Question

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

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

Answer 1

Answer: (A)

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

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

$\therefore T_{r+1}={}^8{C}_r{\left(a{x}^2\right)}^{8-r}{\left(\frac{1}{bx}\right)}^r$

$={}^8{C}_r\frac{a^{8-r}}{b^r}{x}^{16-3r}$

Since, this term contains $x^7$

$\therefore 16-3r=7$

$\Rightarrow 3r=9$

$\Rightarrow r=3$

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

${\left(a{x}^2+\frac{1}{bx}\right)}^8={}^8{C}_3\cdot \frac{a^5}{b^3}$

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

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

$T_{R+1}={}^8{C}_R{\left(ax\right)}^{8-R}{\left(-\frac{1}{b{x}^2}\right)}^R$

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

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

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 ${\left(ax-\frac{1}{b{x}^2}\right)}^8$

$={\left(-1\right)}^5{}^8{C}_5\cdot \frac{a^3}{b^5}$

According to the given condition,

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

$\Rightarrow {}^8{C}_3\cdot \frac{a^5}{b^3}={}^8{C}_5\cdot \frac{a^3}{b^5}$

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

$\Rightarrow ab=1$