If the term independent of x in the expansion of (√a x²+1/(2 x³))¹⁰ is 105 , then a² is equal to :

Solution

Maths sir · 4 steps

Write the general term, set the power of x to zero, and that one value of r does all the work.

1. General term. In (p+q)¹⁰ the term Tᵣ₊₁ = C(10, r) p¹⁰⁻ʳ qʳ. Here p = √a x² and q = 1/2x³.

   Tᵣ₊₁ = C(10, r) (√a)¹⁰⁻ʳ (1/2)ʳ x²⁰⁻⁵ʳ

2. Find r. The power of x is 2(10-r) - 3r = 20 - 5r. For the term independent of x, 20 - 5r = 0, so r = 4.
3. Use the value 105. With r = 4, the term is C(10, 4) (√a)⁶ (1/2)⁴ = 210 × a³ × 1/16. Setting this equal to 105 gives a³ = (105 × 16)/210 = 8, so a = 2.
4. So. a² = 2² = 4, which is option (b).

The idea to keep. For a term independent of x, write the power of x in Tᵣ₊₁ as a function of r and set it to zero. Then read off exactly what the question asks for: a or a².

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