JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q12

MathematicsSingle correct+4 / -1Question 12 of 90

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

3+\dfrac{\sin(1)\cos^2\left(\frac12\right)}2

3+\dfrac12(1+2\cos(1))\sin(1)

5+\dfrac12(\sin(1)+\sin(2))

2+\sin\left(\dfrac12\right)\cos\left(\dfrac12\right)

Answer and worked solution

Official answer: B

Sign of the quadratic. $2x^2+3x-2=(2x-1)(x+2)$ is negative on $[0,\frac12)$ and positive after.

Monotonic pieces. On $[0,\frac12]$ : $f=-(2x^2+3x-2)+\frac12\sin2x$ , $f'=-4x-3+\cos2x<0$ . On $[\frac12,1]$ : $f'=4x+3+\cos2x>0$ .

Extremes. Minimum $f\left(\frac12\right)=\frac12\sin1$ . Maximum is the larger of $f(0)=2$ and $f(1)=3+\frac12\sin2$ , i.e. $3+\frac12\sin2$ .

Sum. $3+\frac12(\sin2+\sin1)=3+\frac12\sin1\,(2\cos1+1)$ .

(B)

Topics:Maxima and minima Inequalities and modulus

JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 12