# If b is very small as compared to the value of a,

Question:

If b is very small as compared to the value of a, so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity

$\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b}+\ldots .+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}$

then the value of $\gamma$ is :

1. $\frac{\mathrm{a}^{2}+\mathrm{b}}{3 \mathrm{a}^{3}}$

2. $\frac{\mathrm{a}+\mathrm{b}}{3 \mathrm{a}^{2}}$

3. $\frac{\mathrm{b}^{2}}{3 \mathrm{a}^{3}}$

4. $\frac{a+b^{2}}{3 a^{3}}$

Correct Option: , 4

Solution:

$\frac{\mathrm{dy}-\mathrm{dx}}{\mathrm{e}^{\mathrm{y}-\mathrm{x}}}=\mathrm{x} \mathrm{dx}$

$\Rightarrow \frac{d y-d x}{e^{y-x}}=x d x$

$\Rightarrow-e^{x-y}=\frac{x^{2}}{2}+c$

At $x=0, y=0 \Rightarrow c=-1$

$\Rightarrow e^{x-y}=\frac{2-x^{2}}{2}$

$\Rightarrow y=x-\ell n\left(\frac{2-x^{2}}{2}\right)$

$\Rightarrow \frac{d y}{d x}=1+\frac{2 x}{2-x^{2}}=\frac{2+2 x-x^{2}}{2-x^{2}}$

So minimum value occurs at $x=1-\sqrt{3}$

$y(1-\sqrt{3})=(1-\sqrt{3})-\ell n\left(\frac{2-(4-2 \sqrt{3})}{2}\right)$

$=(1-\sqrt{3})-\ell \mathrm{n}(\sqrt{3}-1)$