The curve y = ax³ + bx² + cx + d has a tangent at (1, 2) with slope 4, and at (−1, −2) with slope 0. Then the value of a + b + c + d is:

The curve y = ax³ + bx² + cx + d has a tangent at (1, 2) with slope 4, and at (−1, −2) with slope 0. Then the value of a + b + c + d is:

• A. 2
• B. 4
• C. 6
• D. 8

Answer: A) 2

Explanation: y' = 3ax² + 2bx + c. At (1, 2): a + b + c + d = 2. y'(1) = 3a + 2b + c = 4. At (−1, −2): −a + b − c + d = −2. y'(−1) = 3a − 2b + c = 0. Solve: from eq1 and eq3: 2a + 2c = 4 → a + c = 2. From eq2 and eq4: 4b = 4 → b = 1. Then 3a + 2 + c = 4 → 3a + c = 2. With a + c = 2, subtract: 2a = 0 → a = 0, c = 2. From eq1: 0 + 1 + 2 + d = 2 → d = −1. Sum a + b + c + d = 0 + 1 + 2 − 1 = 2.