JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, a,b and c are in AP.
(a,c),(2,b) and (a,b) are vertices of triangle.
Centroid =(‌

10

3

,‌

7

3

)
α and β are the roots of equation ax2+bx+1=0
∵a,b,c are in AP.
∴ 2b=a+c. . . (i)
⇒‌‌

2a+2

3

‌‌=‌

10

3

‌ and ‌‌

c+2b

3

=‌

7

3

‌
a‌‌=4‌
‌ and ‌‌‌c+a+c‌‌=7‌
⇒‌‌2c‌‌=7−4‌‌[∵a=a+c]
c=3∕2
‌ Also, ‌2b=a+c=4+‌

3

2

⇒‌‌b=11∕4
Now, α and β are roots of ax2+bx+1=0
∴‌‌α+β=‌

−b

a

=‌

−11∕4

4

⇒‌‌α+β=‌

−11

16

⇒αβ=‌

1

a

=‌

1

4

⇒‌‌αβ=‌

1

4

Now, α2+β2−αβ
=(α+β)2−3αβ=(‌

−11

16

)2−3×‌

1

4

=‌

121−192

256

=‌

−71

256