Q61 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
If the function $f(x)=\mu \sin x+\dfrac{1}{3} \sin 3 x$ has its derivative equal to zero at $x=\dfrac{\pi}{3}$, then the value of ' $\mu$ ' is
- A$1$
- B$-1$
- Correct answerC$2$
- D$0$
Solution
The given function is $f(x) = \mu \sin x + \dfrac{1}{3} \sin 3x$.
The derivative of the function is $f'(x) = \dfrac{d}{dx} (\mu \sin x + \dfrac{1}{3} \sin 3x) = \mu \cos x + \dfrac{1}{3} \cos 3x \times 3 = \mu \cos x + \cos 3x$.
It is given that $f'(x) = 0$ at $x = \dfrac{\pi}{3}$.
Substituting $x = \dfrac{\pi}{3}$ into the derivative expression:
$f'(\dfrac{\pi}{3}) = \mu \cos(\dfrac{\pi}{3}) + \cos(3 \times \dfrac{\pi}{3}) = 0$.
Since $\cos(\dfrac{\pi}{3}) = \dfrac{1}{2}$ and $\cos(\pi) = -1$, we have:
$\mu(\dfrac{1}{2}) + (-1) = 0$.
$\dfrac{\mu}{2} = 1$.
$\mu = 2$.
Answer: $2$
The derivative of the function is $f'(x) = \dfrac{d}{dx} (\mu \sin x + \dfrac{1}{3} \sin 3x) = \mu \cos x + \dfrac{1}{3} \cos 3x \times 3 = \mu \cos x + \cos 3x$.
It is given that $f'(x) = 0$ at $x = \dfrac{\pi}{3}$.
Substituting $x = \dfrac{\pi}{3}$ into the derivative expression:
$f'(\dfrac{\pi}{3}) = \mu \cos(\dfrac{\pi}{3}) + \cos(3 \times \dfrac{\pi}{3}) = 0$.
Since $\cos(\dfrac{\pi}{3}) = \dfrac{1}{2}$ and $\cos(\pi) = -1$, we have:
$\mu(\dfrac{1}{2}) + (-1) = 0$.
$\dfrac{\mu}{2} = 1$.
$\mu = 2$.
Answer: $2$
Hence, the correct answer is option: 'C'.