Question

The maximum value of $1+\sin (\frac{\pi }{4}+\theta )+2\cos (\frac{\pi }{4}-\theta )$ for real value of $\theta$ is

• A. $3$
• B. $5$
• C. $4$
• D. none of these

Answer 1

The correct option is C $4$

$=1+\frac{\sin \theta +\cos \theta }{\sqrt{2}}+2\frac{\sin \theta +\cos \theta }{\sqrt{2}}$

$=1+3\frac{\sin \theta +\cos \theta }{\sqrt{2}}$

$max(\sin \theta +\cos \theta )=\sqrt{2}$

$f\left(x\right)=\sin \theta +\cos \theta$
$f^{\prime }\left(x\right)=\sin \theta -\cos \theta =0$
So,$\sin \theta =\cos \theta =\frac{1}{\sqrt{2}}$

So, $max\left(f\right(x\left)\right)=\frac{2}{\sqrt{2}}=\sqrt{2}$

So, max of the eqn$=3+1=4$