Question

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

The Correct Option is D

To solve the given problem, we need to analyze and calculate the expressions for \(f(x)\) and its derivatives at specific points, as well as evaluate the integral \( \int_{-2}^{2} f(x) \, dx \). Let's proceed step by step.

Step 1: Evaluate \( f(x) \)

The function is given as \( f(x) = 2 + |x| - |x-1| + |x+1| \).

Step 2: Break the Function into Intervals

Since \(f(x)\) involves absolute values, we need to consider different cases based on the critical points of the absolute functions, which are \(x = -1, 0, 1\).

Step 3: Calculate for Intervals

1. For \(x < -1\):
   • \(f(x) = 2 + (-x) - (-(x-1)) + (-(x+1)) = -3x\)
2. For \(-1 \leq x < 0\):
   • \(f(x) = 2 - x - (-(x-1)) + (x+1) = 4 \)
3. For \(0 \leq x < 1\):
   • \(f(x) = 2 + x - (1-x) + (x+1) = 2x + 2\)
4. For \(x \geq 1\):
   • \(f(x) = 2 + x - (x-1) + (x+1) = 4\)

Step 4: Examine the Derivatives

We need the derivative \(f'(x)\) at points \(x = -\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}\).

1. For \(x = -\frac{3}{2}\): Falls in \(x < -1\), \(f'(x) = -3\).
2. For \(x = -\frac{1}{2}\): Falls in \(-1 \leq x < 0\), \(f'(x) = 0\).
3. For \(x = \frac{1}{2}\): Falls in \(0 \leq x < 1\), \(f'(x) = 2\).
4. For \(x = \frac{3}{2}\): Falls in \(x \geq 1\), \(f'(x) = 0\).

So, \(f'\left(-\frac{3}{2}\right) + f'\left(-\frac{1}{2}\right) + f'\left(\frac{1}{2}\right) + f'\left(\frac{3}{2}\right) = -3 + 0 + 2 + 0 = -1\), not equal to 2. Therefore, statement (S1) is wrong.

Step 5: Evaluate the Integral \(\int_{-2}^{2} f(x) \, dx\)

1. For \(x < -1\) (from \(-2\) to \(-1\)), \(f(x) = -3x\), thus \(\int_{-2}^{-1} -3x \, dx = \left[ -\frac{3x^2}{2} \right]_{-2}^{-1} = -\frac{9}{2} + 6 = \frac{3}{2}\).
2. For \(-1 \leq x < 0\), \(f(x) = 4\), thus \(\int_{-1}^{0} 4 \, dx = 4x \Big|_{-1}^{0} = 0 - (-4) = 4\).
3. For \(0 \leq x < 1\), \(f(x) = 2x + 2\), thus \(\int_{0}^{1} (2x + 2) \, dx = \left[ x^2 + 2x \right]_0^1 = 1 + 2 = 3\).
4. For \(x \geq 1\) (from \(1\) to \(2\)), \(f(x) = 4\), thus \(\int_{1}^{2} 4 \, dx = 4x \Big|_1^2 = 8 - 4 = 4\).

Total area \(= \frac{3}{2} + 4 + 3 + 4 = 12\). So, statement (S2) holds true.

Conclusion

Based on the calculations, the correct answer is: Only (S2) is correct.