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Circle question

2023 · 6 Apr · Shift 1 · Q38
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Circle question

2023 · 6 Apr · Shift 1 · Q38

JEE MainMathematicsCircleNumerical+4 / −1
Let the point (p,p+1)(p, p+1)(p,p+1) lie inside the region E={(x,y):3−x≤y≤9−x2,0≤x≤3}E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^{2}}, 0 \leq x \leq 3\right\}E={(x,y):3−x≤y≤9−x2​,0≤x≤3}. If the set of all values of p\mathrm{p}p is the interval (a,b)(a, b)(a,b), then b2+b−a2b^{2}+b-a^{2}b2+b−a2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3

  1. Understand the region EEE

The region is

E={(x,y):3−x≤y≤9−x2,  0≤x≤3}.E=\{(x,y): 3-x \le y \le \sqrt{9-x^2},\; 0\le x\le 3\}.E={(x,y):3−x≤y≤9−x2​,0≤x≤3}.

So for each x∈[0,3]x\in[0,3]x∈[0,3], the point lies between:

  • the line y=3−xy=3-xy=3−x
  • the upper semicircle y=9−x2y=\sqrt{9-x^2}y=9−x2​ of x2+y2=9x^2+y^2=9x2+y2=9.

Thus EEE is the region in the first quadrant enclosed by the line and the semicircle.

  1. Condition for (p,p+1)(p,p+1)(p,p+1) to lie inside EEE

We substitute

(x,y)=(p,p+1).(x,y)=(p,p+1).(x,y)=(p,p+1).

Then the conditions become:

(i) 0≤x≤30\le x\le 30≤x≤3

0≤p≤3.0\le p\le 3.0≤p≤3.

(ii) Lower bound: 3−x≤y3-x\le y3−x≤y

3−p≤p+1.3-p\le p+1.3−p≤p+1.

Solving,

3−p≤p+1  ⟹  2≤2p  ⟹  p≥1.3-p\le p+1 \implies 2\le 2p \implies p\ge 1.3−p≤p+1⟹2≤2p⟹p≥1.

(iii) Upper bound: y≤9−x2y\le \sqrt{9-x^2}y≤9−x2​

p+1≤9−p2.p+1\le \sqrt{9-p^2}.p+1≤9−p2​.

Since the right side is nonnegative, square both sides:

(p+1)2≤9−p2.(p+1)^2 \le 9-p^2.(p+1)2≤9−p2.

Expand:

p2+2p+1≤9−p2p^2+2p+1 \le 9-p^2p2+2p+1≤9−p2 2p2+2p−8≤02p^2+2p-8\le 02p2+2p−8≤0 p2+p−4≤0.p^2+p-4\le 0.p2+p−4≤0.

The roots of p2+p−4=0p^2+p-4=0p2+p−4=0 are

p=−1±172.p=\frac{-1\pm\sqrt{17}}{2}.p=2−1±17​​.

Hence

−1−172≤p≤−1+172.\frac{-1-\sqrt{17}}{2}\le p\le \frac{-1+\sqrt{17}}{2}.2−1−17​​≤p≤2−1+17​​.

Combining with p≥1p\ge 1p≥1, we get

1≤p≤−1+172.1\le p\le \frac{-1+\sqrt{17}}{2}.1≤p≤2−1+17​​.
  1. Inside the region

The question says the point lies inside the region, so boundary values are excluded.

Therefore,

p∈(a,b)=(1,−1+172).p\in (a,b)=\left(1,\frac{-1+\sqrt{17}}{2}\right).p∈(a,b)=(1,2−1+17​​).

So

a=1,b=−1+172.a=1,\qquad b=\frac{-1+\sqrt{17}}{2}.a=1,b=2−1+17​​.
  1. Compute b2+b−a2b^2+b-a^2b2+b−a2

Since bbb satisfies

b2+b−4=0,b^2+b-4=0,b2+b−4=0,

we have

b2+b=4.b^2+b=4.b2+b=4.

Also,

a2=1.a^2=1.a2=1.

Thus

b2+b−a2=4−1=3.b^2+b-a^2=4-1=3.b2+b−a2=4−1=3.
  1. Final answer
3\boxed{3}3​
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