First slide

Functions (XII)

Question

Consider the real-valued function satisfying $2 f (\sin x) + f (\cos x) = x$ Then the

Difficult

A

$domain of f (x) is R$
B

$domain of f (x) is [- 1, 1]$
C

$range of f (x) is [- \frac{2 π}{3}, \frac{π}{3}]$
D

$range of f (x) is R$

Solution

$\begin{array}{l} Given 2 f (\sin x) + f (\cos x) = x (1) \\ Replacing x by \frac{π}{2} - x, we ge \\ 2 f (\cos x) + f (\sin x) = \frac{π}{2} - x (2) \end{array}$
Eliminating f(cosx) from (l) and (2), we get
$\begin{array}{l} 3 f (\sin x) = 3 x - \frac{π}{2} \\ or f (\sin x) = x - \frac{π}{6} \\ or f (x) = \sin^{- 1} x - \frac{π}{6} \\ f (x) has domain [- 1, 1] . \\ Also, \sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}] or \sin^{- 1} x - \frac{π}{6} \in [- \frac{2 π}{3}, \frac{π}{3} .] \end{array}$

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