Chapter 2 Applications of Derivatives Ex 2.4 Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 2 Applications of Derivatives Ex 2.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

(ii) f(x) = 2 – 3x + 3x²– x³, x ∈ R.
Solution:
f(x) = 2 – 3x + 3x²– x³
∴ f'(x) = (2 – 3x + 3x²– x³)
= 0 – 3 × 1 + 3 × 2x – 3x²
= -3 + 6x – 3x²
= -3(x²– 2x + 1)
= -3(x – 1)²≤ 0 for all x ∈ R
∴ f'(x) ≤ 0 for all x ∈ R
∴ f is decreasing for all x ∈ R.

(iii) f(x) = x – , x ∈ R, x ≠ 0.
Solution:
f(x) = x –
f'(x) =
= > 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.

(ii) f(x) = 3 + 3x – 3x²+ x³
Solution:
f(x) = 3 + 3x – 3x²+ x³
∴ f'(x) = (3 + 3x – 3x²+ x³)
= 0 + 3 × 1 – 3 × 2x + 3x²
= 3 – 6x + 3x²
= 3(x²– 2x + 1)
f is strictly increasing if f'(x) > 0
i.e. if 3(x²– 2x + 1) > 0
i.e. if x²– 2x + 1 > 0
i.e. if (x – 1)²> 0
This is possible if x ∈ R and x ≠ 1
i.e. x ∈ R – {1}
∴ f is strictly increasing if x ∈ R – {1}.

(iii) f(x) = x³– 6x²– 36x + 7
Solution:
f(x) = x³– 6x²– 36x + 7
∴ f'(x) = (x³– 6x²– 36x + 7)
= 3x²– 6 × 2x – 36 × 1 + 0
= 3x²– 12x – 36
= 3(x²– 4x – 12)
f is strictly increasing if f'(x) > 0
i.e. if 3(x²– 4x – 12) > 0
i.e. if x²– 4x – 12 > 0
i.e. if x²– 4x > 12
i.e. if x²– 4x + 4 > 12 + 4
i.e. if (x – 2)²> 16
i.e. if x – 2 > 4 or x – 2 < -4 i.e. if x > 6 or x < -2
∴ f is strictly increasing if x < -2 or x > 6.

(ii) f(x) = x +
Solution:
f(x) = x + , x ≠ 0
∴ f'(x) =
= 1 + 25(-1) x^-2
= 1 –
f is is strictly decreasing if f'(x) < 0
i.e. if 1 – < 0
i.e. if 1 <
i.e. if x²< 25
i.e. if -5 < x < 5, x ≠ 0
i.e. if x ∈ (-5, 5) – {0}
∴ f is strictly decreasing if x ∈ (-5, 5) – {0}.

(iii) f(x) = x³– 9x²+ 24x + 12
Solution:
f(x) = x³– 9x²+ 24x + 12
∴ f'(x) = (x³– 9x²+ 24x + 12)
= 3x²– 9 × 2x + 24 × 1 + 0
= 3x²– 18x + 24
= 3(x²– 6x + 8)
f is strictly decreasing if f'(x) < 0
i.e. if 3(x²– 6x + 8) < 0
i.e. if x²– 6x + 8 < 0
i.e. if x²– 6x < -8
i.e. if x²– 6x + 9 < -8 + 9
i.e. if (x – 3)²< 1
i.e. if -1 < x – 3 < 1
i.e. if -1 + 3 < x – 3 + 3 < 1 + 3
i.e. if 2 < x < 4
i.e., if x ∈ (2, 4)
∴ f is strictly decreasing if x ∈ (2, 4)

(a) f is increasing if f'(x) ≥ 0
i.e. if 3(x²– 8x – 48) ≥ 0
i.e. if x²– 8x – 48 ≥ 0
i.e. if x²– 8x ≥ 48
i.e. if x²– 8x + 16 ≥ 48 + 16
i.e. if (x – 4)²≥ 64
i.e. if x – 4 ≥ 8 or x – 4 ≤ -8
i.e. if x > 12 or x ≤ -4
∴ f is increasing if x ≤ -4 or x ≥ 12,
i.e. x ∈ (-∞, -4] ∪ [12, ∞).

(b) f is decreasing if f'(x) ≤ 0
i.e. if 3(x²– 8x – 48) ≤ 0
i.e. if x²– 8x – 48 ≤ 0
i.e. if x²– 8x ≤ 48
i.e. if x²– 8x + 16 ≤ 48 + 16
i.e. if (x – 4)²≤ 64
i.e. if -8 ≤ x – 4 ≤ 8
i.e. if -4 ≤ x ≤ 12
∴ f is decreasing if -4 ≤ x ≤ 12, i.e. x ∈ [-4, 12].

(b) f is strictly 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 + < 24 +
i.e. if
i.e. if
i.e. if
i.e. if -3 < x < 8
∴ f is strictly decreasing, if -3 < x < 8.

(a) f is strictly increasing if f'(x) > 0
i.e. if > 0
i.e. if 1 – x²> 0 ……..[∵ (x²+ 1)²> 0]
i.e. if 1 > x²
i.e. if x²< 1
i.e. if -1 < x < 1
∴ f is strictly increasing if -1 < x < 1

(b) f is strictly decreasing if f'(x) < 0
i.e. if < 0
i.e. if 1 – x²< 0 ……..[∵ (x²+ 1)²> 0]
i.e. if 1 < x²i.e. if x²> 1
i.e. if x > 1 or x < -1
∴ f is strictly decreasing if x < -1 or x > 1
i.e. x ∈ (-∞, -1) ∪ (1, ∞).

(ii) f(x) = 2x³– 21x²+ 36x – 20
Solution:
f(x) = 2x³– 21x²+ 36x – 20
∴ f'(x) = (2x³– 21x²+ 36x – 20)
= 2 × 3x²– 21 × 2x + 36 × 1 – 0
= 6x²– 42x + 36
and f”(x) = (6x²– 42x + 36)
= 6 × 2x – 42 × 1 + 0
= 12x – 42
f'(x) = 0 gives 6x²– 42x + 36 = 0
∴ x²– 7x + 6 = 0
∴ (x – 1)(x – 6) = 0
the roots of f'(x) = 0 are x₁= 1 and x₂= 6.

Method 1 (Second Derivative Test):
(a) f”(1) = 12(1) – 42 = -30 < 0
∴ by the second derivative test, f has maximum at x = 1
and maximum value of f at x = 1
f(1) = 2(1)³– 21(1)²+ 36(1) – 20
= 2 – 21 + 36 – 20
= -3

(b) f”(6) = 12(6) – 42 = 30 > 0
∴ by the second derivative test, f has minimum at x = 6
and minimum value of f at x = 6
f(6) = 2(6)³– 21(6)²+ 36(6) – 20
= 432 – 756 + 216 – 20
= -128.
Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.

Method 2 (First Derivative Test):
(a) f'(x) = 6(x – 1)(x – 6)
Consider x = 1
Let h be a small positive number. Then
f'(1 – h) = 6(1 – h – 1)(1 – h – 6)
= 6(-h)(-5 – h)
= 6h(5 + h)> 0
and f'(1 + h) = 6(1 + h – 1)(1 + h – 6)
= 6h(h – 5) < 0, as h is small positive number.
∴ by the first derivative test, f has maximum at x = 1 and maximum value of f at x = 1
f(1) = 2(1)³– 21(1)²+ 36(1) – 20
= 2 – 21 + 36 – 20
= -3

(b) f'(x) = 6(x – 1)(x – 6)
Consider x = 6
Let h be a small positive number. Then
f'(6 – h) = 6(6 – h – 1)(6 – h – 6)
= 6(5 – h)(-h)
= -6h(5 – h) < 0, as h is small positive number
and f'(6 + h) = 6(6 + h – 1)(6 + h – 6) = 6(5 + h)(h) > 0
∴ by the first derivative test, f has minimum at x = 6
and minimum value of f at x = 6
f(6) = 2(6)³– 21(6)²+ 36(6) – 20
= 432 – 756 + 216 – 20
= -128
Hence, the function f has maximum value -3 at x = 1
and minimum value -128 at x = 6.

(iii) f(x) = x³– 9x²+ 24x
Solution:
f(x) = x³– 9x²+ 24x
∴ f'(x) = (x³– 9x²+ 24x)
= 3x²– 9 × 2x + 24 × 1
= 3x²– 18x + 24
and f”(x) = (3x²– 18x + 24)
= 3 × 2x – 18 × 1 + 0
= 6x – 18
f'(x) = 0 gives 3x²– 18x + 24 = 0
∴ x²– 6x + 8 = 0
∴ (x – 2)(x – 4) = 0
∴ the roots of f'(x) = 0 are x₁= 2 and x₂= 4.

(a) f”(2) = 6(2) – 18 = -6 < 0
∴ by the second derivative test, f has maximum at x = 2
and maximum value of f at x = 2
f(2) = (2)³– 9(2)²+ 24(2)
= 8 – 36 + 48
= 20

(b) f”(4) = 6(4) – 18 = 6 > 0
∴ by the second derivative test, f has minimum at x = 4
and minimum value of f at x = 4
f(4) = (4)³– 9(4)²+ 24(4)
= 64 – 144 + 96
= 16
Hence, the function f has maximum value 20 at x = 2 and minimum value 16 at x = 4.

(iv) f(x) = x²+
Solution:

(v) f(x) = x log x
Solution:

(vi) f(x) =
Solution: