Q180 / 270 · NDA 2026 (Phase 1) · Mathematics
$\cos x+\sqrt{3}\sin x$ is maximum when $x$ is equal to
- A$\dfrac{\pi}{2}$
- Correct answerB$\dfrac{\pi}{3}$
- C$\dfrac{\pi}{4}$
- D$\dfrac{\pi}{6}$
Solution
Given expression is $f(x) = \cos x + \sqrt{3} \sin x$
Multiplying and dividing by $\sqrt{1^2 + (\sqrt{3})^2} = 2$
$f(x) = 2 \left( \dfrac{1}{2} \cos x + \dfrac{\sqrt{3}}{2} \sin x \right)$
$f(x) = 2 \left( \cos \dfrac{\pi}{3} \cos x + \sin \dfrac{\pi}{3} \sin x \right)$
$f(x) = 2 \cos \left( x - \dfrac{\pi}{3} \right)$
The maximum value of $\cos \left( x - \dfrac{\pi}{3} \right)$ is $1$, which occurs when $x - \dfrac{\pi}{3} = 0$
$x = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
Multiplying and dividing by $\sqrt{1^2 + (\sqrt{3})^2} = 2$
$f(x) = 2 \left( \dfrac{1}{2} \cos x + \dfrac{\sqrt{3}}{2} \sin x \right)$
$f(x) = 2 \left( \cos \dfrac{\pi}{3} \cos x + \sin \dfrac{\pi}{3} \sin x \right)$
$f(x) = 2 \cos \left( x - \dfrac{\pi}{3} \right)$
The maximum value of $\cos \left( x - \dfrac{\pi}{3} \right)$ is $1$, which occurs when $x - \dfrac{\pi}{3} = 0$
$x = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
Hence, the correct answer is option: 'B'.