Let f be an odd function defined on the set of real numbers such that for x ≥ 0, f(x)=3 sin x+4 cos x. Then f( x) at x=- 11 π 6 is equal to:

Given $\mathrm{f}$f be an odd function $f(x)=3 \sin x+4 \cos x$f(x)=3sinx+4cosx Now, $$\begin{array}{r} f\left(\frac{-11 \pi}{6}\right)=3 \sin \left(\frac{-11 \pi}{6}\right)+4 \cos \left(\frac{-11 \pi}{6}\right) \\ f\left(\frac{-11 \pi}{6}\right)=3 \sin \left(-2 \pi+\frac{\pi}{6}\right)+4 \cos \left(-2 \pi+\frac{\pi}{6}\right) \\ f\left(\frac{-11 \pi}{6}\right)=3 \sin \left\{-\left(2 \pi-\frac{\pi}{6}\right)\right\}+4 \cos \left\{-\left(2 \pi-\frac{\pi}{6}\right)\right\} \\ \left\{\text { For odd functions } \begin{array}{l} \sin (-\theta)=-\sin \theta \\ \text { and } \cos (-\theta)=-\cos \theta \end{array}\right\} \end{array}$$f(6−11π​)=3sin(6−11π​)+4cos(6−11π​)f(6−11π​)=3sin(−2π+6π​)+4cos(−2π+6π​)f(6−11π​)=3sin{−(2π−6π​)}+4cos{−(2π−6π​)}{ For odd functions sin(−θ)=−sinθ and cos(−θ)=−cosθ​}​ $$\begin{aligned} \therefore f\left(\frac{-11 \pi}{6}\right)=-3 \sin \left(2 \pi-\frac{\pi}{6}\right)-4 \cos \left(2 \pi-\frac{\pi}{6}\right) \\ \Rightarrow f\left(\frac{-11 \pi}{6}\right) &=+3 \sin \left(\frac{\pi}{6}\right)-4 \cos \frac{\pi}{6} \\ \Rightarrow f\left(\frac{-11 \pi}{6}\right) &=3 \times \frac{1}{2}-4 \times \frac{\sqrt{3}}{2} \\ \text { or } f\left(\frac{-11 \pi}{6}\right) &=\frac{3}{2}-2 \sqrt{3} \end{aligned}$$∴f(6−11π​)=−3sin(2π−6π​)−4cos(2π−6π​)⇒f(6−11π​)⇒f(6−11π​) or f(6−11π​)​=+3sin(6π​)−4cos6π​=3×21​−4×23​​=23​−23​​