Chapter 4 Applications of Derivatives Ex 4.2 Solutions

Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 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

(ii) f(x) = x – , x ∈ R, x ≠ 0
Solution:
f(x) = x –
∴ f'(x) =
= 1 –
= 1 + > 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) = – 3, x ∈ R, x ≠ 0
Solution:
f(x) = – 3
∴ f'(x) =
= < 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.

(ii) f(x) = x²+ 2x – 5
Solution:
f(x) = x²+ 2x – 5
∴ f'(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) = (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 + > 24 +
i.e. if
i.e. if
i.e. if x > 8 or x < -3
i.e. if x ∈ (-∞, -3) ∪ (8, ∞)
∴ f is increasing, if x ∈ (-∞, -3) ∪ (8, ∞).

(ii) f(x) = x⁴– 2x³+ 1
Solution:
f(x) = x⁴– 2x³+ 1
∴ f'(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 <
i.e. -∞ < x <
∴ f is decreasing, if -∞ < x < .

(iii) f(x) = 2x³– 15x²– 84x – 7
Solution:
f(x) = 2x³– 15x²– 84x – 7
∴ f'(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 + < 14 +
i.e. if
i.e. if
i.e. if
i.e. if -2 < x < 7
∴ f is decreasing, if -2 < x < 7.