If f(x) = { ∫₀ˣ ( 5 + | 1 - t | )dt,, x > 2; 5x + 1,, x ≤ 2 , then

Solution

Maths sir · 4 steps

Work out the integral in closed form first, then compare values and slopes on both sides of x = 2.

1. Split the modulus. For x > 2, the modulus changes form at t = 1. On [0, 1] the integrand is 6 - t, and ∫₀¹ (6 - t) dt = 6 - 1/2 = 11/2. On [1, x] it is 4 + t, and ∫₁ˣ (4 + t) dt = 4(x - 1) + (x² - 1)/2.

   f(x) = 11/2 + 4x - 4 + x²/2 - 1/2 = x²/2 + 4x + 1, x > 2

2. Continuity at 2. The right-hand limit is 4/2 + 8 + 1 = 11. The left piece gives f(2) = 5(2) + 1 = 11. The two agree, so f is continuous at x = 2.
3. Slopes at 2. The left derivative is 5. The right derivative is x + 4 = 6 at x = 2. Since 5 ≠ 6, the graph has a corner at x = 2.
4. So. f is continuous but not differentiable at x = 2. Near x = 1, f(x) = 5x + 1 is a straight line and is differentiable there, so the answer is (c).

The idea to keep. When a piece is defined by an integral, evaluate it in closed form first. Then check the joint twice: values must match for continuity, one-sided slopes must match for differentiability.

Answer from the NTA final answer key. Solution by PrepWiser.