Q42 / 90 · JEE Main 2022 (24 Jun Shift 1) · Mathematics

The sum of absolute maximum and absolute minimum values of the function $f (x) = |2 x^{2} + 3 x - 2| + \sin x \cos x$ in the interval $[0, 1]$ is

A
$3 + \frac{\sin (1) \cos^{2} (\frac{1}{2})}{2}$
Correct answer
B
$3 + \frac{1}{2} (1 + 2 \cos (1)) \sin (1)$
C
$5 + \frac{1}{2} (\sin (1) + \sin (2))$
D
$2 + \sin (\frac{1}{2}) \cos (\frac{1}{2})$

Solution

Given $f (x) = |2 x^{2} + 3 x - 2| + \sin x \cos x$

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

Now, $f^{'} (x) = \{\begin{matrix} 4 x + 3 + \frac{\cos 2 x}{4}, & \frac{1}{2} < x < 1 \\ - (4 x + 3) + \frac{\cos 2 x}{4}, & 0 \leq x < \frac{1}{2} \end{matrix}$

For $0 \leq x < \frac{1}{2} \Rightarrow f^{'} (x) < 0$

For $\frac{1}{2} < x \leq 1 \Rightarrow f^{'} (x) > 0$

So $f (x)$ has minima at $x = \frac{1}{2}$ and maxima at $x = 1$

Hence, $f (\frac{1}{2}) + f (1) = 3 + \frac{1}{2} (1 + 2 \cos 1) \sin 1$

Hence, the correct answer is option: 'B'.

The sum of absolute maximum and absolute minimum values of the function fx=2x2+3x-2+sinxcosx in the interval 0,1 is

Solution

Relevant Questions

The sum of absolute maximum and absolute minimum values of the function $f (x) = |2 x^{2} + 3 x - 2| + \sin x \cos x$ in the interval $[0, 1]$ is