Question

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0$, $x=1$, $y^2=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha^2$. Then $(f(0)+f(1))$ is equal to

• A. 12
• B. 9
• C. 7
• D. 14

Hint:

When absolute values are involved, always simplify the expression interval-wise before integrating.

Answer 1

The Correct Option is C

Step 1: Evaluate $f(0)$.
For $\alpha=0$, \[ y=| -5|-|1|=5-1=4 \] Area between $y=4$ and $y^2=x$ from $x=0$ to $x=1$: \[ f(0)=\int_{0}^{1}(4-\sqrt{x})\,dx \] \[ =4-\frac{2}{3}= \frac{10}{3} \] Step 2: Evaluate $f(1)$.
For $\alpha=1$, \[ y=|x-5|-|1-x|+1 \] In $[0,1]$, this simplifies to \[ y=(5-x)-(1-x)+1=5 \] Area becomes \[ f(1)=\int_{0}^{1}(5-\sqrt{x})\,dx \] \[ =5-\frac{2}{3}=\frac{13}{3} \] Step 3: Add the areas.
\[ f(0)+f(1)=\frac{10}{3}+\frac{13}{3}=7 \]