If the coefficient of x^7 in [ax^2 + (1/bx)]^11 equals the coefficient of x^-7 in [ax - (1/bx^2)]^11 , then a and b satisfy the relation: - Mathematics MCQ AIEEE 2026 | ExamTest.live

Solution & Step-by-step Explanation

1. Expansion of

[a x^{2} + \frac{1}{b x}]^{11}

:General term

T_{r + 1} =^{11} C_{r} (a x^{2})^{11 - r} (\frac{1}{b x})^{r} =^{11} C_{r} a^{11 - r} b^{- r} x^{22 - 2 r - r} =^{11} C_{r} a^{11 - r} b^{- r} x^{22 - 3 r}

.For

x^{7}

22 - 3 r = 7 ⟹ 3 r = 15 ⟹ r = 5

.Coefficient

=^{11} C_{5} a^{6} b^{- 5}

.Expansion of

[a x - \frac{1}{b x ^{2}}]^{11}

:General term

T_{r^{'} + 1} =^{11} C_{r^{'}} (a x)^{11 - r^{'}} (- \frac{1}{b x ^{2}})^{r^{'}} =^{11} C_{r^{'}} a^{11 - r^{'}} (- 1)^{r^{'}} b^{- r^{'}} x^{11 - r^{'} - 2 r^{'}} =^{11} C_{r^{'}} a^{11 - r^{'}} (- 1)^{r^{'}} b^{- r^{'}} x^{11 - 3 r^{'}}

.For

x^{- 7}

11 - 3 r^{'} = - 7 ⟹ 3 r^{'} = 18 ⟹ r^{'} = 6

.Coefficient

=^{11} C_{6} a^{5} (- 1)^{6} b^{- 6} =^{11} C_{6} a^{5} b^{- 6}

.Given the coefficients are equal:

^{11} C_{5} a^{6} b^{- 5} =^{11} C_{6} a^{5} b^{- 6}

Since

^{11} C_{5} =^{11} C_{6}

a^{6} b^{- 5} = a^{5} b^{- 6} ⟹ \frac{a ^{6}}{a ^{5}} = \frac{b ^{- 6}}{b ^{- 5}} ⟹ a = b^{- 1} ⟹ ab = 1

If the coefficient of $x^{7}$ in $[a x^{2} + (1/ b x)]^{11}$ equals the coefficient of $x^{- 7}$ in $[a x - (1/ b x^{2})]^{11}$ , then $a$ and $b$ satisfy the relation: