Let a and b be real constants such that the function f defined by f(x)={x²+3 x+a, , x ≤ 1; b x+2, , x>1 be differentiable on ℝ. Then, the value of ∫-2² f(x) d x equals

Solution

Maths sir · 4 steps

Match the slopes at x = 1 to get b, match the values to get a, then split the integral at 1.

1. Match the slopes. For f to be differentiable at x = 1, the left and right derivatives must be equal. On the left, f'(x) = 2x + 3, which gives 5 at x = 1. On the right, f'(x) = b, so b = 5.
2. Match the values. A differentiable function is also continuous, so both pieces must agree at x = 1: 1 + 3 + a = b + 2 = 7. This gives a = 3.

   ∫₋₂² f(x) dx = ∫₋₂¹ (x² + 3x + 3) dx + ∫₁² (5x + 2) dx

3. Evaluate each piece. First piece: [x³/3 + 3x²/2 + 3x]₋₂¹ = 29/6 - (-8/3) = 45/6 = 15/2. Second piece: [5x²/2 + 2x]₁² = 14 - 9/2 = 19/2.
4. So. The integral is 15/2 + 19/2 = 17, which is option (c).

The idea to keep. For a piecewise function to be differentiable at the joint, the values must match and the slopes must match. Two conditions give you two unknowns. When you integrate it, split the integral at the joint.

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