Question

The value of ${\int }_1^2\left(g\right(x)-f(x\left)\right)dx\left)\right)dx$ is

• A. $2$
• B. $3$
• C. $4$
• D. $4\ln 2$

Answer 1

The correct option is B $3$

Given $y=f\left(x\right),y=g\left(x\right)$ are two curves, the equation of tangents to
these curves at points with equal abscissa say $(x=\alpha )$
are given by
$y-f\left(\alpha \right)=f^{\prime }\left(\alpha \right)(x-\alpha )$ & $y-g\left(\alpha \right)=g^{\prime }\left(\alpha \right)(x-\alpha )$...........{1}
These curves (lines) given by {1} 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
$y-f\left(\alpha \right)=-\frac{1}{f^{\prime }\left(\alpha \right)}(x-\alpha )$ & $y-g\left(\alpha \right)=-\frac{1}{g^{\prime }\left(\alpha \right)}(x-\alpha )$.......... {2}
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 $(1,1)$ when $x=1,f\left(x\right)=1$ and
Equation (D) passes through $(2,3)$ when $x=-1,g\left(x\right)=-2$
$\therefore$ $2\times 1=c+\frac{k}{c}$ & $3\times 2=\frac{k}{2c}-2c$
i.e. $4=c+\frac{k}{c}$.........{3}
and $6=\frac{k}{2c}-2c$.........{4}
Solving {3} and {4} for k & c we get $c=-2$, $k=-8$
Now putting the value of $c,k$ in $f\left(x\right)$ & $g\left(x\right)$ i.e. in (C) & (D) we get
$f\left(x\right)=\frac{2}{x}-x$ and $g\left(x\right)=\frac{2}{x}+x$
Now, ${\int }_1^2\left(g\right(x)-f(x\left)\right)dx={\int }_1^2(\frac{2}{x}+x-(\frac{2}{x}-x))dx$
$\Rightarrow {\int }_1^2\left(2x\right)dx=3$