Question

If $f\left(x\right)=\sin x+\cos x$, then the number of solution of $f\left(x\right)=\left[f\left(\frac{\pi }{10}\right)\right]$, for $x\in (0,4\pi )$ is
(where [.] represent greatest integer function)

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3$

We know that,
$f\left(\frac{\pi }{10}\right)=\sin \frac{\pi }{10}+\cos \frac{\pi }{10}\Rightarrow f\left(\frac{\pi }{10}\right)=\sqrt{2}\sin (\frac{\pi }{10}+\frac{\pi }{4})$
Now,
$\sin \frac{\pi }{4}<\sin (\frac{\pi }{10}+\frac{\pi }{4})<\sin \frac{\pi }{2}\frac{1}{\sqrt{2}}<\sin (\frac{\pi }{10}+\frac{\pi }{4})<11<\sqrt{2}\sin (\frac{\pi }{10}+\frac{\pi }{4})<\sqrt{2}$
So,

Now the general solution will be,
$\sin x+\cos x=1\Rightarrow \sin (\frac{\pi }{4}+x)=\frac{1}{\sqrt{2}}\Rightarrow \frac{\pi }{4}+x=2n\pi +\frac{\pi }{4}\Rightarrow x=2n\pi$
Or
$\Rightarrow \frac{\pi }{4}+x=2n\pi +\frac{3\pi }{4}\Rightarrow x=2n\pi +\frac{\pi }{2}$
Hence, $x=\frac{\pi }{2},2\pi ,\frac{5\pi }{2}$