<p>The curve $C$ has equation $y=\frac{{{x}^{2}}+5}{\lambda x+1}$, where $\lambda $ is a non-zero constant.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Obtain the equations of the asymptotes of $C$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Show that $C$ has exactly two points at which $\frac{\mathrm{d}y}{\mathrm{d}x}=0$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Draw a sketch of $C$ to illustrate the case $\lambda >0$, and a separate of $C$ to illustrate the case $\lambda <0$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the set of values of $\lambda $ for which the line $y=4x$ and $C$ has at least one point in common.</span></div></li></ol>

The curve $C$ has equation $y=\frac{{{x}^{2}}+5}{\lambda x+1}$, where $\lambda $ is a non-zero constant.

Obtain the equations of the asymptotes of $C$.

Show that $C$ has exactly two points at which $\frac{\mathrm{d}y}{\mathrm{d}x}=0$.

Draw a sketch of $C$ to illustrate the case $\lambda >0$, and a separate of $C$ to illustrate the case $\lambda <0$.

Find the set of values of $\lambda $ for which the line $y=4x$ and $C$ has at least one point in common.

Applications of Differentiation

International Baccalaureate Higher Level Mathematics

IB HL Math

N1999 P1 Q1 (FM)

By long division of $x^{2}+5$ by $\lambda x + 1$:

$y = \dfrac{x^{2}+5}{\lambda x+1} = \dfrac{1}{\lambda}x - \dfrac{1}{\lambda^{2}} + \dfrac{5+\frac{1}{\lambda^{2}}}{\lambda x+1}$

**Oblique asymptote:** $y = \dfrac{x}{\lambda} - \dfrac{1}{\lambda^{2}}$

**Vertical asymptote:** $\lambda x + 1 = 0 \Rightarrow x = -\dfrac{1}{\lambda}$

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(\lambda x+1)(2x)-(x^{2}+5)(\lambda)}{(\lambda x+1)^{2}}$

$= \dfrac{2\lambda x^{2}+2x-\lambda x^{2}-5\lambda}{(\lambda x+1)^{2}}$

$= \dfrac{\lambda x^{2}+2x-5\lambda}{(\lambda x+1)^{2}}$

When $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: $\lambda x^{2}+2x-5\lambda = 0$

Discriminant $D = 2^{2}-4(\lambda)(-5\lambda) = 4+20\lambda^{2} > 0$ since $\lambda \neq 0$.

$\therefore$ there are two distinct roots for $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$.

$\Rightarrow C$ has exactly two points at which $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$. (shown)

**Case $\lambda > 0$:**
- VA at $x = -\frac{1}{\lambda}$ (in negative $x$ region)
- OA: $y = \frac{x}{\lambda}-\frac{1}{\lambda^{2}}$ (positive slope)
- $y$-intercept at $(0,5)$
- Right branch: above OA, approaches OA from above as $x \to +\infty$
- Left branch: below OA, has a local maximum, approaches OA from below as $x \to -\infty$

**Case $\lambda < 0$:**
- VA at $x = -\frac{1}{\lambda}$ (in positive $x$ region)
- OA: $y = \frac{x}{\lambda}-\frac{1}{\lambda^{2}}$ (negative slope)
- $y$-intercept at $(0,5)$
- Left branch: above OA, passes through $(0,5)$, has a local maximum, approaches VA from the left
- Right branch: below OA, has a local minimum, approaches OA from below as $x \to +\infty$

Substitute $y = 4x$ into $y = \dfrac{x^{2}+5}{\lambda x+1}$:

$4x = \dfrac{x^{2}+5}{\lambda x+1}$ \;\; ($x \neq -\dfrac{1}{\lambda}$)

$4x(\lambda x+1) = x^{2}+5$

$4\lambda x^{2}+4x = x^{2}+5$

$(4\lambda-1)x^{2}+4x-5 = 0$

For at least one common point, we need at least one real solution.

Discriminant $\geq 0$:

$4^{2}-4(4\lambda-1)(-5) \geq 0$

$16+20(4\lambda-1) \geq 0$

$80\lambda - 4 \geq 0$

$\lambda \geq \dfrac{1}{20}$

N1999 P1 Q1 (FM)

N1999 P1 Q1 (FM)

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