Skip to main content
Back to DEFINITE INTEGRATION

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy
1 of 1
Analyzing +4 | -1
Let f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈Rf(x)=2+|x|-|x-1|+|x+1|, x \in R. Consider
(S1):f′(−32)+f′(−12)+f′(12)+f′(32)=2(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(S 2): \int\limits_{ - 2}^2 {f\left( x \right)dx = 12}

Then,
View answer and explanation
Correct option: D
Only (S2) is correct
Only (S2) is correct because the integral of the piecewise function f(x) from -2 to 2 equals 12, while the sum of the derivatives at the specified points equals 4, not 2.

More Questions from DEFINITE INTEGRATION

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.