Question

Let $y=f\left(x\right)$ and $y=g\left(x\right)$ be the pair of curves such that
(i) the tangents at point with equal abscissae intersect on y-axis.
(ii) the normal drawn at points with equal abscissae intersect on x-axis and
(iii) curve $f\left(x\right)$ passes through $(1,1)$ and $g\left(x\right)$ passes through $(2,3)$ then.
On the basis of given information, answer the following question.
The curve $f\left(x\right)$ is given by?

• A. $\frac{2}{x}-x$
• B. $2x^2-\frac{1}{x}$
• C. $\frac{2}{x^2}-x$
• D. None of these

Answer 1

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

let $P$ and $Q$ be the points on $f\left(x\right)$ and $g\left(x\right)$ respectively such that $P(x,f(x\left)\right)$ and $Q(x,g(x\left)\right)$

$y=f\left(x\right)$ and $g\left(x\right)=y$

(i) $\frac{dy}{dx}=d\Rightarrow d_1=1$

for second curve

$\frac{dy}{dx}=d_2$

(ii) $\frac{dy}{dy}=c\Rightarrow c_1=1$

for second curve

$\frac{dy}{dy}=c\Rightarrow c_2=1$

from eq $\left(i\right)$ and $\left(ii\right)$

$d_1=2=d_2$

$\Rightarrow {\int }_1^2\left(g\right(x)-f(x\left)\right)dx$

$\Rightarrow {\int }_1^2(4-2)dx$

$\Rightarrow 2{\int }_1^{2}dx\Rightarrow 2(-1+2)=2$

So, $g\left(x\right)=\frac{2}{x}-x$

Hence, $\left(A\right)$ is the correct option