The availability of step-by-step OP Malhotra Maths Class 12 Solutions Chapter 10 Mean Value Theorems Ex 10(a) can make challenging problems more manageable.
S Chand Class 12 ICSE Maths Solutions Chapter 10 Mean Value Theorems Ex 10(a)
Question 1.
f(x) = x² – x – 6 on [-2, 3].
Solution:
Given f(x) = x² – x – 6 on [-2, 3]
Since f(x) be poly nomial in x and hence continuous and differentiable on R.
Thus f(x) be continuous on [- 2, 3]
and f(x) be differentiable on (- 2, 3).
& f(-2) = (-2)² + 2 – 6 = 0;
f(3) = 3² – 3 – 6 = 0
∴ f(-2) = f(3)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real number c ∈ (-2, 3)
s.t. f(c) = 0
Now, f'(x) = 2x – 1 ⇒ f'(c) = 2c – 1
So f'(c) = 0 ⇒ 2c – 1 = 0
⇒ c = \(\frac { 1 }{ 2 }\) ∈ (- 2, 3)
Hence Rolle’s theorem is verified
Question 2.
f(x) = x²-6x + 5, in the interval [1, 5],
Solution:
Given f(x) = x² – 6x + 5 in [1, 5]
∴ f’ (x) = 2x – 6
which exists for all x ∈ R
Thus f is derivable on (1, 5), Since every derivable function is continuous
f(x) is continuous on [1, 5].
Now f(1) = 1 – 6 + 5 = 0;
f(5) = 25 – 30 + 5 = 0
∴ f(1) = f(5)
So all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real no c ∈ (1, 5)
s.t f(c) = 0 ⇒ 2c – 6 = 0 ⇒ c = 3 ∈ (1, 5)
Thus Rolle’s theorem is verified.
Question 3.
f(x) = x² -5x + 6, 2 ≤ x ≤ 3.
Solution:
Given f(x) = x² – 5x + 6, 2 ≤ x ≤ 3
Since f(x) be polynomial in x and hence continuous and differentiable for all x ∈ R.
Thus f(x) is continuous on [2, 3]
and diffemtiable in (2, 3)
Now, f(2) = 4 – 10 + 6 = 0;
f(3) = 9 – 15 + 6 = 0
∴ f(2) = f(3)
Thus all the three conditions of Roll’s theorem are satisfied So ∃ atleast one real numbers c ∈ (2, 3)
s.t. f'(c) = 0
Now f'(x) = 2x – 5 ∴ f'(c) = 0
⇒ 2c – 5 = 0 ⇒ c = \(\frac { 5 }{ 2 }\) ∈(2, 3)
Thus, Rolle’s theorem is verified.
Question 4.
y = 16 – x², x ∈ [-1, 1].
Solution:
Given y = f(x) = 16 – x², x ∈ [-1, 1]
∴ f'(x) = – 2x which exists eveiywhere on R.
Thus f(x) is continuous on [-1, 1]
Si, f(x) is differentiable on (-1, 1)
Since f(x) be polynomial in x and every polynomial function is differentiable.
Also, f(-1) = f(1)
Thus, all the three conditions of Rolle’s
theorem are satisfied So ∃ atleast one real no. c ∈ (-1, 1)
s.t. f(c) = 0 ⇒ – 2c = 0
⇒ c = 0 ∈ (-1, 1)
Thus, Rolle’s theorem is verified.
Question 5.
f(x) = x(x – 3)² in the interval [0, 3].
Solution:
Given f(x) = x (x – 3)² in [0, 3]
Since f(x) is polynomial in x hence continuous in [0, 3]
Also, f'(x) = (x – 3)² + 2x (x – 3)
= (x – 3) (3x – 3) = 3x² – 12x + 9
which exists for all x ∈ R
Thus f(x) is derivable on (0, 3).
Now f(0) = 0 = f(3)
Now, all the three conditions of Rolle’s
theorem are satisfied so ∃, atleast one real number c ∈ (0, 3)
s.t. f'(c) = 0 ⇒ 3c² – 12c + 9 = 0
⇒ 3(c² – 4c + 3) = 0
⇒ (c – 1) (c -3) = 0
⇒ c = 1, 3
but c = 3 ∉ (0, 3) and c = 1 ∈(0, 3)
Thus, Rolle’s theorem is verified.
Question 6.
f(x) = x(x – 3)² on [2, 4].
Solution:
Given f(x) = (x – 2) (x – 3) (x – 4)
= (x – 2) (x² – 7x + 12)
∴ f(x) = x³ – 9x² + 26x – 24
Since f(x) is polynomial in x and hence continuous on [2, 4]
Now f'(x) = 3x² – 18x + 26 which exists ∀x ∈ R
∴ f(x) be derivable on (2, 4)
Now f(2) = f(4) = 2
Thus, all the three conditions of Rolle’s theorem are satisfied ∃ so atleast one real no. c ∈ (2, 4)
s.t. f(c) = 0 ⇒ 3c² – 18c + 26 = 0
∴ c = \(\frac{18 \pm \sqrt{324-312}}{6}=\frac{18 \pm 2 \sqrt{3}}{6}\)
= 3 ± \(\frac{1}{\sqrt{3}} \in(2,4)\)
Thus Rolle’s theorem is verified.
Question 7.
f(x) = x³ – 7x² + 16x – 12 on [2, 3],
Solution:
Given f(x) = x³ – 7x² + 16x – 12 on [2, 3]
Since f(x) be polynomial in x and hence continuous in [2, 3]
Also, f'(x) = 3x² – 14x + 16 which exists for all x ∈ R
∴ f(x) is differentiable on (2, 3)
Now f(2) = 8 – 28 + 32 – 12 = 0;
f(3) = 27 – 63 + 48 – 12 = 0
∴ f(2) = f(3)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one rreal number c ∈ (2, 3)
s.t. f ‘(c) = 0 ⇒ 3c² – 14c + 16 = 0
∴ c = \(\frac{14 \pm \sqrt{196-192}}{6}\)
= \(\frac{14 \pm 2}{6}\) = \(\frac { 8 }{ 3 }\), 2
Now c = \(\frac { 8 }{ 3 }\) ∈ (2, 3) [∵ c = 2 ∉ (2, 3)]
Therefore Rolle’s theorem is verified.
Question 8.
f(x) = 4sinx, x ∈ [0, π]
Solution:
Given f(x) n since, x ∈ [0, π]
Since stine function is continous and differentiable energy where in its domain.
∴ f(x) is continous on [0, π] and derivable on (0, π)
Now f(0) = 4 x 0 = 0;
f(x) = sin π = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem are satisfied so ∃ one real no. C∈(0, π)
s.t. f'(c) = 0 4 cosx = 0
⇒ cos x = 0 ⇒ x = \(\frac { π }{ 2 }\) ∈ (0, π)
Thus Rolle’s theorem is neasified & c = \(\frac { π }{ 2 }\).
Question 9.
f(x) = sin2x, x ∈ [0, \(\frac { π }{ 2 }\) ].
Solution:
Given f (x) = sin 2x in [0, \(\frac { π }{ 2 }\) ]
Since sine function is continuous and derivable everywhere
∴ f(x) is continuous in [0, \(\frac { π }{ 2 }\) ] and derivable in ([0, \(\frac { π }{ 2 }\)).
Also, f (0) = 0 = f\(\frac { π }{ 2 }\)
∴ All the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no. c ∈(0, \(\frac { π }{ 2 }\)) s.t. f'(c) = 0
i.e. 2 cos 2c = 0 ⇒ cos 2c = 0
⇒ 2c = \(\frac { π }{ 2 }\) ⇒ c = \(\frac { π }{ 4 }\) ∈ (0, \(\frac { π }{ 2 }\))
Hence Rolle’s theorem is verified and c = \(\frac { π }{ 4 }\).
Question 10.
f(x) = sin² x, 0 ≤ x ≤ π.
Solution:
Given f(x) = sin²x; 0 ≤ x ≤ π
since sine function and cosine function both are continuous & differentiable everywhere in its domain.
Also, f'(x) = 2 sin x cos x = sin 2x
which exists for all x ∈ [0, π]
Thus f(x) is continuous on [0, π]
& f(x) is differentiable on (0, π)
Now f(0) = sin² 0 = 0; f(π) sin²π = 0
∴ f(0) = f(π)
Thus, all the theree conditions of Rolle’s theorem are satisfied so ∃ atleast one real so c ∈ (0, π) s.t. f'(c) = 0
⇒ sin 2c = 0
⇒ x = 0, π, 2π ⇒ c = 0, \(\frac { π }{ 2 }\), π
since c = 0, π ∉ (0, π)
but c = \(\frac { π }{ 2 }\) (0, π)
Thus Rolle’s theorem is verified.
Question 11.
f(x) = excosx, x ∈ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Solution:
Given f(x) = excosx, x ∈ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Since cosine and exponential function are continuous everywhere.
∴ f (x) is continuous in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and derivable in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Also, f (-\(\frac { π }{ 2 }\)) = 0 = f(\(\frac { π }{ 2 }\))
Therefore all the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast are real number c ∈ (0, \(\frac { π }{ 2 }\)) s.t. f'(c) = 0 i.e. ec (cos c – sin c) = 0
⇒ cos c – sin c = 0 [∵ ec ≠ 0]
⇒ tan c = 1 ⇒ c = \(\frac { π }{ 4 }\) ∈ (-\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
∴ Rolle’s theorem is verified.
Question 12.
f(x) = \(\frac{\sin x}{e^x}\), x ∈ [0, π].
Solution:
Given f (x) = \(\frac{\sin x}{e^x}\)
since sin x and ex are continuous and differentiable everywhere.
∴ quotient function f (x) is also continuous and differentiable everywhere.
∴ f (x) is continuous on [0, π] and differentiable on (0, π).
Now f(0) = \(\frac{0}{e^0}=\frac{0}{1}\) = 0;
f(π) = \(\frac{\sin \pi}{e^\pi}\) = 0
∴ f(0) = f(π)
Hence, all the three conditions of Rolle’s theorem are satisfied. Now we want to show that ∃ atleast one real number
c ∈ (0, π) s.t. f’ (c) = 0
we have f (x) = \(\frac{\sin x}{e^x}\)
⇒ f'(x) = \(\frac{e^x \cos x-\sin x e^x}{e^{2 x}}\)
∴ f’ (c) = 0 ⇒ ec (cos c – sin c) = 0
⇒ cos c – sin c = 0 [∵ ec > 0]
⇒ tan c = 1 ⇒ c = \(\frac { π }{ 4 }\) ∈ (0, π)
such that f’ (c) = 0.
Hence Rolle’s theorem verified.
Question 13.
f(x) = x(x + 3)e-x/2 defined in the interval [-3, 0].
Solution:
Given f(x) = x (x + 3) e-x/2 on [-3, 0] Since polynomial function x (x + 3) & exponential function are differential everywhere and product of two differential function is differentiable.
Thus f(x) is continuous on [- 3, 0] and differentiable on (-3, 0)
Now f(- 3) = 0 = f(0)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real no. c∈(-3, 0) s.t. f'(c) = 0
Now f'(x) =
Thus Rolle’s theorem is verified.
Question 14.
f(x) = \(\sqrt{4-x^2}\) on [- 2, 2].
Solution:
Given f (x) = \(\sqrt{4-x^2}\) …(1)
Clearly Df = [- 2, 2],
So f(x) is clearly continuous in its domain
i.e. [- 2, 2]
Also, f’ (x) = \(\frac{1}{2 \sqrt{4-x^2}}\) (- 2x)
= \(\frac{-x}{\sqrt{4-x^2}}\)
Clearly f’ (x) exists in (- 2, 2).
So f(x) is clearly derivable in (- 2, 2).
Also f(- 2) = 0 = f(2)
So all the three conditions of Rolle’s theorem are satisfied. So ∃ atleast one real number c ∈(- 2, 2) s.t f’ (c) = 0
Now, f’ (c) = 0 ⇒ \(\frac{-c}{\sqrt{4-c^2}}\) = 0
⇒ c = 0 ∈ (- 2, 2)
So there exists a real number c ∈ (- 2, 2)
s.t f’ (c) = 0
Hence Rolle’s Theorem verified and c = 0
Question 15.
If Rolle’s theorem holds for the function
f(x) = x³ + bx² + ax + 5 on [1, 3] with c = \(\left(2+\frac{1}{\sqrt{3}}\right)\), find the values of a and b.
Solution:
Given f(x) = x³ + bx² + ax + 5
∴ f'(x) = 3x² + 2bx + a
since Rolle’s theorem hold for f(x) on [1, 3]
∴ f(x) = x³
⇒ 1 + b + a + 5 = 27 + 9b + 3a + 5
⇒ b + a + 6 = 9b + 3a + 32
⇒ 2a + 8b = – 26 ⇒ a + 4b = – 13 …(1)
Now f'(c) = 0 ⇒ 3c² + 2bc + a = 0
Question 16.
Apply Rolle’s theorem to find point (or points) on the given curve where the tangent is parallel to the x-axi,
(i) y = x² in [-2, 2]
(ii) y = 16 – x², x ∈ [-1, 1].
Solution:
(i) Given y = x² in [- 2, 2]
Since f (x) is polynomial in x
∴ it is continuous and derivable everywhere
∴ f(x) is continuous in [- 2, 2] and derivable in (-2, 2).
Also f (- 2) = 4 = f( 2)
Hence all the three conditions of Rolle’s theorem are satisfied
∴ ∃ atleast one real no. c ∈ (-2, 2) s.t. f'(c) = 0 i.e. c = 0 ∈ (-2, 2). When x = 0, y = 0² = 0
∴ (0, 0) be the required point at which tangent is || to x-axis.
[By Geomatrical interpretation of Rolle’s Theorem]
(ii) Given y = f(x) = 16 – x²
which is polynomial in x and hence differentiable and continuous everywhere. Thus,f(x) is continuous on [- 1, 1] and differentiable on (- 1, 1).
Here f (- 1) = 16 – 1 = 15
and f(1) = 16 – 1 = 15
∴ f(- 1) = f(0)
Thus, all the three conditions of Rolle’s Theorem are satisfied. So ∃ atleast one real number c ∈ (- 1, 1) s.t.f’ (c) = 0
But f’ (c) = 0 ⇒ – 2c = 0 ⇒ c = 0
∴ f(c) = f(0) = 16
Then by the geometrical interpretation of Rolle’s Theorem, (0, 16) be the point on given curve, where tangent is parallel to x- axis.
Question 17.
Examine the applicability of Rolle’s theorem on the following functions :
(0 f(x) = \(\sqrt{x}\) on [-1, 1]
(ii) f(x) = 1 – (x – 1)2/3 on [0, 2]
(iii) f (x) = x2/3 on [- 1, 1].
Solution:
(i) Given f(x) = \(\sqrt{x}\) in [- 1, 1]
∴ f'(x) = \(\frac{1}{2} x^{\frac{1}{2}-1}=\frac{1}{2 \sqrt{x}}\)
which does not exists at x = 0 ∈ (-1, 1)
Thus f(x) is not differentiable at x = 0 ∈ (- 1, 1)
∴ f(x) is not differentiable on (-1, 1)
Thus Rolle’s theorem is not applicable.
(ii) Given f(x) = 1 – (x – 1)2/3 on [0, 2]
∴ f'(x) = –\(\frac{2}{3}(x-1)^{\frac{2}{3}-1}=-\frac{2}{3(x-1)^{1 / 3}}\)
which does not exists at x = 1 ∈ (0, 2).
Thus function f(x) is not derivable at x = 1 ∈ (0, 2)
(iii) Given f (x) = x2/3
∴ f'(x) = \(\frac{2}{3} x^{-1 / 3}=\frac{2}{3 x^{1 / 3}}\) does not exists at x = 0 ∈ (- 1, 1)
∴function f is not derivable on (0, 2)
Thus, Rolle’s theorem is not applicable.