JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 11)
Let $f(x)=\left|2 x^2+5\right| x|-3|, x \in \mathbf{R}$. If $\mathrm{m}$ and $\mathrm{n}$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $\mathrm{m}+\mathrm{n}$ is equal to :
5
3
2
0
Explanation
$\begin{aligned} & f(x)=\left|2 x^2+5\right| x|-3| \\\\ & \text { Graph of } y=\left|2 x^2+5 x-3\right|\end{aligned}$
_1st_February_Evening_Shift_en_11_1.png)
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Now $f(x)$ is continuous $\forall x \in R$
but non- differentiable at $x=\frac{-1}{2}, \frac{1}{2}, 0$
$$ \begin{aligned} \therefore m & =0 \\\\ n & =3 \\\\ m+n & =3 \end{aligned} $$
Now $f(x)$ is continuous $\forall x \in R$
but non- differentiable at $x=\frac{-1}{2}, \frac{1}{2}, 0$
$$ \begin{aligned} \therefore m & =0 \\\\ n & =3 \\\\ m+n & =3 \end{aligned} $$
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