Question

If a curve passes through the point $(1,0)$ and has slope $(1+\frac{1}{x^2})$ at any point (x, y) on it, then the ordinate of point on the curve whose abscissa is $-3$, is?

• A. $\frac{3}{8}$
• B. $\frac{-8}{3}$
• C. $\frac{-3}{8}$
• D. $\frac{8}{3}$

Answer 1

Answer: (B)

$\frac{-8}{3}$

$\frac{dy}{dx}=1+\frac{1}{x^2}\Rightarrow f\left(x\right)=x-\frac{1}{x}+\lambda$
As, $(1,0)$ satisfy it
$\Rightarrow 0=1-\frac{1}{1}+\lambda \Rightarrow \lambda =0$
So, $f\left(x\right)=x-\frac{1}{x}$
$\Rightarrow f(-3)=-3-\frac{1}{(-3)}=\frac{-3+1}{3}=\frac{-8}{3}$