Question

The function y = 1 + sin x is maximum, when x = ______ 

पर्याय

• ${\displaystyle \frac{3\pi }{2}}$

• ${\displaystyle \frac{\pi }{2}}$

• π

• ${\displaystyle -\frac{\pi }{6}}$

Answer 1

The function y = 1 + sin x is maximum, when x = ${\displaystyle \underline{\frac{\pi }{2}}}$.

Explanation:

y = 1 + sinx

∴ y' = cosx

For maximum or minimum,

y' = 0 ⇒ cosx = 0 ⇒ ${\displaystyle x=\frac{\pi }{2},\frac{3\pi }{2}}$

Now, y'' = -sinx

⇒ ${\displaystyle y^{\prime \prime }\left(\frac{\pi }{2}\right)=-1<0}$ and ${\displaystyle y^{\prime \prime }\left(\frac{3\pi }{2}\right)=1>0}$

∴ y is maximum w en x = ${\displaystyle \frac{\pi }{2}}$.