Question

The minimum & maximum value of $f\left(x\right)=\sin \left(\cos x\right)+\cos \left(\sin x\right)\forall -\frac{\pi }{2}\le x\le \frac{\pi }{2}$ are respective

• A. $\cos 1$ and $1+\sin 1$
• B. $\sin 1$ and $1+\cos 1$
• C. $\cos 1$ and $\cos \left(\frac{1}{\sqrt{2}}\right)+\sin \left(\frac{1}{\sqrt{2}}\right)$
• D. None of these

Answer 1

Answer: (A)

$\cos 1$ and $1+\sin 1$

$f^{\prime }\left(x\right)=cos\left(cosx\right)(-sinx)+-sin\left(sinx\right)cosxas{f}^{\prime }\left(x\right)=0-cos\left(cosx\right)sinx=sin\left(sinx\right)cosxatx=-\left(\frac{\pi }{2}\right),0,\left(\frac{\pi }{2}\right)nowf\left(\frac{\pi }{2}\right)=sin(cos-(\frac{\pi }{2}\left)\right)+cos\left(sin\right(\frac{\pi }{2}\left)\right)=cos1f\left(0\right)=sin\left(cos0\right)+cos\left(sin0\right)=sin1+1f\left(\frac{\pi }{2}\right)=sin\left(cos\right(\frac{\pi }{2}\left)\right)+cos\left(sin\right(\frac{\pi }{2}\left)\right)=cos1maxvalue=1+sin1andminvalue=cos1$