Question

provide answer with explanation

Answer 1

Answer:

(B)

Step-by-step explanation:

We have,

$f(x)=2\sin(x)-x+1$

$\implies\,f^{\prime}(x)=2\cos(x)-1$

For maxima or minima, we have f'(x)=0

So,

$\implies\,2\cos(x)-1=0$

$\implies\,\cos(x)=\dfrac{1}{2}$

Since  $x\in\left(0,\dfrac{\pi}{2}\right)$, so,

$\implies\,x=\dfrac{\pi}{3}$

$\tt{\bullet\,When\,\,x<\dfrac{\pi}{3}\,,\,\,\,}\tt{cos(x)>\dfrac{1}{2}}\\\\\tt{\implies2\,cos(x)-1>0}$

$\tt{\bullet\,When\,\,x>\dfrac{\pi}{3}\,,\,\,\,}\tt{cos(x)<\dfrac{1}{2}}\\\\\tt{\implies2\,cos(x)-1<0}$

So, f'(x) changes its sign from positive to negative

$\sf{f(x)\,\,has\,\,local\,\,maxima\,\,at\,\,x=\dfrac{\pi}{3}}$