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Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 5, “Continuity and Differentiability,” is an essential study material designed to help students understand the concepts of continuity and differentiability in calculus. The chapter covers various topics, including continuity, differentiability, mean value theorem, and Rolle’s theorem. Swastik Classes, a leading coaching institute, has developed comprehensive NCERT solutions that provide step-by-step explanations and solved examples to help students develop a deeper understanding of the subject. These solutions are designed according to the latest CBSE syllabus, making them useful for students preparing for board exams or competitive exams like JEE and NEET. With the help of Swastik Classes’ NCERT solutions, students can improve their problem-solving skills and gain the confidence to tackle complex calculus problems. Overall, Swastik Classes’ NCERT Solution for Class 12 Mathematics Chapter 5 is an essential resource for students who want to excel in mathematics and build a strong foundation in calculus.

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Answers of Physics NCERT solutions for class 12 Chapter 5 Continuity & Differentiability

Chapter 5

Continuity & Differentiability

Exercise-5.1

Question 1:

Prove that the function is continuous at at and 

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6748/Chapter%205_html_a69856a.gif

Therefore, f is continuous at x = 0

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6748/Chapter%205_html_m3baa23bc.gif

Therefore, is continuous at x = −3

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6748/Chapter%205_html_75094717.gif

Therefore, f is continuous at x = 5

Question 2:

Examine the continuity of the function at

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6751/Chapter%205_html_266c3190.gif

Thus, f is continuous at x = 3

Question 3:

Examine the following functions for continuity.

(a) (b)

(c)   (d)

Answer:

(a) The given function is

It is evident that f is defined at every real number k and its value at k is k − 5.

It is also observed that, 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_13958887.gif

Hence, f is continuous at every real number and therefore, it is a continuous function.

(b) The given function is

For any real number k ≠ 5, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_50a3ea5c.gif

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.

(c) The given function is

For any real number c ≠ −5, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_m3f3d37a6.gif

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.

(d) The given function is 

This function f is defined at all points of the real line.

Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5

Case I: c < 5

Then, (c) = 5 − c

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_m1296b13d.gif

Therefore, f is continuous at all real numbers less than 5.

Case II : c = 5

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_m37a44f8.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_c8fc2c0.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_15e897ad.gif

Therefore, is continuous at x = 5

Case III: c > 5

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_m45cd474e.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6754/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all real numbers greater than 5.

Hence, f is continuous at every real number and therefore, it is a continuous function.

Question 4:

Prove that the function  is continuous at x = n, where n is a positive integer.

Answer:

The given function is fx=xn

It is evident that f is defined at all positive integers, n, and its value at n is nn.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6755/Chapter%205_html_75e17cf3.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6755/Chapter%205_html_4dc28689.gif

Therefore, is continuous at n, where n is a positive integer.

Question 5:

Is the function f defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6756/Chapter%205_html_m43306763.gif

continuous at x = 0? At x = 1? At x = 2?

Answer:

The given function f is 

At x = 0,

It is evident that f is defined at 0 and its value at 0 is 0.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6756/Chapter%205_html_1a9d0cd6.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6756/Chapter%205_html_m5835d82a.gif

Therefore, f is continuous at x = 0

At x = 1,

is defined at 1 and its value at 1 is 1.

The left hand limit of f at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6756/Chapter%205_html_58751911.gif

The right hand limit of at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6756/Chapter%205_html_m12aff82a.gif

Therefore, f is not continuous at x = 1

At = 2,

is defined at 2 and its value at 2 is 5.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6756/Chapter%205_html_2480fbf6.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6756/Chapter%205_html_189a3d61.gif

Therefore, f is continuous at = 2

Question 6:

Find all points of discontinuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6758/Chapter%205_html_6f5d7057.gif

Answer:

The given function f is

It is evident that the given function f is defined at all the points of the real line.

Let c be a point on the real line. Then, three cases arise.

(i)  c < 2

(ii)  c > 2

(iii) c = 2

Case (i) c < 2

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6758/Chapter%205_html_m78776ee0.gif

Therefore, f is continuous at all points x, such that x < 2

Case (ii) c > 2

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6758/Chapter%205_html_m5ef9a613.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6758/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x, such that x > 2

Case (iii) c = 2

Then, the left hand limit of at x = 2 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6758/Chapter%205_html_mbe21158.gif

The right hand limit of f at x = 2 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6758/Chapter%205_html_6855250.gif

It is observed that the left and right hand limit of f at x = 2 do not coincide.

Therefore, f is not continuous at x = 2

Hence, x = 2 is the only point of discontinuity of f.

Question 7:

Find all points of discontinuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_m763140dc.gif

Answer:

The given function f is

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_67637c17.gif

Therefore, f is continuous at all points x, such that x < −3

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_6d71b1df.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_727680e3.gif

Therefore, f is continuous at x = −3

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_m2903fe5e.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous in (−3, 3).

Case IV:

If c = 3, then the left hand limit of at x = 3 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_5941cb89.gif

The right hand limit of at x = 3 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_29f0b36.gif

It is observed that the left and right hand limit of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3

Case V:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6759/Chapter%205_html_79947981.gif

Therefore, f is continuous at all points x, such that x > 3

Hence, x = 3 is the only point of discontinuity of f.

Question 8:

Find all points of discontinuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_m4ccee9e7.gif

Answer:

The given function f is

It is known that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_5509af4.gif

Therefore, the given function can be rewritten as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_m292cde6.gif

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_m28f79be1.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x < 0

Case II:

If c = 0, then the left hand limit of at x = 0 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_3fa92674.gif

The right hand limit of at x = 0 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_m4c329f7d.gif

It is observed that the left and right hand limit of f at x = 0 do not coincide.

Therefore, f is not continuous at x = 0

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_m4bef27dd.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6760/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x, such that x > 0

Hence, x = 0 is the only point of discontinuity of f.

Question 9:

Find all points of discontinuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6761/Chapter%205_html_m3ec67b4a.gif

Answer:

The given function f is

It is known that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6761/Chapter%205_html_m27c7ceea.gif

Therefore, the given function can be rewritten as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6761/Chapter%205_html_2ac3ed3e.gif

Let c be any real number. Then, 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6761/Chapter%205_html_3ef9fc07.gif

Also,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6761/Chapter%205_html_378e5a68.gif

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Question 10:

Find all points of discontinuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6762/Chapter%205_html_m5cd13fd3.gif

Answer:

The given function f is

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6762/Chapter%205_html_3501b6a4.gif

Therefore, f is continuous at all points x, such that x < 1

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6762/Chapter%205_html_m4ba2db85.gif

The left hand limit of at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6762/Chapter%205_html_4d9c2e78.gif

The right hand limit of at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6762/Chapter%205_html_m2b97949e.gif

Therefore, f is continuous at x = 1

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6762/Chapter%205_html_m68ca6455.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6762/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x, such that x > 1

Hence, the given function has no point of discontinuity.

Question 11:

Find all points of discontinuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6763/Chapter%205_html_224340ea.gif

Answer:

The given function f is

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6763/Chapter%205_html_m35d1b7b9.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6763/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x, such that x < 2

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6763/Chapter%205_html_m442e0a43.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6763/Chapter%205_html_6003b281.gif

Therefore, f is continuous at x = 2

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6763/Chapter%205_html_m2052425d.gif

Therefore, f is continuous at all points x, such that x > 2

Thus, the given function f is continuous at every point on the real line.

Hence, has no point of discontinuity.

Question 12:

Find all points of discontinuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6765/Chapter%205_html_21ccc991.gif

Answer:

The given function f is

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6765/Chapter%205_html_2b68e9d.gif

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6765/Chapter%205_html_m4a164abb.gif

The right hand limit of f at = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6765/Chapter%205_html_6e433823.gif

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6765/Chapter%205_html_6d5b96ba.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6765/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

Question 13:

Is the function defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_m1d76af7c.gif

a continuous function?

Answer:

The given function is

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_71e1a3a3.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x, such that x < 1

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_m49f14e3.gif

The left hand limit of at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_m33dba451.gif

The right hand limit of f at = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_47b80a2.gif

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_3afae62a.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6767/Chapter%205_html_m3078a2af.gif

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

Question 14:

Discuss the continuity of the function f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_20fb2c5f.gif

Answer:

The given function is

The given function is defined at all points of the interval [0, 10].

Let c be a point in the interval [0, 10].

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_m682944c5.gif

Therefore, f is continuous in the interval [0, 1).

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_m6367ab7.gif

The left hand limit of at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_m7a7b2037.gif

The right hand limit of f at = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_49b439a6.gif

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_m6269866a.gif

Therefore, f is continuous at all points of the interval (1, 3).

Case IV:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_me0f6680.gif

The left hand limit of at x = 3 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_m7ac0d075.gif

The right hand limit of f at = 3 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_m590ecce0.gif

It is observed that the left and right hand limits of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3

Case V:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_m561a1438.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6768/Chapter%205_html_5a589017.gif

Therefore, f is continuous at all points of the interval (3, 10].

Hence, is not continuous at = 1 and = 3

Question 15:

Discuss the continuity of the function f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_m24dbf833.gif

Answer:

The given function is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_964708f.gif

Therefore, f is continuous at all points x, such that x < 0

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_4fe26b04.gif

The left hand limit of at x = 0 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_15c64d62.gif

The right hand limit of f at = 0 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_6c4c75cf.gif

Therefore, f is continuous at x = 0

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_m18bc350f.gif

Therefore, f is continuous at all points of the interval (0, 1).

Case IV:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_6b889dd5.gif

The left hand limit of at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_d825cb9.gif

The right hand limit of f at = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_30c158c5.gif

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case V:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6769/Chapter%205_html_m2bca5b0a.gif

Therefore, f is continuous at all points x, such that x > 1

Hence, is not continuous only at = 1

Question 16:

Discuss the continuity of the function f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_5a22c2c0.gif

Answer:

The given function f is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_2d706456.gif

Therefore, f is continuous at all points x, such that x < −1

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_m5c4dc14e.gif

The left hand limit of at x = −1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_71dd50a6.gif

The right hand limit of f at = −1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_7e7fb68a.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_m2c8a5e24.gif

Therefore, f is continuous at x = −1

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_m1f17b7c9.gif

Therefore, f is continuous at all points of the interval (−1, 1).

Case IV:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_m441906ff.gif

The left hand limit of at x = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_m6ed05fd3.gif

The right hand limit of f at = 1 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_518dec0f.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_2124cbf9.gif

Therefore, f is continuous at x = 2

Case V:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_m56801209.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6773/Chapter%205_html_5a589017.gif

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.

Question 17:

Find the relationship between a and b so that the function f defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6779/Chapter%205_html_7394646a.gif

is continuous at = 3.

Answer:

The given function f is

If f is continuous at x = 3, then

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6779/Chapter%205_html_m33fde032.gif

Therefore, from (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6779/Chapter%205_html_4ee69190.gif

Therefore, the required relationship is given by,

*Question 18:

For what value of https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6784/Chapter%205_html_m11cc021f.gifis the function defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6784/Chapter%205_html_2ac5cb7a.gif

continuous at x = 0? What about continuity at x = 1?

Answer:

The given function f is

If f is continuous at x = 0, then

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6784/Chapter%205_html_5b2103db.gif

Therefore, there is no value of λ for which f is continuous at x = 0

At x = 1,

f (1) = 4x + 1 = 4 × 1 + 1 = 5

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6784/Chapter%205_html_m7a9447a8.gif

Therefore, for any values of λ, f is continuous at x = 1

Question 19:

Show that the function defined by  is discontinuous at all integral point. Here  denotes the greatest integer less than or equal to x.

Answer:

The given function is

It is evident that g is defined at all integral points.

Let n be an integer.

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6788/Chapter%205_html_m207aa0cb.gif

The left hand limit of at x = n is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6788/Chapter%205_html_m1c2be5f0.gif

The right hand limit of f at n is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6788/Chapter%205_html_m172b5fc1.gif

It is observed that the left and right hand limits of f at x = n do not coincide.

Therefore, f is not continuous at x = n

Hence, g is discontinuous at all integral points.

Question 20:

Is the function defined by  continuous at =π?

Answer:

The given function is

It is evident that f is defined at =π.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6792/Item%2020_html_m7c663ad3.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6792/Item%2020_html_m588c68e3.gif

Therefore, the given function f is continuous at = π

Question 21:

Discuss the continuity of the following functions.

(a)  f (x) = sin x + cos x

(b)  f (x) = sin x − cos x

(c)  f (x) = sin x × cos x

Answer:

It is known that if and are two continuous functions, then and g.h are also continuous.

It has to proved first that g (x) = sin and h (x) = cos x are continuous functions.

Let (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let be a real number. Put x = c + h

If x → c, then h → 0

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6796/Chapter%205_html_m40481c7a.gif

Therefore, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let be a real number. Put x = c + h

If x → c, then h → 0

(c) = cos c

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6796/Chapter%205_html_m60ec81a5.gif

Therefore, h is a continuous function.

Therefore, it can be concluded that

(a)  f (x) = g (x) + h (x) = sin x + cos x is a continuous function

(b)  f (x) = g (x) − h (x) = sin x − cos x is a continuous function

(c)  f (x) = g (x) × h (x) = sin x × cos x is a continuous function

Question 22:

Discuss the continuity of the cosine, cosecant, secant and cotangent functions,

Answer:

It is known that if and are two continuous functions, then

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6802/Item%2022_html_b9dd9b.gif

It has to be proved first that g (x) = sin and h (x) = cos x are continuous functions.

Let (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let be a real number. Put x = c + h

If x c, then h → 0

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6802/Item%2022_html_m40481c7a.gif

Therefore, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let be a real number. Put x = c + h

If x → c, then h → 0

(c) = cos c

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6802/Item%2022_html_m60ec81a5.gif

Therefore, h (x) = cos x is continuous function.

It can be concluded that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6802/Item%2022_html_m6892dbfe.gif

Therefore, cosecant is continuous except at np, n ε Z

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6802/Item%2022_html_mb65bf10.gif

Therefore, secant is continuous except at 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6802/Item%2022_html_4fef99da.gif

Therefore, cotangent is continuous except at np, n ε Z

*Question 23:

Find the points of discontinuity of f, where

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6807/Chapter%205_html_m324e9506.gif

Answer:

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6807/Chapter%205_html_m7f62ef9c.gif

Therefore, f is continuous at all points x, such that x < 0

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6807/Chapter%205_html_30eff037.gif

Therefore, f is continuous at all points x, such that x > 0

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6807/Chapter%205_html_2b68db9d.gif

The left hand limit of f at x = 0 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6807/Chapter%205_html_m7ab148d0.gif

The right hand limit of f at x = 0 is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6807/Chapter%205_html_mae478ee.gif

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at all points of the real line.

Thus, f has no point of discontinuity.

*Question 24:

Determine if f defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6812/Chapter%205_html_1bf52e85.gif

is a continuous function?

Answer:

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6812/Chapter%205_html_5690438c.gif

Therefore, f is continuous at all points ≠ 0

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6812/Chapter%205_html_73e8f29.gif
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/Selection_007(10).png
GR0I3EcEkWa3MRVKZwF2bHhur4MtcF IUg1NlZU7IcZE4V daEQwQYXGm5DeeJp 27Gco8pYS1 p4UT1zzXtST9bCQcyF FOPsPx39bjVB9Rkb3Wy2xblirdD c2EMPVbQzXSb8
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/Selection_009(12).png
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6812/Chapter%205_html_m473a2a2d.gif

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

Question 25:

Examine the continuity of f, where f is defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6817/Chapter%205_html_75920e75.gif

Answer:

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6817/Chapter%205_html_31cd584e.gif

Therefore, f is continuous at all points x, such that x ≠ 0

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6817/Chapter%205_html_m14ad73a7.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6817/Chapter%205_html_c367a60.gif

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

*Question 26:

Find the values of so that the function f is continuous at the indicated point.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6819/Item%2026_html_50475cb3.gif

Answer:

The given function f is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6819/Item%2026_html_338e09a5.gif

The given function f is continuous at , if f is defined at  and if the value of the f at  equals the limit of f at

It is evident that is defined at  and

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6819/Item%2026_html_46e34999.gif

Therefore, the required value of k is 6.

Question 27:

Find the values of so that the function f is continuous at the indicated point.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6822/Chapter%205_html_m2d83caea.gif

Answer:

The given function is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6822/Chapter%205_html_38fec36d.gif

The given function f is continuous at x = 2, if f is defined at x = 2 and if the value of f at x = 2 equals the limit of f at x = 2

It is evident that is defined at x = 2 and

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6822/Chapter%205_html_79a9c467.gif

Therefore, the required value of k is

Question 28:

Find the values of so that the function f is continuous at the indicated point.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6826/Item%2028_html_66eb254b.gif

Answer:

The given function is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6826/Item%2028_html_12f6ac2f.gif

The given function f is continuous at x = p, if f is defined at x = p and if the value of f at x = p equals the limit of f at x = p

It is evident that is defined at x = p and

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6826/Item%2028_html_m1ce988b8.gif

Therefore, the required value of k is

Question 29:

Find the values of so that the function f is continuous at the indicated point.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6831/Chapter%205_html_1b0651b3.gif

Answer:

The given function is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6831/Chapter%205_html_m71365e67.gif

The given function f is continuous at x = 5, if f is defined at x = 5 and if the value of f at x = 5 equals the limit of f at x = 5

It is evident that is defined at x = 5 and

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6831/Chapter%205_html_19cdbff6.gif

Therefore, the required value of k is

Question 30:

Find the values of a and b such that the function defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6833/Chapter%205_html_m49453141.gif

is a continuous function.

Answer:

The given function is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6833/Chapter%205_html_m49453141.gif

It is evident that the given function f is defined at all points of the real line.

If f is a continuous function, then f is continuous at all real numbers.

In particular, f is continuous at = 2 and = 10

Since f is continuous at = 2, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6833/Chapter%205_html_m6a67a506.gif

Since f is continuous at = 10, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6833/Chapter%205_html_m62db6148.gif

On subtracting equation (1) from equation (2), we obtain

8a = 16

⇒ a = 2

By putting a = 2 in equation (1), we obtain

2 × 2 + b = 5

⇒ 4 + b = 5

⇒ b = 1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.

 Question 31:

Show that the function defined by f (x) = cos (x2) is a continuous function.

Answer:

The given function is (x) = cos (x2)

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where g (x) = cos x and h (x) = x2

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6835/Chapter%205_html_2414acbb.gif

It has to be first proved that (x) = cos x and h (x) = x2 are continuous functions.

It is evident that g is defined for every real number.

Let c be a real number.

Then, g (c) = cos c

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6835/Chapter%205_html_m6d0b3abd.gif

Therefore, g (x) = cos x is continuous function.

h (x) = x2

Clearly, h is defined for every real number.

Let k be a real number, then h (k) = k2

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6835/Chapter%205_html_m129f05e7.gif

Therefore, h is a continuous function.

It is known that for real valued functions and h, such that (h) is defined at c, if is continuous at and if is continuous at (c), then (g) is continuous at c.

Therefore,  is a continuous function.

Question 32:

Show that the function defined by  is a continuous function.

Answer:

The given function is 

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where and  

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6839/Chapter%205_html_m400b359b.gif

It has to be first proved that  and  are continuous functions.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6839/Chapter%205_html_5c7c3891.gif

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6839/Chapter%205_html_m28e427f3.gif

Therefore, g is continuous at all points x, such that x < 0

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6839/Chapter%205_html_7eaf7bf6.gif

Therefore, g is continuous at all points x, such that x > 0

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6839/Chapter%205_html_m40b8e9ef.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6839/Chapter%205_html_7f384cbb.gif

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

(x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let be a real number. Put x = c + h

If x → c, then h → 0

(c) = cos c

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6839/Chapter%205_html_m60ec81a5.gif

Therefore, h (x) = cos x is a continuous function.

It is known that for real valued functions and h,such that (h) is defined at c, if is continuous at and if is continuous at (c), then (g) is continuous at c.

Therefore,  is a continuous function.

Question 33:

Examine that  is a continuous function.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_4c157e2.gif

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where and

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_4d6be309.gif

It has to be proved first that  and  are continuous functions.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_5c7c3891.gif

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_m28e427f3.gif

Therefore, g is continuous at all points x, such that x < 0

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_7eaf7bf6.gif

Therefore, g is continuous at all points x, such that x > 0

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_m40b8e9ef.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_7f384cbb.gif

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

(x) = sin x

It is evident that h (x) = sin x is defined for every real number.

Let be a real number. Put x = c + k

If x → c, then k → 0

(c) = sin c

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6845/Chapter%205_html_4d11afc.gif

Therefore, h is a continuous function.

It is known that for real valued functions and h,such that (h) is defined at c, if is continuous at and if is continuous at (c), then (g) is continuous at c.

Therefore,  is a continuous function.

Question 34:

Find all the points of discontinuity of defined by

Answer:

The given function is

The two functions, g and h, are defined as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_m5b88ece7.gif

Then, f = − h

The continuity of g and is examined first.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_5c7c3891.gif

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_m28e427f3.gif

Therefore, g is continuous at all points x, such that x < 0

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_7eaf7bf6.gif

Therefore, g is continuous at all points x, such that x > 0

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_m40b8e9ef.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_7f384cbb.gif

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_2697ef2d.gif

Clearly, h is defined for every real number.

Let be a real number.

Case I:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_602d838d.gif

Therefore, h is continuous at all points x, such that x < −1

Case II:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_m7a5a8238.gif

Therefore, h is continuous at all points x, such that x > −1

Case III:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_a45610c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_m56c2d36f.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6849/Chapter%205_html_m2f351985.gif

Therefore, h is continuous at x = −1

From the above three observations, it can be concluded that h is continuous at all points of the real line.

g and h are continuous functions. Therefore, g − is also a continuous function.

Therefore, has no point of discontinuity.

Exercise-3.2

Question 1:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6853/Chapter%205_html_m4d909628.gif

Answer:

Let fx=sin x2+5 , u(x)=x2+5, and v(t)=sint

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6853/Chapter%205_html_m76e301f4.gif

Alternate method

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6853/Chapter%205_html_7e76c191.gif

Question 2:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6856/Chapter%205_html_m73e47a83.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6856/Chapter%205_html_m26cdca88.gif

Thus, is a composite function of two functions.

Put t = u (x) = sin x

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6856/Chapter%205_html_4ceae8bd.gif

By chain rule,https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6856/Chapter%205_html_m119d5d9e.gif

Alternate method

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6856/Chapter%205_html_664cfa69.gif

Question 3:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6858/Chapter%205_html_me9e6b3b.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6858/Chapter%205_html_m4ff198ec.gif

Thus, is a composite function of two functions, u and v.

Put t = u (x) = ax + b

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6858/Chapter%205_html_7838a27b.gif

Hence, by chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6858/Chapter%205_html_e4e4928.gif

Alternate method

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6858/Chapter%205_html_5e6626a8.gif

Question 4:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6860/Chapter%205_html_m39f8d27b.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6860/Chapter%205_html_714e4461.gif

Thus, is a composite function of three functions, u, v, and w.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6860/Chapter%205_html_m27e732b4.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6860/Chapter%205_html_6bc2354c.gif

Hence, by chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6860/Chapter%205_html_21a687a3.gif

Alternate method

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6860/Chapter%205_html_2ba7a7d4.gif

Question 5:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_m74aa0a8b.gif

Answer:

The given function is where g (x) = sin (ax + b) and

h (x) = cos (cx d)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_55c8a329.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_2129c4ad.gif

∴ is a composite function of two functions, u and v.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_56dd424a.gif

Therefore, by chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_m5cf33a7.gif

h is a composite function of two functions, p and q.

Put y = p (x) = cx d

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_71b91de8.gif

Therefore, by chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_7b73ba25.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6862/Chapter%205_html_695dd5a1.gif

Question 6:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6865/Chapter%205_html_5333257e.gif

Answer:

The given function is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6865/Chapter%205_html_2f2abb3d.gif

Question 7:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6866/Chapter%205_html_728931d9.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6866/Chapter%205_html_231350d7.gif

Question 8:

Differentiate the functions with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6869/Chapter%205_html_m69c04209.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6869/Chapter%205_html_38a98fe8.gif

Clearly, is a composite function of two functions, and v, such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6869/Chapter%205_html_m18ee6d2d.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6869/Chapter%205_html_7147c9af.gif

By using chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6869/Chapter%205_html_me24073e.gif

Alternate method

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6869/Chapter%205_html_61c17f97.gif

*Question 9:

Prove that the function given by  is not differentiable at x = 1.

Answer:

The given function is 

It is known that a function f is differentiable at a point x = c in its domain if both

  and are finite and equal.

To check the differentiability of the given function at x = 1,

consider the left hand limit of f at x = 1

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6873/Chapter%205_html_m169cbd00.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6873/Chapter%205_html_m274af491.gif

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

*Question 10:

Prove that the greatest integer function defined by is not

differentiable at x = 1 and x = 2.

Answer:

The given function f is 

It is known that a function f is differentiable at a point x = c in its domain if both and are finite and equal.

To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6875/Chapter%205_html_5dc0d8f0.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6875/Chapter%205_html_7bb8909e.gif

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at

x = 1

To check the differentiability of the given function at x = 2, consider the left hand limit

of f at x = 2

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6875/Chapter%205_html_6d34caac.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6875/Chapter%205_html_14d07761.gif

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2

Exercise-3.2

Question 1:

Find

Answer:

The given relationship is 

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6876/Chapter%205_html_m2ccfaf8f.gif

Question 2:

Find

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6877/Chapter%205_html_m749090b0.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6877/Chapter%205_html_16722ab9.gif

Question 3: 

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6878/Chapter%205_html_63021146.gif

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6878/Chapter%205_html_29ef79fd.gif

Using chain rule, we obtain

 and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6878/Chapter%205_html_m29378c58.gif

From (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6878/Chapter%205_html_62d94b80.gif

Question 4:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6879/Chapter%205_html_md2c130.gif

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6879/Chapter%205_html_2f8e50f7.gif

Question 5:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6880/Chapter%205_html_76c408f6.gif

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6880/Chapter%205_html_m6a27f00e.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6880/Chapter%205_html_6db06843.gif [Derivative of constant function is 0]

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6880/Chapter%205_html_m6a84e470.gif

Question 6:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6881/Chapter%205_html_m32c26aae.gif

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6881/Chapter%205_html_m165a9612.gif

*Question 7:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6882/Item%2051_html_m6df1a6d3.gif

Answer:

The given relationship is 

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6882/Item%2051_html_m38c7e354.gif

Using chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6882/Item%2051_html_773d7508.gif

From (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6882/Item%2051_html_62e6dc51.gif

Question 8:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6883/Chapter%205_html_m50398b58.gif

Answer:

The given relationship is

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6883/Chapter%205_html_410d4b38.gif
b50EXjR30rz28g4Vuht4HYX KwheRGwQKf4o92CIQm9i5FftqT07QXlydGRzj7Z
623zAmGfrD vj3EiRYadEQSLjv3oJ OCGx22NGCWJn3hhND4FfhBKRH8CMBp6JaU7fuV9HVOCW0HC42QEmmGqQTmCT hno4LkStbC2YPsvM OkAwY7oxxNwDe9B 3rXx2xK 1k

Question 9:H9FgC3UbYX8bZPKDrz5o3ERBte2XDGxCL7dYSrkZoIYxWY6JNGtY5Ql 4GEJHzkNXo2lS7v8iQtIU8 N5viQQ RijOMj

Find 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6884/Chapter%205_html_4df2fdc5.gif

Answer:OXJXuAGfJ2WPKs98459c Ie6YaV9KkYF9nY0pqXTALVNILiN3PLZykkcgeYIdm0aElfSGTMz7cxqMLngLKS53XtlvrO67qcmrrrjeiJXTfwUwDaUGiBGg n4g1XM5sfKrO0HDSM

We have

Put

rMvafcIxbRLyYhASWtXgQz4t41M0x
SRnWNjUMoKQ1UkPI5PwRNEMeCYzj CNtbVyZ23nLD6mhRsOxh7GO8v8eLJE0PQhem1QqvNlTMzZAb8k1AQEbAEYjWVpxgwPCjapGkJUelndD5F yCBzuwsLObvkNTB6qcCyQVk
l2XttI8GsPkjPhTWfvBIhVqJpkBCMRSuJCNDn9JoPHt5iG6UJrEb5yjTvasUc8vntWcrZXIOZLFvePuyJGrxU rTYPnb0mEXunUl3nU vN1C0P6I47z kpjgmDUBT 9wcu0TCBM
p0ru1Db8uXILg293K 4hPgDYURNIUqFvUi HLnvGRXIdP2rMYAAB5uDW4rg6gtQYleITFNWFkFkBmk9C041UCWiolWTk6iIJW06IQwzhPQF Y1JSXftA 9i56QIeRIW0rQ19WfM

We have,y = sin-12×1 + x2put x = tan θ ⇒ θ = tan-1xNow,    y = sin-12 tan θ1 + tan2θ⇒y = sin-1sin 2θ, as sin 2θ=2 tan θ1 + tan2θ⇒y = 2θ,  as sin-1sin x = x ⇒ y = 2 tan-1x⇒dydx = 2 × 11 + x2, because dtan-1xdx=11 + x2 ⇒ dy dx = 21 + x2

Question 10:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6885/Chapter%205_html_m5de545ea.gif

Answer:

The given relationship is  

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6885/Chapter%205_html_5fffeb1a.gif

It is known that, 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6885/Chapter%205_html_3ae6bb5d.gif

Comparing equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6885/Chapter%205_html_266bfd1e.gif

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6885/Chapter%205_html_57eda34e.gif

Question 11:

Find 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6886/Chapter%205_html_m693adbde.gif

Answer:

The given relationship is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6886/Chapter%205_html_m526f5daa.gif

On comparing L.H.S. and R.H.S. of the above relationship, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6886/Chapter%205_html_46de67e4.gif

Differentiating this relationship with respect to x, we obtain

lZzjV92ZGjuIM8T8wriZeXSUXGio9BnN6uL3maDVVUCJfWh5HBIiXdCmeeDl9jmJpbrd0d03m7 rk2pJBLxa8HVWwWvZ4XQW UqAmlGq3iwkZYVBj3TFXFjt EJssim7Vq4SspI
5 M1XLcaezx9xLUZ12khYVgD6DRj7ycowy6YQX19m5brL2Cr67qDpBtSs1qTzriRTZgmJkzADmY QG6dYup5vl1NTsKDdIaU31s3AoSiwgU8v3E ptO851vRN9t22bsCWNMwHLw
B8tcF0MESUSDwEC d0Am9Y9SOkwTWock63w4rkjc5Ml3UctTUk7iVJ PDMhNfTKJchvsoGER6YoqDQvZevLX7ofD6rCLQNqYovidcgarWLtz2LnDF88 07IsuJQ3rrw 8ngcSDI
neMyg1UGYPRK YwGWpwcp65BPpswMelsQ1eZneZ7z1QaS Yd24E bQURXKgv8jWFIX9CApiqram39nAoVTJ8L53MBmkfgnNN35wkAgfhN9eWjNghf nMhak8esJPVRukZTRreec

*Question 12:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_18b02256.gif

Answer:

The given relationship is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_md8884c8.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_md8884c8.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_62fdca80.gif

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_m7c7470a4.gif

Using chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_54fd059c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_7aeaeaa4.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_56a3dccc.gif

From (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_m5c82195e.gif

Alternate method

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_md8884c8.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_7ae83ac.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_m7156045c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_534c54ce.gif

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6887/Chapter%205_html_m1e8bf772.gif

Question 13:

Find  

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6888/Chapter%205_html_6cd64398.gif

Answer:

The given relationship is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6888/Chapter%205_html_mdb8500c.gif

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6888/Chapter%205_html_m9d8c768.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6888/Chapter%205_html_m1573b6af.gif

*Question 14:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6892/Chapter%205_html_5110c1c6.gif

Answer:

The given relationship is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6892/Chapter%205_html_m3d245b50.gif

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6892/Chapter%205_html_m3b8e2f0d.gif

Question 15:

Find

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6895/Chapter%205_html_m3100b8d8.gif

Answer:

The given relationship is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6895/Chapter%205_html_m1635bb2.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6895/Chapter%205_html_m1635bb2.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6895/Chapter%205_html_c9b0602.gif

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6895/Chapter%205_html_m4da5753f.gif

Exercise-5.4

Question 1:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6897/Item%2060_html_5a6ef49c.gif

Answer:

Let

By using the quotient rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6897/Item%2060_html_m44826895.gif

Question 2:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6899/Chapter%205_html_10665804.gif

Answer:

Let  

By using the chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6899/Chapter%205_html_m61ef7eca.gif

Question 3:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6902/Chapter%205_html_62b1ecd9.gif

Answer:

Let  

By using the chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6902/Chapter%205_html_m6cd6d4fc.gif

*Question 4:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6904/Chapter%205_html_m417e6043.gif

Answer:

Let

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6904/Chapter%205_html_m85aa204.gif

By using the chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6904/Chapter%205_html_m2beb318a.gif

Question 5:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6906/Chapter%205_html_67353335.gif

Answer:

Let  

By using the chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6906/Chapter%205_html_f778309.gif

*Question 6:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6907/Chapter%205_html_7cef6427.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6907/Chapter%205_html_m6e457112.gif

Question 7:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6908/Chapter%205_html_m289c9b29.gif

Answer:

Let  

Then,  

By differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6908/Chapter%205_html_63fdcfa6.gif

Question 8:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6910/Chapter%205_html_5f9334b3.gif

Answer:

Let  

By using the chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6910/Chapter%205_html_m72ced0c9.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6910/Chapter%205_html_56a4a2d6.gifx > 1

*Question 9:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6912/Chapter%205_html_m1401ed6e.gif

Answer:

Let

By using the quotient rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6912/Chapter%205_html_6256b73f.gif

*Question 10:

Differentiate the following w.r.t. x:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6913/Chapter%205_html_a5d499d.gif

Answer:

Let

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6913/Chapter%205_html_26fadfea.gif

By using the chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6913/Chapter%205_html_m7e614266.gif

Exercise-5.5

Question 1:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6914/Chapter%205_html_5195ccf8.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6914/Chapter%205_html_a90b370.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6914/Chapter%205_html_m659895a.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6914/Chapter%205_html_4069ee03.gif

Question 2:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6915/Chapter%205_html_630e908.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6915/Chapter%205_html_756604ee.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6915/Chapter%205_html_m7d328c07.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6915/Chapter%205_html_m6c60117d.gif

Question 3:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6918/Chapter%205_html_783d3a39.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6918/Chapter%205_html_51ab72c3.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6918/Chapter%205_html_m7f8be8c7.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6918/Chapter%205_html_m37df01e.gif

Question 4:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6922/Chapter%205_html_71de10.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6922/Chapter%205_html_m2a70b6e1.gif

xx

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6922/Chapter%205_html_4a39af9.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6922/Chapter%205_html_m278d76e2.gif

v = 2sin x

Taking logarithm on both the sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6922/Chapter%205_html_md5917e4.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6922/Chapter%205_html_m363ba02e.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6922/Chapter%205_html_63a5305b.gif

Question 5:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6926/Chapter%205_html_m173da24a.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6926/Chapter%205_html_m23f8f68.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6926/Chapter%205_html_4d1e6258.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6926/Chapter%205_html_2e65f58d.gif

Question 6:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6929/Chapter%205_html_6db13fca.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6929/Chapter%205_html_5ce4fc8c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6929/Chapter%205_html_10dec314.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6929/Chapter%205_html_m67161841.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6929/Chapter%205_html_4d0680ca.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6929/Chapter%205_html_m291a6d50.gif

Therefore, from (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6929/Chapter%205_html_1e132b57.gif

Question 7:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_m1b6d35e3.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_m31ae62d8.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_m533a693d.gif

= (log x)x

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_45b75a57.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_m3bd5e52e.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_55f8f753.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_m3e924d53.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_5324cf14.gif

Therefore, from (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6932/Chapter%205_html_m3d5fae4.gif

*Question 8:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6935/Chapter%205_html_1a8d3394.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6935/Chapter%205_html_1596bc50.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6935/Chapter%205_html_34ed9641.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6935/Chapter%205_html_m4cfb3914.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6935/Chapter%205_html_m220de476.gif

Therefore, from (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6935/Chapter%205_html_66e67501.gif

*Question 9:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6937/Chapter%205_html_263eebdd.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6937/Chapter%205_html_731c8f29.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6937/Chapter%205_html_555a94f9.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6937/Chapter%205_html_me8cba15.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6937/Chapter%205_html_m4369494.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6937/Chapter%205_html_m61700c16.gif

From (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6937/Chapter%205_html_54c0cb1e.gif

*Question 10:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6942/Chapter%205_html_m17b64539.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6942/Chapter%205_html_m319891f6.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6942/Chapter%205_html_193f9bb9.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6942/Chapter%205_html_771ab082.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6942/Chapter%205_html_3ad2122b.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6942/Chapter%205_html_m3adb796c.gif

From (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6942/Chapter%205_html_m3d78aab0.gif

*Question 11:

Differentiate the function with respect to x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6949/Chapter%205_html_3e605c11.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6949/Chapter%205_html_6064c284.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6949/Chapter%205_html_m59e0dc54.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6949/Chapter%205_html_33ac2b2.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6949/Chapter%205_html_m6800765d.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6949/Chapter%205_html_760c0f18.gif

From (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6949/Chapter%205_html_md0b5321.gif

Question 12:

Find  of function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_3b66b7c4.gif

Answer:

The given function ishttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_3b66b7c4.gif

Let xy = u and yx = v

Then, the function becomes u v = 1

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_m6e2ab7d2.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_m58067ee2.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_m2c850875.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_m32442ecb.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_m34966692.gif

From (1), (2), and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6955/Chapter%205_html_m51d87f3b.gif

*Question 13:

Find  of function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6959/Chapter%205_html_m243ff079.gif

Answer:

The given function is

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6959/Chapter%205_html_m586a78c.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6959/Chapter%205_html_1c72ea1e.gif

Question 14:

Find  of function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6964/Chapter%205_html_m50f80e68.gif

Answer:

The given function is

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6964/Chapter%205_html_61526371.gif

Differentiating both sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6964/Chapter%205_html_7bed3156.gif

Question 15:

Find  of function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6972/Chapter%205_html_m44b33473.gif

Answer:

The given function is

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6972/Chapter%205_html_m6648756a.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6972/Chapter%205_html_14891e0e.gif

Question 16:

Find the derivative of the function given by  and hence find

Answer:

The given relationship is

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6975/Chapter%205_html_5394fe11.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6975/Chapter%205_html_3196b276.gif

Question 18:

If uv and w are functions of x, then show that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6977/Chapter%205_html_64c95fbc.gif

in two ways-first by repeated application of product rule, second by logarithmic differentiation.

Answer:

Let 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6977/Chapter%205_html_m3be4ac90.gif

By applying product rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6977/Chapter%205_html_m2058d63e.gif

By taking logarithm on both sides of the equation , we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6977/Chapter%205_html_m76a64de7.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6977/Chapter%205_html_e86d78d.gif

Exercise-5.6

Question 1:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6979/Chapter%205_html_40388a4c.gif

Answer:

The given equations arehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6979/Chapter%205_html_2586dc24.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6979/Chapter%205_html_689f5229.gif

Question 2:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

x = a cos θy = b cos θ

Answer:

The given equations are x = a cos θ and y = b cos θ

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6980/Chapter%205_html_298f8226.gif

*Question 3:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

x = sin ty = cos 2t

Answer:

The given equations are x = sin t and y = cos 2t

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6981/Chapter%205_html_3ed7b1fd.gif

Question 4:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6982/Chapter%205_html_cf72344.gif

Answer:

The given equations are and  

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6982/Chapter%205_html_4ae74387.gif

Question 5:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6983/Chapter%205_html_m61204350.gif

Answer:

The given equations are

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6983/Chapter%205_html_7c434e2.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6983/Chapter%205_html_88d11c7.gif

Question 6:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6984/Chapter%205_html_53f8470a.gif

Answer:

The given equations are

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6984/Chapter%205_html_m3709b23c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6984/Chapter%205_html_65b29f2c.gif

Question 7:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6985/Chapter%205_html_m33c9ca76.gif

Answer:

The given equations are

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6985/Chapter%205_html_m3de5aa4.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6985/Chapter%205_html_m63668abf.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6985/Chapter%205_html_1d45c44d.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6985/Chapter%205_html_530f4da5.gif

Question 8:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6986/Chapter%205_html_201f59d8.gif

Answer:

The given equations are

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6986/Chapter%205_html_eb65268.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6986/Chapter%205_html_m180a8a0b.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6986/Chapter%205_html_1e944e50.gif

Question 9:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

nlxoYSbWi54wJ06mbmCMKB0piiSjiF0FAHnvGIq Bv41LcRCyHNq3xDnwdVcX5fFMLKquzA gc3xxxihdksl it wEOV1uMpq15j I1kklFDzKw5rDN7kvpoEARs l HGw0qVmI

Answer:

The given equations are

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6987/Chapter%205_html_4ec41ea3.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6987/Chapter%205_html_m426d257d.gif

*Question 10:

If x and y are connected parametrically by the equation, without eliminating the parameter, find .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6988/Chapter%205_html_m4c97a497.gif

Answer:

The given equations are

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6988/Chapter%205_html_11d3141d.gif
PAd7 FVliZ8I5hk6DfyK4U8r0HO2yJm9Nbu 2YXT05hSdPeEFKh303kzsXBtGtKiONueaNyzLccdGXeLMIbiqBIo6hRgl2N3uIuWGhl4UiXvjWR6rhGPQzg67bFKbIHD6FbqQdk
udEL7EnMmUzLldzWx8tnRvqBQDwURgPngwFssZUODqhOEjHaZsUqQTvvAmClCmT3TNSd4 gs

*Question 11:

If

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6989/Chapter%205_html_2d605598.gif

Answer:

The given equations are

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6989/Chapter%205_html_60faf63f.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6989/Chapter%205_html_m211dfd34.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6989/Chapter%205_html_643d06c3.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6989/Chapter%205_html_m39cfcd1.gif

Hence, proved.

Exercise-5.7

Question 1:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6990/Chapter%205_html_m7a59faf5.gif

Answer:

Let

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6990/Chapter%205_html_41f03fdb.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6990/Chapter%205_html_140e559d.gif

Question 2:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6991/Chapter%205_html_m2468ff53.gif

Answer:

Lethttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6991/Chapter%205_html_m7bd30f86.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6991/Chapter%205_html_m57b41e71.gif

Question 3:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6992/Chapter%205_html_m23f4c69b.gif

Answer:

Let

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6992/Chapter%205_html_2aec51b6.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6992/Chapter%205_html_71a5f565.gif

Question 4:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6993/Chapter%205_html_727b0d3e.gif

Answer:

Lethttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6993/Chapter%205_html_m5d5bb1c3.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6993/Chapter%205_html_4c379a84.gif

Question 5:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6994/Chapter%205_html_33c4eae6.gif

Answer:

Lethttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6994/Chapter%205_html_m34818b33.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6994/Chapter%205_html_192f1554.gif

Question 6:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6995/Chapter%205_html_480384f8.gif

Answer:

Lethttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6995/Chapter%205_html_m338bf3ed.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6995/Chapter%205_html_mf68e5a3.gif

Question 7:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6996/Chapter%205_html_47bf6e60.gif

Answer:

Lethttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6996/Chapter%205_html_6f6eedb7.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6996/Chapter%205_html_m72401aa9.gif

Question 8:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6997/Chapter%205_html_6b1d3f6d.gif

Answer:

Let

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6997/Chapter%205_html_m5f1b07e9.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6997/Chapter%205_html_m1e4696a0.gif

*Question 9:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6998/Chapter%205_html_f8f0eb6.gif

Answer:

Let

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6998/Chapter%205_html_6451572a.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6998/Chapter%205_html_m49c47554.gif

Question 10:

Find the second order derivatives of the function.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6999/Chapter%205_html_m4ebb87cd.gif

Answer:

Let

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6999/Chapter%205_html_5f4c3a91.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/6999/Chapter%205_html_203b096f.gif

*Question 11:

If , prove that

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7000/Chapter%205_html_m287fb11e.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7000/Chapter%205_html_4a272fe7.gif

Hence, proved.

Question 12:

If  find in terms of y alone.

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7001/Chapter%205_html_27f47e6b.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7001/Chapter%205_html_500433ae.gif

Question 13:

If show that  

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7002/Chapter%205_html_6a6aad23.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7002/Chapter%205_html_m1d04d99e.gif

Hence, proved.

*Question 14:

If show that

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7003/Chapter%205_html_m3ce836af.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7003/Chapter%205_html_m1a6b17e8.gif

Hence, proved.

Question 15:

If , show that

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7005/Chapter%205_html_m173671fd.gif

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7005/Chapter%205_html_m12fbda7f.gif

Hence, proved.

Question 16:

If show that

Answer:

The given relationship is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7006/Chapter%205_html_4513d279.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7006/Chapter%205_html_m7a51c783.gif

Differentiating this relationship with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7006/Chapter%205_html_4686e8e2.gif

Hence, proved.

Question 17:

If , show that

Answer:

The given relationship is

Then,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7007/Chapter%205_html_m10ec30e3.gif

Hence, proved.

Exercise-5.8

Question 1:

Verify Rolle’s Theorem for the function

Answer:

The given function,, being a polynomial function, is continuous in [−4, 2] and is differentiable in (−4, 2).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_m7310293a.gif

∴ f (−4) = f (2) = 0

⇒ The value of f (x) at −4 and 2 coincides.

Rolle’s Theorem states that there is a point c ∈ (−4, 2) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7008/Chapter%205_html_7b1d23d4.gif

Hence, Rolle’s Theorem is verified for the given function.

Question 2:

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?

(i) for

(ii)  for

(iii) for  

Answer:

By Rolle’s Theorem, for a function , if

(a) f is continuous on [ab]

(b) f is differentiable on (ab)

(c)  (a) = f (b)

then, there exists some c ∈ (ab) such that 

Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

(i)  for

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = 5 and = 9

⇒ f (x) is not continuous in [5, 9].

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_7c07b588.gif

The differentiability of f in (5, 9) is checked as follows.

Let be an integer such that n ∈ (5, 9).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_332814a3.gif

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for for

(ii)  for 

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = −2 and = 2

⇒ f (x) is not continuous in [−2, 2].

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_4dac9e05.gif

The differentiability of f in (−2, 2) is checked as follows.

Let be an integer such that n ∈ (−2, 2).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_332814a3.gif

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for for .

(iii) for

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7009/Chapter%205_html_1291b89d.gif

(1) ≠ f (2)

It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for for

*Question 3:

If  is a differentiable function and if  does not vanish anywhere, then prove that

Answer:

It is given that is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) f is continuous on [−5, 5].

(b) is differentiable on (−5, 5).

Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_413b28d7.gif

It is also given that  does not vanish anywhere.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7010/Chapter%205_html_m53324de.gif

Hence, proved.

Question 4:

Verify Mean Value Theorem, if   in the interval where and, b = 4

Answer:

The given function is

f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x − 4.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_26322854.gif

Mean Value Theorem states that there is a point c ∈ (1, 4) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7011/Chapter%205_html_74f0b6d.gif

Hence, Mean Value Theorem is verified for the given function.

*Question 5:

Verify Mean Value Theorem, if  in the interval [ab], where a = 1 and b = 3. Find all  for which  

Answer:

The given function f is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_245fb984.gif

f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x2 − 10x − 3.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_m6eec5660.gif

Mean Value Theorem states that there exist a point c ∈ (1, 3) such thathttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_33edd1e0.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7012/Chapter%205_html_23b1964d.gif

Hence, Mean Value Theorem is verified for the given function and  is the only point for which  

Question 6:

Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Answer:

Mean Value Theorem states that for a function if

(a)  f is continuous on [ab]

(b) f is differentiable on (ab)

then, there exists some c ∈ (ab) such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_10edd9c.gif

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

(i) 

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = 5 and = 9

⇒ f (x) is not continuous in [5, 9].

The differentiability of f in (5, 9) is checked as follows.

Let be an integer such that n ∈ (5, 9).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_332814a3.gif

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_1e3439b6.gif.

(ii) 

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = −2 and = 2

⇒ f (x) is not continuous in [−2, 2].

The differentiability of f in (−2, 2) is checked as follows.

Let be an integer such that n ∈ (−2, 2).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_332814a3.gif

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_97a178a.gif.

(iii) 

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for .

It can be proved as follows.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_m6ac1a3cd.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7013/Chapter%205_html_7f2b4507.gif

Miscellaneous exercise

Differentiate wrt x the function in question 1 to 11

Question 1:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7014/Chapter%205_html_d961562.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7014/Chapter%205_html_m4d94c2f0.gif

Using chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7014/Chapter%205_html_m3c95ac0f.gif

Question 2:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7015/Chapter%205_html_m3549a5f.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7015/Chapter%205_html_6b8a7ae0.gif

Question 3:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7016/Chapter%205_html_5c8f433.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7016/Chapter%205_html_m4e5ec0bf.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7016/Chapter%205_html_2f8798fc.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7016/Chapter%205_html_m76406458.gif

Question 4:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7017/Chapter%205_html_m51f28dd5.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7017/Chapter%205_html_m656ffe8.gif

Using chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7017/Chapter%205_html_m5eefc5ae.gif

Question 5:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7018/Chapter%205_html_2aa0355a.gif

Answer:

gaNpkPD83zv2iFkQzqfCMtqz30SwhYtMgQwe09s5YoT0gS6NgXKcCiP2pdHWeoHI1hABAdPN aa6i6lZa82SRSpXwY2SKAs1uZSD2KKdVlySQskpnfX9yNd1NP
5h xw r586GsDo b5NEfu1Z0DsN9xtTCUDRzGn6ndvDpIWiHUvieNIDuLm NuqM6FieeslunTnnIKyAsVkqh87cuJwXH8n4BwC6CnIaqq2V8Y3re8 bHupadAE7NiIDknH2t14

*Question 6:

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7019/Chapter%205_html_m3ee6a316.gif

Therefore, equation (1) becomes

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7019/Chapter%205_html_17aff96e.gif

*Question 7:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7020/Chapter%205_html_m4826e034.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7020/Chapter%205_html_m297fcb9d.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7020/Chapter%205_html_m7b8e6006.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7020/Chapter%205_html_m33b8f35d.gif

Question 8:

 for some constant a and b.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7021/Chapter%205_html_m76132c97.gif

By using chain rule, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7021/Chapter%205_html_454748c.gif

Question 9:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7022/Chapter%205_html_m1fa12acd.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7022/Chapter%205_html_m640ad7d0.gif

Taking logarithm on both the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7022/Chapter%205_html_m7c35ee16.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7022/Chapter%205_html_1ad53114.gif

Question 10:

for some fixed  and 

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7025/Chapter%205_html_1b6ce38c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7025/Chapter%205_html_m4e23c929.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7025/Chapter%205_html_m2b571ec3.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7025/Chapter%205_html_22731031.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7025/Chapter%205_html_m29814724.gif

s = aa

Since a is constant, aa is also a constant.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7025/Chapter%205_html_m28f31a96.gif

From (1), (2), (3), (4), and (5), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7025/Chapter%205_html_m32f7549d.gif

Question 11:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_m17e50c14.gif, for https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_5f6ad269.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_31a56b1e.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_3c96230d.gif

Differentiating with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_4c578ae1.gif

Also,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_7d588169.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_1f5c0947.gif

Substituting the expressions of https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_5c6433d9.gifin equation (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7029/Chapter%205_html_a5b5dce.gif

Question 12:

Find if 

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7033/Chapter%205_html_6ee472a6.gif

*Question 13:

Find if 

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7037/Chapter%205_html_4ecf365e.gif

*Question 14:

If for, −1 < x <1, prove that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7060/Chapter%205_html_m224d4bdb.gif

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7060/Chapter%205_html_3daa2805.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7060/Chapter%205_html_426367e2.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7060/Chapter%205_html_m32da6f71.gif

Hence, proved.

*Question 15:

If , for some  prove that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7062/Chapter%205_html_37a69f32.gif

is a constant independent of a and b.

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7062/Chapter%205_html_m7e6932d6.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7062/Chapter%205_html_m2884619c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7062/Chapter%205_html_m6c362b98.gif

Hence, proved.

Question 16:

If  with  prove that

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7065/Chapter%205_html_m44367968.gif

Then, equation (1) reduces to

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/Selection_014(11).png
UpP5F9WzHPL26bWL0BY2VypnqIta5x20FEH bRoCHtg5Y cNEjUNfQ11o2N4 ttTqFxNIP4zZXA91N2XcgaNJYkrL7dWNzOorq2vPRP2uAzD4SUY8kg 5G2wiZaO28 pO3hVELo
999QtoK PfrwdtJP1T1e1AU2EvOBJT 5W2 GUtTH26xBw2JuQhsmeicy1sVWnNEIhpLlvkW3o1n3DAG7Iy4uLvm1RTjmzi8 LGUYLgcnVEQa1IKCu6O8kIoEqp9QR6zlOKchOKw

Hence, proved.

*Question 17:5ySHiEhFxhA ooI2AQKLgXOGwX bhsekRsPDmPI9wquKQqLHGAi

If  and , find 

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7067/Chapter%205_html_2d91050f.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7067/Chapter%205_html_mbbd3588.gif

Question 18:

If show that  exists for all real x, and find it.

Answer:

It is known that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7070/Chapter%205_html_m4f621101.gif

Therefore, when x ≥ 0, 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7070/Chapter%205_html_m111c14a3.gif

In this case,  and hence, 

When x < 0, 

In this case,  and hence, 

Thus, for exists for all real x and is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7070/Chapter%205_html_534c3f5e.gif

Question 19:

Using mathematical induction prove that  for all positive integers n.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7072/Chapter%205_html_m265b040e.gif

For n = 1,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7072/Chapter%205_html_m2a97324c.gif

∴P(n) is true for n = 1

Let P(k) is true for some positive integer k.

That is,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7072/Chapter%205_html_19a5767d.gif

It has to be proved that P(k + 1) is also true.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7072/Chapter%205_html_m364fa7b2.gif

Thus, P(k + 1) is true whenever P (k) is true.

Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.

Hence, proved.

Question 20:

Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7073/Chapter%205_html_16c8c39e.gif

Differentiating both sides with respect to x, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7073/Chapter%205_html_3406b25.gif

Question 21:

Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer ?

Answer:

y=x           -∞<x≤1    2-x         1≤x≤∞https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/3(389).png It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.

*Question 22:

GORYUuNFihnQ6xVWkeBeXyfw YBPvAouehrtcbHqzy4oBZlMQyMxDxrYeZt 21ASMt7Uacv2Hp6Il32PCJnJQ2UlYZdfr PbvKdgim9PaS7 m6WvUquEXMD3vCF591IqjtlLa1I

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7074/Chapter%205_html_m25dad79b.gif

Thus,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7074/Chapter%205_html_5c68ffc9.gif

Question 23:

If show that 

Answer:

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7077/Chapter%205_html_m613dfb4.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/234/7077/Chapter%205_html_m41b41031.gif

Conclusion

NCERT solutions for Class 12 Math Chapter 5 “Continuity and Differentiability” are designed to provide a deeper understanding of the concepts of continuity and differentiability of functions. These concepts play an essential role in calculus and are fundamental to understand higher-level mathematics.
The chapter begins with the definition of continuity and its different types, such as pointwise continuity and uniform continuity. The chapter then proceeds to discuss differentiability and the relationship between continuity and differentiability.
NCERT solutions for Class 12 Math Chapter 5 provide detailed explanations of the concepts along with a variety of solved examples and exercises to reinforce the learning. Additionally, the solutions offer step-by-step guidance on how to approach different types of problems and help students to develop problem-solving skills.
Overall, NCERT solutions for Class 12 Math Chapter 5 are an essential resource for students who want to master the concepts of continuity and differentiability and develop their skills in calculus.

Topics to study in NCERT Class 12 Math Chapter 5

Section noTopics
5.1Introduction to Continuity and Differentiability
5.2Exponential and Logarithmic Functions
5.3Logarithmic Differentiation
5.4Derivatives of Functions in Parametric Forms
5.5Second order Derivatives
5.6Mean Value Theorem

Weightage of Math class 12 Chapter 5 in exams

ChaptersMarks
Continuity and Differentiability4 Marks

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Videos on Class 12 Mathematics Chapter 5

Related Links

NCERT Solution for Class XIIth Maths Chapter 6 Applications of DerivativesNCERT Solution for Class XIIth Maths Chapter 7 Integrals
NCERT Solution for Class XIIth Maths Chapter 8 Application of IntegralsNCERT Solution for Class XIIth Maths Chapter 9 Differential Equations
NCERT Solution for Class XIIth Maths Chapter 10 VectorNCERT Solution for Class XIIth Maths Chapter 11 Three Dimensional Geometry

FAQS on NCERT Solutions for Class 12 Math Chapter 5

What is continuity and differentiation?

One of the most essential subjects in mathematics is continuity and differentiability, which helps students grasp concepts like continuity at a point, continuity on an interval, derivative of functions, and many more. Continuity and Differentiability of functional parameters, on the other hand, are extremely difficult to achieve.

What are the four relationships between differentiability and continuity?

If a function has a derivative, it is differentiable. A function’s derivative can be thought of as its slope. Continuous functions and differentiability have the following relationship: all differentiable functions are continuous, but not all continuous functions are differentiable.

Is differentiation and differentiability same?

Differentiability relates to the presence of a derivative, whereas differentiation refers to the act of taking the derivative. As a result, we may state that any function can only be differentiated if it is differentiable.

How is continuity and differentiability related?

The relationship between continuity and differentiability is that all differentiable functions are also continuous, but not all continuous functions are.

Who discovered continuous function?

Bolzano discovered continuous function.

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