Question

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

• A. 62.0
• B. 62
• C. 62.00

Answer 1

Answer: (A), (B), (C)

$f\left(g\right(x\left)\right)=\{\begin{array}{c}\left[2\right(2x-3{)}^2]+1,x<0\\ \left[2\right(2x+3{)}^2]+1,x\ge 0\end{array}.$
The possible points where $\text{fog}\left(x\right)$ may be discontinuous are
$2(2x-3{)}^2\in I\text{\&}x\in (-1,0)$
$2(2x+3{)}^2\in I\text{\&}x\in [0,1)$

for $x\in (-1,0)$
$\Rightarrow 2x-3\in (-5,-3)$
$\Rightarrow 2(2x-3{)}^2\in (18,50)$
$\therefore$ number of points of discontinuity $=31$
for $x\in [0,1)$
$\Rightarrow 2x+3\in [3,5)$
$\Rightarrow 2(2x+3{)}^2\in [18,50)$
It is discontinuous at all points except at $x=0$
so, number of discontinuous points in this case $=31$
$\therefore$ Total number of points where $fog$ is discontinuous is $=31+31=62$