Question

Let $R$ be the region in the first quadrant bounded by the $x$ and $y$ axis and the graphs of $f(x)=\frac{9}{25}x+b$ and $y=f^{-1}(x)$. If the area of $R$ is 49 , then the value of $b$, is

• A. $\frac{18}{5}$
• B. $\frac{22}{5}$
• C. $\frac{28}{5}$
• D. none

Answer 1

Answer: (C)

If $f(x)=mx+b$, then $f^{-1}(x)=\frac{x-b}{m}$ and their point of intersection can be found by setting $x=my+b$ since they intersect on $y=x$.
Thus $x=\frac{b}{1-m}$ and the point of intersection is $\left(\frac{b}{1-m},\frac{b}{1-m}\right)$.
Region $R$ can be broken up into congruent triangles $PAB$ and $PCB$ which both have a base of $b$ and a height of $\frac{b}{1-m}$.
The area of $R$ is $2\left(\frac{b}{2}\right)\left(\frac{b}{1-m}\right)=\frac{b^2}{1-m}=49$. For $m=\frac{9}{25},b^2=\frac{16}{25}\cdot 49\Rightarrow b=\frac{28}{5}$