Q84 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
If $f(x) = \cos x\,\cos 2x\,\cos 4x\,\cos 8x\,\cos 16x$, then $f'\left(\dfrac{\pi}{4}\right) =$
- Correct answerA$\text{cosec}\left(\dfrac{\pi}{4}\right)$
- B$\cos\left(\dfrac{\pi}{4}\right)$
- C$\tan\left(\dfrac{\pi}{4}\right)$
- D$0$
Solution
Using the trigonometric identity $\cos x \cos 2x \cos 2^2x \dots \cos 2^{n-1}x = \dfrac{\sin(2^n x)}{2^n \sin x}$
For $n=5$, the given function can be written as:
$f(x) = \dfrac{\sin(32x)}{32 \sin x}$
Differentiating with respect to $x$ using the quotient rule:
$f'(x) = \dfrac{32 \sin x \cdot 32 \cos(32x) - \sin(32x) \cdot 32 \cos x}{(32 \sin x)^2}$
$f'(x) = \dfrac{32 \cos(32x) \sin x - \sin(32x) \cos x}{32 \sin^2 x}$
Substituting $x = \dfrac{\pi}{4}$:
$32x = 32 \times \dfrac{\pi}{4} = 8\pi$
Since $\cos(8\pi) = 1$ and $\sin(8\pi) = 0$, we get:
$f'\left(\dfrac{\pi}{4}\right) = \dfrac{32(1)\sin\left(\dfrac{\pi}{4}\right) - 0}{32 \sin^2\left(\dfrac{\pi}{4}\right)}$
$f'\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sin\left(\dfrac{\pi}{4}\right)}$
$f'\left(\dfrac{\pi}{4}\right) = \text{cosec}\left(\dfrac{\pi}{4}\right)$
Answer: $\text{cosec}\left(\dfrac{\pi}{4}\right)$
For $n=5$, the given function can be written as:
$f(x) = \dfrac{\sin(32x)}{32 \sin x}$
Differentiating with respect to $x$ using the quotient rule:
$f'(x) = \dfrac{32 \sin x \cdot 32 \cos(32x) - \sin(32x) \cdot 32 \cos x}{(32 \sin x)^2}$
$f'(x) = \dfrac{32 \cos(32x) \sin x - \sin(32x) \cos x}{32 \sin^2 x}$
Substituting $x = \dfrac{\pi}{4}$:
$32x = 32 \times \dfrac{\pi}{4} = 8\pi$
Since $\cos(8\pi) = 1$ and $\sin(8\pi) = 0$, we get:
$f'\left(\dfrac{\pi}{4}\right) = \dfrac{32(1)\sin\left(\dfrac{\pi}{4}\right) - 0}{32 \sin^2\left(\dfrac{\pi}{4}\right)}$
$f'\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sin\left(\dfrac{\pi}{4}\right)}$
$f'\left(\dfrac{\pi}{4}\right) = \text{cosec}\left(\dfrac{\pi}{4}\right)$
Answer: $\text{cosec}\left(\dfrac{\pi}{4}\right)$
Hence, the correct answer is option: 'A'.