Question

Let $y=f\left(x\right)$ & $y=g\left(x\right)$ be two curves satisfying the following conditions

(1) Tangents at points with equal abscissa intersect on y-axis.

(2) The normal drawn at points with equal abscissa intersect on x-axis.

(3) One curve passes through (2, 2) & other passes through (-1, -2 )

On the basis of above information answer the following

question

The value of ${\int }_1^4\{f\left(x\right)-g\left(x\right)\}dx$ is?

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is D 6

Given y=f(x), y=g(x) are two curves, the equation of tangents to these curves at points with equal abscissa say $(x=\alpha )$ are given by
$y-f\left(\alpha \right)=f^{\prime }\left(\alpha \right)(x-\alpha )$ & $y-g\left(\alpha \right)=g^{\prime }\left(\alpha \right)(x-\alpha )$ (*)
These curves (lines) given by (*) intersect at a point on y-axis
$\therefore$ $y-f\left(\alpha \right)=-\alpha f^{\prime }\left(\alpha \right)$ & $y-g\left(\alpha \right)=-\alpha g^{\prime }\left(\alpha \right)$
$\Rightarrow$ $y=f\left(\alpha \right)-\alpha f^{\prime }\left(\alpha \right)$ & $y=g\left(\alpha \right)-\alpha g^{\prime }\left(\alpha \right)$
$\Rightarrow$ $f\left(\alpha \right)-\alpha f^{\prime }\left(\alpha \right)=g\left(\alpha \right)-\alpha g^{\prime }\left(\alpha \right)$
$\Rightarrow$ $f\left(\alpha \right)-g\left(\alpha \right)=\alpha \{f^{\prime }\left(\alpha \right)-g^{\prime }\left(\alpha \right)\}$ replacing $\alpha =x$
$\therefore$ $f\left(x\right)-g\left(x\right)=x\{f^{\prime }\left(x\right)-g^{\prime }\left(x\right)\}$
$\Rightarrow$ $f\left(x\right)-g\left(x\right)=x\frac{d}{dx}\{f\left(x\right)-g\left(x\right)\}$
$\Rightarrow$ $\frac{d\{f\left(x\right)-g\left(x\right)\}}{f\left(x\right)-g\left(x\right)}=\frac{dx}{x}$
$\Rightarrow$ $\log \{f\left(x\right)-g\left(x\right)\}=\log x+\log c=\log \left(xc\right)$
$\Rightarrow$ $f\left(x\right)-g\left(x\right)=xc$ (A)
Again, the equations ofthe normals at points with equal abscissa say $x=\alpha$ the two curves are
$y-f\left(\alpha \right)=-\frac{1}{f^{\prime }\left(\alpha \right)}(x-\alpha )$ & $y-g\left(\alpha \right)=-\frac{1}{g^{\prime }\left(\alpha \right)}(x-\alpha )$
These lines intersect at a point on x-axis
$\therefore$ y=0
$0-f\left(\alpha \right)=-\frac{1}{f^{\prime }\left(\alpha \right)}(x-\alpha )$ & $0-g\left(\alpha \right)=-\frac{1}{g^{\prime }\left(\alpha \right)}(x-\alpha )$
$\Rightarrow$ $x=\alpha +f^{\prime }\left(\alpha \right)f\left(\alpha \right)$ & $x=\alpha +g^{\prime }\left(\alpha \right)g\left(\alpha \right)$
$\Rightarrow$ $\alpha +f^{\prime }\left(\alpha \right)f\left(\alpha \right)=\alpha +g^{\prime }\left(\alpha \right)g\left(\alpha \right)$
$\Rightarrow$ $f^{\prime }\left(\alpha \right)f\left(\alpha \right)=g^{\prime }\left(\alpha \right)g\left(\alpha \right)$ replacing $\alpha =x$ we get
$\Rightarrow$ $f^{\prime }\left(x\right)f\left(x\right)=g^{\prime }\left(x\right)g\left(x\right)$ $\Rightarrow$ ${\left(f\left(x\right)\right)}^2={\left\{g\left(x\right)\right\}}^2+k$
$\Rightarrow$ ${\left\{f\left(x\right)\right\}}^2-{\left\{g\left(x\right)\right\}}^2=k$
$\Rightarrow$ $\{f\left(x\right)+g\left(x\right)\}\{f\left(x\right)-g\left(x\right)\}=k$
$\Rightarrow$ $f\left(x\right)+g\left(x\right)=\frac{k}{cx}$ (using A)(B)
Now solving (A) & (B) we get
$f\left(x\right)=\frac{1}{2}(cx+\frac{k}{cx})$ (C)
and $g\left(x\right)=\frac{1}{2}(\frac{k}{cx}-cx)$ (D)
Now equation (C) passes through (2, 2) when x=2,
f(x)=2 and equation (D) passes through (-1, -2) when
$x=-1$, $g\left(x\right)=-2$
$\therefore$ $2\times 2=2c+\frac{k}{2c}$ & $-2\times 2=\frac{k}{-c}+c$
i.e. $4=2c+\frac{k}{2c}$ (*)
and $-2=\frac{c}{2}+\frac{k}{-2c}$ (**)
Solvmg (*}and (**) for k & c we get $c=\frac{4}{5}$, $k=\frac{96}{25}$
Now putting the value of c, k in f(x) & g(x) i.e. in (C) & (D) we get
$f\left(x\right)-g\left(x\right)=(\frac{2}{5}x+\frac{12}{5x})-(\frac{12}{5x}-\frac{2}{5}x)=\frac{4}{5}x$
$\therefore$ ${\int }_1^4\{f\left(x\right)-g\left(x\right)\}dx=\frac{4}{5}{\int }_1^{4}xdx=\frac{4}{5}{\left(\frac{x^2}{2}\right)}_1^4=\frac{2}{5}\left(15\right)=6$