Let H(x)=3x^4+6x^3-2x^2+1 and g(x) be a polyn | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$Multiplying both sides by $(x-1)(x+1)(x-2)$, we get:$H(x)=(x-1

Vedclass · Accepted Answer

$g(-1)+2g(2)-3g(1)$

Vedclass · Answer

(C) Given the equation: $\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$Multiplying both sides by $(x-1)(x+1)(x-2)$, we get:$H(x)=(x-1)(x+1)(x-2)f(x)+g(x)$Now, we evaluate $H(x)$ at $x=-1, 2, 1$:For $x=-1$: $H(-1)=( -1-1)( -1+1)( -1-2)f(-1)+g(-1) = 0+g(-1) = g(-1)$For $x=2$: $H(2)=(2-1)(2+1)(2-2)f(2)+g(2) = 0+g(2) = g(2)$For $x=1$: $H(1)=(1-1)(1+1)(1-2)f(1)+g(1) = 0+g(1) = g(1)$Substituting these into the expression $H(-1)+2H(2)-3H(1)$:$H(-1)+2H(2)-3H(1) = g(-1)+2g(2)-3g(1)$

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

Similar Questions

The fraction $\frac{x^2}{(x-a)(x-b)}$ is

If $\frac{x^2-3}{(x+2)(x^2+1)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}$ then $3A+2B-C=$

If $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A + \frac{B}{2 x-1} + \frac{C}{x+2} + \frac{D}{x-3}$,then $A$ is equal to

If $\frac{ax+5}{(x^2+b)(x+3)}=\frac{x+21}{12(x^2+b)}+\frac{c}{12(x+3)}$,then $b^2=$

If $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$, then $2(A-C+B+D) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module