Question

The maximum and minimum value of $ab\sin \left(x\right)+b\sqrt{1-a^2}\cos \left(x\right)+c$ lie in the interval where $\left|a\right|<1$, $b>0$

• A. $\left[b-c,b+c\right]$
• B. $\left(b-c,b+c\right)$
• C. $\left[c-b,b+c\right]$
• D. None of these

Answer 1

The correct option is C

$\left[c-b,b+c\right]$

Explanation for the correct option

Step 1: Solve for the critical points

Given functions is $ab\sin \left(x\right)+b\sqrt{1-a^2}\cos \left(x\right)+c$
We can assume that $a=\cos \left(\theta \right)$ since $\left|a\right|<1$

So,
$\begin{array}{rcl}f\left(x\right)& =& b\left(\cos \left(\theta \right)\sin \left(x\right)+\cos \left(x\right)\sin \left(\theta \right)\right)+c\left[\because \sqrt{1-{\cos }^2\left(\theta \right)}=\sin \left(\theta \right)\right]\\ & =& b\sin \left(x+\theta \right)+c\left[\because \sin \left(x+y\right)=\sin \left(x\right)\cos \left(y\right)+\cos \left(x\right)\sin \left(y\right)\right]\end{array}$

A function has local minima at $a$ if $f\text{'}\left(a\right)=0$ and $f\text{'}\text{'}\left(a\right)>0$ and has local maxima if $f\text{'}\text{'}\left(a\right)<0$

So,
$f\text{'}\left(x\right)=b\cos \left(x+\theta \right)$

To have $f\text{'}\left(x\right)=0$,
$\begin{array}{cc}& b\cos \left(x+\theta \right)=0\\ \Rightarrow & \cos \left(x+\theta \right)=0\\ \Rightarrow & x=\frac{\mathrm{\pi }}{2}-\theta ,\frac{3\mathrm{\pi }}{2}-\theta \end{array}$

Step 2: Solve for the required interval
$f\text{'}\text{'}\left(x\right)=-b\sin \left(x+\theta \right)$

At $x=\frac{\mathrm{\pi }}{2}-\theta$,
$\begin{array}{rcl}f\left(\frac{\mathrm{\pi }}{2}-\theta \right)& =& b\sin \left(\frac{\mathrm{\pi }}{2}-\theta +\theta \right)+c\\ & =& b+c\end{array}$

At $x=\frac{3\mathrm{\pi }}{2}-\theta$,
$\begin{array}{rcl}f\left(\frac{3\mathrm{\pi }}{2}-\theta \right)& =& b\sin \left(\frac{3\mathrm{\pi }}{2}-\cancel{\theta }+\cancel{\theta }\right)+c\\ & =& b\sin \left(\frac{\mathrm{\pi }}{2}+\mathrm{\pi }\right)+c\\ & =& c-b\left[\because \sin \left(\mathrm{\pi }+x\right)=-\sin \left(x\right)\right]\end{array}$

Therefore, the interval in which the given function has a maximum and minimum value is $\left[c-b,b+c\right]$

Hence, option(C) i.e. $\left[c-b,b+c\right]$ is correct.