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Mathematics Question on Definite Integral

Let
f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈R.f(x) = 2+|x|-|x-1|+|x+1|,x∈R.
Consider f′(−32)+f′(−12)+f′(12)+f′(32)=2f'\left(-\frac{3}{2}\right) + f'\left(-\frac{1}{2}\right) + f'\left(\frac{1}{2}\right) + f'\left(\frac{3}{2}\right) = 2
((S2):∫−22f(x) dx=12(S2):\int_{-2}^{2} f(x) \,dx = 12
Then,

A

Both (S1) and (S2) are correct

B

Both (S1) and (S2) are wrong

C

Only (S1) is correct

D

Only (S2) is correct

Answer

Only (S2) is correct

Explanation

Solution

The correct answer is (D) : Only (S2) is correct
f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈Rf(x) = 2+|x|-|x-1|+|x+1|,x∈R
∴f(x)={−xx<−1 x+2−1≤x<0 3x+20≤x<1 x+4x≥1 \therefore f(x) = \begin{cases} -x & x < -1 \\\ x + 2 & -1 \leq x < 0 \\\ 3x + 2 & 0 \leq x < 1 \\\ x + 4 & x \geq 1 \\\ \end{cases}
∴f′(−32)+f′(−12)+f′(12)+f′(32)\therefore f'\left(-\frac{3}{2}\right) + f'\left(-\frac{1}{2}\right) + f'\left(\frac{1}{2}\right) + f'\left(\frac{3}{2}\right)
=−1+1+3+1=4= -1+1+3+1 = 4
and ∫−22f(x) dx=∫−2−1f(x) dx+∫−10f(x) dx+∫01f(x) dx+∫12f(x) dx\int_{-2}^{2} f(x) \,dx = \int_{-2}^{-1} f(x) \,dx + \int_{-1}^{0} f(x) \,dx + \int_{0}^{1} f(x) \,dx + \int_{1}^{2} f(x) \,dx
=[−x22]−2−1+[(x+2)22]−10+[(3x+2)26]01+[(x+4)22]12= \left[ -\frac{x^2}{2} \right]_{-2}^{-1} + \left[ \frac{(x+2)^2}{2} \right]_{-1}^{0} + \left[ \frac{(3x+2)^2}{6} \right]_{0}^{1} + \left[ \frac{(x+4)^2}{2} \right]_{1}^{2}
=32+32+72+112=\frac{3}{2}+\frac{3}{2}+\frac{7}{2}+\frac{11}{2}
=242=12=\frac{24}{2}=12
∴ Only (S2) is correct