JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 12)
Let ,$$f:[-1,2] \rightarrow \mathbf{R}$$ be given by $$f(x)=2 x^2+x+\left[x^2\right]-[x]$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. The number of points, where $$f$$ is not continuous, is :
5
6
4
3
Explanation
$$\begin{aligned} & f(x)=2 x^2+x+\left[x^2\right]-[x]=2 x^2+\left[x^2\right]+\{x\} \\ & f(-1)=2+1+0=3 \\ & f\left(-1^{+}\right)=2+0+0=2 \\ & f\left(0^{-}\right)=0+1=1 \\ & f\left(0^{+}\right)=0+0+0=0 \\ & f\left(1^{+}\right)=2+1+0=3 \end{aligned}$$
$$\begin{aligned} & f\left(1^{-}\right)=2+0+1=3 \\ & f\left(2^{-}\right)=8+3+1=12 \\ & f\left(2^{+}\right)=8+4+0=12 \end{aligned}$$
$$\therefore$$ discontinuous at $$x=0, \sqrt{2}, \sqrt{3},-1$$
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