FPM(2023) January Paper 02R Q4 and solution

Problem and Solution

We are given the equation of the curve $C$ as

$$ y = \frac{2x + q}{x + r} $$

where $q$ and $r$ are integers. The asymptote parallel to the $y$-axis is $x = -4$, and the asymptote parallel to the $x$-axis is $y = p$. Additionally, the curve crosses the $y$-axis at the point $$ \left(0,\frac{3}{2}\right) $$ and the curve crosses the $x$-axis at the point $$ (s,0) $$

Part (a)

• (i) The value of $p$:
  
  As $x \to \infty$, the curve tends to the horizontal asymptote, where the ratio of the leading terms in the numerator and denominator gives the asymptote. Therefore, for large values of $x$:

$$ y \approx \frac{2x}{x} = 2 $$

Thus, the horizontal asymptote is $$ y = 2 $$ so the value of $p$ is

$$ p = 2 $$

• (ii) The value of $r$:
  
  The vertical asymptote occurs when the denominator is zero, i.e. when $$ x + r = 0 $$ which gives $$ x = -r $$

We are told that the vertical asymptote is at $$ x = -4 $$ so

$$ r = 4 $$

Part (b)

We are given that the curve crosses the $y$-axis at the point $$ \left(0,\frac{3}{2}\right) $$ To find $q$, substitute $x = 0$ and $$ y = \frac{3}{2} $$ into the equation of the curve:

$$ \frac{3}{2} = \frac{2(0) + q}{0 + 4} $$

This simplifies to:

$$ \frac{3}{2} = \frac{q}{4} $$

Multiplying both sides by $4$:

$$ q = 6 $$

Thus, the value of $q$ is $6$.

Part (c)

Given that the curve crosses the $x$-axis at the point $$ (s,0) $$ substitute $y = 0$ into the equation of the curve:

$$ 0 = \frac{2x + 6}{x + 4} $$

For the fraction to be zero, the numerator must be zero. Therefore:

$$ 2x + 6 = 0 $$

Solving for $x$:

$$ 2x = -6 \quad \Rightarrow \quad x = -3 $$

Thus, the value of $s$ is $-3$.

Final Answers

$$ \boxed{ p = 2, \quad r = 4, \quad q = 6, \quad s = -3 } $$