STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

$\therefore $  $ \mathrm{f}^{\prime}(\theta) =12 \cos \theta-18 \sin \theta \cos \theta $

$=6 \cos \theta(2-3) \sin \theta) $

Now $f^{\prime}(\theta)=0$ gives $\cos \theta=0$ or $\sin \theta=\frac{2}{3}$

$\Rightarrow \quad \sin \theta=1$ or $\sin \theta=\frac{2}{3}$

$\mathrm{f}^{\prime \prime}(\theta)=-12 \sin \theta-18\left[\cos ^{2} \theta-\sin ^{2} \theta\right]$

when $\sin \theta=1$

$f^{\prime \prime}(\theta)=-12-18[1-2]=+v e$

and when $\sin \theta=2 / 3$

$f^{\prime \prime}(\theta)=-8-18\left[1-\frac{4}{9}\right]=-v e$

$\therefore $   $f(\theta)$ is Max. when $\sin \theta=2 / 3$

$\therefore $    Max. $\mathrm{f}(\theta)=8-4=4$