Question

Let $\alpha$ be a fixed constant number such that $0<\alpha< \frac{\pi}{2}$. The function $F$ is defined by $F(\theta)=\int\limits_0^\theta x \cos (x+\alpha) d x$. 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 $\theta \in\left[0, \frac{\pi}{2}-\alpha\right) \left[\right.$ As $0<\theta+\alpha<\pi$, so $\left.\theta+\alpha=\frac{\pi}{2} \Rightarrow \theta=\left(\frac{\pi}{2}-\alpha\right)\right]$ and $F$ is decreasing on $\theta \in\left(\frac{\pi}{2}-\alpha, \frac{\pi}{2}\right]$
$\text { we have } F (\theta)=\int\limits_0^\theta \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 $
$\Rightarrow F _{\max }= F \left(\frac{\pi}{2}-\alpha\right)=\frac{\pi}{2}-\alpha-\cos \alpha$