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JEE Main 2021
Definite Integration
Definite Integration
Hard

Question

Let f(x)=2+xx1+x+1,xRf(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}. Consider (S1):f(32)+f(12)+f(12)+f(32)=2(\mathrm{S} 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2 (S2):22f(x)dx=12(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12 Then,

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Solution

This problem requires us to work with a piecewise function defined using absolute values. We need to evaluate derivatives and a definite integral of this function. The core skills tested are:

  1. Understanding Absolute Value Functions: How to rewrite a function involving absolute values as a piecewise function.
  2. Differentiation of Piecewise Functions: Finding the derivative of each piece and evaluating it at specific points.
  3. Definite Integration of Piecewise Functions: Splitting the integral into sub-intervals based on the function's definition and integrating each piece.

1. Defining the Piecewise Function f(x)f(x)

The given function is f(x)=2+xx1+x+1f(x)=2+|x|-|x-1|+|x+1|. To work with this function, especially for differentiation and integration, we must first express it as a piecewise function by removing the absolute value signs. The critical points where the expressions inside the absolute values change sign are x=0x=0 (for x|x|), x=1x=1 (for x1|x-1|), and x=1x=-1 (for x+1|x+1|). These points divide the real number line into four intervals: x<1x < -1, 1x<0-1 \le x < 0, 0x<10 \le x < 1, and x1x \ge 1.

Let's analyze f(x)f(x) in each interval:

  • Case 1: x<1x < -1 In this interval: x=x|x| = -x (since xx is negative) x1=(x1)=x+1|x-1| = -(x-1) = -x+1 (since x1x-1 is negative) x+1=(x+1)=x1|x+1| = -(x+1) = -x-1 (since x+1x+1 is negative) Substituting these into f(x)f(x): $f(x) =

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