Let the point (p, p+1) lie inside the region E=\(x, y): 3-x <= y <= sqrt9-x^2, 0 <= x <= 3\. If the set of all values of p is the interval (a, b), then b^2+b-a^2 is equal to ___________.

Given region, E=\(x, y): 3-x 2 x=2 => x=1 and y=2 and point of intersection of y=x+1 and y=sqrt(9-x^2) i.e., (x+1)^2=9-x^2 \lc => x^2+1+2 x=9-x^2 \ => 2 x^2+2 x-8=0 \ => x^2+x-4=0 \ => x=frac-1 ± sqrt(1+4(1)(4))2 \ => x=frac-1 ± sqrt(17)2\ \lll => x=frac-1+sqrt(17)2, (Since, x in [0,3]) \ Therefore p in (1, frac-1+sqrt(17)2)\ => a=1, b=frac-1+sqrt(17)2 Therefore b^2+b-a \ = frac1+17-2 sqrt(17)4+frac(-1+sqrt(17))2-1 \ = frac18-2 sqrt(17)-2+2 sqrt(17)-44 \ = (12 / 4)=3\

Let the point p p1 lie inside the region Ex y 3x leq y ... | null (Mathematics)

Given region,

E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\right\}

Since, point

(p, p+1)

lie on line

y=x+1

\therefore

Point of intersection of

y=x+1

and

y=3-x

i.e.,

x+1=3-x

\Rightarrow

2 x=2 \Rightarrow x=1

and

y=2

and point of intersection of

y=x+1

and

y=\sqrt{9-x^2}

i.e.,

(x+1)^2=9-x^2

\begin{array}{lc} &\Rightarrow x^2+1+2 x=9-x^2 \\\\ &\Rightarrow 2 x^2+2 x-8=0 \\\\ &\Rightarrow x^2+x-4=0 \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{1+4(1)(4)}}{2} \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{17}}{2}\end{array}

\begin{array}{lll}\Rightarrow & x=\frac{-1+\sqrt{17}}{2}, & \text { (Since, } x \in[0,3]) \\\\ \therefore & p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)\end{array}

\Rightarrow a=1, b=\frac{-1+\sqrt{17}}{2}

\begin{aligned} & \therefore b^2+b-a \\\\ &= \frac{1+17-2 \sqrt{17}}{4}+\frac{(-1+\sqrt{17})}{2}-1 \\\\ &= \frac{18-2 \sqrt{17}-2+2 \sqrt{17}-4}{4} \\\\ &= \frac{12}{4}=3\end{aligned}

Explanation

The minimum number of elements that must be added to the relation R = {(a,b),(b,c),(b,d)} on the set { a,b,c,d} so that it is an equivalence relation is

For integers n and r let n r nCr ifn ge r ge 0 0 otherwise

The integral int_0^1 \frac{1}{7^{\left[\frac{1}{x}\right]}} {dx}, where [.] denotes the greatest integer function is equal to

The area of the region enclosed by the parabolas y=4x-x^2 and 3y=(x-4)^2 is equal to

For (2 leq r leq n,binom{n}{r} + 2binom{n}{r - 1} + binom{n}{r - 2} =)

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have atleast one letter repeated, is

Area of the region (x,y):x^2+(y-2)^2≤4,x^2≥2y is

Let R be the set of real numbers. end{enumerate} Statement-1: A = {(x,y) in R X R:y - x is an integer } is an equivalence . relation on R. Statement-2: B = {(x,y) in R X R:x = alpha y for some rational number } is . an equivalence relation on R.

In a college of 300 students, every student reads 5 newspapers and every newspaper is read by 60 students. The number of newspaper is

Download Divine JEE App

Quick Links

Exam Categories

Contact Info

Divine JEE