Question

Let `R` be the region in the first quadrant bounded by the `x` and `y` axis and the graphs of `f(x) = 9/25 x +b` and `y = f^-1 (x)`. If the area of `R` is 49, then the value of `b`, is ? A) `18/5` B) `22/5` C) `28/5` D) none

Answer 1

C) `28/5`

Explanation:

Let's first imagine what the region would look like. Inverse functions are reflections of themselves over the line `x=y`, so the lines would resemble the following (ignore any numbers, this is just a speculation):

We see that the top line will be given by `L_1(x)=f(x)=9/25x+b`.

The inverse of this is given by:

`y=9/25x+b" "=>" "x=9/25y+b`

`color(white)(y=9/25x+b)" "=>" "y=25/9(x-b)=f^-1(x)`

Since this is the lower line we say that `L_2(x)=25/9(x-b)`.

The area of the region is given by two distinct parts: the first is the trapezoid that lies under `L_1`. This extends from `x=0` to the `x` intercept of `L_2`.

The second part is a triangle that extends from the `x` intercept of `L_2` to the intersection point of `L_1` and `L_2`. Its height is the value of `L_1` at the `x` intercept of `L_2`.

Let's find these important points first:

`mathbf(L_2)` `mathbfx`-intercept

`L_2(x)=9/25(x-b)=0" "=>" "x=b`

Intersection Point

`L_1(x)=L_2(x)" "=>" "9/25x+b=25/9(x-b)`

`color(white)(L_1(x)=L_2(x))" "=>" "81/625x+9/25b=x-b`

`color(white)(L_1(x)=L_2(x))" "=>" "34/25b=544/625x`

`color(white)(L_1(x)=L_2(x))" "=>" "x=25/16b`

We can now, without any integration necessary (since we're dealing with lines) find these areas.

Trapezoid Area

One base is `b`, which is the `y` intercept of `L_1`. The other base is `L_1(b)=34/25b`. The height is `b` as well, so the trapezoid's area is

`A_1=1/2b(b+34/25b)=59/50b^2`

Triangle Area

The base is again `L_1(b)=34/25b`. The height (horizontal) is `25/16b-b=9/16b`. The area of the triangle is then

`A_2=1/2(34/25b)(9/16b)=153/400b^2`

The total region `R` has area `49`.

`R=A_1+A_2=59/50b^2+153/400b^2=49`

`25/16b^2=49`

`b=sqrt((49(16))/25)=28/5`