Question

If $\frac{3ax}{4}+\sqrt{4+ax}-\frac{2}{\sqrt{1-ax}}=-x^2+b{x}^3+...$ to $\infty$, then
Note: ( $|ax|<1$)

• A. $a=\frac{8}{7},b=-\frac{319}{343}$
• B. $a=-\frac{8}{7},b=\frac{319}{343}$
• C. $a=\frac{8}{7},b=\frac{319}{343}$
• D. $a=-\frac{8}{7},b=-\frac{319}{343}$

Answer 1

Answer: (A), (B)

The correct options are
A $a=\frac{8}{7},b=-\frac{319}{343}$
B $a=-\frac{8}{7},b=\frac{319}{343}$

$\sqrt{4+ax}=2{(1+\frac{a}{4}x)}^{1/2}=2[1+\frac{a/4}{2}x+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{{\left(a/4\right)}^2{x}^2}{2!}+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{-3}{2}\frac{{\left(a/4\right)}^3{x}^3}{3!}+...]..........\left(1\right)$
$-2{(1-ax)}^{-1/2}=-2[1+\frac{a}{2}x+\frac{1}{2}\times \frac{3}{2}\times \frac{1}{2}{a}^2{x}^2+\frac{1\times 3\times 5}{2^3\times 3!}{a}^3{x}^3+...].....\left(2\right)$
Adding $\left(1\right)$, $\left(2\right)$ and $\frac{3ax}{4}$, we get
$\frac{3ax}{4}+\sqrt{4+ax}-\frac{2}{\sqrt{1-ax}}$
$=2[\frac{-a^2}{128}-\frac{3}{8}{a}^2]x^2+2[\frac{3a^3}{64\times 8\times 6}-\frac{3\times 5a^3}{8\times 6}]x^3+....$
Comparing the coefficient of $x^2$ with right hand side of the equation in the question we get
$2[\frac{-a^2}{128}-\frac{3}{8}{a}^2]=-1\Rightarrow a^2=\frac{64}{49}\Rightarrow a=\pm \frac{8}{7}......\left(3\right)$
Now comparing the coefficient of $x^3$, we get
$2[\frac{3a^3}{64\times 8\times 6}-\frac{3\times 5a^3}{8\times 6}]=b\Rightarrow \frac{a^3}{8}[\frac{1}{64}-5]=b\Rightarrow -\frac{319}{8^3}{a}^3=b$
Using the value of $a$ from $\left(3\right)$, we get
$b=-\frac{319}{343}$, when $a=\frac{8}{7}$
and,
$b=\frac{319}{343}$, when $a=-\frac{8}{7}$