Home / 9709_s23_qp_12

Question 1

Topic – ALV: 3.8

The equation of a curve is such that $ \frac{dy}{dx} = \frac{4}{(x-3)^{3}} $ for $x > 3$. The curve passes through the point $(4,5)$. Find the equation of the curve.

▶️Answer/Explanation

Solution: –

$y=\frac{4}{-2}(x-3)^{-3+1}+c~~or~~\frac{4}{-2{(x-3)}^{2}}+c$

$5=\frac{4}{-2}(4-3)^{-2}+c~~or~~5=\frac{4}{-2(4-3)^{2}}+c~~leading~~to~~c=7$

$y=\frac{-2}{(x-3)^{2}}+7~~or~~y=-2(x-3)^{-2}+7$

Question 2

Topic – ALV: 1.6

The coefficient of x in the expansion of (x + a)⁶ is p and the coefficient of x² in the expansion of (ax + 3)⁴ is q. It is given that p + q = 276. Find the possible values of the constant a.

▶️Answer/Explanation

Solution :-

[Coefficient of x⁴ = p] = 15a²

[Coefficient of x² = q] = 54a²

Equating their p + their q to 276 leading to an equation in a² only

a = ±2

Question 3

Topic – ALV: 3.1

(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

Solution: –

(a) 4(x – 3)² seen or a = 4 and b = -3

-36 + p or p – 36 seen or c = p – 36

(b) p – 36 > 0 leading to p > 36 or 24² – 4 x 4 x p/0 => p > 36 or 36 < p

Question 4

Topic – ALV: 3.1

 Solve the equation 8x⁶ + 215x³ – 27 = 0.

▶️Answer/Explanation

Solution: –

$[8x⁶ + 215x³ – 27 = 0]$ leading to $(8x³ – 1)(x³ + 27) = 0$

or

$\frac{-215 \pm \sqrt{215² – 4 \cdot 8 \cdot (-27)}}{2 \cdot 8}$

or

$\frac{-215 \pm \sqrt{47089}}{2 \cdot 8}$

$\frac{1}{8}$, -27 or 0.5, -3

Question 5

Topic – ALV: 1.8

The diagram shows the curve with equation \(y=10x^{\frac{1}{2}}-\frac{5}{2}x^{\frac{3}{2}}\) for \(x>0\). The curve meets the x-axis at the points (0, 0) and (4, 0).

Find the area of the shaded region.

▶️Answer/Explanation

Solution: –

$\left[\int(10x^{\frac{1}{2}}-\frac{5}{2}x^{\frac{3}{2}})dx\right]=\left[10\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{5}{2}\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]=\left[\frac{20}{3}x^{\frac{3}{2}}-x^{\frac{5}{2}}\right]$

$=\left[\text{their }\frac{20}{3}\times8-32\right]-[0]$

[Area of shaded region] = $\frac{64}{3}$, 21$\frac{1}{3}$ or 21.3[333…]

Question 6

Topic – ALV: 1.4

The diagram shows a sector OAB of a circle with centre O. Angle AOB = \theta radians and OP = AP = x.

(a) Show that the arc length AB is \(2~\theta~cos~\theta\)

(b) Find the area of the shaded region APB in terms of x and θ.

▶️Answer/Explanation

Solution: –

(a) $\frac{1}{2}OA=x~cos~\theta$ or $\frac{OA}{sin(\pi-2\theta)}=\frac{x}{sin~\theta}$

$OA^{2}=x^{2}+x^{2}-2x^{2}cos(\pi-2\theta)$ or $x^{2}=r^{2}+x^{2}-2rx~cos~\theta$ or other valid method.

$OA=2x~cos~\theta$ leading to Arc length = $2~\theta~x~cos~\theta$

(b) Sector area = $\frac{1}{2}(2x~cos~\theta)^{2}\times\theta$

Triangle area = $\frac{1}{2}\times2x~cos~\theta\times x~sin~\theta$ OR $\frac{1}{2}x^{2}sin(\pi-2\theta)$

[Area APB =] Their sector area – their triangle area

[Area APB =] $\frac{1}{2}(2x~cos~\theta)^{2}\times\theta-\frac{1}{2}x^{2}sin(\pi-2\theta)$

[= $x^{2}(2\theta~cos^{2}\theta-\frac{1}{2}sin~2\theta)$] or [$x^{2}cos~\theta(2\theta~cos~\theta-sin~\theta)$]

Question 7

Topic – ALV: 1.5

(a) (i) By first expanding (cos θ + sin θ)², find the three solutions of the equation \[(cos~\theta+sin~\theta)^{2}=1\]

for 0 ≤ θ ≤ π.

(a) (ii) Hence verify that the only solutions of the equation cos θ + sin θ = 1 for 0 ≤ θ ≤ π are 0 and \(\frac{1}{2}\pi\).

(b) Prove the identity

\[\frac{sin~\theta}{cos~\theta+sin~\theta}+\frac{1-cos~\theta}{cos~\theta-sin~\theta}\equiv\frac{cos~\theta+sin~\theta-1}{1-2~sin^{2}\theta}.\]

(c) Using the results of (a) (ii) and (b), solve the equation

\[\frac{sin~\theta}{cos~\theta+sin~\theta}+\frac{1-cos~\theta}{cos~\theta-sin~\theta}=2(cos~\theta+sin~\theta-1)\]

for 0 ≤ θ ≤ π.

▶️Answer/Explanation

Solution: –

(a)(i) $cos²θ + 2sinθcosθ + sin²θ = 1$ leading to $2sinθcosθ = 0$ or $sin2θ = 0$
$[θ] = 0, π/2, π$

(a)(ii) $cos0+sin0 = [1+0] = 1$ and $cosπ/2+sinπ/2 = [0+1] = 1$
$cosπ+sinπ = [-1+0] = -1$ or ≠ 1

(b) $\frac{(cos~\theta-sin~\theta)sin~\theta+(cos~\theta+sin~\theta)(1-cos~\theta)}{(cos~\theta+sin~\theta)(cos~\theta-sin~\theta)}$

$=\frac{cos~\theta~sin~\theta-sin^{2}\theta+cos~\theta-cos^{2}\theta+sin~\theta-sin~\theta~cos~\theta}{cos^{2}\theta-sin^{2}\theta}$

$=\frac{sin~\theta+cos~\theta-cos^{2}\theta-sin^{2}\theta}{cos^{2}\theta-sin^{2}\theta}=\frac{cos~\theta+sin~\theta-1}{1-2~sin^{2}\theta}$

(c) $\frac{cos~\theta+sin~\theta-1}{1-2sin^{2}\theta}=2(cos~\theta+sin~\theta-1)$
leading to $1 = 2(1 – 2sin²θ)$

k sin²θ = 1 or 3 leading to $sinθ = [±]\sqrt{\frac{1 or~3}{k}}$
[4sin²θ = 1] leading to $sinθ = ±1/2$
Solutions 0, 1/6π, 1/2π, 5/6π

Question 8

Topic – ALV: 1.2

The diagram shows the graph of y = f(x) where the function f is defined by

\[f(x) = 3 + 2\sin\left(\frac{1}{4}x\right)\] for \(0 \leq x \leq 2\pi\).

(a) On the diagram above, sketch the graph of y = f⁻¹(x).

(b) Find an expression for f⁻¹(x).

(c)

The diagram above shows part of the graph of the function \(g(x) = 3 + 2\sin\left(\frac{1}{4}x\right)\) for \(-2\pi \leq x \leq 2\pi\).

Complete the sketch of the graph of \(g(x)\) on the diagram above and hence explain whether the function g has an inverse.

(d) Describe fully a sequence of three transformations which can be combined to transform the graph of \(y = \sin x\) for \(0 \leq x \leq \frac{1}{2}\pi\) to the graph of \(y = f(x)\), making clear the order in which the transformations are applied.

▶️Answer/Explanation

Solution: –

(a)

(b) $y=3+2sin\frac{1}{4}x$ leading to $sin\frac{1}{4}x=\frac{y+3}{2}$

$x=4sin^{-1}\left(\frac{y-3}{2}\right)$ leading to $[f^{-1}(x)]or~y=4sin^{-1}\left(\frac{x-3}{2}\right)$

(c)

Yes it does have an inverse, because the graph is always increasing
OR because it is one-one OR because it passes the horizontal line test OR
it is not a many to one function.

(d)$1. \text{Stretch by a factor of 4 in the x-direction}$

$2. \text{Stretch by a factor of 2 in the y-direction}$

$3. \text{Translation by the vector } \begin{pmatrix} 0 \\ 3 \end{pmatrix}$

Question 9

Topic – ALV: 1.6

9. The second term of a geometric progression is 16 and the sum to infinity is 100.

(a) Find the two possible values of the first term.

(b) Show that the nth term of one of the two possible geometric progressions is equal to 4^{n-2} multiplied by the nth term of the other geometric progression.

▶️Answer/Explanation

Solution: –

(a) $\begin{cases}
ar=16, \\
\frac{a}{1-r}=100
\end{cases}$

leading to $a=\frac{16}{r}$ and $a=100(1-r)$

$100(1-r)r=16$ leading to $100r^{2}-100r+16=0$

$(5r-4)(5r-1)=0$

OR

$\frac{25\pm\sqrt{25^{2}-4\cdot25\cdot4}}{2\cdot25}$

leading to $r=\frac{4}{5}$ or $r=\frac{1}{5}$

$a=20$, $a=80$

Alternative Method

$\begin{cases}
ar=16, \\
\frac{a}{1-r}=100
\end{cases}$

leading to $r=\frac{16}{a}$ and $r=\frac{100-a}{100}$

$1600=100a-a^{2}$ leading to $a^{2}-100a+1600=0$

$(a-20)(a-80)=0~OR~\frac{100\pm\sqrt{100^{2}-4\cdot1600}}{2}$

$a=20$, $a=80$

(b) $r=\frac{4}{5}, \frac{1}{5}$

$u_{n}=their~20\times\left(\frac{4}{5}\right)^{n-1}$

$v_{n}=their~80\times\left(\frac{1}{5}\right)^{n-1}$

Method 1

$20\times\left(\frac{1}{5}\right)^{n-1}\times4^{n-1}$

$u_{n}=\frac{1}{4}\times80\times\left(\frac{1}{5}\right)^{n-1}\times4^{n-1}=4^{n-2}\times80\times\left(\frac{1}{5}\right)^{n-1}=4^{n-2}\times v_{n}$

Method 2

$20\times0.8^{n-1}$

$80\times0.2^{n-1}=\frac{1}{4}\times4^{n-1}$

$=4^{-1}\times4^{n-1}=4^{n-2}$

Question 10

Topic – ALV: 1.3

The equation of a circle is \((x-a)^{2}+(y-3)^{2}=20.\) The line \(y=\frac{1}{2}x+6\) is a tangent to the circle at the point P.

(a) Show that one possible value of a is 4 and find the other possible value.

(b) For \(a=4\), find the equation of the normal to the circle at P.

(c) For \(a=4\), find the equations of the two tangents to the circle which are parallel to the normal found in (b).

▶️Answer/Explanation

Solution: –

(a) $(x-a)^{2}+\left(\frac{1}{2}x+6-3\right)^{2}=20$ or using $x=2y-12$

$\frac{5}{4}x^{2}+(3-2a)x+a^{2}-11=0$

$(3-2a)^{2}-4\times\frac{5}{4}(a^{2}-11)=0$

Method 1

Using $a=4$: $(3-8)^{2}-5(5)=0$

$a=-16$

Method 2

$-a^{2}-12a+64=0 \Rightarrow (a-4)(a+16)=0 \Rightarrow a=4$

$a=-16$

(b) Centre (4, 3) identified or used or the point P is (2, 7)

gradient of normal = -2

Forming normal equation using their gradient (not 0.5) and their centre or P

$\frac{y-3}{(x-4)}=-2$ or $y-7=-2(x-2)$

(c) Method 1 

Diameter:

$y-3=\frac{1}{2}(x-4)$ [leading to $y=\frac{1}{2}x+1$]

Or

$2(x-4)+2(y-3)\frac{dy}{dx}=0$ [leading to $y=\frac{1}{2}x+1$]

$(x-4)^{2}+\left(\frac{1}{2}x+1-3\right)^{2}=20$ [$\frac{5}{4}x^{2}-10x=0$]

$x=0$ or 8, $y=1$ or $5$ [(0,1) and (8,5)]

Equations are $y-1=-2x$ and $y-5=-2(x-8)$

Method 2

Coordinates of points at which tangents meet curve are

(4+4, 3+2) = (8,5) and (4-4, 3-2) = (0,1)

Equations are $y-5=-2(x-8)$ and $y-1=-2x$

Method 3

$(x-4)^{2}+(-2x+c-3)^{2}=20$

$[5x^{2}+(4-4c)x+(c-3)^{2}-4=0]$

$(4-4c)^{2}-20((c-3)^{2}-4)=0$

[leading to $-4c^{2}-32c+120c+16-100=0$]

$4c^{2}-88c+84=0$ [leading to $c^{2}-22c+21=0$]

$c=21$ and $c=1$ or $y=-2x+21$ and $y=-2x+1$

Question 11

Topic – ALV: 1.7

The equation of a curve is

\[y=k\sqrt{4x+1}-x+5.\]

where k is a positive constant.

(a) Find \(\frac{dy}{dx}\)

(b) Find the x-coordinate of the stationary point in terms of k.

(c) Given that \(k = 10.5\), find the equation of the normal to the curve at the point where the tangent to the curve makes an angle of \(tan^{-1}(2)\) with the positive x-axis.

▶️Answer/Explanation

Solution: –

(a) $\frac{dy}{dx}=\left[\frac{1}{2}k(4x+1)^{-\frac{1}{2}}\right]\left\{\times4\right\}-1$

(b) $2k(4x+1)^{-\frac{1}{2}}-1=0$ leading to $(4x+1)^{\frac{1}{2}}=2k$ or $\frac{2k}{(4x+1)^{\frac{1}{2}}}=1$

$x=\frac{4k^{2}-1}{4}$

(c) $2\times10.5(4x+1)^{-\frac{1}{2}}-1=2$

$7=(4x+1)^{\frac{1}{2}}$ leading to $4x+1=49$ leading to $x=12$

$y=[10.5\sqrt{4x+1}-x+5]=66.5$ [leading to $(12,66.5)]$

$y-66.5=-\frac{1}{2}(x-12)$

Scroll to Top