NTSE Stage 2 SAT 2017 Question Paper

Reason :
Let ${\mathrm{q}}_1$ $\left(\mathrm{x}\right)$ and ${\mathrm{q}}_2$ $\left(\mathrm{x}\right)$ be the quotients when $\mathrm{p}\left(\mathrm{x}\right)$ is divided by $\left(\mathrm{x}\mathit{–}1\right)\mathrm{and}\left(\mathrm{x}\mathit{–}3\right)$ respectively.
Also asume that ${\mathrm{q}}_3\left(\mathrm{x}\right)$ is quotient when $\mathrm{p}\left(\mathrm{x}\right)$ is divided by $\left(\mathrm{x}\mathit{–}1\right)\left(\mathrm{x}\mathit{–}3\right).$
Now A.G.C I, $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{q}}_1\left(\mathrm{x}\right)\times \left(\mathrm{x}\mathit{–}1\right)+3$...(1)
and A.G.C II, $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{q}}_2\left(\mathrm{x}\right)\times \left(\mathrm{x}\mathit{–}3\right)+5$ ...(2)
Also A.G.C III, $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{q}}_3\left(\mathrm{x}\right)\left(\mathrm{x}\mathit{–}1\right)\left(\mathrm{x}\mathit{–}3\right)+\mathrm{r}\left(\mathrm{x}\right)$ ...(3)
Now at x = 1, from eqn no. (1), p(1) = 3...(4)
at x = 3, from eqn no. (2), p(3) = 5...(5)
Now put x = 1 in equation no. (3) we get $\mathrm{p}\left(1\right)=\mathrm{r}\left(1\right)\Rightarrow \mathrm{r}\left(1\right)=3$ [using equation (4)]...(6)
Similarly from (3) & (5) we get r(3) = 5...(7)
from equation no.(3) it is clear that divisor $\left(\mathrm{x}\mathit{–}1\right)\left(\mathrm{x}\mathit{–}3\right)$ is of degree two so its remainder r(x) will be of degree one or zero so
Let $\mathrm{r}\left(\mathrm{x}\right)=\mathrm{Ax}+\mathrm{B}$ put r = 1 & r = 3 is equation (8) we get $\mathrm{r}\left(1\right)=\mathrm{A}+\mathrm{B}\Rightarrow \mathrm{A}+\mathrm{B}=3$ [using (6)]...(8)
similarly $\mathrm{r}\left(3\right)=3\mathrm{A}+\mathit{B}\Rightarrow 3\mathrm{A}+\mathrm{B}=5$ [using 7] ...(9)
on solving equation no.(8) & (9) we getA = 1 & B = 2⇒ $\mathrm{r}\left(\mathrm{x}\right)=1\mathrm{x}+2$ ...(10)Now put x = –2 in equation no.(10) we get $\mathrm{r}\left(\mathit{–}2\right)=\mathit{–}2+2=0$
Hence option no.(3) is correct.