Question

If the magnitude of the coefficient of $ x^7$ in the expansion of $\left(ax^{2} +\frac{1}{bx}\right)^{8}$ where $a, b$ are positive numbers, is equal to the magnitude of the coefficient of $x^{-7}$ in the expansion of $\left(ax -\frac{1}{bx^{2}}\right)^{8}$, then $a$ and $b$ are connected by the relation

• A. $ab = 1$
• B. $ ab = 2$
• C. $ a^2b = 1$
• D. $ ab^2 = 2$

Answer 1

Answer: (A)

Let the term containing $x^ 7$ in the expansion of

$\left(ax^{2} +\frac{1}{bx}\right)^{8} $ is $T_{r+1}$

$ \therefore T_{r+1} =\,{}^{8}C_{r}\left(ax^{2}\right)^{8-r} \left(\frac{1}{bx}\right)^{r} $

$= \,{}^{8}C_{r} \frac{a^{8-r}}{b^{r}} x^{16-3r} $

Since, this term contains $x^{7} $

$ \therefore 1 6 - 3 r = 7$

$\Rightarrow 3r = 9 $

$\Rightarrow r = 3 $

$\therefore $ Coefficient of $x^7$ in the expansion of

$\left(ax^{2} +\frac{1}{bx}\right)^{8} = \,{}^{8}C_{3} \cdot\frac{a^{5}}{b^{3}} $

Also, the term containing $x^{-7}$ in the expansion of

$\left(ax-\frac{1}{bx^{2}}\right)^{8}$ is $T_{R+1}$.

$ T_{R+1} =\,{}^{8}C_{R}\left(ax\right)^{8-R} \left(-\frac{1}{bx^{2}}\right)^{R} $

$ =\,{}^{8}C_{R} \frac{a^{8-R} x^{8-R}\left(-1\right)^{R}}{b^{R} x^{2R}} $

$= \left(-1\right)^{R} \,{}^{8}C_{R} \frac{a^{8-R}}{b^{R}} \cdot x^{8-3R}$

Since, this term contains $x^{-7} $

$ \therefore 8-3R=-7 $

$ \Rightarrow 3R = 15 $

$ \Rightarrow R =5 $

$ \therefore $ coefficient of $x^{-7}$ in the expansion of $ \left(ax - \frac{1}{bx^{2}}\right)^{8}$

$ = \left(-1\right)^{5} \,{}^{8}C_{5} \cdot \frac{a^{3}}{b^{5}} $

According to the given condition,

$ \left|T_{r+1}\right| = \left|T_{R+1}\right| $

$ \Rightarrow \,{}^{8}C_{3} \cdot \frac{a^{5}}{b^{3}} = \,{}^{8}C_{5} \cdot\frac{a^{3}}{b^{5}} $

$\Rightarrow a^{2}b^{2} = 1 $

$\Rightarrow ab = 1$