MHT CET · Maths · Trigonometric Equations
The length of the longest interval, in which the function \(3 \sin x-4 \sin ^3 x\) is increasing, is
- A \(\frac{\pi}{3}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{3 \pi}{2}\)
- D \(\pi\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{f}(x)=3 \sin x-4 \sin ^3 x \\
& \Rightarrow \mathrm{f}(x)=\sin 3 x \\
& \Rightarrow \mathrm{f}^{\prime}(x)=3 \cos 3 x
\end{aligned}\)
For \(\mathrm{f}(x)\) to be increasing,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x) \geq 0 \\
& \Rightarrow 3 \cos 3 x \geq 0 \\
& \Rightarrow \cos 3 x \geq 0 \\
& \Rightarrow \frac{-\pi}{2} \leq 3 x \leq \frac{\pi}{2} \\
& \Rightarrow \frac{-\pi}{6} \leq x \leq \frac{\pi}{6}
\end{aligned}\)
\(\therefore \quad\) largest length in which \(\mathrm{f}(x)\) is increasing
\(=\frac{\pi}{6}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{3}\)
& \mathrm{f}(x)=3 \sin x-4 \sin ^3 x \\
& \Rightarrow \mathrm{f}(x)=\sin 3 x \\
& \Rightarrow \mathrm{f}^{\prime}(x)=3 \cos 3 x
\end{aligned}\)
For \(\mathrm{f}(x)\) to be increasing,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x) \geq 0 \\
& \Rightarrow 3 \cos 3 x \geq 0 \\
& \Rightarrow \cos 3 x \geq 0 \\
& \Rightarrow \frac{-\pi}{2} \leq 3 x \leq \frac{\pi}{2} \\
& \Rightarrow \frac{-\pi}{6} \leq x \leq \frac{\pi}{6}
\end{aligned}\)
\(\therefore \quad\) largest length in which \(\mathrm{f}(x)\) is increasing
\(=\frac{\pi}{6}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{3}\)
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