Question

If a curve passes through the point $(2,\frac{7}{2})$ and has slope $(1-\frac{1}{x^2})$ at any point $(x,y)$ on it, then

the ordinate of the point on the curve whose abscissa is $-2$ is:

• A. $-\frac{3}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $-\frac{5}{2}$

Answer 1

The correct option is A $-\frac{3}{2}$

Let $y=f\left(x\right)$ be the curve.

Now, the slope of the curve at $(x,y)$ is $\frac{dy}{dx}$

According to the problem,

$\frac{dy}{dx}=1-\frac{1}{x^2}$

Integrating we get,

$y=x+\frac{1}{x}+c$ [where 'x' is integrating constant]

Gives the curve passes the $(2,\frac{7}{2})$

$\frac{7}{2}=2+\frac{1}{2}+c$

or $\frac{7}{2}-\frac{5}{2}=c$

or, $c=1$

So, the curve is $y=(x=\frac{1}{x}+1)$

For $x=-2$; $y=-2-\frac{1}{2}+1$

$=-1-\frac{1}{2}=-\left(\frac{3}{2}\right)$

Required coordinate is $(-\frac{3}{2})$.