Area Under the Curve
Area with Absolute Value Curves — f(0)+f(1)
nta_pyq_2026_jan
Grade 12
Question:
Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0$, $x=1$, $y^2=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^2$. Then $f(0)+f(1)$ is equal to
Step-by-Step Solution
Key Concept: For $\alpha=0$: $y=|-5|-|1|+0=5-1=4$ (constant). Area $=\int_0^1(4-\sqrt{x})dx=4-\frac{2}{3}=\frac{10}{3}$. For $\alpha=1$, $0\leq x\leq1$: $|x-5|=5-x$, $|1-x|=1-x$, $y=4+x^2$. Area $=\int_0^1(4+x^2-\sqrt{x})dx=4+\frac{1}{3}-\frac{2}{3}=\frac{11}{3}$.
$f(0)=10/3$, $f(1)=11/3$. $f(0)+f(1)=7$.
Correct Answer: 3