TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given, f(x)=ax2+bx+c and GCD of a,b,c is 1.
Also given, ‌

−7+√11i

6

is a root of f(x)=0, then
‌

−7−√11i

6

must be another root of f(x)=0.
So, sum of roots =−‌

b

a

⇒‌

−7+√11i−7−√11i

6

=−‌

b

a

⇒‌‌‌

b

a

=‌

7

3

. . . (i)
and product of roots =‌

c

a

⇒(‌

−7+√11i

6

)(‌

−7−√11i

6

)=‌

c

a

⇒‌‌‌

49+11

6

=‌

c

a

⇒‌‌‌

c

a

=‌

5

3

. . . (ii)
From Eqs. (i) and (ii),
a=3,b=7 and c=5 and GCD(a,b,c)=1
∴f(x)=3x2+7x+5
Now, f(‌

x

k

)−L=(x+4)(3x−5)
⇒3(‌

x

k

)2+7(‌

x

k

)+5−L=3x2+7x−20
⇒3x2+7kx+5k2−Lk2=3k2x2+7k2x−20k2
On comparing the coefficients, we get

3k2=3	7k=7k2	5k2−Lk2=−20k2
k2=1	7k(k−1)=0	25k2=Lk2
k=−1... (iii)	k=0,1... (iv)	L=25

From Eqs. (iii) and (iv),
k=1‌ and ‌L=25