Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Ex 8(c) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Exercise 8(c)

Question 1.

Find the intervals where the following functions are (a) increasing and (b) decreasing.

(i) y = sin x, x ∈ [0, 2π]

Solution:

(ii) y = In x, x ∈ R_{+}

Solution:

⇒ \(\frac{d y}{d x}\) = \(\frac{1}{x}\) > 0

for all x ∈ R_{+}.

Thus y = In x is increasing in (0, ∞) and decreasing nowhere.

(iii) y = a^{x}, a > 0, x ∈ R

Solution:

⇒ \(\frac{d y}{d x}\) = a^{x} In a > 0 for all x ∈ R provided a > 1.

The function is increasing in (-∞, ∞) is a > 1.

Again \(\frac{d y}{d x}\) = a^{x} In a < 0 if 0 < a < 1.

Thus the function is decreasing in (-∞, ∞) if (0 < a < 1)

(iv) y = sin x + cos x, x ∈ [0, 2π]

Solution:

(v) y = 2x^{3} + 3x^{2} – 36x – 7

Solution:

\(\frac{d y}{d x}\) = 6x^{2} – 6x – 36

y is increasing if \(\frac{d y}{d x}\) > 0

⇒ x^{2} + x – 6>0

⇒ x^{2} + 3x – 2x – 6>0

⇒ x (x + 3) -2 (x + 3) > 0

⇒ (x + 3) (x – 2) > 0

for x > 2 or x < -3 and (x + 3 ) (x – 2) < 0 for -3 < x < 2

∴ The function is increasing in (-∞, -3) ∪ (2, ∞) and is decreasing in (-3, 2).

(vi) y = \(\frac{1}{x-1^{\prime}}\) x ≠ 1

Solution:

\(\frac{d y}{d x}\) = \(\frac{-1}{(x-1)^2}\)

The function is increasing if \(\frac{d y}{d x}\) > 0

⇒ \(\frac{-1}{(x-1)^2}\) > 0 which is impossible because \(\frac{-1}{(x-1)^2}\) is always -ve for all x ≠ 1. So the function is increasing nowhere. It is decreasing in R – {1}

(vii) y = \(\left\{\begin{array}{cc}

x^2+1, & x \leq-3 \\

x^3-8 x+13, & x>-3

\end{array}\right.\)

Solution:

(viii) y = 4x^{2} + \(\frac{1}{x}\)

Solution:

(ix) y = (x – 1)^{2} (x + 2)

Solution:

\(\frac{d y}{d x}\) = 2 (x – 1) (x + 2) + (x – 1)^{2}

= (x – 1) (2x + 4 + x – 1)

= (x – 1 ) (3x + 3)

= 3 (x – 1) (x + 1 )

The function is increasing if \(\frac{d y}{d x}\) > 0.

⇒ (x + 1) (x – 1) > 0

⇒ x < -1 or x > 1

∴ The function is increasing in (-∞, -1) ∪ (1, ∞)

It is decreasing in (-1, 1).

(x) y = \(\frac{\ln x}{x}\), x > 0

Solution:

(xi) y = tan x – 4 (x – 2), x ∈ (-\(\frac{\pi}{2}\), \(\frac{\pi}{2}\))

Solution:

(xii) y = sin 2x – cos 2x, x ∈ [0, 2π]

Solution:

Question 2.

Give a rough sketch of the functions given in question 1.

Solution:

Do yourself.

Question 3.

Show that the function \(\frac{e^x}{x^p}\) is strictly increasing for x > p > 0.

Solution:

Question 4.

Show that 2 sin x + tan x ≥ 3x for all x ∈ (0, \(\frac{\pi}{2}\)).

Solution:

Let f(x) = 2 sin x + tan x – 3x

Then f(x) = 2 cos x + sec^{2} x – 3