Question

The constant $c$ of Lagrange's mean value theorem for $f(x)=\cos x-\sin 2x$ in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ is

• A. 0
• B. ${\sin }^{-1}\left(\frac{1\pm \sqrt{33}}{8}\right)$
• C. ${\cos }^{-1}\left(\frac{1\pm \sqrt{33}}{8}\right)$
• D. $\pm \frac{\pi }{4}$

Answer 1

Answer: (B)

The given function $f(x)=\cos x-\sin 2x$, is a
continuous function in interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$
and differentiable in interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.
Now, according to Lagrange's mean value theorem, there exist
$c\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, such that
$f^{\prime }(c)=\frac{f\left(\frac{\pi }{2}\right)-f\left(-\frac{\pi }{2}\right)}{\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)}$
$\Rightarrow -\sin c-2\cos (2c)=0$
$\Rightarrow \sin c+2-4{\sin }^{2}c=0$
$\Rightarrow 4{\sin }^{2}c-\sin c-2=0$
So, $\sin c=\frac{1\pm \sqrt{1+32}}{8}$
$\Rightarrow \sin c=\frac{1\pm \sqrt{33}}{8}$
$\Rightarrow c={\sin }^{-1}\left(\frac{1\pm \sqrt{33}}{8}\right)$