Q81 / 200 · WBJEE 2014 · Mathematics
Let $f(x)=2 x^{2}+5 x+1$. If we write $f(x)$ as $f(x)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)$
for real numbers $a, b, c,$ then
- Athere are infinite number of choices for $a, b, c$
- Bonly one choice for a but infinite number of choices for $b$ and $c$
- Correct answerCexactly one choice for each of $a, b, c$
- Dmore than one but finite number of choices for $a, b, c$
Solution
$$
\begin{array}{l}
\text { Given, } f(x)=2 x^{2}+5 x+1 \\
\text { Also, } f(x)=a(x+1)(x-2)+b(x-2)(x-1) \\
+c(x-1)(x+1) \\
\qquad \begin{array}{r}
=a\left(x^{2}-x-2\right)+b\left(x^{2}-3 x+2\right) \\
+c\left(x^{2}-1\right) \\
f(x)=(a+b+c) x^{2}+(-a-3 b) x \\
+(-2 a+2 b-c)
\end{array}
\end{array}
$$
On equating the coefficients of $x^{2}$, $x$ and constant term we get
$$
\begin{array}{l}
a+b+c=2,-a-3 b=5 \\
\text { and }-2 a+2 b-c=1
\end{array}
$$
On solving above equations, we get $a=-\frac{35}{4}, b=\frac{5}{4}$ and $c=\frac{38}{4}$
Hence, exactly one choice for each of b and c
Hence, the correct answer is option: 'C'.