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JEE Main 20241 FebruaryEvening Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2024 · 1 February · Evening Shift · Mathematics · Continuity and Differentiability

Let f(x) = |2 x ^ 2 + 5|x| - 3|, x ∈ R . If m and n denote the...

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

Options

  1. A55
  2. B22
  3. C00
  4. D33

Correct answer

D. 33

Step-by-step solution

Given: f(x)=∣2x2+5∣x∣−3∣f\left(x\right) = \left|2{x}^{2} + 5\left|x\right| - 3\right|

Now, let y=∣2x2+5x−3∣y = \left|2{x}^{2} + 5x - 3\right|

⇒y=∣2x2+6x−x−3∣⇒ y = \left|2{x}^{2} + 6x - x - 3\right|

⇒y=∣2x(x+3)−(x+3)∣⇒ y = \left|2x\left(x + 3\right) - \left(x + 3\right)\right|

⇒y=∣(2x−1)(x+3)∣⇒ y = \left|\left(2x - 1\right)\left(x + 3\right)\right|

Graph of ⇒y=∣(2x−1)(x+3)∣⇒ y = \left|\left(2x - 1\right)\left(x + 3\right)\right| is given below:

Let f(x) = |2 x ^ 2 + 5|x| - 3|, x ∈ R . If m and n denote the number of points where f is not continuous and not

Graph of f(x)=∣2x2+5∣x∣−3∣f\left(x\right) = \left|2{x}^{2} + 5\left|x\right| - 3\right| is given below:

Let f(x) = |2 x ^ 2 + 5|x| - 3|, x ∈ R . If m and n denote the number of points where f is not continuous and not

So, f(x)f\left(x\right) is continuous for x∈Rx \in R and non-differentiable at x=±12,0x = \pm \dfrac{1}{2}, 0

So, number of points of discontinuity=0=m= 0 = m

Also, number of points of non-discontinuity=3=n= 3 = n

⇒m+n=3⇒ m + n = 3

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