Question

Let $H (x)=3 x^{4}+6 x^{3}-2 x^{2}+1$ and $g(x)$ be a polynomial of degree one. If $H \frac{(x)}{(x-1) x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$ then $H (-1)+2 H (2)-3 H (1)=$

• A. $f(-1)+2 f(2)-3 f(1)$
• B. $hH (-1)+f(2)+g(3)$
• C. $g(-1)+2 g(2)-3 g(1)$
• D. $H (1)+2 f(2)-g(1)$

Answer 1

Answer: (C)

We have,
$\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$
$\Rightarrow H(x)=(x-1)(x+1)(x-2) f(x)+g(x)$
$\Rightarrow H(-1)=0+g(-1)=g(-1)$
$H(2)=0+g(2)=g(2)$
$H(1)=0+g(1)=g(1)$
$\therefore H(-1)+2 H(2)-3 H(1)$
$=g(-1)+2 g(2)-3 g(1)$