JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q22

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MathematicsInteger+4 / -1Question 22 of 90

Let $f(x)=\left[2x^2+1\right]$ and $g(x)=\begin{cases}2x-3, & x<0\\2x+3, & x\ge0\end{cases}$ , where $[t]$ is the greatest integer $\le t$ . Then, in the open interval $(-1,1)$ , the number of points where $f\circ g$ is discontinuous is equal to ______.

Your answer

Answer and worked solution

Official answer: 62

For $x\in(-1,0)$ : $g=2x-3\in(-5,-3)$ , and $2g^2+1$ runs strictly through $(19,51)$ ; $f\circ g$ jumps each time it passes an integer from 20 to 50: 31 points.

For $x\in[0,1)$ : $g=2x+3\in[3,5)$ , and $2g^2+1$ runs through $[19,51)$ : again 31 jumps (20 to 50).

At $x=0$ . Both sides give $[19^+]=19$ , so it is continuous there.

Total 62.

Answer: 62

Topics:Continuity Composition and inverse functions

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JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 22