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  • The number of points where the function <img alt="\mathrm{f(x)= \begin{cases}\left|2 x^{2}-3 x-7\right| & \text { if } x \leq-1 \\ {\left[4 x^{2}-1\right]} & \text { if }-1<x<1

The number of points where the function
\mathrm{f(x)= \begin{cases}\left|2 x^{2}-3 x-7\right| & \text { if } x \leq-1 \\ {\left[4 x^{2}-1\right]} & \text { if }-1<x<1 \\ |x+1|+|x-2| & \text { if } x \geqslant 1,\end{cases}}
\mathrm{[t]} denotes the greatest integer \mathrm{\leqslant t}, is discontinuous is ________.

Option: 1

7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{f(x)= \begin{cases}\left|2x^{2}-3x-7\right| & x \leq-1 \\ {\left[4 x^{2}-1\right]} &-1<x<1 \\ |x+1|+|x-2| & x \geqslant 1\end{cases}}

As modulus function is continuous.
\mathrm{\therefore \; f(x)} can be discontinuous at \mathrm{x= -1,1}  and at all those points where  \mathrm{4x^{2}-1} is integer

\mathrm{\begin{matrix} 4x^{2}-1= 0 & ;4x^{2}-1= -1 &,4x^{2}-1= 1 ,&4x^{2}-1= 2 \\ x= \pm \frac{1}{2}& x= 0 & x= \pm \frac{1}{\sqrt{2}} & x= \pm \frac{\sqrt{3}}{2} \end{matrix}}

Hence discontinuous at
\mathrm{x= \frac{1}{2},\frac{-1}{2},\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},\frac{\sqrt{3}}{2},\frac{-\sqrt{3}}{2},1}
    \mathrm{= 7}

Posted by

Gaurav

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