Question

Consider the real-valued function satisfying

2 f (\sin x) + f (\cos x) = x .

Find the domain and range of

f (x)

, using the Principal domain and range.

Accepted Answer

It is given that $2 f (\sin x) + f (\cos x) = x \dots (i)$

$x$ is replace by $\frac{π}{2} - x$

$2 f (\sin (\frac{π}{2} - x)) + f (\cos (\frac{π}{2} - x)) = \frac{π}{2} - x$

$\Rightarrow$ $2 f (\cos x) + f (\sin x) = \frac{π}{2} - x \dots (ii)$

Multiply by $2$ in equation $(ii)$

$\Rightarrow 4 f (\cos x) + 2 f (\sin x) = π - 2 x \dots (iii)$

Now, subtract equation $(iii)$ from $(i)$

$\Rightarrow f (\cos x) - 4 f (\cos x) = x - (π - 2 x)$

$\Rightarrow - 3 f (\cos x) = 3 x - π$

$\Rightarrow f (\cos x) = - (x - \frac{π}{3})$

Let, $\cos x = θ \Rightarrow x = \cos^{- 1} θ, θ \in [- 1, 1]$ and $x \in [0, π]$

Then, $f (\cos (\cos^{- 1} θ)) = - \cos^{- 1} θ + \frac{π}{3}$

$\Rightarrow f (θ) =$ $- \cos^{- 1} θ + \frac{π}{3}$

Domain $θ \in [- 1, 1]$

For range $,$ $0 \leq \cos^{- 1} θ \leq π$ $\Rightarrow - π \leq - \cos^{- 1} θ \leq 0$

$\Rightarrow - π + \frac{π}{3} \leq - \cos^{- 1} θ + \frac{π}{3} \leq \frac{π}{3}$

$\Rightarrow - \frac{2 π}{3} \leq f (θ) \leq \frac{π}{3}$

$∴ R_{f} = [- \frac{2 π}{3}, \frac{π}{3}]$

Then, Domain of $f$ is $[- 1, 1]$ and range is $[- \frac{2 π}{3}, \frac{π}{3}]$

Amit M Agarwal Solutions for Chapter: Functions, Exercise 6: Target Exercise 3.6

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Functions, Exercise 6: Target Exercise 3.6

Questions from Amit M Agarwal Solutions for Chapter: Functions, Exercise 6: Target Exercise 3.6 with Hints & Solutions

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