The range of the function f(x)=log_53+cos(3π)/4+x+cos(π)/4+x+cos(π)/4-x-cos(3π)/4-x is :

Correct answer: (B) [0, 2]. f(x)=log_5(3-√2sinx+√2cosx) -2≤(-√2sinx+√2cosx)≤2 ⇒1≤(3-√2sinx+√2cosx)≤5 ⇒log_51≤f(x)≤log_55 ⇒0≤f(x)≤2

Q34 / 90 · JEE Main 2021 (01 Sep Shift 2) · Mathematics

The range of the function $f (x) = \log_{\sqrt{5}} (3 + \cos (\frac{3 π}{4} + x) + \cos (\frac{π}{4} + x) + \cos (\frac{π}{4} - x) - \cos (\frac{3 π}{4} - x))$ is :

$f (x) = \log_{\sqrt{5}} (3 - \sqrt{2} \sin x + \sqrt{2} \cos x)$

$- 2 \leq (- \sqrt{2} \sin x + \sqrt{2} \cos x) \leq 2$

$\Rightarrow 1 \leq (3 - \sqrt{2} \sin x + \sqrt{2} \cos x) \leq 5$

$\Rightarrow \log_{\sqrt{5}} 1 \leq f (x) \leq \log_{\sqrt{5}} 5$

$\Rightarrow 0 \leq f (x) \leq 2$

Hence, the correct answer is option: 'B'.