JEE Main 2023 — Mathematics Coordinate Geometry

Given,

The point $(p,p+1)$(p,p+1) lie inside the region $E={(x,y):3-x\leq y\leq \sqrt{9-{x}^{2}},0\leq x\leq 3}$E=(x,y):3−x≤y≤9−x2​,0≤x≤3,

And the set of all values of $p$p is the interval $(a,b),$(a,b),

Now by observation $(p,p+1)$(p,p+1) lies on $y-x=1$y−x=1,

Also point $(p,p+1)$(p,p+1) lies on $x+y=3$x+y=3

So, solving $y-x=1$y−x=1 and $x+y=3$x+y=3 we get, $P(1,2)$P(1,2)

Again given point $(p,p+1)$(p,p+1) lies on ${x}^{2}+{y}^{2}=9$x2+y2=9

So, solving $y-x=1$y−x=1 and ${x}^{2}+{y}^{2}=9$x2+y2=9, we get

$${x}^{2}+(1+x{)}^{2}=9$$x2+(1+x)2=9

$$\Rightarrow 2{x}^{2}+2x-8=0$$⇒2x2+2x−8=0

$$\Rightarrow {x}^{2}+x-4=0.........(i)$$⇒x2+x−4=0.........(i)

$$\Rightarrow x=\frac{-1\pm \sqrt{1+4\cdot 4}}{2}=\frac{\sqrt{17}-1}{2}$$⇒x=2−1±1+4⋅4​​=217​−1​

Hence, $p\in (1,\frac{\sqrt{17}-1}{2})$p∈(1,217​−1​)

Now on comparing with $p\in (a,b)$p∈(a,b) we get,

$\Rightarrow a=1,{b}^{2}+b=4$⇒a=1,b2+b=4 [using $(i)$(i)]

Hence, the value of ${b}^{2}+b-{a}^{2}=4-1=3$b2+b−a2=4−1=3