Verified Solution • 2025–26 Session moderate

QJEEM24Ma806f: JEE Main 2024 Mathematics - Class 12 JEEM24Ma806f Solution: &nbsp;Let&nbsp; f ( x ) = { − 2 , − 2 ≤ x ≤ 0 x − 2 , 0 < x ≤ 2 …

 Let f(x)={−2,−2≤x≤0x−2,0<x≤2 and h(x)=f(|x|)+|f(x)|. Then ∫−22 h(x)dx is equal to: 

Multiple Choice Options

  • A 2
  • B 6
  • C 4
  • D 1

Step-by-Step Verified Solution

f(x)={−2amp;−2≤x≤0x−2amp;0lt;x≤2h(x)=f∣x∣+∣f(x)∣f(x)=\left\{\begin{array}{cc}-2 &amp; -2 \leq x \leq 0 \\x-2 &amp; 0&lt; x \leq 2\end{array} h(x)=f|x|+|f(x)|\right.

JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Definite Integration Question 26 English Explanation

h(x)={−x−2+2=−x−2≤x≤000<x≤2∴∫−22h(x)dx=∫−20−xdx+∫020dxx22|−20=42=2

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