Question

f(x) = $\frac { x } { 1 + x \tan x }$, x ∈$\left( 0 , \frac { \pi } { 2 } \right)$then

• A. f(x) has exactly one point of minima
• B. f(x) has exactly one point of maxima
• C. f(x) is increasing in$\left( 0 , \frac { \pi } { 2 } \right)$
• D. f(x) is decreasing in$\left( 0 , \frac { \pi } { 2 } \right)$

Answer 1

Answer: (B)

f(x) has exactly one point of maxima

Answer 2

f '(x) = $\frac { 1 - x ^ { 2 } \sec ^ { 2 } x } { ( 1 + x \tan x ) ^ { 2 } } = \frac { \sec ^ { 2 } x ( \cos x + x ) ( \cos x - x ) } { ( 1 + x \tan x ) ^ { 2 } }$

clearly f '(x₀) = 0 and f'(x) > 0 ∀ x ∈ (0, x₀)

f '(x) <0 ∀ x ∈ (x₀, π/2). Thus x = x₀ is the only point of local maxima for y = f(x).