If the coefficient of x¹⁵ in the expansion of (a x³+1/(b x^(1 / 3)))¹⁵ is equal to the coefficient of x⁻¹⁵ in the expansion of (a x^(1 / 3)-1/(b x³))¹⁵, where a and b are positive real numbers, then for each such ordered pair (a, b) :

Solution

Maths sir · 4 steps

Find which term carries the required power in each expansion, then set the two coefficients equal.

1. First expansion. The general term is C(15, r)(ax³)¹⁵⁻ʳ(1/bx^(1/3))ʳ, so the power of x is 3(15-r)-r/3 = 45-10r/3. Setting this equal to 15 gives r = 9, so the coefficient is C(15, 9)a⁶/b⁹.
2. Second expansion. The general term is C(15, r)(ax^(1/3))¹⁵⁻ʳ(-1/bx³)ʳ, so the power of x is (15-r)/3-3r = 5-10r/3. Setting this equal to -15 gives r = 6. Since (-1)⁶ = 1, the coefficient is C(15, 6)a⁹/b⁶.
3. Equate them. Because C(15, 9) = C(15, 6), the binomial coefficients cancel and only the powers of a and b are left.

   a⁶/b⁹ = a⁹/b⁶ ⇒ a³b³ = 1

4. So. Since a and b are positive, taking the cube root gives ab = 1, which is option (b).

The idea to keep. For a specific power of x, write the exponent in the general term as a function of r and solve for r first. Then check the sign: (-1)ʳ matters only when r is odd.

Answer from the NTA final answer key. Solution by PrepWiser.