# Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

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

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.2

Question 1.
Test whether the following functions are increasing and decreasing:
(i) f(x) = x3 – 6x2 + 12x – 16, x ∈ R
Solution:
f(x) = x3 – 6x2 + 12x – 16
∴ f'(x) = $$\frac{d}{d x}$$(x3 – 6x2 + 12x – 16)
= 3x2 – 6 × 2x + 12 × 1 – 0
= 3x2 – 12x + 12
= 3(x2 – 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) = 2x3 – 15x2 + 36x + 1
Solution:
f(x) = 2x3 – 15x2 + 36x + 1
∴ f'(x) = $$\frac{d}{d x}$$(2x3 – 15x2 + 36x + 1)
= 2 × 3x2 – 15 × 2x + 36 × 1 + 0
= 6x2 – 30x + 36
= 6(x2 – 5x + 6)
f is increasing, if f'(x) > 0
i.e. if 6(x2 – 5x + 6) > 0
i.e. if x2 – 5x + 6 > 0
i.e. if x2 – 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) = x2 + 2x – 5
Solution:
f(x) = x2 + 2x – 5
∴ f'(x) = $$\frac{d}{d x}$$(x2 + 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) = 2x3 – 15x2 – 144x – 7
Solution:
f(x) = 2x3 – 15x2 – 144x – 7
∴ f'(x) = $$\frac{d}{d x}$$ (2x3 – 15x2 – 144x – 7)
= 2 × 3x2 – 15 × 2x – 144 × 1 – 0
= 6x2 – 30x – 144
= 6(x2 – 5x – 24)
f is increasing if, f'(x) > 0
i.e. if 6(x2 – 5x – 24) > 0
i.e. if x2 – 5x – 24 > 0
i.e. if x2 – 5x > 24
i.e. if x2 – 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) = 2x3 – 15x2 – 144x – 7
Solution:
f(x) = 2x3 – 15x2 – 144x – 7
∴ f'(x) = $$\frac{d}{d x}$$(2x3 – 15x2 – 144x – 7)
= 2 × 3x2 – 15 × 2x – 144 × 1 – 0
= 6x2 – 30x – 144
= 6(x2 – 5x – 24)
f is decreasing, if f'(x) < 0
i.e. if 6(x2 – 5x – 24) < 0
i.e. if x2 – 5x – 24 < 0
i.e. if x2 – 5x < 24
i.e. if x2 – 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) = x4 – 2x3 + 1
Solution:
f(x) = x4 – 2x3 + 1
∴ f'(x) = $$\frac{d}{d x}$$(x4 – 2x3 + 1)
= 4x3 – 2 × 3x2 + 0
= 4x3 – 6x2
f is decreasing, if f'(x) < 0
i.e. if 4x3 – 6x2 < 0
i.e. if x2(4x – 6) < 0
i.e. if 4x – 6 < 0 …….[∵ x2 > 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) = 2x3 – 15x2 – 84x – 7
Solution:
f(x) = 2x3 – 15x2 – 84x – 7
∴ f'(x) = $$\frac{d}{d x}$$(2x3 – 15x2 – 84x – 7)
= 2 × 3x2 – 15 × 2x – 84 × 1 – 0
= 6x2 – 30x – 84
= 6(x2 – 5x – 14)
f is decreasing, if f'(x) < 0
i.e. if 6(x2 – 5x – 14) < 0
i.e. if x2 – 5x – 14 < 0
i.e. if x2 – 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.