Question

The roll'd then is applied to $f\left(x\right)=sinx-cosx-1;xϵ[\frac{\pi }{2},\pi ]thenfin{d}^{-}{c}^{{}^{-}}inusalϵ[\frac{\pi }{2},\pi ]$

Answer 1

We have,

Given that

A real valued function

$f$ is continuous on $[\frac{\pi }{2},\pi ]$

$f$ is differentiable on $(\frac{\pi }{2},\pi )$

$f\left(x\right)=\sin x-\cos x-1$

$f\left(\frac{\pi }{2}\right)=\sin \frac{\pi }{2}-\cos \frac{\pi }{2}-1$

$=1-0-1$

$f\left(\frac{\pi }{2}\right)=0$

$f\left(\pi \right)=\sin \pi -\cos \pi -1$

$=0-(-1)-1$

$=1-1$

$f\left(\pi \right)=0$

Now,

$f\left(a\right)=f\left(b\right)$

$f\left(\frac{\pi }{2}\right)=f\left(\pi \right)$

Now, $f^{\prime }\left(c\right)=0$

So,$f\left(x\right)=\sin x-\cos x-1$

On differentiating this and we get,

$f^{\prime }\left(x\right)=\cos x-(-\sin x)-0$

$=\cos x+\sin x$

$f^{\prime }\left(C\right)=\cos C+\sin C$

Now,

$f^{\prime }\left(C\right)=0$

$0=\cos C+\sin C$

$\cos C+\sin C=0$

$\cos C=-\mathrm{sinC}$

$-\tan C=1$

$\tan C=-1$

$\tan C=-\tan \frac{\pi }{4}$

$\tan C=\tan (\pi -\frac{\pi }{4})$

$C=\frac{3\pi }{4}$

Hence, this is the answer.