Question 1
Find the exact values of $k$ such that the straight line $y=1-k-x$ is a tangent to the curve $y=kx^2+x+2k$.
[4]
Step 1: Set up the problem
We are given the straight line:
$$ y = 1 - k - x $$and the curve:
$$ y = kx^2 + x + 2k $$We need to find the values of $k$ such that the line is tangent to the curve. This means the line and the curve should intersect at exactly one point, and their slopes should be equal at that point.
Step 2: Solve for the intersection points
Equating the two equations for $y$, we get:
$$ 1 - k - x = kx^2 + x + 2k $$Simplify and rearrange:
$$ kx^2 + 2x + 3k - 1 = 0 $$This is a quadratic equation in $x$. For the line to be tangent to the curve, this quadratic equation must have exactly one solution, which occurs when the discriminant is zero.
Step 3: Find the discriminant
For a quadratic equation of the form $ax^2 + bx + c = 0$, the discriminant is given by:
$$ \Delta = b^2 - 4ac $$Here, $a = k$, $b = 2$, and $c = 3k - 1$. Substituting these values:
$$ \Delta = 2^2 - 4(k)(3k - 1) $$Simplify:
$$ \Delta = 4 - 12k^2 + 4k $$ $$ \Delta = -12k^2 + 4k + 4 $$Step 4: Set the discriminant to zero
For the line to be tangent, $\Delta = 0$. Thus:
$$ -12k^2 + 4k + 4 = 0 $$Divide through by $-4$ to simplify:
$$ 3k^2 - k - 1 = 0 $$Step 5: Solve for $k$
Solve the quadratic equation $3k^2 - k - 1 = 0$ using the quadratic formula:
$$ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$where $a = 3$, $b = -1$, and $c = -1$. Substituting these values:
$$ k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-1)}}{2(3)} $$ $$ k = \frac{1 \pm \sqrt{1 + 12}}{6} $$ $$ k = \frac{1 \pm \sqrt{13}}{6} $$Step 6: Verify the solutions
For each value of $k$, check that the discriminant is zero and that the line is tangent to the curve.
- For $k = \frac{1 + \sqrt{13}}{6}$, substitute into the original equations to verify tangency.
- For $k = \frac{1 - \sqrt{13}}{6}$, substitute into the original equations to verify tangency.
Both values of $k$ satisfy the tangency condition.
Final Answer
The exact values of $k$ such that the line is tangent to the curve are:
$$ k = \frac{1 + \sqrt{13}}{6} \quad \text{and} \quad k = \frac{1 - \sqrt{13}}{6} $$Question 2
A curve has equation $y=(5-x)(x+2)^2$.
- (a) Find the $x$-coordinates of the stationary points on the curve. [4]
- (b) On the axes below, sketch the graph of $y=(5-x)(x+2)^2$, stating the coordinates of the points where the curve meets the axes. [3]
- (c) Find the values of $k$ for which the equation $k=(5-x)(x+2)^2$ has one distinct root only. [3]
Step (a): Finding the $x$-coordinates of the stationary points
The given curve is defined by the equation:
$$ y = (5-x)(x+2)^2 $$First, expand $y$:
$$ y = (5-x)(x^2 + 4x + 4) $$ $$ y = 5x^2 + 20x + 20 - x^3 - 4x^2 - 4x $$ $$ y = -x^3 + x^2 + 16x + 20 $$Now, compute the derivative $\frac{dy}{dx}$:
$$ \frac{dy}{dx} = \frac{d}{dx}[-x^3 + x^2 + 16x + 20] $$ $$ \frac{dy}{dx} = -3x^2 + 2x + 16 $$To find the stationary points, set $\frac{dy}{dx} = 0$:
$$ -3x^2 + 2x + 16 = 0 $$Solve this quadratic equation using the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$where $a = -3$, $b = 2$, and $c = 16$.
$$ x = \frac{-2 \pm \sqrt{2^2 - 4(-3)(16)}}{2(-3)} $$ $$ x = \frac{-2 \pm \sqrt{4 + 192}}{-6} $$ $$ x = \frac{-2 \pm \sqrt{196}}{-6} $$ $$ x = \frac{-2 \pm 14}{-6} $$This gives:
$$ x = \frac{-2 + 14}{-6} = \frac{12}{-6} = -2 $$ $$ x = \frac{-2 - 14}{-6} = \frac{-16}{-6} = \frac{8}{3} $$The $x$-coordinates of the stationary points are $x=-2$ and $x=\frac{8}{3}$.
Step (b): Sketching the graph of $y=(5-x)(x+2)^2$
First, find the intercepts:
- $y$-intercept: Substitute $x=0$ into $y$:
The $y$-intercept is $(0,20)$.
- $x$-intercepts: Solve $y=0$:
The $x$-intercepts are $(5,0)$ and $(-2,0)$.
Using these points and the stationary points $x=-2$ and $x=\frac{8}{3}$, sketch the curve $y=(5-x)(x+2)^2$. The curve passes through $(0,20)$, $(5,0)$, and $(-2,0)$, with stationary points at $x=-2$ and $x=\frac{8}{3}$.
Step (c): Finding $k$ for which $k=(5-x)(x+2)^2$ has one distinct root
The equation $k=(5-x)(x+2)^2$ can be rewritten as:
$$ y=(5-x)(x+2)^2 \qquad \text{and} \qquad y=k $$Calculate the corresponding $y$-values:
- At $x=-2$:
- At $x=\frac{8}{3}$:
From Step (b), the turning points occur at $x=-2$ and $x=\frac{8}{3}$. We can check that $(-2,0)$ is a minimum point and $\left(\frac{8}{3}, \frac{1372}{27}\right)$ is a maximum point.
If we take $0 \le k \le \frac{1372}{27}$, the given equation has at least two roots.
For $k$ to have one distinct real root, we must have:
$$ k > \frac{1372}{27} \quad \text{or} \quad k \lt 0 $$
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