Consider and x. The value of x for which f (g(x)) is minimum is where k, then find k+ .

JEE MainMathematicsToughFunctions

Consider $f(x) = |x - 1| + |2x - \pi| + |x - 3|$ and
$g(x) = \sin^{- 1}x + \tan^{- 1}$ x. The value of x for which f (g(x))
is minimum is $\sqrt{k\sin\frac{\pi}{\lambda}}$ where k, $\lambda \in N$
then find k+ $\lambda$ .

Show answer and solution

f (x) is minimum at

\ x = \frac{\pi}{2}

\Rightarrow \sin^{- 1}x + \tan^{- 1}x = \frac{\pi}{2}

\tan^{- 1}x = \cos^{- 1}x

\tan^{- 1}x = \tan^{- 1}\left( \frac{\sqrt{1 - x^{2}}}{x} \right)

= x^{2} = \frac{1 - x^{2}}{x^{2}}

x^{4} + x^{2} - 1 = 0

x^{2} = \frac{- 1 + \sqrt{5}}{2}

x = \sqrt{2sin18^{0}} = \sqrt{2sin\frac{\pi}{10}}

\Rightarrow 12 = k + \lambda

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

Scan any doubt from your notes and get a whiteboard explanation.