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If $f(x)=x^4-2 x^3+3 x^2-a x+b$ is divided by $x-1$ and $x+1$, the remainders are 5 and 19, respectively. If $f(x)$ is divided by $x-2$. The remainder is
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Verified Answer
The correct answer is:
10
$f(x)=x^4-2 x^3+3 x^2-a x+b$
Given that, $f(1)=5$ and $f(-1)=19$, then
$$
\begin{aligned}
& f(1)=1-2+3-a+b=2-a+b=5 \\
& \therefore \quad b-a=3 \\
& \text { and } f(-1)=1+2+3+a+b \\
& =b+a+6=19 \\
& \therefore \quad b+a=13 \\
&
\end{aligned}
$$
adding Eqs. (i) and (ii), we obtain
$$
\begin{aligned}
& \qquad 2 b=16 \Rightarrow b=8 \\
& \text { and } a=13-b \Rightarrow a=5 \\
& \therefore f(x)=x^4-2 x^3+3 x^2-5 x+8
\end{aligned}
$$
Thus,
$$
\begin{aligned}
f(2) & =(2)^4-2(2)^3+3(2)^2-5(2)+8 \\
& =16-16+12-10+8 \\
& =10
\end{aligned}
$$
$\therefore$ If $f(x)$ is divided by $x-2$, then remainder is 10 .
Given that, $f(1)=5$ and $f(-1)=19$, then
$$
\begin{aligned}
& f(1)=1-2+3-a+b=2-a+b=5 \\
& \therefore \quad b-a=3 \\
& \text { and } f(-1)=1+2+3+a+b \\
& =b+a+6=19 \\
& \therefore \quad b+a=13 \\
&
\end{aligned}
$$
adding Eqs. (i) and (ii), we obtain
$$
\begin{aligned}
& \qquad 2 b=16 \Rightarrow b=8 \\
& \text { and } a=13-b \Rightarrow a=5 \\
& \therefore f(x)=x^4-2 x^3+3 x^2-5 x+8
\end{aligned}
$$
Thus,
$$
\begin{aligned}
f(2) & =(2)^4-2(2)^3+3(2)^2-5(2)+8 \\
& =16-16+12-10+8 \\
& =10
\end{aligned}
$$
$\therefore$ If $f(x)$ is divided by $x-2$, then remainder is 10 .
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