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

JEE · Math · previous-year question

  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 correctcorrect

Answer

D. only (S2) is correct

Explanation

$$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$$ $$\therefore$$ $$f(x) = \left\{ {\matrix{ { - x} & , & {x $$\therefore$$ $$f' ( { - {3 \over 2}} ) + f' ( { - {1 \over 2}} ) + f' ( {{1 \over 2}} ) + f' ( {{3 \over 2}} ) = - 1 + 1 + 3 + 1 = 4$$ and $$\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } } $$ $$ = [ { - {{{x^2}} \over 2}} ]_2^{ - 1} + [ {{{{{(x + 2)}^2}} \over 2}} ]_{ - 1}^0 + [ {{{{{(3x + 2)}^2}} \over 6}} ]_0^1 + [ {{{{{(x + 4)}^2}} \over 2}} ]_1^2$$ $$ = {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12$$ $$\therefore$$ Only (S2) is correct

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