Let f(x) = [ 2x^2 + 1 ] and g(x) = \ \beginmatrix 2x - 3, & x \lt 0 2x + 3, & x \ge 0 \endmatrix . , where [t] is the greatest integer \le t. Then, in the open interval ( - 1, 1), the number of points where fog is discontinuous is equal to \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2022 · 25 Jun · Shift 2 · Q38 with Solution

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Correct answer: 63

We need the number of points in $(-1,1)$ where $(f\circ g)(x)=f(g(x))$ is discontinuous.

1. Given functions

$f(x)=[2x^2+1]$

and

2x-3, & x<0\\ 2x+3, & x\ge 0 \end{cases}$$ So $$ (f\circ g)(x) = [2(g(x))^2+1]. $$ --- ## 2. Write $(f\circ g)(x)$ explicitly ### For $x<0$: $$g(x)=2x-3$$ Hence $$ (f\circ g)(x)=[2(2x-3)^2+1]. $$ Now, $$2(2x-3)^2+1=2(4x^2-12x+9)+1=8x^2-24x+19.$$ So for $x<0$, $$ (f\circ g)(x)=[8x^2-24x+19]. $$ ### For $x\ge 0$: $$g(x)=2x+3$$ Hence $$ (f\circ g)(x)=[2(2x+3)^2+1]. $$ Now, $$2(2x+3)^2+1=2(4x^2+12x+9)+1=8x^2+24x+19.$$ So for $x\ge 0$, $$ (f\circ g)(x)=[8x^2+24x+19]. $$ Therefore $$ (f\circ g)(x)=\begin{cases} [8x^2-24x+19], & x<0\\ [8x^2+24x+19], & x\ge 0 \end{cases}$$ --- ## 3. Simplify using symmetry Observe that for both sides, $$8x^2\pm 24x+19 = 8x^2+24|x|+19.$$ Indeed, - if $x<0$, then $|x|=-x$, so $8x^2+24|x|+19=8x^2-24x+19$; - if $x\ge0$, then $|x|=x$, so $8x^2+24x+19$. Thus $$ (f\circ g)(x)=[8x^2+24|x|+19]. $$ Let $$h(x)=8x^2+24|x|+19.$$ Then $(f\circ g)(x)=[h(x)]$. Since $h(x)$ is continuous on $(-1,1)$, the greatest integer function $[h(x)]$ is discontinuous exactly at those points where $h(x)$ is an integer. --- ## 4. Range of $h(x)$ on $(-1,1)$ Because $|x|\in[0,1)$, $$h(x)=8x^2+24|x|+19.$$ For $x\ge0$, $$h(x)=8x^2+24x+19,$$ which is increasing on $[0,1)$ since derivative is $$16x+24>0.$$ For $x<0$, $$h(x)=8x^2-24x+19,$$ and as $x$ increases from $-1$ to $0$, this decreases from near $51$ to $19$. Hence overall minimum occurs at $x=0$: $$h(0)=19.$$ As $x\to 1^-$ or $x\to -1^+$, $$h(x)\to 8+24+19=51,$$ but $x=\pm1$ are not included. So $$h((-1,1))=[19,51).$$ --- ## 5. Count points where $h(x)$ is an integer We need integer values attained by $h(x)$ in $[19,51)$, i.e. $$19,20,21,\dots,50.$$ This is $$50-19+1=32$$ integers. Now solve $$h(x)=n$$ for each such integer $n$. Since $$h(x)=8x^2+24|x|+19,$$ put $y=|x|$ with $0\le y<1$: $$8y^2+24y+19=n.$$ For a fixed integer $n$: - if $n=19$, then $$8y^2+24y=0 \implies y=0,$$ so only $$x=0.$$ Thus 1 point. - if $19<n<51$, then the equation has exactly one solution with $y\in(0,1)$ because $8y^2+24y+19$ is strictly increasing for $y\ge0$. Then each such $y$ gives two $x$ values: $$x=\pm y.$$ Thus 2 points for each integer $n=20,21,\dots,50$. There are $$50-20+1=31$$ such integers. Hence total number of discontinuity points is $$1+2\cdot 31=63.$$ --- ## 6. Check at $x=0$ At $x=0$, $$h(0)=19,$$ so near $0$, $h(x)>19$ for $x\ne0$, hence $$[h(x)]\ge 20$$ for nearby nonzero $x$, while $$[h(0)]=19.$$ Thus $x=0$ is indeed a discontinuity point. --- ## 7. Final answer The number of points in $(-1,1)$ where $f\circ g$ is discontinuous is $$\boxed{63}.$$ --- ## 8. Comparison with stored answer Stored correct answer = $62$. Our derived answer is $63$, so they do **not** agree. The likely omission in the stored answer is the point $x=0$, where $h(0)=19$ is an integer and the floor function has a jump.

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1. Given functions

f(x)=[2x^2+1]

and

2x-3, & x<0\\ 2x+3, & x\ge 0 \end{cases}$$ So $$ (f\circ g)(x) = [2(g(x))^2+1]. $$ --- ## 2. Write $(f\circ g)(x)$ explicitly ### For $x<0$: $$g(x)=2x-3$$ Hence $$ (f\circ g)(x)=[2(2x-3)^2+1]. $$ Now, $$2(2x-3)^2+1=2(4x^2-12x+9)+1=8x^2-24x+19.$$ So for $x<0$, $$ (f\circ g)(x)=[8x^2-24x+19]. $$ ### For $x\ge 0$: $$g(x)=2x+3$$ Hence $$ (f\circ g)(x)=[2(2x+3)^2+1]. $$ Now, $$2(2x+3)^2+1=2(4x^2+12x+9)+1=8x^2+24x+19.$$ So for $x\ge 0$, $$ (f\circ g)(x)=[8x^2+24x+19]. $$ Therefore $$ (f\circ g)(x)=\begin{cases} [8x^2-24x+19], & x<0\\ [8x^2+24x+19], & x\ge 0 \end{cases}$$ --- ## 3. Simplify using symmetry Observe that for both sides, $$8x^2\pm 24x+19 = 8x^2+24|x|+19.$$ Indeed, - if $x<0$, then $|x|=-x$, so $8x^2+24|x|+19=8x^2-24x+19$; - if $x\ge0$, then $|x|=x$, so $8x^2+24x+19$. Thus $$ (f\circ g)(x)=[8x^2+24|x|+19]. $$ Let $$h(x)=8x^2+24|x|+19.$$ Then $(f\circ g)(x)=[h(x)]$. Since $h(x)$ is continuous on $(-1,1)$, the greatest integer function $[h(x)]$ is discontinuous exactly at those points where $h(x)$ is an integer. --- ## 4. Range of $h(x)$ on $(-1,1)$ Because $|x|\in[0,1)$, $$h(x)=8x^2+24|x|+19.$$ For $x\ge0$, $$h(x)=8x^2+24x+19,$$ which is increasing on $[0,1)$ since derivative is $$16x+24>0.$$ For $x<0$, $$h(x)=8x^2-24x+19,$$ and as $x$ increases from $-1$ to $0$, this decreases from near $51$ to $19$. Hence overall minimum occurs at $x=0$: $$h(0)=19.$$ As $x\to 1^-$ or $x\to -1^+$, $$h(x)\to 8+24+19=51,$$ but $x=\pm1$ are not included. So $$h((-1,1))=[19,51).$$ --- ## 5. Count points where $h(x)$ is an integer We need integer values attained by $h(x)$ in $[19,51)$, i.e. $$19,20,21,\dots,50.$$ This is $$50-19+1=32$$ integers. Now solve $$h(x)=n$$ for each such integer $n$. Since $$h(x)=8x^2+24|x|+19,$$ put $y=|x|$ with $0\le y<1$: $$8y^2+24y+19=n.$$ For a fixed integer $n$: - if $n=19$, then $$8y^2+24y=0 \implies y=0,$$ so only $$x=0.$$ Thus 1 point. - if $19<n<51$, then the equation has exactly one solution with $y\in(0,1)$ because $8y^2+24y+19$ is strictly increasing for $y\ge0$. Then each such $y$ gives two $x$ values: $$x=\pm y.$$ Thus 2 points for each integer $n=20,21,\dots,50$. There are $$50-20+1=31$$ such integers. Hence total number of discontinuity points is $$1+2\cdot 31=63.$$ --- ## 6. Check at $x=0$ At $x=0$, $$h(0)=19,$$ so near $0$, $h(x)>19$ for $x\ne0$, hence $$[h(x)]\ge 20$$ for nearby nonzero $x$, while $$[h(0)]=19.$$ Thus $x=0$ is indeed a discontinuity point. --- ## 7. Final answer The number of points in $(-1,1)$ where $f\circ g$ is discontinuous is $$\boxed{63}.$$ --- ## 8. Comparison with stored answer Stored correct answer = $62$. Our derived answer is $63$, so they do **not** agree. The likely omission in the stored answer is the point $x=0$, where $h(0)=19$ is an integer and the floor function has a jump.

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