Basic Mathematics & Logarithm
Point Inside Region — Inequalities and Intervals
nta_pyq_2023_apr
Grade 11
Question:
Let the point $(p,p+1)$ lie inside the region $E=\{(x,y):\ 3-x\leq y\leq\sqrt{9-x^2},\ 0\leq x\leq 3\}$. If the set of all values of $p$ is the interval $(a,b)$, then $b^2+b-a^2$ is equal to ________.
Step-by-Step Solution
Key Concept: The point $(p,p+1)$ lies on the line $y=x+1$. The region $E$ is bounded below by $y=3-x$ and above by the quarter-circle $y=\sqrt{9-x^2}$. Find where $y=x+1$ enters and exits $E$.
Intersection with $x+y=3$: $p=1\Rightarrow a=1$. Intersection with $x^2+y^2=9$: $b^2+b-4=0$ from $2b^2+2b-8=0$, so $b^2+b=4$. $b^2+b-a^2=4-1=3$.
Correct Answer: 3