The range of the function $f(x)=\log _{\sqrt{5}}\left(3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\right)$ is:

(B) $\left[\frac{1}{\sqrt{5}}, \sqrt{5}\right]$

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Sets Relations and Functions

The range of the function

f (x) = lo g_{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:

Answer: (D)

Simplify the argument using sum-to-product formulas:

cos (\frac{π}{4} + x) + cos (\frac{π}{4} - x) = 2 cos \frac{π}{4} cos x = 2 cos x

cos (\frac{3 π}{4} + x) - cos (\frac{3 π}{4} - x) = - 2 sin \frac{3 π}{4} sin x = - 2 sin x

So the argument becomes:

3 + 2 cos x - 2 sin x = 3 + 2 (cos x - sin x)

Since

- 2 \leq cos x - sin x \leq 2

, we have:

3 - 2 \leq 3 + 2 (cos x - sin x) \leq 3 + 2

1 \leq argument \leq 5

Therefore:

lo g_{5} 1 \leq f (x) \leq lo g_{5} 5

0 \leq f (x) \leq 2

The range is

[0, 2]

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option