Question

Let $\alpha$ be a fixed constant number such that $0<\alpha <\frac{\pi }{2}$. The function $F$ is defined by $F(\theta )=\underset{0}{\overset{\theta }{\int }}x\cos (x+\alpha )dx$. If $\theta$ lies in the range of $\left[0,\frac{\pi }{2}\right]$, then the maximum value of $F(\theta )$, is

• A. $\frac{\pi }{2}-\alpha +\sin \alpha$
• B. $\frac{\pi }{2}-\alpha +\cos \alpha$
• C. $\frac{\pi }{2}-\alpha -\sin \alpha$
• D. $\frac{\pi }{2}-\alpha -\cos \alpha$

Answer 1

Answer: (D)

As $F^{\prime }(\theta )=\theta \cos (\theta +\alpha )$, where $\theta +\alpha \in (0,\pi )$
For maximum $F^{\prime }(\theta )=0\Rightarrow \theta =0$ or $\theta =\frac{\pi }{2}-\alpha$
$F^{\prime \prime }(\theta )=\cos (\theta +\alpha )-\theta \sin (\theta +\alpha )$
$F^{\prime \prime }(\theta )>0\text{ and }{F}^{\prime \prime }\left(\frac{\pi }{2}-\alpha \right)<0\Rightarrow \text{ Maxima at }x=\frac{\pi }{2}-\alpha$
Now, $F$ is increasing on $\theta \in \left[0,\frac{\pi }{2}-\alpha \right)[$ As $0<\theta +\alpha <\pi$, so $\theta +\alpha =\frac{\pi }{2}\Rightarrow \theta =\left(\frac{\pi }{2}-\alpha \right)]$ and $F$ is decreasing on $\theta \in \left(\frac{\pi }{2}-\alpha ,\frac{\pi }{2}\right]$
$\text{ we have }F(\theta )=\underset{0}{\overset{\theta }{\int }}{\cos }_{\text{(I.B.P) }}^x\cos \left(x\frac{11}{4}\frac{\alpha }{3}\right)dx=\theta \sin (\theta +\alpha )+\cos (\theta +\alpha )-\cos \alpha$
$\Rightarrow F_{\max }=F\left(\frac{\pi }{2}-\alpha \right)=\frac{\pi }{2}-\alpha -\cos \alpha$