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Chapter 2 Applications of Derivatives Ex 2.3 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 2 Applications of Derivatives Ex 2.3. Step-by-step solved exercises, numerical problems, and digest answers.

7 Solved Questions• 13 Diagrams• 1996 words

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 2 Applications of Derivatives Ex 2.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

Question 1 Maharashtra Board Solution
Check the validity of the Rolle’s theorem for the following functions. (i) f(x) = x2 – 4x + 3, x ∈ [1, 3]
Solution & Step-by-Step Answer:
The function f given as f(x) = x2 – 4x + 3 is polynomial function. Hence, it is continuous on [1, 3] and differentiable on (1, 3). Now, f(1) = 12 – 4(1) + 3 = 1 – 4 + 3 = 0 and f(3) = 32 – 4(3) + 3 = 9 – 12 + 3 = 0 ∴ f(1) = f(3) Thus, the function f satisfies all the conditions of Rolle’s theorem.

(ii) f(x) = e-xsin x, x ∈ [0, π].
Solution:
The functions e-xand sin x are continuous and differentiable on their domains.
∴ f(x) = e-xsin x is continuous on [0, π] and differentiable on (0, π).
Now, f(0) = e0sin 0 = 1 × 0 = 0
and f(π) = e-πsin π = e-π× 0 = 0
∴ f(0) = f(π)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.

(iii) f(x) = 2x2– 5x + 3, x ∈ [1, 3].
Solution:
The function f given as f(x) = 2x2– 5x + 3 is a polynomial function.
Hence, it is continuous on [1, 3] and differentiable on (1, 3).
Now, f(1) = 2(1)2– 5(1) + 3 = 2 – 5 + 3 = 0
and f(3) = 2(3)2– 5(3) + 3 = 18 – 15 + 3 = 6
∴ f(1) ≠ f(3)
Hence, the conditions of Rolle’s theorem are not satisfied.

(iv) f(x) = sin x – cos x + 3, x ∈ [0, 2π].
Solution:
The functions sin x, cos x and 3 are continuous and differentiable on their domains.
∴ f(x) = sin x – cos x + 3 is continuous on [0, 2π] and differentiable on (0, 2π).
Now, f(0) = sin 0 – cos 0 + 3 = 0 – 1 + 3 = 2
and f(2π) = sin 2π – cos 2π + 3 = 0 – 1 + 3 = 2
∴ f(0) = f(2π)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.

(v) f(x) = x2, if 0 ≤ x ≤ 2
= 6 – x, if 2 < x ≤ 6.
Solution:
f(x) = x2, if 0 ≤ x ≤ 2
= 6 – x, if 2 < x ≤ 6
∴ f(x) = = 2x, if 0 ≤ x ≤ 2
= = -1, if 2 < x ≤ 6
∴ Lf'(2) = 2(2) = 4 and Rf'(2) = -1
∴ Lf'(2) ≠ Rf'(2)
∴ f is not differentiable at x = 2 and 2 ∈ (0, 6).
∴ f is not differentiable at all the points on (0, 6).
Hence, the conditions of Rolle’s theorem are not satisfied.

(vi) f(x) = , x ∈ [-1, 1].
Solution:
f(x) =
∴ =
This does not exist at x = 0 and 0 ∈ (-1, 1)
∴ f is not differentiable on the interval (-1, 1).
Hence, the conditions of Rolle’s theorem are not satisfied.

Question 2 Maharashtra Board Solution
Given an interval [a, b] that satisfies hypothesis of Rolle’s theorem for the function f(x) = x4 + x2 – 2. It is known that a = -1. Find the value of b.
Solution & Step-by-Step Answer:
f(x) = x4 + x2 – 2 Since the hypothesis of Rolle’s theorem are satisfied by f in the interval [a, b], we have f(a) = f(b), where a = -1 Now, f(a) = f(-1) = (-1)4 + (-1)2 – 2 = 1 + 1 – 2 = 0 and f(b) = b4 + b2 – 2 ∴ f(a) = f(b) gives 0 = b4 + b2 – 2 i.e. b4 + b2 – 2 = 0. Since, b = 1 satisfies this equation, b = 1 is one of the roots of this equation. Hence, b = 1.
Question 3 Maharashtra Board Solution
Verify Rolle’s theorem for the following functions. (i) f(x) = sin x + cos x + 7, x ∈ [0, 2π]
Solution & Step-by-Step Answer:
The functions sin x, cos x and 7 are continuous and differentiable on their domains. ∴ f(x) = sin x + cos x + 7 is continuous on [0, 2π] and differentiable on (0, 2π) Now, f(0) = sin 0 + cos 0 + 7 = 0 + 1 + 7 = 8 and f(2π) = sin 2π + cos 2π + 7 = 0 + 1 + 7 = 8 ∴ f(0) = f(2π) Thus, the function f satisfies all the conditions of Rolle’s theorem. ∴ there exists c ∈ (0, 2π) such that f'(c) = 0. Now, f(x) = sin x + cos x + 7 ∴ f'(x) = (sin x + cos x + 7) = cos x – sin x + 0 = cos x – sin x ∴ f'(c) = cos c – sin c ∴ f'(c) = 0 gives, cos c – sin c = 0 ∴ cos c = sin c ∴ c = But ∈ (0, 2π) ∴ c = Hence, the Rolle’s theorem is verified.

(ii) f(x) = sin(), x ∈ [0, 2π]
Solution:
The function f(x) = sin() is continuous on [0, 2π] and differentiable on (0, 2π).
Now, f(0) = sin 0 = 0
and f(2π) = sin π = 0
∴ f(0) = f(2π)
Thus, the function f satisfies all the conditions of Rolle’s theorem.
∴ there exists c ∈ (0, 2π) such that f'(c) = 0.


Hence, Rolle’s theorem is verified.

(iii) f(x) = x2– 5x + 9, x ∈ [1, 4].
Solution:
The function f given as f(x) = x2– 5x + 9 is a polynomial function.
Hence it is continuous on [1, 4] and differentiable on (1, 4).
Now, f(1) = 12– 5(1) + 9 = 1 – 5 + 9 = 5
and f(4) = 42– 5(4) + 9 = 16 – 20+ 9 = 5
∴ f(1) = f(4)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.
∴ there exists c ∈ (1, 4) such that f'(c) = 0.
Now, f(x) = x2 – 5x + 9
∴ f'(x) = (x2– 5x + 9)
= 2x – 5 × 1 + 0
= 2x – 5
∴ f'(c) = 2c – 5
∴ f'(c) = 0 gives, 2c – 5 = 0
∴ c = 5/2 ∈ (1, 4)
Hence, the Rolle’s theorem is verified.

Question 4 Maharashtra Board Solution
If Rolle’s theorem holds for the function f(x) = x3 + px2 + qx + 5, x ∈ [1, 3] with c = 2 + , find the values of p and q.
Solution & Step-by-Step Answer:
The Rolle’s theorem holds for the function f(x) = x3 + px2 + qx + 5, x ∈ [1, 3] ∴ f(1) = f(3) ∴ 13 + p(1)2 + q(1) + 5 = 33 + p (3)2 + q(3) + 5 ∴ 1 + p + q + 5 = 27 + 9p + 3q + 5 ∴ 8p + 2q = -26 ∴ 4p + q = -13 ….. (1) Also, there exists at least one point c ∈ (1, 3) such that f'(c) = 0. Now, f'(x) = (x3 + px2 + qx + 5) = 3x2 + p × 2x + q × 1 + 0 = 3x2 + 2px + q But f'(c) = 0 ∴ ∴ (4√3 + 2)p + √3q + (13√3 + 12) = 0 ∴ (4√3 + 2)p + √3q = -13√3 – 12 ……. (2) Multiplying equation (1) by √3, we get 4√3p + √3q= -13√3 Subtracting this equation from (2), we get 2p = -12 ⇒ p= -6 ∴ from (1), 4(-6) + q = -13 ⇒ q = 11 Hence, p = -6 and q = 11.

Question 5 Maharashtra Board Solution
If Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2], show that the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).
Solution & Step-by-Step Answer:
The Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2]. ∴ there exists at least one real number c ∈ (1, 2) such that f'(c) = 0. Now, f(x) = (x – 2) log x ∴ f'(c) = 0 gives 1 – + log c = 0 ∴ c – 2 + c log c = 0 ∴ c log c = 2 – c, where c ∈ (1, 2) ∴ c satisfies the equation x log x = 2 – x, c ∈ (1, 2). Hence, the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).

Question 6 Maharashtra Board Solution
The function f(x) = satisfies all the conditions of Rolle’s theorem on [-3, 0]. Find the value of c such that f'(c) = 0.
Solution & Step-by-Step Answer:
The function f(x) satisfies all the conditions of Rolle’s theorem, therefore there exist c ∈ (-3, 0) such that f'(c) = 0.

Question 7 Maharashtra Board Solution
Verify Lagrange’s mean value theorem for the following functions: (i) f(x) = log x on [1, e].
Solution & Step-by-Step Answer:
The function f given as f(x) = log x is a logarithmic function that is continuous for all positive real numbers. Hence, it is continuous on [1, e] and differentiable on (1, e). Thus, the function f satisfies the conditions of Lagrange’s mean value theorem. ∴ there exists c ∈ (1, e) such that Hence, Lagrange’s mean value theorem is verified.

(ii) f(x) = (x – 1)(x – 2)(x – 3) on [0, 4].
Solution:
The function f given as
f(x) = (x – 1)(x – 2)(x – 3)
= (x – 1)(x2– 5x + 6)
= x3– 5x2+ 6x – x2+ 5x – 6
= x3– 6x2+ 11x – 6 is a polynomial function.
Hence, it is continuous on [0, 4] and differentiable on (0, 4).
Thus, the function f satisfies the conditions of Lagrange’s, mean value theorem.
∴ there exists c ∈ (0, 4) such that


Hence, Lagrange’s mean value theorem is verified.

(iii) f(x) = x2– 3x – 1, x ∈
Solution:
The function f given as f(x) = x2– 3x – 1 is a polynomial function.
Hence, it is continuous on and differentiable on .
Thus, the function f satisfies the conditions of LMVT.
∴ there exists c ∈ such that


Hence, Lagrange’s mean value theorem is verified.

(iv) f(x) = 2x – x2, x ∈ [0, 1].
Solution:
The function f given as f(x) = 2x – x2is a polynomial function.
Hence, it is continuous on [0, 1] and differentiable on (0, 1).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (0, 1) such that

Hence, Lagrange’s mean value theorem is verified.

(v) f(x) = on [4, 5].
Solution:
The function f given as
f(x) = is a rational function which is continuous except at x = 3.
But 3 ∉ [4, 5]
Hence, it is continuous on [4, 5] and differentiable on (4, 5).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (4, 5) such that
f'(c) = ……..(1)

Hence, Lagrange’s mean value theorem is verified.

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