JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 18)
The number of points where the function
$$f(x) = \left\{ {\matrix{ {|2{x^2} - 3x - 7|} & {if} & {x \le - 1} \cr {[4{x^2} - 1]} & {if} & { - 1 < x < 1} \cr {|x + 1| + |x - 2|} & {if} & {x \ge 1} \cr } } \right.$$
[t] denotes the greatest integer $$\le$$ t, is discontinuous is _____________.
Answer
7
Explanation
$\because f(-1)=2$ and $f(1)=3$
For $x \in(-1,1),\left(4 x^{2}-1\right) \in[-1,3)$
hence $f(x)$ will be discontinuous at $x=1$ and also
whenever $4 x^{2}-1=0,1$ or 2
$$ \Rightarrow x=\pm \frac{1}{2}, \pm \frac{1}{\sqrt{2}} \text { and } \pm \frac{\sqrt{3}}{2} $$
So there are total 7 points of discontinuity.
For $x \in(-1,1),\left(4 x^{2}-1\right) \in[-1,3)$
hence $f(x)$ will be discontinuous at $x=1$ and also
whenever $4 x^{2}-1=0,1$ or 2
$$ \Rightarrow x=\pm \frac{1}{2}, \pm \frac{1}{\sqrt{2}} \text { and } \pm \frac{\sqrt{3}}{2} $$
So there are total 7 points of discontinuity.
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