Tag: Miscellaneous Exercise on Chapter 6 Question 11 NCERT Class 12th Maths Solution

Class 12th   Class 12th Maths

Question 11.
Find the absolute maximum and minimum values of the function

$$f(x)={\cos }^{2}x+\sin x,x\in [0,\pi ]$$

Solution

Given:

$$f(x)={\cos }^{2}x+\sin x=1-{\sin }^{2}x+\sin x$$
$$f(x)=1+\sin x-{\sin }^{2}x$$

Step 1: First derivative

$$f^{\mathrm{\prime }}(x)=\cos x-2\sin x\cos x$$
$$f^{\mathrm{\prime }}(x)=\cos x(1-2\sin x)$$

Set $f^{\mathrm{\prime }}(x)=0$:

$$\cos x=0\text{or}1-2\sin x=0$$

1. $\cos x=0\Rightarrow x=\frac{\pi }{2}$
   

2. $1-2\sin x=0\Rightarrow \sin x=\frac{1}{2}$

$$x=\frac{\pi }{6},\frac{5\pi }{6}$$

So critical points:

$$x=\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}$$

Also evaluate function at end points:

$$x=0,x=\pi$$

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Step 2: Evaluate $f(x)$ at critical and end points

Final Answer

Absolute Maximum value

$$\boxed{{\displaystyle \frac{5}{4}}}\text{at }x=\frac{\pi }{6},\frac{5\pi }{6}$$

Absolute Minimum value

$$\boxed{{\displaystyle 1}}\text{at }x=0,\frac{\pi }{2},\pi$$

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Extras :-

$$\mathbf{\text{Graph}}\mathbf{\text{of}}f(x)={\cos }^{2}x+\sin x\mathbf{\text{on}}[0,\pi ]$$

The graph now properly displays the exact function heading centered at the top.
The curve and marked points show clearly:

• Absolute Maximum Value

  $$f\left(\frac{\pi }{6}\right)=f\left(\frac{5\pi }{6}\right)=\frac{5}{4}$$

• Absolute Minimum Value

  $$f(0)=f\left(\frac{\pi }{2}\right)=f(\pi )=1$$