RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives Ex 8.2 is part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE 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

Question 1.

Prove that f(x) = x^{2} is increasing in interval (0, ∞) and decreasing in interval (-∞,0).

Solution:

Let x_{1}, x_{2} ∈ [0,∞] is such that

x_{1} < x_{2}

∴ x_{1} < x_{2} ⇒ x_{1}^{2}< x_{1}x_{2} …..(i)

and x_{1} < x_{2} ⇒ x_{1}x_{2} < x_{2}^{2}……(i)

From (i) and (ii),

x_{1} < x_{2} ⇒ x_{1}^{2 }< X_{2}^{2}

⇒ f(x_{1}) < f(x_{2})

∴ X_{1} < X_{2} ⇒ f(x_{1}) < f(x_{2})

where x_{1}, x_{2} ∈ [0, ∞]

Hence, f(x) is continuously increasing in [0, ∞)

Again, let x_{1}, x_{2} ∈ (-∞, 0) is such that

x_{1} < x_{2}

Then

x_{1} < x_{2} ⇒ x_{1}^{2}> x_{1}x_{2}

[∵ – 3 < – 2, (- 3) (- 3) = 9

(- 3) × (- 2) = 6

∴ 9 > 6

x_{1}^{2} > x_{1} x_{2}]

Again

x_{1} < x_{2} ⇒ x_{1}x_{2} > x_{2}^{2}

Again – 3 < – 2

(- 3) × (- 2) = 6

(- 2) × (- 2) = 4

6 > 4

∴ x_{1}x_{2} > x_{2}^{2}]

From (i) and (ii),

x_{1} < x_{2} ⇒ X_{1}^{2} > x_{2}^{2}

⇒ f(x_{1}) > f(x_{2})

∴ x_{1} < x_{2} ⇒ f(x_{1}) > f(x_{2})

Hence, f(x) is continuously decreasing in (-∞, 0).

Question 2.

Prove that f(x) = a^{x}, 0 < a < 1, is decreasing in R.

Solution:

Let x_{1},x_{2} ∈R is such that

x_{1} < x_{2}

Then, x_{1} < x_{2}

⇒ a^{x1} > a^{x2}

[∵ 0 < a < 1 and x < x_{2} ⇒ a^{x1} > a^{x2}]

⇒ f(x_{1}) > f(x_{2})

∴ x_{1} < x_{2}

⇒ f(x_{1}) > f(x_{2}) ∀ x_{1}, x_{2} ∈ R

Hence, function f(x) = a^{x}, 0 < a < 1, R is decreasing in R.

Question 3.

f(x) = log sin x, x ∈ (0,\(\frac { \pi }{ 2 } \))

Solution:

Hence, in interval (0,\(\frac { \pi }{ 2 } \)) function is continuously increasing

Question 4.

f(x) = x^{100} + sin x + 1- x ∈ (0,\(\frac { \pi }{ 2 } \))

Solution:

f(x) = x^{100} + sin x + 1

Diff. w.r.t. x,

f'(x) = 100x^{99} + cos x

In interval (0,\(\frac { \pi }{ 2 } \))

f'(x) = 100x^{99} + cos x > 0

[∵ cos x > 0 and 100x^{99} > 0]

⇒ f'(x) > 0

Hence, function in interval (0,\(\frac { \pi }{ 2 } \)) is increasing.

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.

Question 6.

f(x) = x^{3} – 6x^{2} + 12x – 1,x ∈ R.

Solution:

f(x) = x^{3} – 6x_{6}^{2} + 12x – 1

DifF. w.r.t x,

f'(x) = 3x^{2} – 12x + 12

⇒ f'(x) = 3(x^{2} – 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.

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.

Question 8.

f(x) = sin^{4} x + cos^{4} x, x ∈(0,\(\frac { \pi }{ 4 } \))

Solution:

f(x) = sin^{4} x + cos^{4} x

Diff. w.r.t x,

f'(x) = 4 sin^{3} x cos x + 4 cos^{3} x(-sin x)

⇒ f'(x) = 4 sin x cos x (sin^{2} x – cos^{2} x)

⇒ f'(x) = -2.2 sin x cos x (cos^{2} x – sin^{2} x)

⇒ f'(x) = – 2sin 2x cos 2x

⇒ f'(x) = – sin 4x

Question 9.

f(x) = \(\frac { 3 }{ x } \) + 5, x ∈ R, x ≠ 0.

Solution:

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

Question 10.

f(x) = x^{2} – 2x + 3, x < 1.

Solution:

f(x) = x^{2} – 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.

Question 11.

f(x) = 2x^{3} – 3x^{2} – 36x + 7

Solution:

Given

f(x) = 2x^{3} – 3x^{2} – 36x + 7

⇒ f'(x) = 6x^{2} – 6x – 36

f'(x) = o

⇒ 6x^{2} – 6x – 36 = 0

⇒ 6(x^{2} – 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) = 6x^{2} – 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) = 6x^{2} – 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) = 6x^{2} – 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).

Question 12.

f(x) = x^{4} – 2x^{2}.

Solution:

Given

f(x) = x^{4} – 2x^{2}

f'(x) = 4x_{6}^{3} – 4x (∵f(x) = 0)

⇒ 4x^{3} – 4x = 0

⇒ 4x(x^{2} – 1) = 0

⇒ 4x = 0 or x^{2} – 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 x^{3} – 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) = 4x^{3} – 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) = 4x^{3} – 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) = 4x^{3} – 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) = 2x^{3} – 9x^{2} + 12x + 5.

Solution:

Given

f(x) = 2x^{3} – 9x^{2} + 12x + 15

f'(x) = 6x^{2} – 18x + 12

= 6(x^{2 }– 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) = – 2x^{3} + 3×2 + 12x + 5.

Solution:

Given, f(x) = -2×3 + 3×2 + 12x + 5

f'(x) = – 6x^{2} + 6x + 12

= – 6(x^{2} – 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) = x^{2} + ax + 5, is increasing in interval [1, 2]

Solution:

f(x) = x^{2} + 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,\(\frac { \pi }{ 4 } \))

Solution:

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

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