Question

The function $f ( x ) = \frac { x } { 1 + x \tan x }$has

• A. One point of minimum in the interval (0, π/2)
• B. One point of maximum in the interval (0, π/2)
• C. No point of maximum, no point of minimum in the interval (0, π/2)
• D. ) Two points of maxima in the interval (0, π/2)

Answer 1

Answer: (B)

One point of maximum in the interval (0, π/2)

Answer 2

$g ( x ) = \frac { 1 } { f ( x ) } = \frac { 1 } { x } + \tan x$

When $g ( x )$ has minimum $f ( x )$ has maximum $g ^ { \prime } ( x ) = - \frac { 1 } { x ^ { 2 } } + \sec ^ { 2 } x$

For maxima & minima

$g ^ { \prime } ( x ) = 0$ ⇒ $\sec ^ { 2 } x = \frac { 1 } { x ^ { 2 } }$ ⇒ $\cos ^ { 2 } x = x ^ { 2 }$

⇒ $\cos x = x$ or $\cos x = - x$

$g ^ { \prime \prime } ( x ) = \frac { 2 } { x ^ { 3 } } + 2 \sec ^ { 2 } x \tan x$

$g ^ { \prime \prime } ( x ) = + v e$ so, only one minimum for $g ( x )$ in the interval $\left( 0 , \frac { \pi } { 2 } \right)$

⇒ only one maximum for $f ( x )$ in the interval $\left( 0 , \frac { \pi } { 2 } \right)$ .

So, 'b' is correct.