JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q8

MathematicsSingle correct+4 / -1Question 8 of 90

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

\dfrac{\sqrt{17}+3}2

\dfrac{\sqrt{17}+5}2

\dfrac{9-\sqrt{17}}2

Answer and worked solution

Official answer: A

Sign change. $3x-x^2+2=0$ at $x_0=\frac{3-\sqrt{17}}2\approx-0.56$ , which is in $[-1,2]$ .

Pieces. On $[-1,x_0]$ : $f=x^2-4x-2$ , decreasing from $f(-1)=3$ . On $[x_0,2]$ : $f=-x^2+2x+2$ , rising to $f(1)=3$ , then $f(2)=2$ .

Extremes. Maximum 3; minimum $f(x_0)=-x_0=\frac{\sqrt{17}-3}2$ . Sum $\dfrac{\sqrt{17}+3}2$ .

(A)

Topics:Maxima and minima Inequalities and modulus

JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 8