JEE Advanced 2020 Full Test 7 Paper 2

I−T;II−P;III−S;IV−Q
(I) Let the sides a and b are the roots of x2−5x+3=0, then a+b=5 and ab=3 Alos, ∠C=‌

π

3

∴‌

1

2

=‌

a2+b2−c2

2ab

⇒ab=(a+b)2−2ab−c2⇒c=4
Now, rR=‌

abc

2(a+b+c)

=‌

3×4

2(5+4)

=‌

2

3

(II) We have a=‌

r1(r2+r3)

√r1r2+r2r3+r3r1

=‌

8×36

24

=12
(III) Give 1≤a≤10
1 Also, for roots of opposite sign 2a2−7a+3<0
or ‌

1

2

<a<3⇒1 or 2
∴ Desired probability =‌

2

10

=‌

1

5

(IV) (α,β) lies on the director circle of the ellipse i. e. on x2+y2=9
So, we can assume α=3‌cos‌θ,β=3‌sin‌θ
Thus F=12‌cos‌θ+9‌sin‌θ=3(4‌cos‌θ+3‌sin‌θ)
⇒−15≤F≤15