CONTINUITY AND DIFFERENTIABILITY
5.1.16 Mean Value Theorem (Lagrange)
Let f : [a, b] R be a continuous function on [a, b] and differentiable on (a, b). Then there exists at least one point c in (a, b) such that f ′ (c) = f b( ) f a( )
b a .
Geometrically, Mean Value Theorem states that there exists at least one point c in (a, b) such that the tangent at the point (c, f (c)) is parallel to the secant joining the points (a, f (a) and (b, f (b)).
5.2 Solved Examples Short Answer (S.A.)
Example 1 Find the value of the constant k so that the function f defined below is
continuous at x = 0, where ( ) 1 – cos 42 , 0
Solution It is given that the function f is continuous at x = 0. Therefore,lim0
x→ f (x) = f (0)
Example 2 Discuss the continuity of the function f(x) = sin x . cos x.
Solution Since sin x and cos x are continuous functions and product of two continuous function is a continuous function, therefore f(x) = sin x . cos x is a continuous function.
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Example 4 Show that the function f defined by sin ,1 0
x . Find the points of discontinuity of the composite function y = f [f(x)].
Solution We know that f (x) = 1 –1
x is discontinuous at x = 1 Now, for x 1,
CONTINUITY AND DIFFERENTIABILITY 93 which is discontinuous at x = 2.
Hence, the points of discontinuity are x = 1 and x = 2.
Example 6 Let f(x) = x x , for all x ∈ R. Discuss the derivability of f(x) at x = 0
Solution We may rewrite f as
2
Since the left hand derivative and right hand derivative both are equal, hence f is differentiable at x = 0.
Example 7 Differentiate tan x w.r.t. x
Solution Let y = tan x . Using chain rule, we have
Solution Given y = tan (x + y). differentiating both sides w.r.t. x, we have
94 MATHEMATICS which implies that –
–
CONTINUITY AND DIFFERENTIABILITY 95 Differentiating w.r.t. x, we get
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Differentiating both sides w.r.t. x, we get
2 2 Solution We have y = tanx + secx. Differentiating w.r.t. x, we get
dy
Now, differentiating again w.r.t. x, we get
2
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Solution When
2 < x < π, cosx < 0 so that |cos x| = – cos x, i.e., f (x) = – cos x f ′ (x) = sin x.
Hence, f ′ 3
4 = sin 3
4 = 1 2
Example 16 If f (x) = |cos x – sinx|, find f ′ 6 .
Solution When 0 < x <
4
π, cos x > sin x, so that cos x – sin x > 0, i.e., f (x) = cos x – sin x
f ′ (x) = – sin x – cos x Hence f ′ 6
= – sin 6 – cos
6 = 1
(1 3)
−2 + .
Example 17 Verify Rolle’s theorem for the function, f (x) = sin 2x in 0, 2 .
Solution Consider f (x) = sin 2x in 0, 2 . Note that:
(i) The function f is continuous in 0, 2 , as f is a sine function, which is always continuous.
(ii) f ′ (x) = 2cos 2x, exists in 0, 2 , hence f is derivable in 0, 2
⎛ π⎞
⎜ ⎟
⎝ ⎠. (iii) f (0) = sin0 = 0 and f 2
= sinπ = 0 ⇒ f (0) = f 2 .
Conditions of Rolle’s theorem are satisfied. Hence there exists at least one c ∈ 0, 2 such that f ′(c) = 0. Thus
2 cos 2c = 0 ⇒ 2c = 2 ⇒ c = 4.
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Example 18 Verify mean value theorem for the function f (x) = (x – 3) (x – 6) (x – 9) in [3, 5].
Solution (i) Function f is continuous in [3, 5] as product of polynomial functions is a polynomial, which is continuous.
(ii) f ′(x) = 3x2 – 36x + 99 exists in (3, 5) and hence derivable in (3, 5).
Thus conditions of mean value theorem are satisfied. Hence, there exists at least one c ∈ (3, 5) such that
Long Answer (L.A.)
Example 19 If f (x) = 2 cos 1
cos sin cos sin 2 cos 1
x
x x x x
x x x x x
x
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Example 20 Show that the function f given by
1
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≠ does not exist. Hence f is discontinuous at x = 0.
Example 21 Let
2
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Example 22 Examine the differentiability of the function f defined by
2 3, if 3 2
Solution The only doubtful points for differentiability of f (x) are x = – 2 and x = 0.
Differentiability at x = – 2.
Now L f ′ (–2) = 0 (–2 ) (–2)
Similarly, for differentiability at x = 0, we have
L (f ′(0)= 0 (0 ) (0)
which does not exist. Hence f is not differentiable at x = 0.
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Example 23 Differentiate tan-1
1 x2
x with respect to cos-1 2 1x x2 , where 1 ,1
x 2 .
Solution Let u = tan-1
1 x2
x and v = cos-1 2 1x x2 .
We want to find
du du dx dv dv
dx
Now u = tan-1
1 x2
x . Put x = sinθ. 4 2
π π
⎛ <θ< ⎞
⎜ ⎟
⎝ ⎠.
Then u = tan-1
1 sin2
sin = tan-1 (cot θ)
= tan-1 tan 2 2
⎧ ⎛π− θ⎞⎫= − θπ
⎨ ⎜⎝ ⎟⎠⎬
⎩ ⎭
sin–1
2 x
Hence 2
1 1 du
dx x .
Now v = cos–1 (2x 1 x2 )
= 2– sin–1 (2x 1 x2 )
= 2– sin–1 (2sinθ 1 sin2 ) sin (sin 2 )–1 2
− θ = −π θ
= 2 – sin–1 {sin (π – 2θ)} [since 2
π < 2 θ < π]
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Choose the correct answer from the given four options in each of the Examples 24 to 35.
Example 24 The function f (x) =
Solution (B) is the Correct answer.
Example 25 The function f (x) = [x], where [x] denotes the greatest integer function, is continuous at
(A) 4 (B) – 2
(C) 1 (D) 1.5
Solution (D) is the correct answer. The greatest integer function[x] is discontinuous at all integral values of x. Thus D is the correct answer.
Example 26 The number of points at which the function f (x) = 1
x–[ ]x is not continuous is
(A) 1 (B) 2
(C) 3 (D) none of these
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Solution (D) is the correct answer. As x – [x] = 0, when x is an integer so f (x) is discontinuous for all x ∈ Z.
Example 27 The function given by f (x) = tanx is discontinuous on the set
(A) n :n Z (B) 2n :n Z
(C) (2 1) :
n 2 n Z (D) :
2
n n Z
Solution C is the correct answer.
Example 28 Let f (x)= |cosx|. Then,
(A) f is everywhere differentiable.
(B) f is everywhere continuous but not differentiable at n = nπ, n Z. (C) f is everywhere continuous but not differentiable at x = (2n + 1)
2 π,
n∈ Z. (D) none of these.
Solution C is the correct answer.
Example 29 The function f (x) = |x| + |x – 1| is (A) continuous at x = 0 as well as at x = 1.
(B) continuous at x = 1 but not at x = 0.
(C) discontinuous at x = 0 as well as at x = 1.
(D) continuous at x = 0 but not at x = 1.
Solution Correct answer is A.
Example 30 The value of k which makes the function defined by sin ,1 if 0
( )
, if 0
f x x x
k x
, continuous at x = 0 is
(A) 8 (B) 1
(C) –1 (D) none of these
Solution(D) is the correct answer. Indeed
0
lim sin1
x→ xdoes not exist.
Example 31 The set of points where the functions f given by f (x) = |x – 3| cosx is differentiable is
CONTINUITY AND DIFFERENTIABILITY 105
(A) R (B) R – {3}
(C) (0, ∞) (D) none of these
Solution B is the correct answer.
Example 32 Differential coefficient of sec (tan–1x) w.r.t. x is
(A) 1 2 Solution (A) is the correct answer.
Example 33 If u = –1 2
Solution (D) is the correct answer.
Example 34 The value of c in Rolle’s Theorem for the function f (x) = ex sinx, [0, ]
Solution (D) is the correct answer.
Example 35 The value of c in Mean value theorem for the function f (x) = x (x – 2), Solution (A) is the correct answer.
Example 36 Match the following
COLUMN-I COLUMN-II
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(B) Every continuous function is differentiable (b) True (C) An example of a function which is continuous (c) 6
everywhere but not differentiable at exactly one point
(D) The identity function i.e. f (x) = x ∀ ∈x Ris a (d) False continuous function
Solution A → c, B → d, C → a, D → b Fill in the blanks in each of the Examples 37 to 41.
Example 37 The number of points at which the function f (x) = 1 log | |x is discontinuous is ________.
Solution The given function is discontinuous at x = 0, ± 1 and hence the number of points of discontinuity is 3.
Example 38 If
Example 39 The derivative of log10x w.r.t. x is ________.
Solution
(
10)
dxis equal to ______.
Solution 0.
Example 41 The deriative of sin x w.r.t. cos x is ________.
Solution – cot x
State whether the statements are True or False in each of the Exercises 42 to 46.
Example 42 For continuity, at x = a, each of xlim→a+ f x( )and lim ( )–
x a f x
→ is equal to f (a).
Solution True.
Example 43 y = |x – 1| is a continuous function.
Solution True.
Example 44 A continuous function can have some points where limit does not exist.
Solution False.
Example 45 |sinx| is a differentiable function for every value of x.
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Solution False.
Example 46 cos |x| is differentiable everywhere.
Solution True.
5.3 EXERCISE Short Answer (S.A.)
1. Examine the continuity of the function f (x) = x3 + 2x2 – 1 at x = 1
Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:
2. 2
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Find the value of k in each of the Exercises 11 to 14 so that the function f is continuous at the indicated point:
11.