For $x \in\left(0, \frac{3}{2}\right)$, let $f(x)=\sqrt{x}, g(x)=\tan x$ and $h(x)=\frac{1-x^{2}}{1+x^{2}}$. If $\quad \phi(x)=((h \operatorname{of}) \operatorname{og})(x)$, then $\phi\left(\frac{\pi}{3}\right)$ is equal to :

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Sets Relations and Functions

For

x \in (0, \frac{3}{2})

, let

f (x) = x, g (x) = tan x

and

h (x) = \frac{1 - x ^{2}}{1 + x ^{2}}

. If

ϕ (x) = ((h of) og) (x)

, then

ϕ (\frac{π}{3})

is equal to :

Answer: (A)

Build the composite from the inside out:

(f \circ g) (x) = tan x

, and then

ϕ (x) = h (tan x) = \frac{1 - tan x}{1 + tan x}

. Recognising the tangent-difference pattern,

\frac{1 - tan x}{1 + tan x} = tan (\frac{π}{4} - x)

. At

x = \frac{π}{3}

this is

tan (\frac{π}{4} - \frac{π}{3}) = tan (- \frac{π}{12}) = - tan \frac{π}{12}

, and since

- tan θ = tan (π - θ)

, this equals

tan \frac{11 π}{12}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option