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Definite Integration question

2022 · 27 Jul · Shift 2 · Q26
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  5. /2022 · 27 Jul · Shift 2 · Q26

Definite Integration question

2022 · 27 Jul · Shift 2 · Q26

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈Rf(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈R. Consider (S1):f′(−32)+f′(−12)+f′(12)+f′(32)=2(S2):∫−22f(x)dx=12(\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(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12(S1):f′(−23​)+f′(−21​)+f′(21​)+f′(23​)=2(S2):−2∫2​f(x)dx=12 Then,
  1. A
    both (S1) and (S2) are correct
  2. B
    both (S1) and (S2) are wrong
  3. C
    only (S1) is correct
  4. D
    only (S2) is correct
View written solutionFree

Correct answer: D

  1. Given function

We have f(x)=2+∣x∣−∣x−1∣+∣x+1∣.f(x)=2+|x|-|x-1|+|x+1|.f(x)=2+∣x∣−∣x−1∣+∣x+1∣. The critical points for absolute values are where the expressions inside modulus become zero: x=−1,  0,  1.x=-1,\;0,\;1.x=−1,0,1. So we split the real line into intervals:

  • x<−1x<-1x<−1
  • −1≤x<0-1\le x<0−1≤x<0
  • 0≤x<10\le x<10≤x<1
  • x≥1x\ge 1x≥1
  1. Find piecewise form of f(x)f(x)f(x)

Case 1: x<−1x<-1x<−1

Then ∣x∣=−x,∣x−1∣=−(x−1)=−x+1,∣x+1∣=−(x+1)=−x−1.|x|=-x,\quad |x-1|=-(x-1)=-x+1,\quad |x+1|=-(x+1)=-x-1.∣x∣=−x,∣x−1∣=−(x−1)=−x+1,∣x+1∣=−(x+1)=−x−1. So, f(x)=2+(−x)−(−x+1)+(−x−1)=2−x+x−1−x−1=−x.f(x)=2+(-x)-(-x+1)+(-x-1)=2-x+x-1-x-1=-x.f(x)=2+(−x)−(−x+1)+(−x−1)=2−x+x−1−x−1=−x.

Case 2: −1≤x<0-1\le x<0−1≤x<0

Then ∣x∣=−x,∣x−1∣=−x+1,∣x+1∣=x+1.|x|=-x,\quad |x-1|=-x+1,\quad |x+1|=x+1.∣x∣=−x,∣x−1∣=−x+1,∣x+1∣=x+1. So, f(x)=2−x−(−x+1)+(x+1)=2−x+x−1+x+1=x+2.f(x)=2-x-(-x+1)+(x+1)=2-x+x-1+x+1=x+2.f(x)=2−x−(−x+1)+(x+1)=2−x+x−1+x+1=x+2.

Case 3: 0≤x<10\le x<10≤x<1

Then ∣x∣=x,∣x−1∣=−x+1,∣x+1∣=x+1.|x|=x,\quad |x-1|=-x+1,\quad |x+1|=x+1.∣x∣=x,∣x−1∣=−x+1,∣x+1∣=x+1. So, f(x)=2+x−(−x+1)+(x+1)=2+x+x−1+x+1=3x+2.f(x)=2+x-(-x+1)+(x+1)=2+x+x-1+x+1=3x+2.f(x)=2+x−(−x+1)+(x+1)=2+x+x−1+x+1=3x+2.

Case 4: x≥1x\ge 1x≥1

Then ∣x∣=x,∣x−1∣=x−1,∣x+1∣=x+1.|x|=x,\quad |x-1|=x-1,\quad |x+1|=x+1.∣x∣=x,∣x−1∣=x−1,∣x+1∣=x+1. So, f(x)=2+x−(x−1)+(x+1)=2+x−x+1+x+1=x+4.f(x)=2+x-(x-1)+(x+1)=2+x-x+1+x+1=x+4.f(x)=2+x−(x−1)+(x+1)=2+x−x+1+x+1=x+4.

Hence,

f(x)={−x,x<−1,x+2,−1≤x<0,3x+2,0≤x<1,x+4,x≥1.f(x)= \begin{cases} -x, & x<-1,\\[4pt] x+2, & -1\le x<0,\\[4pt] 3x+2, & 0\le x<1,\\[4pt] x+4, & x\ge 1. \end{cases}f(x)=⎩⎨⎧​−x,x+2,3x+2,x+4,​x<−1,−1≤x<0,0≤x<1,x≥1.​
  1. Check (S1)

We need f′(−32)+f′(−12)+f′(12)+f′(32).f'\left(-\frac32\right)+f'\left(-\frac12\right)+f'\left(\frac12\right)+f'\left(\frac32\right).f′(−23​)+f′(−21​)+f′(21​)+f′(23​).

From the piecewise linear form, slopes are:

  • for x<−1x<-1x<−1, f′(x)=−1f'(x)=-1f′(x)=−1
  • for −1<x<0-1<x<0−1<x<0, f′(x)=1f'(x)=1f′(x)=1
  • for 0<x<10<x<10<x<1, f′(x)=3f'(x)=3f′(x)=3
  • for x>1x>1x>1, f′(x)=1f'(x)=1f′(x)=1

Thus, f′(−32)=−1,f'\left(-\frac32\right)=-1,f′(−23​)=−1, f′(−12)=1,f'\left(-\frac12\right)=1,f′(−21​)=1, f′(12)=3,f'\left(\frac12\right)=3,f′(21​)=3, f′(32)=1.f'\left(\frac32\right)=1.f′(23​)=1.

Their sum is −1+1+3+1=4.-1+1+3+1=4.−1+1+3+1=4. But (S1) claims this sum is 222.

So, (S1) is wrong.

  1. Check (S2)

We need ∫−22f(x) dx.\int_{-2}^{2} f(x)\,dx.∫−22​f(x)dx. Split according to intervals:

∫−2−1(−x) dx+∫−10(x+2) dx+∫01(3x+2) dx+∫12(x+4) dx.\int_{-2}^{-1}(-x)\,dx+ \int_{-1}^{0}(x+2)\,dx+ \int_{0}^{1}(3x+2)\,dx+ \int_{1}^{2}(x+4)\,dx.∫−2−1​(−x)dx+∫−10​(x+2)dx+∫01​(3x+2)dx+∫12​(x+4)dx.

First integral

\left[-\frac{x^2}{2}\right]_{-2}^{-1} =-\frac{1}{2}-\left(-2\right)=\frac32.$$ ### Second integral $$\int_{-1}^{0}(x+2)\,dx= \left[\frac{x^2}{2}+2x\right]_{-1}^{0} =0-\left(\frac12-2\right)=\frac32.$$ ### Third integral $$\int_{0}^{1}(3x+2)\,dx= \left[\frac{3x^2}{2}+2x\right]_{0}^{1} =\frac32+2=\frac72.$$ ### Fourth integral $$\int_{1}^{2}(x+4)\,dx= \left[\frac{x^2}{2}+4x\right]_{1}^{2} =(2+8)-\left(\frac12+4\right)=10-\frac92=\frac{11}{2}.$$ Now add: $$\frac32+\frac32+\frac72+\frac{11}{2} =3+9=12.$$ So, **(S2) is correct**. --- 5. **Final conclusion** - (S1) is wrong - (S2) is correct Therefore, the correct option is: $$\boxed{\text{D: only (S2) is correct}}$$ --- 6. **Comparison with stored answer** Stored correct answer: **D** My derived answer: **D** So they agree.
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