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:
Therefore, f is continuous at x = 0
Therefore, f is continuous at x = −3
Therefore, f is continuous at x = 5
Question 2:
Examine the continuity of the function at
Answer:
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,
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
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
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, f (c) = 5 − c
Therefore, f is continuous at all real numbers less than 5.
Case II : c = 5
Then,
Therefore, f is continuous at x = 5
Case III: c > 5
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.
Therefore, f is continuous at n, where n is a positive integer.
Question 5:
Is the function f defined by
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.
Therefore, f is continuous at x = 0
At x = 1,
f is defined at 1 and its value at 1 is 1.
The left hand limit of f at x = 1 is,
The right hand limit of f at x = 1 is,
Therefore, f is not continuous at x = 1
At x = 2,
f is defined at 2 and its value at 2 is 5.
Therefore, f is continuous at x = 2
Question 6:
Find all points of discontinuity of f, where f is defined by
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
Therefore, f is continuous at all points x, such that x < 2
Case (ii) c > 2
Therefore, f is continuous at all points x, such that x > 2
Case (iii) c = 2
Then, the left hand limit of f at x = 2 is,
The right hand limit of f at x = 2 is,
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
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:
Therefore, f is continuous at all points x, such that x < −3
Case II:
Therefore, f is continuous at x = −3
Case III:
Therefore, f is continuous in (−3, 3).
Case IV:
If c = 3, then the left hand limit of f at x = 3 is,
The right hand limit of f at x = 3 is,
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:
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
Answer:
The given function f is
It is known that,
Therefore, the given function can be rewritten as
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:
Therefore, f is continuous at all points x < 0
Case II:
If c = 0, then the left hand limit of f at x = 0 is,
The right hand limit of f at x = 0 is,
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:
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
Answer:
The given function f is
It is known that,
Therefore, the given function can be rewritten as
Let c be any real number. Then,
Also,
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
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:
Therefore, f is continuous at all points x, such that x < 1
Case II:
The left hand limit of f at x = 1 is,
The right hand limit of f at x = 1 is,
Therefore, f is continuous at x = 1
Case III:
Therefore, f is continuous at all points x, such that x > 1
Hence, the given function f has no point of discontinuity.
Question 11:
Find all points of discontinuity of f, where f is defined by
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:
Therefore, f is continuous at all points x, such that x < 2
Case II:
Therefore, f is continuous at x = 2
Case III:
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, f has no point of discontinuity.
Question 12:
Find all points of discontinuity of f, where f is defined by
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:
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,
The right hand limit of f at x = 1 is,
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:
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
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:
Therefore, f is continuous at all points x, such that x < 1
Case II:
The left hand limit of f at x = 1 is,
The right hand limit of f at x = 1 is,
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:
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
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:
Therefore, f is continuous in the interval [0, 1).
Case II:
The left hand limit of f at x = 1 is,
The right hand limit of f at x = 1 is,
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:
Therefore, f is continuous at all points of the interval (1, 3).
Case IV:
The left hand limit of f at x = 3 is,
The right hand limit of f at x = 3 is,
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:
Therefore, f is continuous at all points of the interval (3, 10].
Hence, f is not continuous at x = 1 and x = 3
Question 15:
Discuss the continuity of the function f, where f is defined by
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:
Therefore, f is continuous at all points x, such that x < 0
Case II:
The left hand limit of f at x = 0 is,
The right hand limit of f at x = 0 is,
Therefore, f is continuous at x = 0
Case III:
Therefore, f is continuous at all points of the interval (0, 1).
Case IV:
The left hand limit of f at x = 1 is,
The right hand limit of f at x = 1 is,
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:
Therefore, f is continuous at all points x, such that x > 1
Hence, f is not continuous only at x = 1
Question 16:
Discuss the continuity of the function f, where f is defined by
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:
Therefore, f is continuous at all points x, such that x < −1
Case II:
The left hand limit of f at x = −1 is,
The right hand limit of f at x = −1 is,
Therefore, f is continuous at x = −1
Case III:
Therefore, f is continuous at all points of the interval (−1, 1).
Case IV:
The left hand limit of f at x = 1 is,
The right hand limit of f at x = 1 is,
Therefore, f is continuous at x = 2
Case V:
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
is continuous at x = 3.
Answer:
The given function f is
If f is continuous at x = 3, then
Therefore, from (1), we obtain
Therefore, the required relationship is given by,
*Question 18:
For what value of 
continuous at x = 0? What about continuity at x = 1?
Answer:
The given function f is
If f is continuous at x = 0, then
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
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,
The left hand limit of f at x = n is,
The right hand limit of f at x = n is,
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 x =π?
Answer:
The given function is
It is evident that f is defined at x =π.
Therefore, the given function f is continuous at x = 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 g and h are two continuous functions, then and g.h are also continuous.
It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.
Let g (x) = sin x
It is evident that g (x) = sin x is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
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 c be a real number. Put x = c + h
If x → c, then h → 0
h (c) = cos c
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 g and h are two continuous functions, then
It has to be proved first that g (x) = sin x and h (x) = cos x are continuous functions.
Let g (x) = sin x
It is evident that g (x) = sin x is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
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 c be a real number. Put x = c + h
If x → c, then h → 0
h (c) = cos c
Therefore, h (x) = cos x is continuous function.
It can be concluded that,
Therefore, cosecant is continuous except at x = np, n ε Z
Therefore, secant is continuous except at
Therefore, cotangent is continuous except at x = np, n ε Z
*Question 23:
Find the points of discontinuity of f, where
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:
Therefore, f is continuous at all points x, such that x < 0
Case II:
Therefore, f is continuous at all points x, such that x > 0
Case III:
The left hand limit of f at x = 0 is,
The right hand limit of f at x = 0 is,
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
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:
Therefore, f is continuous at all points x ≠ 0
Case II:
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
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:
Therefore, f is continuous at all points x, such that x ≠ 0
Case II:
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 k so that the function f is continuous at the indicated point.
Answer:
The given function f is
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 f is defined at and
Therefore, the required value of k is 6.
Question 27:
Find the values of k so that the function f is continuous at the indicated point.
Answer:
The given function is
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 f is defined at x = 2 and
Therefore, the required value of k is
Question 28:
Find the values of k so that the function f is continuous at the indicated point.
Answer:
The given function is
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 f is defined at x = p and
Therefore, the required value of k is
Question 29:
Find the values of k so that the function f is continuous at the indicated point.
Answer:
The given function f is
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 f is defined at x = 5 and
Therefore, the required value of k is
Question 30:
Find the values of a and b such that the function defined by
is a continuous function.
Answer:
The given function f is
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 x = 2 and x = 10
Since f is continuous at x = 2, we obtain
Since f is continuous at x = 10, we obtain
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 f (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
It has to be first proved that g (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
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
Therefore, h is a continuous function.
It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o 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
It has to be first proved that and are continuous functions.
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
Therefore, g is continuous at all points x, such that x < 0
Case II:
Therefore, g is continuous at all points x, such that x > 0
Case III:
Therefore, g is continuous at x = 0
From the above three observations, it can be concluded that g is continuous at all points.
h (x) = cos x
It is evident that h (x) = cos x is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
h (c) = cos c
Therefore, h (x) = cos x is a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, is a continuous function.
Question 33:
Examine that is a continuous function.
Answer:
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
It has to be proved first that and are continuous functions.
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
Therefore, g is continuous at all points x, such that x < 0
Case II:
Therefore, g is continuous at all points x, such that x > 0
Case III:
Therefore, g is continuous at x = 0
From the above three observations, it can be concluded that g is continuous at all points.
h (x) = sin x
It is evident that h (x) = sin x is defined for every real number.
Let c be a real number. Put x = c + k
If x → c, then k → 0
h (c) = sin c
Therefore, h is a continuous function.
It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Therefore, is a continuous function.
Question 34:
Find all the points of discontinuity of f defined by
Answer:
The given function is
The two functions, g and h, are defined as
Then, f = g − h
The continuity of g and h is examined first.
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
Therefore, g is continuous at all points x, such that x < 0
Case II:
Therefore, g is continuous at all points x, such that x > 0
Case III:
Therefore, g is continuous at x = 0
From the above three observations, it can be concluded that g is continuous at all points.
Clearly, h is defined for every real number.
Let c be a real number.
Case I:
Therefore, h is continuous at all points x, such that x < −1
Case II:
Therefore, h is continuous at all points x, such that x > −1
Case III:
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, f = g − h is also a continuous function.
Therefore, f has no point of discontinuity.
Exercise-3.2
Question 1:
Differentiate the functions with respect to x.
Answer:
Let fx=sin x2+5 , u(x)=x2+5, and v(t)=sint
Alternate method
Question 2:
Differentiate the functions with respect to x.
Answer:
Thus, f is a composite function of two functions.
Put t = u (x) = sin x
By chain rule,
Alternate method
Question 3:
Differentiate the functions with respect to x.
Answer:
Thus, f is a composite function of two functions, u and v.
Put t = u (x) = ax + b
Hence, by chain rule, we obtain
Alternate method
Question 4:
Differentiate the functions with respect to x.
Answer:
Thus, f is a composite function of three functions, u, v, and w.
Hence, by chain rule, we obtain
Alternate method
Question 5:
Differentiate the functions with respect to x.
Answer:
The given function is where g (x) = sin (ax + b) and
h (x) = cos (cx + d)
∴ g is a composite function of two functions, u and v.
Therefore, by chain rule, we obtain
∴h is a composite function of two functions, p and q.
Put y = p (x) = cx + d
Therefore, by chain rule, we obtain
Question 6:
Differentiate the functions with respect to x.
Answer:
The given function is
Question 7:
Differentiate the functions with respect to x.
Answer:
Question 8:
Differentiate the functions with respect to x.
Answer:
Clearly, f is a composite function of two functions, u and v, such that
By using chain rule, we obtain
Alternate method
*Question 9:
Prove that the function f 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
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
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
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
Question 2:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Question 3:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Using chain rule, we obtain
and
From (1) and (2), we obtain
Question 4:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Question 5:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Question 6:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
*Question 7:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Using chain rule, we obtain
From (1), (2), and (3), we obtain
Question 8:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Question 9:
Find
Answer:
We have
Put
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
Answer:
The given relationship is
It is known that,
Comparing equations (1) and (2), we obtain
Differentiating this relationship with respect to x, we obtain
Question 11:
Find
Answer:
The given relationship is,
On comparing L.H.S. and R.H.S. of the above relationship, we obtain
Differentiating this relationship with respect to x, we obtain
*Question 12:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Using chain rule, we obtain
From (1), (2), and (3), we obtain
Alternate method
⇒
Differentiating this relationship with respect to x, we obtain
Question 13:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
*Question 14:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Question 15:
Find
Answer:
The given relationship is
Differentiating this relationship with respect to x, we obtain
Exercise-5.4
Question 1:
Differentiate the following w.r.t. x:
Answer:
Let
By using the quotient rule, we obtain
Question 2:
Differentiate the following w.r.t. x:
Answer:
Let
By using the chain rule, we obtain
Question 3:
Differentiate the following w.r.t. x:
Answer:
Let
By using the chain rule, we obtain
*Question 4:
Differentiate the following w.r.t. x:
Answer:
Let
By using the chain rule, we obtain
Question 5:
Differentiate the following w.r.t. x:
Answer:
Let
By using the chain rule, we obtain
*Question 6:
Differentiate the following w.r.t. x:
Answer:
Question 7:
Differentiate the following w.r.t. x:
Answer:
Let
Then,
By differentiating this relationship with respect to x, we obtain
Question 8:
Differentiate the following w.r.t. x:
Answer:
Let
By using the chain rule, we obtain
*Question 9:
Differentiate the following w.r.t. x:
Answer:
Let
By using the quotient rule, we obtain
*Question 10:
Differentiate the following w.r.t. x:
Answer:
Let
By using the chain rule, we obtain
Exercise-5.5
Question 1:
Differentiate the function with respect to x.
Answer:
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 2:
Differentiate the function with respect to x.
Answer:
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 3:
Differentiate the function with respect to x.
Answer:
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 4:
Differentiate the function with respect to x.
Answer:
u = xx
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
v = 2sin x
Taking logarithm on both the sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
Question 5:
Differentiate the function with respect to x.
Answer:
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 6:
Differentiate the function with respect to x.
Answer:
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
Therefore, from (1), (2), and (3), we obtain
Question 7:
Differentiate the function with respect to x.
Answer:
u = (log x)x
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
Therefore, from (1), (2), and (3), we obtain
*Question 8:
Differentiate the function with respect to x.
Answer:
Differentiating both sides with respect to x, we obtain
Therefore, from (1), (2), and (3), we obtain
*Question 9:
Differentiate the function with respect to x.
Answer:
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
From (1), (2), and (3), we obtain
*Question 10:
Differentiate the function with respect to x.
Answer:
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
From (1), (2), and (3), we obtain
*Question 11:
Differentiate the function with respect to x.
Answer:
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
From (1), (2), and (3), we obtain
Question 12:
Find of function.
Answer:
The given function is
Let xy = u and yx = v
Then, the function becomes u + v = 1
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
From (1), (2), and (3), we obtain
*Question 13:
Find of function.
Answer:
The given function is
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 14:
Find of function.
Answer:
The given function is
Taking logarithm on both the sides, we obtain
Differentiating both sides, we obtain
Question 15:
Find of function.
Answer:
The given function is
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
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
Differentiating both sides with respect to x, we obtain
Question 18:
If u, v and w are functions of x, then show that
in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Answer:
Let
By applying product rule, we obtain
By taking logarithm on both sides of the equation , we obtain
Differentiating both sides with respect to x, we obtain
Exercise-5.6
Question 1:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are
Question 2:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
x = a cos θ, y = b cos θ
Answer:
The given equations are x = a cos θ and y = b cos θ
*Question 3:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
x = sin t, y = cos 2t
Answer:
The given equations are x = sin t and y = cos 2t
Question 4:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are and
Question 5:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are
Question 6:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are
Question 7:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are
Question 8:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are
Question 9:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are
*Question 10:
If x and y are connected parametrically by the equation, without eliminating the parameter, find .
Answer:
The given equations are
*Question 11:
If
Answer:
The given equations are
Hence, proved.
Exercise-5.7
Question 1:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 2:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 3:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 4:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 5:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 6:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 7:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 8:
Find the second order derivatives of the function.
Answer:
Let
Then,
*Question 9:
Find the second order derivatives of the function.
Answer:
Let
Then,
Question 10:
Find the second order derivatives of the function.
Answer:
Let
Then,
*Question 11:
If , prove that
Answer:
It is given that,
Then,
Hence, proved.
Question 12:
If find in terms of y alone.
Answer:
It is given that,
Then,
Question 13:
If show that
Answer:
It is given that,
Then,
Hence, proved.
*Question 14:
If show that
Answer:
It is given that,
Then,
Hence, proved.
Question 15:
If , show that
Answer:
It is given that,
Then,
Hence, proved.
Question 16:
If show that
Answer:
The given relationship is
Taking logarithm on both the sides, we obtain
Differentiating this relationship with respect to x, we obtain
Hence, proved.
Question 17:
If , show that
Answer:
The given relationship is
Then,
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).
∴ 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
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 [a, b]
(b) f is differentiable on (a, b)
(c) f (a) = f (b)
then, there exists some c ∈ (a, b) 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 x = 5 and x = 9
⇒ f (x) is not continuous in [5, 9].
The differentiability of f in (5, 9) is checked as follows.
Let n be an integer such that n ∈ (5, 9).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f 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 x = −2 and x = 2
⇒ f (x) is not continuous in [−2, 2].
The differentiability of f in (−2, 2) is checked as follows.
Let n be an integer such that n ∈ (−2, 2).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f 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).
∴f (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) f is differentiable on (−5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that
It is also given that does not vanish anywhere.
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.
Mean Value Theorem states that there is a point c ∈ (1, 4) such that
Hence, Mean Value Theorem is verified for the given function.
*Question 5:
Verify Mean Value Theorem, if in the interval [a, b], where a = 1 and b = 3. Find all for which
Answer:
The given function f is
f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x2 − 10x − 3.
Mean Value Theorem states that there exist a point c ∈ (1, 3) such that
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 [a, b]
(b) f is differentiable on (a, b)
then, there exists some c ∈ (a, b) such that
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 x = 5 and x = 9
⇒ f (x) is not continuous in [5, 9].
The differentiability of f in (5, 9) is checked as follows.
Let n be an integer such that n ∈ (5, 9).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f 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
(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 x = −2 and x = 2
⇒ f (x) is not continuous in [−2, 2].
The differentiability of f in (−2, 2) is checked as follows.
Let n be an integer such that n ∈ (−2, 2).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f 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
(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.
Miscellaneous exercise
Differentiate wrt x the function in question 1 to 11
Question 1:
Answer:
Using chain rule, we obtain
Question 2:
Answer:
Question 3:
Answer:
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 4:
Answer:
Using chain rule, we obtain
Question 5:
Answer:
*Question 6:
Answer:
Therefore, equation (1) becomes
*Question 7:
Answer:
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 8:
for some constant a and b.
Answer:
By using chain rule, we obtain
Question 9:
Answer:
Taking logarithm on both the sides, we obtain
Differentiating both sides with respect to x, we obtain
Question 10:
for some fixed and
Answer:
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
s = aa
Since a is constant, aa is also a constant.
∴
From (1), (2), (3), (4), and (5), we obtain
Question 11:
Answer:
Differentiating both sides with respect to x, we obtain
Differentiating with respect to x, we obtain
Also,
Differentiating both sides with respect to x, we obtain
Substituting the expressions of 
Question 12:
Find if
Answer:
*Question 13:
Find if
Answer:
*Question 14:
If for, −1 < x <1, prove that
Answer:
It is given that,
Differentiating both sides with respect to x, we obtain
Hence, proved.
*Question 15:
If , for some prove that
is a constant independent of a and b.
Answer:
It is given that,
Differentiating both sides with respect to x, we obtain
Hence, proved.
Question 16:
If with prove that
Answer:
Then, equation (1) reduces to
Hence, proved.
*Question 17:
If and , find
Answer:
Question 18:
If show that exists for all real x, and find it.
Answer:
It is known that,
Therefore, when x ≥ 0,
In this case, and hence,
When x < 0,
In this case, and hence,
Thus, for exists for all real x and is given by,
Question 19:
Using mathematical induction prove that for all positive integers n.
Answer:
For n = 1,
∴P(n) is true for n = 1
Let P(k) is true for some positive integer k.
That is,
It has to be proved that P(k + 1) is also true.
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:
Differentiating both sides with respect to x, we obtain
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≤∞
*Question 22:
Answer:
Thus,
Question 23:
If show that
Answer:
It is given that,
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 no | Topics |
| 5.1 | Introduction to Continuity and Differentiability |
| 5.2 | Exponential and Logarithmic Functions |
| 5.3 | Logarithmic Differentiation |
| 5.4 | Derivatives of Functions in Parametric Forms |
| 5.5 | Second order Derivatives |
| 5.6 | Mean Value Theorem |
Weightage of Math class 12 Chapter 5 in exams
| Chapters | Marks |
| Continuity and Differentiability | 4 Marks |
Why do students prefer SWC’s NCERT Solutions for math class 12?
One of the top IIT JEE coaching institutes is Swastik Classes. Shobhit Bhaiya and Alok Bhaiya, pioneering mentors of IIT JEE Coaching Classes, started Swastik Classes in Anand Vihar. Over the last 15 years, they have educated and sent over 2000+ students to IITs and 5000+ students to different famous universities such as BITS, NITs, DTU, and NSIT. When it comes to coaching programmes for IIT JEE, Swastik Classes is the top IIT JEE Coaching in Delhi, favoured by students from all over India.
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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.
