Q31 / 90 · JEE Main 2014 (11 Apr Online) · Mathematics
Let $\mathrm{f}$ be an odd function defined on the set of real numbers such that for $\mathrm{x} \geq 0$, $f(x)=3 \sin x+4 \cos x$.
Then $\mathrm{f}(\mathrm{x})$ at $\mathrm{x}=-\frac{11 \pi}{6}$ is equal to:
- A$\frac{3}{2}+2 \sqrt{3}$
- B$-\frac{3}{2}+2 \sqrt{3}$
- Correct answerC$\frac{3}{2}-2 \sqrt{3}$
- D$-\frac{3}{2}-2 \sqrt{3}$
Solution
Given $\mathrm{f}$ be an odd function $f(x)=3 \sin x+4 \cos x$
Now,
$$
\begin{array}{r}
f\left(\frac{-11 \pi}{6}\right)=3 \sin \left(\frac{-11 \pi}{6}\right)+4 \cos \left(\frac{-11 \pi}{6}\right) \\
f\left(\frac{-11 \pi}{6}\right)=3 \sin \left(-2 \pi+\frac{\pi}{6}\right)+4 \cos \left(-2 \pi+\frac{\pi}{6}\right) \\
f\left(\frac{-11 \pi}{6}\right)=3 \sin \left\{-\left(2 \pi-\frac{\pi}{6}\right)\right\}+4 \cos \left\{-\left(2 \pi-\frac{\pi}{6}\right)\right\} \\
\left\{\text { For odd functions } \begin{array}{l}
\sin (-\theta)=-\sin \theta \\
\text { and } \cos (-\theta)=-\cos \theta
\end{array}\right\}
\end{array}
$$
$$
\begin{aligned}
\therefore f\left(\frac{-11 \pi}{6}\right)=-3 \sin \left(2 \pi-\frac{\pi}{6}\right)-4 \cos \left(2 \pi-\frac{\pi}{6}\right) \\
\Rightarrow f\left(\frac{-11 \pi}{6}\right) &=+3 \sin \left(\frac{\pi}{6}\right)-4 \cos \frac{\pi}{6} \\
\Rightarrow f\left(\frac{-11 \pi}{6}\right) &=3 \times \frac{1}{2}-4 \times \frac{\sqrt{3}}{2} \\
\text { or } f\left(\frac{-11 \pi}{6}\right) &=\frac{3}{2}-2 \sqrt{3}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.