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(x) passes through $(1,1)$ and $g\left(x\right)$ passes through $(2,3)$ then: The curve g(x) is given by.

• A. $x-\frac{1}{x}$
• B. $x+\frac{2}{x}$
• C. $x^2-\frac{1}{x^2}$
• D. $(x+\frac{1}{x})$$(x+\frac{2}{x})$

Answer 1

The correct option is A $x-\frac{1}{x}$

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

(i)

$\frac{dy}{dy}=c\Rightarrow c_1=1$

for second curve

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

(ii)

$\frac{dy}{dx}=d_1---\left(1\right)$

for second curve

$\frac{dy}{dx}=d_2----\left(2\right)$

(iii)

From eq (1) and (2)

$d_1=2=d_2$

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

${\int }_1^2(4-2)d\left(x\right)$

$2{\int }_1^{2}d\left(x\right)$

$2(2-1)=2$

SO $g\left(x\right)=x-\frac{1}{x}$