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QJEEM24Ma488f: JEE Main 2024 Mathematics - Class 12 JEEM24Ma488f Solution: Let . If and denote the number of points where is not…

Let f(x)=∣2x2+5∣x∣−3∣,x∈Rf(x)=\left|2 x^2+5\right| x|-3|, x \in \mathbf{R} . If m\mathrm{m} and n\mathrm{n} denote the number of points where ff is not continuous and not differentiable respectively, then m+n\mathrm{m}+\mathrm{n} is equal to :

Multiple Choice Options

  • A 5
  • B 3
  • C 2
  • D 0

Step-by-Step Verified Solution

amp;f(x)=∣2x2+5∣x∣−3∣amp; Graph of y=∣2x2+5x−3∣\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}

JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Limits, Continuity and Differentiability Question 41 English Explanation 1
JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Limits, Continuity and Differentiability Question 41 English Explanation 2

Now f(x)f(x) is continuous ∀x∈R\forall x \in R

but non- differentiable at x=−12,12,0x=\frac{-1}{2}, \frac{1}{2}, 0

∴mamp;=0namp;=3m+namp;=3\begin{aligned}\therefore m & =0 \\\\n & =3 \\\\m+n & =3\end{aligned}

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