CIE 2023 june 0606/11 Question 1/2 and solution

Question 1

1. Write $5x^2 - 14x + 8$ in the form $a(x+b)^2 + c$, where $a$, $b$ and $c$ are constants to be found. [3]
2. Hence write down the coordinates of the stationary point on the curve $y = 5x^2 - 14x + 8$. [2]
3. On the axes below, sketch the graph of $y = |5x^2 - 14x + 8|$, stating the coordinates of the points where the graph meets the coordinate axes. [3]
   
   
4. Write down the range of values of $k$ for which the equation $|5x^2 - 14x + 8| = k$ has $4$ distinct roots. [2]

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Solution

Step (a): Completing the square

We aim to write $5x^2 - 14x + 8$ in the form $a(x+b)^2 + c$, where $a$, $b$, and $c$ are constants.

• Start with $5x^2 - 14x + 8$ and factor out $5$ from the quadratic terms: $$ 5x^2 - 14x + 8 = 5\left(x^2 - \frac{14}{5}x\right) + 8 $$
• Complete the square inside the parentheses: $$ x^2 - \frac{14}{5}x = \left(x - \frac{7}{5}\right)^2 - \left(\frac{7}{5}\right)^2 $$ $$ = \left(x - \frac{7}{5}\right)^2 - \frac{49}{25} $$
• Substitute back and simplify: $$ 5\left(x^2 - \frac{14}{5}x\right) + 8 = 5\left[ \left(x - \frac{7}{5}\right)^2 - \frac{49}{25} \right] + 8 $$ $$ = 5\left(x - \frac{7}{5}\right)^2 - \frac{245}{25} + 8 $$ $$ = 5\left(x - \frac{7}{5}\right)^2 - 9.8 + 8 $$ $$ = 5\left(x - \frac{7}{5}\right)^2 - 1.8 $$

Thus, $$ 5x^2 - 14x + 8 = 5\left(x - \frac{7}{5}\right)^2 - 1.8 $$

Step (b): Stationary point

The curve is given by $y = 5x^2 - 14x + 8$, and its stationary point corresponds to the vertex of the parabola.

• From the completed square form $$ y = 5\left(x - \frac{7}{5}\right)^2 - 1.8 $$ the vertex occurs at $$ x = \frac{7}{5} $$
• Substituting $x = \frac{7}{5}$ into the equation gives: $$ y = 5\left( \frac{7}{5} - \frac{7}{5} \right)^2 - 1.8 = -1.8 $$

Thus, the stationary point is $$ \left( \frac{7}{5}, -1.8 \right) $$

Step (c): Sketching the graph of $y = |5x^2 - 14x + 8|$

• The graph of $y = 5x^2 - 14x + 8$ is a parabola opening upwards with vertex at $$ \left( \frac{7}{5}, -1.8 \right) $$
• The effect of the absolute value is to reflect the portion of the graph below the $x$-axis above the $x$-axis.
• The points of intersection with the $x$-axis can be found by solving $$ 5x^2 - 14x + 8 = 0 $$
• Using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(5)(8)}}{2(5)} $$ $$ = \frac{14 \pm \sqrt{196 - 160}}{10} $$ $$ = \frac{14 \pm 6}{10} $$ $$ = \{2,\ 0.8\} $$
• Thus, the graph meets the $x$-axis at $(0.8, 0)$ and $(2, 0)$.
• The graph meets the $y$-axis when $x = 0$: $$ y = 8 $$ so the point is $(0, 8)$.

Key points on the graph:

• Reflected vertex: $$ \left( \frac{7}{5}, 1.8 \right) $$
• $x$-intercepts: $(0.8, 0)$ and $(2, 0)$
• $y$-intercept: $(0, 8)$

Step (d): Range of $k$ for four distinct roots

The equation $$ |5x^2 - 14x + 8| = k $$ has four distinct roots if the horizontal line $y = k$ intersects the graph of $y = |5x^2 - 14x + 8|$ in four distinct places.

• The graph has minimum value $$ y = 0 $$ at the points $(0.8,0)$ and $(2,0)$.
• The original parabola has vertex value $$ y = -1.8 $$ so after reflection the maximum turning value becomes $$ y = 1.8 $$

Therefore, $k$ must satisfy $$ 0 \lt k \lt 1.8 $$ for the equation to have four distinct roots.