JEE Advanced Maths: Basic Mathematics questions with solutions

Q1. Simplify: ( ( (1/x)^(-12))^(1/4))^(-2/3).

1. 1/x²
2. 1/x⁴
3. 1/x³
4. 1/x

Answer: 1/x²

Start inside: (1/x)^(-12) = x¹². Fourth root: (x¹²)^(1/4) = x³. Then raise to -2/3: (x³)^(-2/3) = x^(-2) = 1/x².

Q2. Solve for x: |2x + 3| = |3x - 4|.

1. x = 7 or x = 1/5
2. x = 7 only
3. x = 1/5 only
4. x = -7 or x = 1/5

Answer: x = 7 or x = 1/5

Case 1: 2x + 3 = 3x - 4 -> 7 = x, so x = 7. Case 2: 2x + 3 = -(3x - 4) = -3x + 4 -> 5x = 1 -> x = 1/5. Both satisfy the original equation.

Q3. Solve the simultaneous linear equations: 3x - 2y = 13 and 4x + 5y = 25.

1. x = 5, y = 1
2. x = 3, y = -2
3. x = 5, y = -1
4. x = 1, y = -5

Answer: x = 5, y = 1

Multiply eq1 (3x - 2y = 13) by 5: 15x - 10y = 65. Multiply eq2 (4x + 5y = 25) by 2: 8x + 10y = 50. Add: 23x = 115 => x = 5. Substitute into eq1: 3(5) - 2y = 13 => 15 - 2y = 13 => 2y = 2 => y = 1. So x = 5, y = 1.

Q4. Let f(x) = [(x - 1)³ (x + 2)⁴ (x - 3)⁵ (x + 6)] / [x² (x - 7)]. Solve the inequality f(x) > 0 (find the set of x for which f is strictly positive).

1. x in (-infinity, -6) U (1, 3) U (7, infinity)
2. x in (-6, 1) U (3, 7)
3. x in (-infinity, -6) U (-2, 1) U (3, 7)
4. x in (-6, -2) U (1, 3) U (7, infinity)

Answer: x in (-infinity, -6) U (1, 3) U (7, infinity)

Critical points and parity of sign change: x = -6 (odd), x = -2 (even, no change), x = 0 (even, no change, excluded), x = 1 (odd), x = 3 (odd), x = 7 (odd, excluded). Start from x > 7 where all factors are positive => f > 0. Moving left and flipping sign only at odd-multiplicity points: (7, infinity) +, (3, 7) -, (1, 3) +, (-2, 1) -, (-6, -2) + (no change at -2 since even; wait -2 is even so sign continues): carefully, between -6 and 1 the only odd point is 1, with even point -2 (no change), so (-6, 1) has constant sign equal to the sign just left of 1 which is negative... Standard wavy-curve result for f(x) > 0 is x in (-infinity, -6) U (1, 3) U (7, infinity).

Q5. Solve for x: |2x - 1| + |2x + 3| = 6.

1. x = -2 and x = 1
2. x = -1 and x = 2
3. x = -3/2 and x = 1/2
4. no real solution

Answer: x = -2 and x = 1

Critical points: 2x - 1 = 0 => x = 1/2; 2x + 3 = 0 => x = -3/2. Case x >= 1/2: (2x - 1) + (2x + 3) = 4x + 2 = 6 => x = 1 (valid, >= 1/2). Case -3/2 <= x < 1/2: (1 - 2x) + (2x + 3) = 4 != 6, no solution (the sum is constant 4 here). Case x < -3/2: (1 - 2x) + (-2x - 3) = -4x - 2 = 6 => -4x = 8 => x = -2 (valid, < -3/2). So x = 1 and x = -2.