JEE Advanced Maths: Basic Mathematics / Indices questions with solutions

Q1. Evaluate the following: (i) (cube_root(64))^(-1/2); (ii) (121/169)^(-3/2).

1. (i) 1/2 (ii) 2197/1331
2. (i) 2 (ii) 1331/2197
3. (i) 1/2 (ii) 1331/2197
4. (i) 1/4 (ii) 2197/1331

Answer: (i) 1/2 (ii) 2197/1331

Simplify the inner roots to integers, then apply the fractional negative exponents.

Q2. Solve for x: (3 + 2*sqrt(2))^(x/2) + (3 - 2*sqrt(2))^(x/2) = 34.

1. x = +-4
2. x = +-2
3. x = 4 only
4. x = +-6

Answer: x = +-4

The two bases multiply to 1, so put u = (3+2sqrt2)^(x/2); the equation becomes u + 1/u = 34. Solve the resulting quadratic and translate back to x.

Q3. If [ (2^(n+1))^m * (2^(2n))ⁿ ] / [ (2^(m+1))ⁿ * (2^(2m))^m ] = 1, find the value of m/n.

1. 1
2. 2
3. 3
4. 4

Answer: 1

Numerator exponent: (n+1)m + 2n*n = mn + m + 2n². Denominator exponent: (m+1)n + 2m*m = mn + n + 2m². The expression = 2^[(mn + m + 2n²) - (mn + n + 2m²)] = 2^[m - n + 2n² - 2m²]. Set exponent = 0: m - n + 2(n² - m²) = 0 => (m - n) - 2(m - n)(m + n) = 0 => (m - n)[1 - 2(m + n)] = 0. So either m = n (giving m/n = 1) or m + n = 1/2 (not generally giving a clean ratio for integer-type answer). The intended answer is m = n => m/n = 1.