Question

If the Co-efficient of x⁷ in (ax² + a/bx)¹¹ equals the co-efficients of x^-7 in (ax - 1/bx²)¹¹ then 

(a) a/b = 1

(b) ab = 1

(c) a - b = 1

(d) a + b = 1

Answer 1

Correct option: (b) ab = 1

Explanation:

given: coefficient of x⁷ of [ax² + (1 / bx)]¹¹ = coefficient of x^–7 of [ax – (1 / bx²)]¹¹ 

Now T_r+1 = ¹¹c_r (ax²)^11–r (1 / bx)^r and T_r+1' = ¹¹c_r (ax)^11–r [– (1 / bx²)]^r 

∴ For 1^st expression, T_r+1 = ¹¹c_r a^11–r x^22–2r–r ∙ (1 / b^r)

For 2^nd expression T_r+1' = ¹¹c_r ∙ a^11–r ∙ x^11–r–2r ∙ [(– 1) / b^r]

In T_r+1 for coefficient of x^–7, put 22 – 3r = 7 ⇒ r = 5

In T_r+1', for coefficient of x^–7, put 11 – 3r = – 7 ⇒ r = 6

∴  Coefficient of x⁷ = ¹¹c₅ (a⁵ / b⁵) ------ for r = 5  (1)

Coefficient of x^–7 = ¹¹c₆ (a⁵ / b⁶) ------ for r = 6    (2)

as (1) & (2) are equal

∴  ¹¹c₅ (a⁶ / b⁵) = ¹¹c₆ (a⁵ / b⁶) 

⇒ ab = 1