Question

Let $f(x)=2x^2+5x+1$. 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

• A. there are infinite number of choices for a,b,c
• B. only one choice for a but infinite number of choices for b and c
• C. exactly one choice for each of a,b,c
• D. more than one but finite number of choices for a,b,c

Answer 1

Answer: (C)

Given, $f(x)=2x^2+5x+1...(i)$
Also, $f(x)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)$
$=a\left(x^2-x-2\right)+b\left(x^2-3x+2\right)+c\left(x^2-1\right)$
$\Rightarrow f(x)=(a+b+c)x^2+(-a-3b)x+(-2a+2b-c)...(ii)$
On equating the coefficients of $x^2,x$ and
constant term in Eqs. (i) and (ii), we get
$a+b+c=2,-a-3b=5$
and-$2a+2b-c=1$
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 $c.$