AP EAPCET 24-Aug-2021 Shift 1 Paper

Q(x) be a polynomial of degree n
Q(1)=1 and

Q(2x)

Q(x+1)

+

56

x+7

−8=0 ...(i)
Put x=0 in Eq. (i), we get

Q(0)

Q(1)

+

56

7

−8=0
⇒

Q(0)

1

+8−8=0
Q(0)=0
Solving Eq. (i), we get

Q(2x)

Q(x+1)

+

56−8x−56

x+7

=0
⇒

Q(2x)

Q(x+1)

=

8x

x+7

=0
⇒

Q(2x)

Q(x+1)

=

8x

x+7

...(ii)
∵Q(0)=0
∴ x is a factor of Q(x).
Let Q(x)=xP(x)
∴

Q(2x)

Q(x+1)

=

8x

x+7

2x.P(2x)

(x+1)P(x+1)

=

8x

x+7

P(2x)

P(x+1

=

4(x+1)

x+7

...(ii)
Put x=−1

P(−2)

P(0)

=0
⇒P(−2)=0
⇒x+2 is factor of P(x)
P(x)=(x+2)R(x)
Now, in Eq, (ii), we get

4(x+1)

x+7

=

P(2x)

P(x+1)

=

(2x+2)R(2x)

(x+1+2)R(x+1)

⇒

4(x+1)

x+7

=

2x+2

x+3

R(2x)

R(x+1)

2(x+3)

x+7

=

R(2x)

R(x+1)

∵x=−3
⇒0=

R(−6)

R(−2)

⇒R(−6)=0
=x+6 is a factor of R(x).
So, R(x)=(x+6)S(x)

2(x+3)

x+7

=

R(2x)

R(x+1)

=

(2x+6)S(2x)

(x+6+1)S(x+1)

S(2x)

S(x+1)

=1
⇒S(2x)=S(x+1) for all x
S(x) is constant function. ∴Q(x) is cubic polynomial.
∴n=3
Required value =‌3C0+‌3C1+‌3C2+‌3C3
=1+3+3+1=8