JEE Advanced Maths: Polynomials / Theory of Equations questions with solutions

Q1. When a polynomial f(x) is divided by x² - 3x + 2, the remainder is a*x + b. Given f(1) = 4 and f(2) = 7, find the values of a and b.

1. a = 3, b = 1
2. a = 1, b = 3
3. a = 2, b = 2
4. a = 4, b = 0

Answer: a = 3, b = 1

Since the divisor factors as (x-1)(x-2), substituting its roots makes the quotient term vanish, leaving the linear remainder equal to f at those points.

Q2. A polynomial f(x) of degree 4 satisfies f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4, and f(0) = 1. Find f(5).

1. 5
2. 6
3. 7
4. 8

Answer: 6

Since f(x) - x is zero at x = 1, 2, 3, 4 and has degree 4, it equals a(x-1)(x-2)(x-3)(x-4). The condition f(0) = 1 fixes a. Then f(5) = g(5) + 5.

Q3. A cubic polynomial P(x) has leading coefficient 3 and satisfies P(1) = 1, P(2) = 2, P(3) = 3. Find P(4).

1. 4
2. 22
3. 28
4. 7

Answer: 22

Let Q(x) = P(x) - x. Since P(1)=1, P(2)=2, P(3)=3, Q has roots 1, 2, 3. P is cubic with leading coefficient 3, and subtracting x (degree 1) does not change the leading term, so Q(x) = 3*(x-1)*(x-2)*(x-3). Then P(x) = 3*(x-1)*(x-2)*(x-3) + x, so P(4) = 3*(3)*(2)*(1) + 4 = 18 + 4 = 22.