JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q7

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MathematicsSingle correct+4 / -1Question 7 of 90

Let $[t]$ denote the greatest integer less than or equal to $t$ . Then, the value of the integral $\displaystyle\int_0^1\left[-8x^2+6x-1\right]dx$ is equal to

-1

\dfrac{-5}4

\dfrac{\sqrt{17}-13}8

\dfrac{\sqrt{17}-16}8

Answer and worked solution

Official answer: C

Value of $g=-8x^2+6x-1$ on [0,1]. It is in $[0,\frac18]$ on $[\frac14,\frac12]$ , in $[-1,0)$ on $[0,\frac14)\cup(\frac12,\frac34]$ , in $[-2,-1)$ up to $x=\frac{3+\sqrt{17}}8$ , and in $[-3,-2)$ after that.

Integral.
$-\tfrac12-2\left(\tfrac{3+\sqrt{17}}8-\tfrac34\right)-3\left(1-\tfrac{3+\sqrt{17}}8\right)=\frac{\sqrt{17}-13}8.$

(C)

Topics:Properties of definite integrals Quadratic expressions

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 7