Question

If $x$ be very small compared to unity such that $\frac{\sqrt{1+x}+\sqrt[3]{3(1-x{)}^2}}{\sqrt{1+x}+(1+x)}\approx a+bx$, then value of $a+b$ is

• A. $\frac{11}{6}$
• B. $\frac{1}{6}$
• C. $\frac{8}{3}$
• D. $-\frac{2}{3}$

Answer 1

Answer: (B)

$\frac{1}{6}$

As $x<<1$, So, while expanding we can ignore from 3rd term onwards.

So, $\frac{{(1+x)}^{1/2}+{(1-x)}^{2/3}}{{(1+x)}^{1/2}+(1+x)}=\frac{1+\frac{1}{2}x+...+1-\frac{2}{3}x+...}{1+\frac{1}{2}x+...+1+x+...}=d\frac{1+\frac{1}{2}x+1-\frac{2}{3}x}{1+\frac{1}{2}x+1+x}(\because x<<1)=\frac{2-\frac{x}{6}}{2+\frac{3x}{2}}=\frac{12-x}{12+9x}$

This is equal to $a+bx$

Hence,

$\frac{12+9x-10x}{12+9x}=a+bx\Rightarrow 1-\frac{10x}{12+9x}=a-bx\Rightarrow 1-\frac{10x}{12(1+\frac{9}{12}x)}=a-bx\Rightarrow 1-\frac{10}{12}x=a+bx\{\because x<<1\Rightarrow \frac{9}{12}x<<<1\}\Rightarrow a=1,b=-\frac{10}{12}\Rightarrow a+b=\frac{1}{6}$