JEE Advanced Maths: Number System / Surds questions with solutions

Q1. Rationalize the denominator of 1/(3*sqrt(2) + sqrt(5)) and choose the simplified form.

1. (3*sqrt(2) - sqrt(5))/13
2. (3*sqrt(2) + sqrt(5))/13
3. (3*sqrt(2) - sqrt(5))/7
4. (sqrt(2) - sqrt(5))/13

Answer: (3*sqrt(2) - sqrt(5))/13

Multiplying top and bottom by the conjugate removes the radicals from the denominator using the difference-of-squares identity.

Q2. Simplify: (i) ((5*sqrt(3) + sqrt(50))*(5 - sqrt(24))) / (sqrt(75) - 5*sqrt(2)) (ii) (3*sqrt(2))/(sqrt(6) + sqrt(3)) - (4*sqrt(3))/(sqrt(6) + sqrt(2)) + (sqrt(6))/(sqrt(3) + sqrt(2))

1. (i) 1; (ii) 0
2. (i) 0; (ii) 1
3. (i) sqrt(2); (ii) sqrt(3)
4. (i) 5; (ii) -1

Answer: (i) 1; (ii) 0

Part (i) simplifies because (5*sqrt(3) + 5*sqrt(2)) is a multiple of (sqrt(75) - 5*sqrt(2))'s structure once you account for sqrt(24) = 2*sqrt(6). Part (ii) telescopes to zero after rationalizing.

Q3. Find rational numbers a and b such that (2 + 3 sqrt(5)) / (1 - 3 sqrt(5)) = a + b sqrt(5).

1. a = -47/44, b = -9/44
2. a = 47/44, b = -9/44
3. a = -47/44, b = 9/44
4. a = 47/44, b = 9/44

Answer: a = -47/44, b = -9/44

Multiply by (1 + 3 sqrt5): denominator = 1 - (3 sqrt5)² = 1 - 45 = -44. Numerator = (2 + 3 sqrt5)(1 + 3 sqrt5) = 2 + 6 sqrt5 + 3 sqrt5 + 45 = 47 + 9 sqrt5. So expression = (47 + 9 sqrt5)/(-44) = -47/44 - (9/44) sqrt5. Hence a = -47/44, b = -9/44.