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  • Let  <img alt="\mathrm{f(x)=\left[2 x^{2}+1\right] \text { and } g(x)= \begin{cases}2 x-3, & x<0 \\ 2 x+3, & x \geq 0\end{cases}}" src="https://entrancecorner.oncodecogs.com/gif.lat

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

Option: 1

62


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{f(g(x))=\left[2 g^{2}(x)\right]+1}

\mathrm{=\left\{\begin{array}{l} {\left[2(2 x-3)^{2}\right]+1, x<0} \\ {\left[2(2 x+3)^{2}\right]+1, x \geq 0} \end{array}\right.}

\therefore Fog is discontinuous whenever \mathrm{2\left ( 2x-3 \right )^{2}\: or\: 2\left ( 2x+3 \right )^{2}} belongs to integer except \mathrm{x=0}

\mathrm{\therefore 62} points of discontinuity.

Hence the answer is \mathrm{ 62}.

Posted by

Riya

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