12th Commerce Maths 1 Chapter 4 Exercise 4.2 Answers Maharashtra Board

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.2 Questions and Answers.

Std 12 Maths 1 Exercise 4.2 Solutions Commerce Maths

Question 1.
Test whether the following functions are increasing and decreasing:
(i) f(x) = x ³ – 6x ² + 12x – 16, x ∈ R
Solution:
f(x) = x ³ – 6x ² + 12x – 16
∴ f'(x) = \(\frac{d}{d x}\)(x ³ – 6x ² + 12x – 16)
= 3x ² – 6 × 2x + 12 × 1 – 0
= 3x ² – 12x + 12
= 3(x ² – 4x + 4)
= 3(x – 2) ² > 0 for all x ∈ R, x ≠ 2
∴ f'(x) > 0 for all x ∈ R – {2}
∴ f is increasing for all x ∈ R – {2}.

(ii) f(x) = x – \(\frac{1}{x}\), x ∈ R, x ≠ 0
Solution:
f(x) = x – \(\frac{1}{x}\)
∴ f'(x) = \(\frac{d}{d x}\left(x-\frac{1}{x}\right)\)
= 1 – \(\left(-\frac{1}{x^{2}}\right)\)
= 1 + \(\frac{1}{x^{2}}\) > 0 for all x ∈ R, x ≠ 0
∴ f'(x) > 0 for all x ∈ R, where x ≠ 0
∴ f is increasing for all x > R, where x ≠ 0.

(iii) f(x) = \(\frac{7}{x}\) – 3, x ∈ R, x ≠ 0
Solution:
f(x) = \(\frac{7}{x}\) – 3
∴ f'(x) = \(\frac{d}{d x}\left(\frac{7}{x}-3\right)=7\left(-\frac{1}{x^{2}}\right)-0\)
= \(-\frac{7}{x^{2}}\) < 0 for all x ∈ R, x ≠ 0
∴ f'(x) < 0 for all x ∈ R, where x ≠ 0.
∴ f is decreasing for all x ∈ R, where x ≠ 0.

Question 2.
Find the values of x, such that f(x) is increasing function:
(i) f(x) = 2x ³ – 15x ² + 36x + 1
Solution:
f(x) = 2x ³ – 15x ² + 36x + 1
∴ f'(x) = \(\frac{d}{d x}\)(2x ³ – 15x ² + 36x + 1)
= 2 × 3x ² – 15 × 2x + 36 × 1 + 0
= 6x ² – 30x + 36
= 6(x ² – 5x + 6)
f is increasing, if f'(x) > 0
i.e. if 6(x ² – 5x + 6) > 0
i.e. if x ² – 5x + 6 > 0
i.e. if x ² – 5x > -6
i.e. if x – 5x + \(\frac{25}{4}\) > -6 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{1}{4}\)
i.e. if x – \(\frac{5}{2}\) > \(\frac{1}{2}\) or x – \(\frac{5}{2}\) < –\(\frac{1}{2}\)
i.e. if x > 3 or x < 2
i.e. if x ∈ (-∞, 2) ∪ (3, ∞)
∴ f is increasing, if x ∈ (-∞, 2) ∪ (3, ∞).

(ii) f(x) = x ² + 2x – 5
Solution:
f(x) = x ² + 2x – 5
∴ f'(x) = \(\frac{d}{d x}\)(x ² + 2x – 5)
= 2x + 2 × 1 – 0
= 2x + 2
f is increasing, if f'(x) > 0
i.e. if 2x + 2 > 0
i.e. if 2x > -2
i.e. if x > -1, i.e. x ∈ (-1, ∞)
∴ f is increasing, if x > -1, i.e. x ∈ (-1, ∞)

(iii) f(x) = 2x ³ – 15x ² – 144x – 7
Solution:
f(x) = 2x ³ – 15x ² – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\) (2x ³ – 15x ² – 144x – 7)
= 2 × 3x ² – 15 × 2x – 144 × 1 – 0
= 6x ² – 30x – 144
= 6(x ² – 5x – 24)
f is increasing if, f'(x) > 0
i.e. if 6(x ² – 5x – 24) > 0
i.e. if x ² – 5x – 24 > 0
i.e. if x ² – 5x > 24
i.e. if x ² – 5x + \(\frac{25}{4}\) > 24 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{121}{4}\)
i.e. if \(x-\frac{5}{2}>\frac{11}{2} \text { or } x-\frac{5}{2}<-\frac{11}{2}\)
i.e. if x > 8 or x < -3
i.e. if x ∈ (-∞, -3) ∪ (8, ∞)
∴ f is increasing, if x ∈ (-∞, -3) ∪ (8, ∞).

Question 3.
Find the values of x such that f(x) is decreasing function:
(i) f(x) = 2x ³ – 15x ² – 144x – 7
Solution:
f(x) = 2x ³ – 15x ² – 144x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x ³ – 15x ² – 144x – 7)
= 2 × 3x ² – 15 × 2x – 144 × 1 – 0
= 6x ² – 30x – 144
= 6(x ² – 5x – 24)
f is decreasing, if f'(x) < 0
i.e. if 6(x ² – 5x – 24) < 0
i.e. if x ² – 5x – 24 < 0
i.e. if x ² – 5x < 24
i.e. if x ² – 5x + \(\frac{25}{4}\) < \(\frac{121}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{121}{4}\)
i.e. if \(-\frac{11}{2}<x-\frac{5}{2}<\frac{11}{2}\)
i.e. if \(-\frac{11}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{11}{2}+\frac{5}{2}\)
i.e. if -3 < x < 8
∴ f is decreasing, if -3 < x < 8.

(ii) f(x) = x ⁴ – 2x ³ + 1
Solution:
f(x) = x ⁴ – 2x ³ + 1
∴ f'(x) = \(\frac{d}{d x}\)(x ⁴ – 2x ³ + 1)
= 4x ³ – 2 × 3x ² + 0
= 4x ³ – 6x ²
f is decreasing, if f'(x) < 0
i.e. if 4x ³ – 6x ² < 0
i.e. if x ² (4x – 6) < 0
i.e. if 4x – 6 < 0 …….[∵ x ² > 0]
i.e. if x < \(\frac{3}{2}\)
i.e. -∞ < x < \(\frac{3}{2}\)
∴ f is decreasing, if -∞ < x < \(\frac{3}{2}\).

(iii) f(x) = 2x ³ – 15x ² – 84x – 7
Solution:
f(x) = 2x ³ – 15x ² – 84x – 7
∴ f'(x) = \(\frac{d}{d x}\)(2x ³ – 15x ² – 84x – 7)
= 2 × 3x ² – 15 × 2x – 84 × 1 – 0
= 6x ² – 30x – 84
= 6(x ² – 5x – 14)
f is decreasing, if f'(x) < 0
i.e. if 6(x ² – 5x – 14) < 0
i.e. if x ² – 5x – 14 < 0
i.e. if x ² – 5x < 14
i.e. if x – 5x + \(\frac{25}{4}\) < 14 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{81}{4}\)
i.e. if \(-\frac{9}{2}<x-\frac{5}{2}<\frac{9}{2}\)
i.e. if \(-\frac{9}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{9}{2}+\frac{5}{2}\)
i.e. if -2 < x < 7
∴ f is decreasing, if -2 < x < 7.