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Question 1

(a) Topic 1.5 – Trigonometry

(b) Topic 1.5 – Trigonometry

The diagram shows the curve with equation $y = a\sin(bx) + c$ for $0 \leq x \leq 2\pi$, where $a$, $b$, and $c$ are positive constants.

(a) State the values of $a$, $b$, and $c$.

(b) For these values of $a$, $b$, and $c$, determine the number of solutions in the interval $0 \leq x \leq 2\pi$ for each of the following equations:

(i) $a\sin(bx) + c = 7 – x$

(ii) $a\sin(bx) + c = 2\pi(x – 1)$

▶️Answer/Explanation

Solution :-

(a) $a = 4$

$b = 2$

$c = 3$

(b)(i) 5

(b)(ii) 1

Question 2

(a) Topic 1.6 – Series

(b) Topic 1.6 – Series

The first term of an arithmetic progression is -20 and the common difference is 5.

(a)Find the sum of the first 20 terms of the progression.

It is given that the sum of the first 2k terms is 10 times the sum of the first k terms.

(b)Find the value of k.

▶️Answer/Explanation

Solution :-

(a) Find the sum of the first 20 terms of the progression.

For an arithmetic progression, the sum of the first $n$ terms is given by:
\[
S_n = \frac{n}{2} (2a + (n-1)d)
\]
where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms.

$a = -20$, $d = 5$, $n = 20$
\[
S_{20} = \frac{20}{2} (2(-20) + (20-1) \cdot 5)
\]
\[
= 10 (-40 + 19 \cdot 5)
\]
\[
= 10 (-40 + 95) = 10 \cdot 55 = 550
\]

Sum of the first 20 terms: 550

(b) Find the value of $k$.

Given: The sum of the first $2k$ terms is 10 times the sum of the first $k$ terms.
\[
S_{2k} = 10 \cdot S_k
\]

Using the sum formula:
\[
S_k = \frac{k}{2} (2a + (k-1)d)
\]
\[
S_{2k} = \frac{2k}{2} (2a + (2k-1)d) = k (2a + (2k-1)d)
\]

Substitute $a = -20$ and $d = 5$:
\[
S_k = \frac{k}{2} (2(-20) + (k-1) \cdot 5) = \frac{k}{2} (-40 + 5k – 5) = \frac{k}{2} (5k – 45) = \frac{k(5k – 45)}{2}
\]
\[
S_{2k} = k (2(-20) + (2k-1) \cdot 5) = k (-40 + (2k-1) \cdot 5) = k (-40 + 10k – 5) = k (10k – 45)
\]

Now apply the given condition:
\[
k (10k – 45) = 10 \cdot \frac{k(5k – 45)}{2}
\]
Assuming $k \neq 0$ (since $k$ must be a positive integer for the number of terms), divide both sides by $k$:
\[
10k – 45 = 10 \cdot \frac{5k – 45}{2}
\]
\[
10k – 45 = 5(5k – 45)
\]
\[
10k – 45 = 25k – 225
\]
\[
-45 + 225 = 25k – 10k
\]
\[
180 = 15k
\]
\[
k = \frac{180}{15} = 12
\]

**Value of $k= 12$

Question 3

(a) Topic 1.7 – Differentiation

(b) Topic 1.7 – Differentiation

The equation of a curve is $y = 2x^2 – 3$. Two points A and B with x-coordinates 2 and (2 + h) respectively
lie on the curve.

(a) Find and simplify an expression for the gradient of the chord AB in terms of h.

(b) Explain how the gradient of the curve at the point A can be deduced from the answer to part (a),
and state the value of this gradient.

▶️Answer/Explanation

Solution :-

(a) To find the gradient of the chord AB, we need the coordinates of points A and B on the curve $y = 2x^2 – 3$.

– For point A, where $x = 2$:
\[y = 2(2)^2 – 3 = 2 \times 4 – 3 = 8 – 3 = 5.\]
So, A is $(2, 5)$.
– For point B, where $x = 2 + h$:
\[y = 2(2 + h)^2 – 3 = 2(4 + 4h + h^2) – 3 = 8 + 8h + 2h^2 – 3 = 2h^2 + 8h + 5.\]
So, B is $(2 + h, 2h^2 + 8h + 5)$.
The gradient of the chord AB is the slope between A and B:
\[\text{Gradient} = \frac{(2h^2 + 8h + 5) – 5}{(2 + h) – 2} = \frac{2h^2 + 8h}{h} = \frac{2h(h + 4)}{h} = 2(h + 4),\]
for $h \neq 0$. So, the gradient of the chord AB in terms of $h$ is $2h + 8$.

(b) The gradient of the curve at point A can be deduced from the gradient of the chord AB by taking the limit as $h$ approaches 0. This limit represents the derivative of the curve at $x = 2$, which is the instantaneous rate of change (or slope of the tangent) at point A.

From part (a), the gradient of the chord AB is $2h + 8$. Now, take the limit as $h \to 0$:
\[\lim_{h \to 0} (2h + 8) = 2(0) + 8 = 8.\]

Alternatively, we can find the gradient of the curve directly by differentiating $y = 2x^2 – 3$:
\[\frac{dy}{dx} = 4x.\]
At $x = 2$:
\[\frac{dy}{dx} = 4(2) = 8.\]

Thus, the gradient of the curve at point A $(2, 5)$ is 8.

Question 4

(a) Topic 1.6 – Series

(b) Topic 1.6 – Series

Find the term independent of x in the expansion of each of the following:

(a) $\left(x + \frac{3}{x^2}\right)^6$

(b) $(4x^3 – 5)\left(x + \frac{3}{x^2}\right)^6$

▶️Answer/Explanation

Solution :-

(a) $\left(x + \frac{3}{x^2}\right)^6$
This is a binomial expansion: $(a + b)^n$, where $a = x$, $b = \frac{3}{x^2}$, and $n = 6$. The general term is:
– $\binom{6}{r} x^{6-r} \left(\frac{3}{x^2}\right)^r = \binom{6}{r} x^{6-r} \cdot 3^r \cdot x^{-2r} = \binom{6}{r} 3^r x^{6-r-2r} = \binom{6}{r} 3^r x^{6-3r}$.

For the term to be independent of $x$, the exponent of $x$ must be 0:
– $6 – 3r = 0$
– $3r = 6$
– $r = 2$.

Now, substitute $r = 2$:
– Term = $\binom{6}{2} 3^2 x^{6-6} = 15 \cdot 9 \cdot x^0 = 135$.

So, the term independent of $x$ is 135.

(b) $(4x^3 – 5)\left(x + \frac{3}{x^2}\right)^6$
First, consider the expansion of $\left(x + \frac{3}{x^2}\right)^6$ (from part (a)). The general term is $\binom{6}{r} 3^r x^{6-3r}$. Now multiply by $(4x^3 – 5)$ and find the constant term.

Distribute:
1. $4x^3 \cdot \binom{6}{r} 3^r x^{6-3r}$:
– $= 4 \cdot \binom{6}{r} 3^r x^{3 + 6 – 3r} = 4 \cdot \binom{6}{r} 3^r x^{9-3r}$.
– Set exponent to 0: $9 – 3r = 0$
– $3r = 9$
– $r = 3$.
– Term = $4 \cdot \binom{6}{3} 3^3 = 4 \cdot 20 \cdot 27 = 2160$.

2. $-5 \cdot \binom{6}{r} 3^r x^{6-3r}$:
– Set exponent to 0: $6 – 3r = 0$
– $3r = 6$
– $r = 2$.
– Term = $-5 \cdot \binom{6}{2} 3^2 = -5 \cdot 15 \cdot 9 = -675$.

Total constant term = $2160 + (-675) = 1485$.

So, the term independent of $x$ is 1485.

Question 5

(a) Topic 1.2 – Functions

(b) Topic 1.2 – Functions

The function f is defined by $f(x) = \frac{2x + 1}{2x – 1}$ for $x < \frac{1}{2}$.

(a) (i) State the value of f(-1).

(ii)

The diagram shows the graph of y = f(x). Sketch the graph of y = f⁻¹(x) on this diagram.
Show any relevant mirror line.

(iii)Topic – ALV: 1.2-Find an expression for f⁻¹(x) and state the domain of the function f⁻¹.

The function g is defined by $g(x) = 3x + 2$ for $x \in \mathbb{R}$.

(b)Topic – ALV: 1.2-Solve the equation $f(x) = g(f(\frac{1}{4}))$.

▶️Answer/Explanation

Solution :-

$(a)(i) [f(-1)=] \frac{1}{3}$

(a)(ii)

(a)(iii) $\frac{2x+1}{2x-1}=y \Rightarrow 2x+1=y(2x-1)$

$2xy-2x=y+1$

$\frac{x+1}{2(x-1)}, \frac{-x-1}{2-2x}$

$Domain of f^{-1} is$ $x<1$

Alternative Method for Question 5(a)(iii)

$y=1+\frac{2}{2x-1} \Rightarrow y-1=\frac{2}{2x-1}$

$2x=\frac{2}{y-1}+1$

$\frac{1}{x-1}+\frac{1}{2}$

(b) $gf\left(\frac{1}{4}\right)=-7$

$\frac{2x+1}{2x-1}=-7$

$[x=] \frac{3}{8}$

Alternative solution for Question 5(b)

$gf\left(\frac{1}{4}\right)=-7$

$x=f^{-1}(-7)$

$[x=] \frac{3}{8}$

Question 6

Topic 1.4 – Circular measure

The diagram shows a metal plate OABCDEF consisting of sectors of two circles, each with centre O.
The radii of sectors AOB and EOF are $r$ cm and the radius of sector COD is $2r$ cm.

Angle AOB = angle EOF = $\theta$ radians and angle COD = $2\theta$ radians.

It is given that the perimeter of the plate is 14 cm and the area of the plate is 10 cm².

Given that $r > \frac{3}{2}$ and $\theta < \frac{\pi}{4}$, find the values of $r$ and $\theta$.

▶️Answer/Explanation

Solution :-

$Perimeter = r+r\theta+r+2r \times 2\theta+r+r\theta+r$ $[=4r+6r\theta]$ $Area =\frac{1}{2}r^{2}\theta+\frac{1}{2}(2r)^{2} \times 2\theta+\frac{1}{2}r^{2}\theta$ $[=5r^{2}\theta]$

$4r+6r\theta=14$ and $5r^{2}\theta=10$

EITHER

$5r^{2}\frac{14-4r}{6r}=10$ or $4r+6r(\frac{10}{5r^{2}})=14$

$\Rightarrow2r^{2}-7r+6=0\Rightarrow](r-2)(2r-3)=0$

$5(\frac{14}{4+6\theta})^{2}\theta=10$ or $4(\sqrt{\frac{10}{5\theta}})+6(\sqrt{\frac{10}{5\theta}})\theta=14$

$\Rightarrow18\theta^{2}-25\theta+8=0\Rightarrow](9\theta-8)(2\theta-1)=0$

Then

$r=2$ and $\theta=0.5$

Question 7

(a) Topic 1.1 – Quadratics

(b) Topic 1.8 – Integration

(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

$\left\{ \frac{-2}{3}(x-2)^{3}+19x \right\} – \left\{ \frac{8}{2}x^{2}+9x \right\}$

$\left\{ \left(\frac{2}{3}+19\right)-\left(18-19\right) \right\} – \left\{ (4+9)-(4-9) \right\}$

= $\frac{8}{3}, 2\frac{2}{3}$

Question 8

(a) Topic 1.3 – Coordinate geometry

(b) Topic 1.3 – Coordinate geometry

The equation of a circle is $x^2 + y^2 + px + 2y + q = 0$, where $p$ and $q$ are constants.

(a)Topic – ALV: 1.3-Express the equation in the form $(x – a)^2 + (y – b)^2 = r^2$, where $a$ is to be given in terms of $p$ and $r^2$ is to be given in terms of $p$ and $q$.

The line with equation $x + 2y = 10$ is the tangent to the circle at the point $A(4,3)$.

(b) (i)Topic – ALV: 1.3-Find the equation of the normal to the circle at the point $A$.

(ii)Topic – ALV: 1.3-Find the values of $p$ and $q$.

▶️Answer/Explanation

Solution :-

(a) $\left(x-\left(-\frac{1}{2}p\right)\right)^{2}+(y-(-1))^{2}$ OE

$\left(x-\left(-\frac{1}{2}p\right)\right)^{2}+(y-(-1))^{2}=-q+1+\left(-\frac{1}{2}P\right)^{2}$ OE

(b)(i) [Gradient of tangent =] $-\frac{1}{2}$

[Gradient of normal =] 2

$\frac{y-3}{x-4}=2$ [y=2x-5]

(b)(ii) Method 1 for the first two marks:

$-1-3=2\left(-\frac{1}{2}p-4\right) \text{ or } -1=-p-5$

p=-4

Method 2 for the first two marks:

$-1=2x-5\Rightarrow x=2 \Rightarrow -\frac{1}{2}p=2$

p=-4

Method 3 for the first two marks:

$2x+2y\frac{dy}{dx}+p+2\frac{dy}{dx}=0 [\Rightarrow p=-8-8\frac{dy}{dx}]$

$\left[\frac{dy}{dx}=-\frac{1}{2} \Rightarrow\right] p=-4$

(b)(ii) Method 1 for the last 3 marks:

$r^{2}=(4-2)^{2}+(3-(-1))^{2}[=20]$

$-q+1+\frac{1}{4}p^{2}=20$

$q=-15$

Method 2 for the last 3 marks:

$r=|\frac{2-2-10}{\sqrt{5}}|[\frac{10}{\sqrt{5}}]$

$-q+1+\frac{1}{4}p^{2}=(\frac{10}{\sqrt{5}})^{2}$

$q=-15$

Method 3 for the last 3 marks:

$4^{2}+3^{2}+4p+6+q=0[\Rightarrow4p+q+31=0]$

$(4-(-\frac{1}{2}p))^{2}+(3-(-1))^{2}=-q+1+(-\frac{1}{2}p)^{2}$

$4(-4)+q+31=0$

$q=-15$

(b)(ii) Alternative Method for Question 8(b)(ii)

$4^{2}+3^{2}+4p+6+q=0$

$x^{2}+(2x-5)^{2}+px+2(2x-5)+q=0$ with $x=4$

$x^{2}+\left(\frac{10-x}{2}\right)^{2}+px+\left(\frac{10-x}{2}\right)+q=0$ with $x=4$

$\left(\frac{y+5}{2}\right)^{2}+y^{2}+p\left(\frac{y+5}{2}\right)+2y+q=0$ with $y=3$

$(10-2y)^{2}+y^{2}+p(10-2y)+2y+q=0$ with $y=3$

[Each of these $\Rightarrow 4p+q+31=0$]

$\frac{5}{4}x^{2}+(p-6)x+35+q=0 \Rightarrow (p-6)^{2}-4\times\frac{5}{4}\times(35+q)=0$

$5y^{2}-y(38+2p)+100+10p+q=0 \Rightarrow (38+2p)^{2}-4\times5\times(100+10p+q)=0$

[Each of these $\Rightarrow p^{2}-12p-139-5q=0$]

Solving the equations simultaneously to find p or q

$p=-4$

$q=-15$

Question 9

(a) Topic 1.1 – Quadratics

(b) Topic 1.3 – Coordinate geometry

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 10

(a) Topic 1.7 – Differentiation

(b) Topic 1.8 – Integration

A function $f$ with domain $x > 0$ is such that $f'(x) = 8(2x – 3)^{\frac{1}{3}} – 10x^{\frac{2}{3}}$. It is given that the curve with equation $y = f(x)$ passes through the point $(1, 0)$.

(a) Find the equation of the normal to the curve at the point $(1, 0)$.

(b) Find $f(x)$.

It is given that the equation $f'(x) = 0$ can be expressed in the form

\[125x^{2}-128x+192=0\]

▶️Answer/Explanation

Solution :-

(a) -18

$\frac{1}{18}$

$\frac{y[=0]}{x-1}=\frac{1}{18}$

$10(b) [f(x)=]\left\{8(2x-3)^{\frac{4}{3}}.\frac{1}{\frac{4}{3}}\right\}-\left\{10x^{\frac{5}{3}}.\frac{1}{\frac{5}{3}}\right\}[+c]$

$[3(2x-3)^{\frac{4}{3}}-6x^{\frac{5}{3}}+c]$

$0=3(2(1)-3)^{\frac{4}{3}}-6(1)^{\frac{5}{3}}+c$
$[0=3-6+c]$

$[f(x)or~y=]3(2x-3)^{\frac{4}{3}}-6x^{\frac{5}{3}}+3$

OR use of the quadratic formula and stating “No solutions”

OR completing the square for the given quadratic and stating positive or $>0.$

OR sketch of the given quadratic and stating positive.

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