Let \(\mathrm{f}(\mathrm{x})=2+|\mathrm{x}|-|\mathrm{x}-1|+|\mathrm{x}+1|, \mathrm{x} \in \mathbf{R}\). Consider (Sl): \(\mathrm{f}^{\prime}\left(-\frac{3}{2}\right)+\mathrm{f}^{\prime}\left(-\frac{1}{2}\right)+\mathrm{f}^{\prime}\left(\frac{1}{2}\right)+\mathrm{f}^{\prime}\left(\frac{3}{2}\right)=2\) (S2) : \(\int_{-2}^{2} f(x) d x=12\) Then,

(A) both (S1) and (S2) are correct

(B) both (S1) and (S2) are wrong

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Continuity and Differentiability

Let

f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣, x \in R

. Consider
(Sl):

f^{'} (- \frac{3}{2}) + f^{'} (- \frac{1}{2}) + f^{'} (\frac{1}{2}) + f^{'} (\frac{3}{2}) = 2

(S2) :

\int_{- 2}^{2} f (x) d x = 12

Then,

Answer: (D)

We need to analyze

f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣

by breaking it into regions based on critical points

x = - 1, 0, 1

.

For

x < - 1

f (x) = 2 - x + (x - 1) - (x + 1) = - x

For

- 1 \leq x < 0

f (x) = 2 - x + (x - 1) + (x + 1) = x + 2

For

0 \leq x < 1

f (x) = 2 + x + (x - 1) + (x + 1) = 3 x + 2

For

x \geq 1

f (x) = 2 + x - (x - 1) + (x + 1) = x + 4

The derivatives in each region are:

- 1, 1, 3, 1

respectively.

Statement S1:

f^{'} (- 3/2) + f^{'} (- 1/2) + f^{'} (1/2) + f^{'} (3/2) = - 1 + 1 + 3 + 1 = 4 \neq = 2

. So S1 is false.

Statement S2: Computing

\int_{- 2}^{2} f (x) d x

by integrating over each region gives 12. So S2 is true.

Therefore, only (S2) is correct.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option