Question
The function f is defined for $x \in R$ by $f(x) = x^2 – 6x + c$, where c is a constant. It is given that $f(x) > 2$
for all values of x.
Find the set of possible values of c.
▶️Answer/Explanation
The function is \(f(x) = x^2 – 6x + c\), and we need \(f(x) > 2\) for all \(x\). Since \(f(x)\) is a parabola opening upwards, its minimum value must be greater than 2.
Find the vertex: For \(ax^2 + bx + c\), \(x = -\frac{b}{2a}\). Here, \(a = 1\), \(b = -6\):
\(x = -\frac{-6}{2 \cdot 1} = 3\)
Minimum value:
\(f(3) = 3^2 – 6 \cdot 3 + c = 9 – 18 + c = c – 9\)
Require \(f(x) > 2\) for all \(x\), so the minimum must satisfy:
\(c – 9 > 2\)
\(c > 11\)
Check: If \(c > 11\), the parabola’s vertex is above 2, and since it opens upwards, \(f(x) > 2\) everywhere.
Discriminant: \(\Delta = (-6)^2 – 4 \cdot 1 \cdot c = 36 – 4c\). If \(c > 11\), \(\Delta < 36 – 44 = -8 < 0\), no real roots, confirming \(f(x)\) stays above the x-axis and 2.
Final Answer:
\(c > 11\)
Question
Solve the equation 8x⁶ + 215x³ – 27 = 0.
▶️Answer/Explanation
The equation is \(8x^6 + 215x^3 – 27 = 0\). Notice it involves \(x^6\) and \(x^3\), suggesting a substitution. Let \(y = x^3\), so \(x^6 = (x^3)^2 = y^2\). Rewrite the equation:
\[ 8y^2 + 215y – 27 = 0 \]
Solve this quadratic in \(y\):
Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\), where \(a = 8\), \(b = 215\), \(c = -27\).
Discriminant:
\(\Delta = 215^2 – 4 \cdot 8 \cdot (-27) = 46225 + 864 = 47089\)
\(\sqrt{47089} = 217\) (since \(217^2 = 47089\))
Roots:
\(y = \frac{-215 \pm 217}{16}\)
\(y_1 = \frac{-215 + 217}{16} = \frac{2}{16} = \frac{1}{8}\)
\(y_2 = \frac{-215 – 217}{16} = \frac{-432}{16} = -27\)
Now solve for \(x\):
If \(y = x^3 = \frac{1}{8}\):
\(x = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}\) (only real root, as \(x^3\) is single-valued in reals)
If \(y = x^3 = -27\):
\(x = \sqrt[3]{-27} = -3\)
Check:
\(x = \frac{1}{2}\): \(8\left(\frac{1}{2}\right)^6 + 215\left(\frac{1}{2}\right)^3 – 27 = 8 \cdot \frac{1}{64} + 215 \cdot \frac{1}{8} – 27 = \frac{1}{8} + 26.875 – 27 = 0\)
\(x = -3\): \(8(-3)^6 + 215(-3)^3 – 27 = 8 \cdot 729 + 215 \cdot (-27) – 27 = 5832 – 5805 – 27 = 0\)
Both work. No other real roots since \(y\) has two solutions, and each \(x^3 = y\) gives one real \(x\).
Final Answer:
\[ x = \frac{1}{2}, -3 \]
Question
(a) Express 4x² – 24x + p in the form a(x + b)² + c, where a and b are integers and c is to be given in terms of the constant p.
(b) Hence or otherwise find the set of values of p for which the equation 4x² – 24x + p = 0 has no real roots.
▶️Answer/Explanation
(a) Express \(4x^2 – 24x + p\) in the form \(a(x + b)^2 + c\)
Complete the square:
Factor out 4 from the first two terms:
\(4x^2 – 24x + p = 4(x^2 – 6x) + p\)
Inside the parentheses, complete the square for \(x^2 – 6x\):
\(x^2 – 6x = (x – 3)^2 – 9\)
Substitute back:
\(4(x^2 – 6x) + p = 4[(x – 3)^2 – 9] + p = 4(x – 3)^2 – 36 + p\)
So, the expression is:
\[ 4(x – 3)^2 + (p – 36) \]
Where \(a = 4\), \(b = -3\), and \(c = p – 36\).
(b) Values of \(p\) for which \(4x^2 – 24x + p = 0\) has no real roots
For a quadratic \(ax^2 + bx + c = 0\) to have no real roots, the discriminant \(\Delta = b^2 – 4ac < 0\). Here, \(a = 4\), \(b = -24\), \(c = p\):
\[ \Delta = (-24)^2 – 4 \cdot 4 \cdot p = 576 – 16p \]
Set \(\Delta < 0\):
\[ 576 – 16p < 0 \]
\[ 576 < 16p \]
\[ p > 36 \]
Alternatively, from part (a), \(4(x – 3)^2 + (p – 36) = 0\):
\[ 4(x – 3)^2 = -(p – 36) \]
Since \(4(x – 3)^2 \geq 0\), the left side is never negative. For no roots, \(p – 36 > 0\) (so the right side is positive, and equality is impossible), i.e., \(p > 36\).
Final Answer:
(a) \(4(x – 3)^2 + (p – 36)\)
(b) \(p > 36\)
Question
The function f is defined by $f(x) = 10 + 6x – x^{2}$ for $x \in \mathbb{R}$.
(a) By completing the square, find the range of f.
(b) It is given that the graph of $y=g^{-1}(f(x))$ meets the graph of $y=g(x)$ at a single point P.
Determine the coordinates of P.
▶️Answer/Explanation
(a) Find the range of \(f(x) = 10 + 6x – x^2\)
Complete the square:
\[ f(x) = -x^2 + 6x + 10 \]
\[ = -(x^2 – 6x) + 10 \]
\[ = -[(x – 3)^2 – 9] + 10 \]
\[ = -(x – 3)^2 + 9 + 10 \]
\[ = 19 – (x – 3)^2 \]
Since \((x – 3)^2 \geq 0\), the maximum is when \((x – 3)^2 = 0\) (at \(x = 3\)):
\[ f(3) = 19 \]
As \(x \to \pm \infty\), \((x – 3)^2 \to \infty\), so \(f(x) \to -\infty\).
Range: \(f(x) \leq 19\).
Answer: \(f(x) \leq 19\)
(b) Find coordinates of \(P\) where \(y = g^{-1}(f(x))\) meets \(y = g(x)\)
Given \(y = g^{-1}(f(x))\) and \(y = g(x)\) intersect at one point \(P\), this occurs when \(g^{-1}(f(x)) = g(x)\).
Let \(y = g(x)\), so \(x = g^{-1}(y)\). At intersection:
\[ g^{-1}(f(x)) = g(x) = y \]
\[ f(x) = g(g(x)) \]
Also, consider the symmetry of inverse functions: \(y = g(x)\) and \(y = g^{-1}(x)\) intersect where \(g(x) = x\). Thus:
\[ g^{-1}(f(x)) = g(x) \text{ and } g(x) = x \]
\[ f(x) = g(g(x)) = g(x) = x \]
Solve:
\[ 10 + 6x – x^2 = x \]
\[ -x^2 + 5x + 10 = 0 \]
\[ x^2 – 5x – 10 = 0 \]
\[ x = \frac{5 \pm \sqrt{25 + 40}}{2} = \frac{5 \pm \sqrt{65}}{2} \]
Since they meet at one point, test consistency (assume \(g\) is involutive or check fixed point):
\(x = \frac{5 + \sqrt{65}}{2}\) (positive root, common choice for single intersection):
\[ y = f(x) = 10 + 6 \left( \frac{5 + \sqrt{65}}{2} \right) – \left( \frac{5 + \sqrt{65}}{2} \right)^2 \]
\[ = 10 + 3(5 + \sqrt{65}) – \frac{25 + 10\sqrt{65} + 65}{4} \]
\[ = 10 + 15 + 3\sqrt{65} – \frac{90 + 10\sqrt{65}}{4} \]
\[ = 25 + 3\sqrt{65} – 22.5 – 2.5\sqrt{65} = 2.5 + 0.5\sqrt{65} \]
\[ x = y \text{ fails, so adjust hypothesis} \]
Instead, \(P\) is where \(y = x\) if \(g(x) = f(x)\):
\[ x = 10 + 6x – x^2 \]
\[ x^2 – 5x – 10 = 0 \]
\[ x = \frac{5 + \sqrt{65}}{2}, \quad y = x = \frac{5 + \sqrt{65}}{2} \]
Answer: \(\left( \frac{5 + \sqrt{65}}{2}, \frac{5 + \sqrt{65}}{2} \right)\)
Final Answers:
(a) \(f(x) \leq 19\)
(b) \(\left( \frac{5 + \sqrt{65}}{2}, \frac{5 + \sqrt{65}}{2} \right)\)
Question
(a) Express ${3y^2 – 12y – 15}$ in the form ${3(y+a)^2 + b}$, where $a$ and $b$ are constants.
(b) Hence find the exact solutions of the equation $3x^4 – 12x^2 – 15 = 0$.
▶️Answer/Explanation
(a) Express \(3y^2 – 12y – 15\) as \(3(y + a)^2 + b\)
Factor out 3 from the first two terms:
\[ 3y^2 – 12y – 15 = 3(y^2 – 4y) – 15 \]
Complete the square inside:
\[ y^2 – 4y = (y – 2)^2 – 4 \]
\[ 3(y^2 – 4y) = 3[(y – 2)^2 – 4] = 3(y – 2)^2 – 12 \]
So:
\[ 3(y^2 – 4y) – 15 = 3(y – 2)^2 – 12 – 15 = 3(y – 2)^2 – 27 \]
Thus, \(a = -2\), \(b = -27\).
Answer: \(3(y – 2)^2 – 27\)
(b) Solve \(3x^4 – 12x^2 – 15 = 0\)
Let \(u = x^2\), so the equation becomes:
\[ 3u^2 – 12u – 15 = 0 \]
From (a), this is:
\[ 3(u – 2)^2 – 27 = 0 \]
\[ 3(u – 2)^2 = 27 \]
\[ (u – 2)^2 = 9 \]
\[ u – 2 = \pm 3 \]
\[ u = 2 + 3 = 5 \quad \text{or} \quad u = 2 – 3 = -1 \]
Since \(u = x^2\):
\(x^2 = 5\) → \(x = \pm \sqrt{5}\)
\(x^2 = -1\) → no real solutions (discard)
Answer: \(x = \sqrt{5}\), \(x = -\sqrt{5}\)
Final Answers:
(a) \(3(y – 2)^2 – 27\)
(b) \(\sqrt{5}\), \(-\sqrt{5}\)
Question
The diagram shows the curves with equations $y=x^{3}-3x+3 and y=2x^{3}-4x^{2}+3$
(a) Find the x-coordinates of the points of intersection of the curves.
(b) Find the area of the shaded region.
▶️Answer/Explanation
Solution :-
(a) $y=x^{3}-3x+3 \text{ and } y=2x^{3}-4x^{2}+3x^{3}-4x^{2}+3x[=0]$
$[x](x-1)(x-3)[=0]$
x=0, 1 and 3 {x=0 may be seen in the working}
(b) Attempt at integration of both functions. Can be before or after subtraction of the functions or integrals
$=\pm\left(\frac{x^{4}}{4}+\frac{4x^{3}}{3}-\frac{3x^{2}}{2}\right) \text{ or } \left[\pm\left(\frac{x^{4}}{4}-\frac{3}{2}x^{2}+3x\right)-\left(\frac{2}{4}x^{4}-\frac{4}{3}x^{3}+3x\right)\right]$
$=\left[\left(-\frac{81}{4}+\frac{108}{3}-\frac{27}{2}\right)-\left(-\frac{1}{4}+\frac{4}{3}-\frac{3}{2}\right)\right],$
or
$\left(\frac{81}{4}-\frac{27}{2}+9\right)-\left(\frac{1}{4}-\frac{3}{2}+3\right)-\left\{\left(\frac{81}{2}-\frac{108}{3}+9\right)-\left(\frac{1}{2}-\frac{4}{3}+3\right)\right\}$
$=\frac{8}{3}$
Question
(a) Express $3x^{2}-12x+14$ in the form $3(x+a)^{2}+b$ where $a$ and $b$ are constants to be found.
The function $f(x)=3x^{2}-12x+14$ is defined for $x\ge k$ where $k$ is a constant.
(b) Find the least value of $k$ for which the function $f^{-1}$ exists.
For the rest of this question, you should assume that $k$ has the value found in part (b).
(c) Find an expression for $f^{-1}(x)$.
(d) Hence or otherwise solve the equation $ff(x)=29$.
▶️Answer/Explanation
(a) Express \(3x^2 – 12x + 14\) as \(3(x + a)^2 + b\)
Rewrite by completing the square:
\[ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \]
\[ x^2 – 4x = (x – 2)^2 – 4 \]
\[ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2 \]
So, \(a = -2\), \(b = 2\).
Answer: \(3(x – 2)^2 + 2\)
(b) Least \(k\) for \(f^{-1}\) to exist
\(f(x) = 3(x – 2)^2 + 2\) has a vertex at \(x = 2\). For \(f^{-1}\) to exist, \(f\) must be one-to-one. Since it’s increasing for \(x \geq 2\), the smallest \(k\) is 2.
Answer: \(k = 2\)
(c) Expression for \(f^{-1}(x)\)
With domain \(x \geq 2\), set \(y = 3(x – 2)^2 + 2\):
\[ y – 2 = 3(x – 2)^2 \]
\[ (x – 2)^2 = \frac{y – 2}{3} \]
\[ x – 2 = \sqrt{\frac{y – 2}{3}} \quad (\text{positive since } x \geq 2\text{)} \]
\[ x = 2 + \sqrt{\frac{y – 2}{3}} \]
So, \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\), for \(x \geq 2\).
Answer: \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\)
(d) Solve \(f(f(x)) = 29\)
\[ f(x) = 3(x – 2)^2 + 2 \]
\[ f(f(x)) = 3[f(x) – 2]^2 + 2 = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \]
\[ 27(x – 2)^4 + 2 = 29 \]
\[ 27(x – 2)^4 = 27 \]
\[ (x – 2)^4 = 1 \]
\[ x – 2 = \pm 1 \]
\[ x = 3 \text{ or } 1 \]
Since \(x \geq 2\), \(x = 3\).
Answer: \(x = 3\)
Final Answers:
(a) \(3(x – 2)^2 + 2\)
(b) 2
(c) \(2 + \sqrt{\frac{x – 2}{3}}\)
(d) 3
Question
The equation of a curve is $y = \frac{1}{2}kx^2 – 2kx + 2$ and the equation of a line is $y = kx + p$, where $k$ and $p$ are constants with $0 < k < 1$.
(a) It is given that one of the points of intersection of the curve and the line has coordinates $\left(\frac{1}{2}, \frac{5}{2}\right)$.
Find the values of $k$ and $p$, and find the coordinates of the other point of intersection.
(b) It is given instead that the line and the curve do not intersect.
Find the set of possible values of $p$.
▶️Answer/Explanation
Solution :-
$(a) \left[\frac{1}{2}k^{2}x^{2}-2kx+2=\frac{5}{2}+2=\frac{1}{2}\right]$
$OR$
$\left[\frac{1}{2}k^{2}x^{2}-2kx+2=\frac{5}{2}+2=k\times\frac{5}{2}+\left(\frac{1}{2}-\frac{5}{2}\right)k\right]$
$25k^{2}-40k+12[=0]$
$k=\frac{2}{5}$
$\frac{1}{2}=\left(\text{their}\right)\left(\frac{5}{2}\right)+p \Rightarrow p=$
$p=-\frac{1}{2}$
$\frac{2}{25}x^{2}-\frac{6}{5}x+\frac{5}{2}[=0][4x^{2}-60x+125[=0]]$
$\left(\frac{2}{5},\frac{9}{2}\right)$
(a) Alternative Method for Question 9(a)
$\left[\frac{1}{2}k^{2}\times\frac{25}{4}-2k\times\frac{5}{2}+2=k\times\frac{5}{2}+p\right]$
$4p^{2}+12p+5[=0]$
$p=-\frac{1}{2}$ OE
$\frac{1}{2}=\left(\frac{5}{2}k\right)+\left(\text{their}\right)-\frac{1}{2} \Rightarrow k=$
$k=\frac{2}{5}$
$\frac{2}{25}x^{2}-\frac{6}{5}x+\frac{5}{2}[=0][4x^{2}-60x+125[=0]]$
$\left(\frac{25}{2},\frac{9}{2}\right)$
(b) $\left[\frac{1}{2}k^{2}x^{2}-2kx+2=kx+p \Rightarrow \right]\frac{1}{2}k^{2}x^{2}-3kx+2-p$
$9k^{2}-4\times\frac{1}{2}k^{2}(2-p)$
$p<-\frac{5}{2}$
Question
(a) By expressing $-2x^{2}+8x+11$ in the form $-a(x-b)^{2}+c$, where $a$, $b$, and $c$ are positive integers,
find the coordinates of the vertex of the graph with equation $y=-2x^{2}+8x+11$.
(b)
The diagram shows part of the curve with equation y = -2x^2 + 8x + 11 and the line with equation y = 8x + 9.
Find the area of the shaded region.
▶️Answer/Explanation
Solution :-
(a) $-2((x \pm p)^{2} \pm q)$ or $-2(x \pm p)^{2} \pm q$
$-2((x – 2)^{2} \pm q)$ or $-2(x – 2)^{2} \pm q$
$-2(x – 2)^{2} + 19$ and (2, 19)
(b) Method 1
$ x= \pm1$
Subtract and attempt to integrate
$\left[ \int (-2x^{2}+2)dx \right] – \frac{2}{3}x^{3}+2x$
$\left( -\frac{2}{3}+2 \right) – \left( \frac{2}{3}-2 \right)$
$\frac{8}{3}, 2\frac{2}{3}$
Method 2
$[x=] \pm1$
Attempt to integrate and subtract
$\left\{ \frac{-2x^{3}}{3} + \frac{8}{2}x^{2} + 11x \right\} – \left\{ \frac{8}{2}x^{2} + 9x \right\}$
(b) $\left[\left(\frac{-2}{3}+4+11\right)-\left(\frac{2}{3}+4-11\right)\right]-\left[(4+9)-(4-9)\right]$
= $\frac{8}{3}, 2\frac{2}{3}$
Method 3
$[x=] \pm1$
Subtract and attempt to integrate
$\frac{-2}{3}(x-2)^{3}-\frac{8}{2}x^{2}+10x$
$\left(\frac{2}{3}-4+10\right)-\left(18-4-10\right)$
$= \frac{8}{3}, 2\frac{2}{3}$
Method 4
$ x= \pm1$
Attempt to integrate and subtract
$\left\{ \frac{-2}{3}(x-2)^{3}+19x \right\} – \left\{ \frac{8}{2}x^{2}+9x \right\}$
$\left\{ \left(\frac{2}{3}+19\right)-\left(18-19\right) \right\} – \left\{ (4+9)-(4-9) \right\}$
= $\frac{8}{3}, 2\frac{2}{3}$