Question

If ƒ′′(x) > 0 and ƒ′(1) = 0 such that

g(x) = ƒ (cot² x + 2 cot x + 2), where 0 < x < π then the interval in which g(x) is decreasing is –

• A. (0, π)
• B. (\frac{\pi}{2},\pi)
• C. (\frac{3\pi}{4},\pi)
• D. (0,\frac{3\pi}{4})

Answer 1

Answer: (D)

(0,\frac{3\pi}{4})

Answer 2

Here, g(x) = ƒ(cot² x + 2 cot x + 2)

⇒ g′(x) = ƒ′(cot² x + 2 cot x + 2)

{–2 cot x cosec² x – 2cosec² x}

For g(x) to be decreasing, g′(x) < 0

⇒ ƒ′{(cot x + 1)² + 1} · (–2 cosec² x) (cot x + 1) < 0

⇒ ƒ′{(cot x + 1)² + 1} · (cot x + 1) > 0 … (i)

{as ƒ′′(x) > 0 ⇒ ƒ′(x) is increasing, then

ƒ′{(cot x + 1)² + 1} > ƒ′(1) = 0

∀ x ∈$\left( 0,\frac{3\pi}{4} \right)$ ∪ $\left( \frac{3\pi}{4},\pi \right)$

Thus, equation (i) holds, if cot x + 1 > 0

⇒ cot x > –1 ∀ x ∈$\left( 0,\frac{3\pi}{4} \right)$

Hence, (4) is the correct answer.