Continuity and Differentiability
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Class XII Chapter 5 – Continuity and Differentiability Maths
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Exercise 5.1
Question 1:
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Prove that the function
is continuous at 
Answer
Therefore, f is continuous at x = 0
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Therefore, f is continuous at x = −3
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Therefore, f is continuous at x = 5
Question 2:
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Examine the continuity of the function
. Answer
Thus, f is continuous at x = 3
Page 1 of 144
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.
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It is also observed that,
Hence, f is continuous at every real number and therefore, it is a continuous function.
(b)
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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
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Therefore, f is continuous at all real numbers less than 5. Case II : c = 5
Then, 
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Therefore, f is continuous at x = 5 Case III: c > 5
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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 f (x) = xn
It is evident that f is defined at all positive integers, n, and its value at n is nn.
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Therefore, f is continuous at n, where n is a positive integer.
Question 5:
Is the function f defined by
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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.
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Therefore, f is continuous at x = 0 At x = 1,
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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.
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Therefore, f is continuous at x = 2
Question 6:
Find all points of discontinuity of f, where f is defined by
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Answer
 |
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)
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c = 2 Case (i) c < 2
Therefore, f is continuous at all points x, such that x < 2 Case (ii) c > 2
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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,
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The right hand limit of f at x = 2 is,
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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 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,
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The right hand limit of f at x = 3 is,
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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:
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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
 |
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,
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The right hand limit of f at x = 0 is,
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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
|
It is known that,
Therefore, the given function can be rewritten as
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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 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,
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The right hand limit of f at x = 1 is,
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Therefore, f is continuous at x = 1 Case III:
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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 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 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,
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The right hand limit of f at x = 1 is,
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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:
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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,
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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
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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:
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Therefore, f is continuous in the interval [0, 1). Case II:
The left hand limit of f at x = 1 is,
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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,
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The right hand limit of f at x = 3 is,
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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:
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Therefore, f is continuous at all points x, such that x < 0
Case II:
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The left hand limit of f at x = 0 is,
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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,
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The right hand limit of f at x = 1 is,
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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:
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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
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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,
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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,
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The right hand limit of f at x = 1 is,
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Therefore, f is continuous at x = 2 Case V:
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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
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is continuous at x = 3. Answer
If f is continuous at x = 3, then
Therefore, from (1), we obtain
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Therefore, the required relationship is given by, 
Question 18:
For what value of is
the function defined
by
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continuous at x = 0? What about continuity at x = 1? Answer
The given function is
If f is continuous at x = 0, then
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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
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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,
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The left hand limit of f at x = n is,
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The right hand limit of f at x = n is,
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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 = p? Answer
The given function is
It is evident that f is defined at x = p
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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
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 Î Z
 |
 |
Therefore, secant is continuous except at
Therefore, cotangent is continuous except at x = np, n Î Z
Question 23:
Find the points of discontinuity of f, where
 |
Answer
 |
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
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
|
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 continuous at 
, if f is defined
at and if the value of
the f
at equals the limit of f at .
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 .
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
Question 29:
Find the values of k so that the function f is continuous at the indicated point.
 |
Answer
 |
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 
Question 30:
Find the values of a and b such that the function defined by
 |
is a continuous function.
Answer
|
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 
 |
It has to be first proved that 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 
 |
It has to be proved
first that 
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 5.2
Question 1:
Differentiate the functions with respect to x.
 |
Answer
 |
Thus, f is a composite of two functions.
 |
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
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∴h is a composite function of two functions, p and q.
Put y = p (x) = cx + d
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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
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Alternate method
 |
Question 9:
Prove that the function f given by
is notdifferentiable 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
 |
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
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 5.3
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
The given relationship is 
 |
Differentiating this relationship with respect to x, we obtain
 |
The function, 
, is of the form of 
. Therefore, by quotient rule, we obtain
Also,
 |
From (1), (2), and (3), we obtain
 |
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 2:
Show that the function given by f(x) = e2x is strictly increasing on R. Answer
 |
Let
be any two numbers
in R. Then, we have:
Hence, f is strictly increasing on R.
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
 |
, x >
1
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 17:
Differentiate 
in three ways mentioned below
(i) By using product rule.
(ii) By expanding the product to obtain a single polynomial. (iii By logarithmic differentiation.
Do they all give the same answer?
Answer (i)
(ii)
(iii)
Taking logarithm on both the sides, we obtain
 |
Differentiating both sides with respect to x, we obtain
 |
From the above three
observations, it can be concluded that all the results of are same.
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
.
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
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) 
(ii) 
(iii) 
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)
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 
. (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 Rolle’s Theorem.
Hence, Rolle’s 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).
∴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 .
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
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 Solutions
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:
 |
Answer
,
for some fixed 
and 
 |
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:
, for 
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 
in equation
(1), we obtain
 |
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 22:
If , prove that
Answer
Thus,
Question 23:
If , show that 
Answer
It is given that,
Class XII Chapter 5 – Continuity and Differentiability Maths
 |
Page 144 of 144