Let P(x) = x² + bx + c be a quadratic polynomial with real coefficients such that ∫₀¹ P(x)dx = 1 and P(x) leaves remainder 5 when it is divided by (x - 2). Then the value of 9(b + c) is equal to :

Solution

Maths sir · 4 steps

Each condition gives one linear equation in b and c. Two equations, two unknowns, and you are done.

1. Integral condition. Integrate term by term: ∫₀¹ (x² + bx + c) dx = 1/3 + b/2 + c. Setting this equal to 1 gives b/2 + c = 2/3, that is, 3b + 6c = 4.
2. Remainder theorem. The remainder on dividing by (x - 2) is P(2) = 4 + 2b + c. Setting this equal to 5 gives 2b + c = 1, so c = 1 - 2b.

   3b + 6(1 - 2b) = 4 ⇒ -9b = -2 ⇒ b = 2/9

3. Find c. c = 1 - 2 × 2/9 = 5/9. So b + c = 2/9 + 5/9 = 7/9.
4. So. 9(b + c) = 9 × 7/9 = 7, which is option (c).

The idea to keep. When a polynomial has unknown coefficients, turn every given condition into one equation. An integral gives a linear equation, and a remainder on division by (x - a) is just P(a).

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