Exams › JEE Advanced › Maths

Find the remainder when f(x) = x⁵ - x³ + 3x² + 3x + 1 is divided by (x² - 1).

1. 3x + 4
2. x - 1
3. 2x + 1
4. 2x - 1

Correct answer: 3x + 4

Solution

Divisor x² - 1 = (x-1)(x+1), so remainder = p x + q. f(1) = 1 - 1 + 3 + 3 + 1 = 7 = p + q. f(-1) = -1 + 1 + 3 - 3 + 1 = 1 = -p + q. Adding: 2q = 8 => q = 4; subtracting: 2p = 6 => p = 3. Remainder = 3x + 4.

Related JEE Advanced Maths questions

• Let P(x) = x¹⁰ + a*x⁸ + b*x⁶ + c*x⁴ + d*x² - 1 be a polynomial with real coefficients. Given P(1) = 1 and P(2) = -1, find the minimum number of real zeroes of P(x).
• Let f(x) be a polynomial of degree 3 such that f(k) = -2/k for k = 2, 3, 4, 5. Find the value of 52 - 10 f(10).
• When (x⁴ + 5x² + 7x - 8) is divided by (x - 1) the remainder is a, and when divided by (x + 2) the remainder is b. Find the value of (b - a).
• The polynomial P(x) = 7x⁴ + 8x² - 3x + lambda gives a remainder of 3 when divided by (x + 1). Find the value of lambda.
• Let f(x) be a polynomial of degree 5 with leading coefficient 1 such that f(1) = 5, f(2) = 4, f(3) = 3, f(4) = 2, f(5) = 1. Find f(6).
• Let P(x) = x⁴ + a x³ + b x² + c x + d, where a, b, c, d are constants. Given P(1) = 10, P(2) = 20, P(3) = 30, compute P(4) + P(0).

⚔️ Practice JEE Advanced Maths free + battle 1v1 →