Question

If $\underset{x\to \infty }{lim}\frac{a(2x^3-x^2)+b(x^3-1)-c(3x^3+x^2)}{a(5x^4-x)-b{x}^4+c(4x^4+1)+2x^2+5x}=1,$ then the value of $(a-b-c)$ can be expressed in the lowest form as $\frac{p}{q}$ where $p,q\in N.$ The value of $p+q$ is

• A. 34.0
• B. 34
• C. 34.00

Answer 1

Answer: (A), (B), (C)

$\underset{x\to \infty }{lim}\frac{a(2x^3-x^2)+b(x^3-1)-c(3x^3+x^2)}{a(5x^4-x)-b{x}^4+c(4x^4+1)+2x^2+5x}=1$

$\underset{x\to \infty }{lim}\frac{(2a+b-3c)x^3+(-a-c)x^2-b}{(5a-b+4c)x^4+2x^2+(-a+5)x+c}=1$

For limit to exist and equal to $1,$
coefficient of $x^4$ in denominator $=0$
$\Rightarrow 5a-b+4c=0\cdots \left(1\right)$
and coefficient of $x^3$ in numerator $=0$
$\Rightarrow 2a+b-3c=0\cdots \left(2\right)$
and $\frac{\text{coefficient of}{x}^2\text{in numerator}}{\text{coefficient of}{x}^2\text{in denominator}}=1$
$\Rightarrow \frac{-a-c}{2}=1$
$\Rightarrow -a-c=2\cdots \left(3\right)$

Solving $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
$a=\frac{1}{3},b=-\frac{23}{3}$ and $c=-\frac{7}{3}$
$\Rightarrow a-b-c=\frac{31}{3}=\frac{p}{q}$
$\therefore p+q=34$