TS EAMCET 6-Aug-2021 Shift 2 Question Paper

From given question, we observe that α occurs at x=‌

−b

2a

i.e. x=‌

−b

2

and β occurs at x=‌

−b

2a

, i.e. x=‌

+a

2

So, ‌‌α=(‌

−b

2

)2+b⋅(‌

−b

2

)+5
⇒‌‌α=5−‌

b2

4

‌‌... (i)
and ‌‌β=−(‌

+a

2

)2+a(‌

+a

2

)+5
=−‌

a2

4

+‌

a2

2

+5
⇒‌‌β=5−‌

a2

4

. . . (ii)
Given, x2−10x+24≤0
⇒‌‌(x−4)(x−6)≤0
⇒‌‌4≤x≤6
∴‌‌α=4 and β=6
From Eqs. (i) and (ii), we get
∴b2=4‌ and ‌a2=4
∴‌‌a2b2=16