RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives Ex 8.2

RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives Ex 8.2

Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.2

RBSE Solutions For Class 12 Maths Chapter 8.2 Question 1.
Prove that f(x) = x² is increasing in interval (0, ∞) and decreasing in interval (-∞,0).
Solution:
Let x₁, x₂ ∈ [0,∞] is such that
x₁ < x₂
∴ x₁ < x₂ ⇒ x₁²< x₁x₂ …..(i)
and x₁ < x₂ ⇒ x₁x₂ < x₂²……(i)
From (i) and (ii),
x₁ < x₂ ⇒ x₁² < X₂²
⇒ f(x₁) < f(x₂)
∴ X₁ < X₂ ⇒ f(x₁) < f(x₂)
where x₁, x₂ ∈ [0, ∞]
Hence, f(x) is continuously increasing in [0, ∞)
Again, let x₁, x₂ ∈ (-∞, 0) is such that
x₁ < x₂
Then
x₁ < x₂ ⇒ x₁²> x₁x₂
[∵ – 3 < – 2, (- 3) (- 3) = 9
(- 3) × (- 2) = 6
∴ 9 > 6
x₁² > x₁ x₂]
Again
x₁ < x₂ ⇒ x₁x₂ > x₂²
Again – 3 < – 2
(- 3) × (- 2) = 6
(- 2) × (- 2) = 4
6 > 4
∴ x₁x₂ > x₂²]
From (i) and (ii),
x₁ < x₂ ⇒ X₁² > x₂²
⇒ f(x₁) > f(x₂)
∴ x₁ < x₂ ⇒ f(x₁) > f(x₂)
Hence, f(x) is continuously decreasing in (-∞, 0).

Exercise 8.2 Class 12 Maths RBSE Question 2.
Prove that f(x) = a^x, 0 < a < 1, is decreasing in R.
Solution:
Let x₁,x₂ ∈R is such that
x₁ < x₂
Then, x₁ < x₂
⇒ a^x1 > a^x2
[∵ 0 < a < 1 and x < x₂ ⇒ a^x1 > a^x2]
⇒ f(x₁) > f(x₂)
∴ x₁ < x₂
⇒ f(x₁) > f(x₂) ∀ x₁, x₂ ∈ R
Hence, function f(x) = a^x, 0 < a < 1, R is decreasing in R.

RBSE Solutions For Class 12 Maths Chapter 8 Question 3.
f(x) = log sin x, x ∈ (0,π/2)
Solution:

Hence, in interval (0,π/2) function is continuously increasing

RBSE Solutions For Class 12 Maths Chapter 8 Exercise 8.2 Question 4.
f(x) = x¹⁰⁰ + sin x + 1- x ∈ (0,π/2)
Solution:
f(x) = x¹⁰⁰ + sin x + 1
Diff. w.r.t. x,
f'(x) = 100x⁹⁹ + cos x
In interval (0,π/2)
f'(x) = 100x⁹⁹ + cos x > 0
[∵ cos x > 0 and 100x⁹⁹ > 0]
⇒ f'(x) > 0
Hence, function in interval (0,π/2) is increasing.

RBSE Solutions For Class 12 Maths Chapter 8 Miscellaneous Question 5.
f(x) = (x – 1)e^x + 1, x > 0.
Solution:
f(x) = (x – 1)e^x+1
Diff. w.r.t. x,

Hence, at x = 0, function is increasing.

12th Class RBSE Solution Question 6.
f(x) = x³ – 6x² + 12x – 1,x ∈ R.
Solution:
f(x) = x³ – 6x₆² + 12x – 1
DifF. w.r.t x,
f'(x) = 3x² – 12x + 12
⇒ f'(x) = 3(x² – 4x + 4)
⇒ f'(x) = 3(x – 2)²
⇒ f'(x) = 3(x – 2)² ≥ 0
⇒ f'(x) ≥ 0 [∵(x – 2)² > 0]
Hence, function f(x) is increasing in R.

RBSE Class 12 Maths Chapter 8 Question 7.
f(x) = tan^-1 x – x, x ∈ R.
Solution:
f(x) = tan^-1 x – x
Diff. w.r.t. x,

Hence, function fix) decreasing in R.

12th Maths RBSE Solution Question 8.
f(x) = sin⁴ x + cos⁴ x, x ∈(0,π/4)
Solution:
f(x) = sin⁴ x + cos⁴ x
Diff. w.r.t x,
f'(x) = 4 sin³ x cos x + 4 cos³ x(-sin x)
⇒ f'(x) = 4 sin x cos x (sin² x – cos² x)
⇒ f'(x) = -2.2 sin x cos x (cos² x – sin² x)
⇒ f'(x) = – 2sin 2x cos 2x
⇒ f'(x) = – sin 4x

Class 12 Maths 8.2 Solutions Question 9.
f(x) = 3x + 5, x ∈ R, x ≠ 0.
Solution:

Hence, function f(x) is decreasing in x ∈ R when x ≠ 0.

Exercise 8.2 Class 12 Maths Question 10.
f(x) = x² – 2x + 3, x < 1.
Solution:
f(x) = x² – 2x + 3
Diff. w.r.t. x,
f'(x) = 2x – 2
⇒ f'(x) = 2(x – 1)
⇒ f'(x) = 2(x – 1) < 0 [When x < 1]
⇒ f'(x) < 0
Hence function f(x) is decreasing in interval x < 1.

Class 10 Maths Ex 8.2 Question 11.
f(x) = 2x³ – 3x² – 36x + 7
Solution:
Given
f(x) = 2x³ – 3x² – 36x + 7
⇒ f'(x) = 6x² – 6x – 36
f'(x) = o
⇒ 6x² – 6x – 36 = 0
⇒ 6(x² – x – 6) = 0
⇒ 6(x – 3) (x + 2) = 0
⇒ x – 3 = 0 or x + 2 = 0
⇒ x = 3 or x = -2
Hence, point x = – 2, x = 3 divides the real number line into three intervals (-∞,-2), (-2, 3) and (3,∞).

(a) For (-∞, -2) interval.
f'(x) = 6x² – 6x – 36 > 0
∵ At x = – 3
f'(x) = 6(- 3)² – 6(- 3) – 36
= 6 × 9 + 6 × 3 – 36
= 54 + 18 – 36 = 36 >0
So,f'(x) > 0 can be shown like this by taking other points. Hence, function is continuously increasing in interval (-∞, – 2).

(b) For (-2, 3)
f'(x) = 6x² – 6x – 36 < 0
∴ At x = 1
f'(x) = 6(1)² – 6 × 1 – 36
= 6 – 6 – 36 = -36 < 0
At x = 0, f'(x) = 6(0)² – 6 × 0 – 36
= – 36 < 0
So,f'(x) < 0 can be shown like this by taking other points.
Hence, function is continuously decreasing for x ∈ (- 2, 3).

(c) For (3, ∞)
f'(x) = 6x² – 6x – 36 > 0
∵ At x = 4,
f'(x) = 6 × (4)² – 6 × 4 – 36
= 96 – 24 – 36
= 96 – 60 = 36 > 0
At x = 5, f'(x)= 6(5)² – 6 × 5 – 36
= 6 × 25 – 30 – 36
= 150 – 30 – 36
= 84 > 0
So,f'(x) > 0 can be shown like this for other points.
Hence, function f(x) is continuous increasing in interval
(-∞, -2) ∪ (3,∞). {∵ f'(x) > 0}
Function is continuously decreasing in interval (- 2, 3).

8.2 Class 10 Question 12.
f(x) = x⁴ – 2x².
Solution:
Given
f(x) = x⁴ – 2x²
f'(x) = 4x₆³ – 4x (∵f(x) = 0)
⇒ 4x³ – 4x = 0
⇒ 4x(x² – 1) = 0
⇒ 4x = 0 or x² – 1 = 0
⇒ x = 0 or x = ± 1
Here, point x = 0, 1, – 1 divides real number line into four interval (-∞, -1), (-1, 0), (0, 1) and (1,∞).

(a) For interval (-∞, -1)
f'(x) = 4 x³ – 4x
At x = – 2, f'(x) = 4 x (- 2)³ – 4(- 2)
= -32 + 8 = -24 < 0
Like this it can be shown for other points (f'(x) <0).

(b) For interval (-1,0)
f'(x) = 4x³ – 4x
For x = -0.5
f'(x) = 4 × (- 0.5)³ – 4 × (- 0.5)
= -0.5 + 2.0 = 1.5 >0
Like this it can be shown for other points {f'(x) > 0}

(c) For interval (0, 1)
f'(x) = 4x³ – 4x
For x = 0.5
f'(x) = 4 × (0.5)³ – 4 × (0.5)
= 4 × 0.125 – 2.0
= 0.5 – 2.0 = – 1.5 < 0
Like this it can be shown for other points {f'(x)< 0}

(d) For interval (1, ∞)
f'(x) = 4x³ – 4x
For x = 2 f'(x) = 4 × (2)³ – 4 × 2
= 32 – 8 = 24 > 0
Like this it can be shown for other points {f'(x) > 0}
Hence, function is decreasing in interval (-∞, – 1) ∪ (0, 1) and function is increasing in interval (-1, 0) ∪ (1, ∞).

Question 13.
f(x) = 2x³ – 9x² + 12x + 5.
Solution:
Given
f(x) = 2x³ – 9x² + 12x + 15
f'(x) = 6x² – 18x + 12
= 6(x² – 3x + 2)
x = 1 and x = 2 divides the real number line into three intervals (-∞, -1), (1, 2) and (2,∞).
when x ∈ (-∞, 1) then f'(x) = + ve
when x ∈ (1,2) then f'(x) = – ve
when x ∈ (2,∞) then f'(x) = + ve
So, function is increasing in interval (-∞, 1) ∪ (2,∞) and decreasing in interval (4, 2).

Question 14.
f(x) = – 2x³ + 3×2 + 12x + 5.
Solution:
Given, f(x) = -2×3 + 3×2 + 12x + 5
f'(x) = – 6x² + 6x + 12
= – 6(x² – x – 2)
= – 6(x – 2)(x + 1)
x = – 1 and x = 2 divides the real number line into three intervals (-∞, -1), (- 1, 2) and (2,∞).
when x ∈ (-∞,-1), then f'(x) = -ve
when x ∈ (-1,2), then f'(x) = +ve
when x ∈ (2,∞), then f'(x) = -ve
So, function is increasing in interval (-1, 2) and decreasing in interval (-∞, -1) ∪ (2,∞).

Question 15.
Find the minimum value of a, such that function f(x) = x² + ax + 5, is increasing in interval [1, 2]
Solution:
f(x) = x² + ax + 5
⇒ f'(x) = 2x + a
x ∈ [1, 2]
⇒ 1 ≤ x ≤ 2
⇒ 2 ≤ 2x ≤ 4
⇒ 2 + a ≤ 2x + a ≤ 4 + a
⇒ 2 + a ≥ 0
⇒ a ≥ -2
Hence, minimum value of a is – 2.

Question 16.
Prove that, function f(x) = tan-1 (sin x + cos x), is increasing function in interval (0,π/4)
Solution:

Hence, function is continuously increasing in interval (0, π/4).

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