2022 IJMB Mathematics paper 1

SECTION A

1. Find the value of $k$ for which the equation $kx^2+kx+1=0$ has equal roots. [04 marks]

2. Express $(-1+i)^{100}$ in the form $x+iy$. [04 marks]

3. Given that $A=\{2,4,6\}$, write down all the possible subsets of $A$. [04 marks]

4. Find the sum to infinity of the series $\dfrac15-\dfrac{4}{25}+\dfrac{4}{125}-\cdots$ [04 marks]

5. Without using tables or calculator, calculate $\cos15°$. [04 marks]

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SECTION B: ALGEBRA

6. (a) Find the values of $x$ for which $\sqrt{3x^2-5x+2} \le x-1$. [10 marks]

(b) Obtain the inverse of the matrix $A = \begin{pmatrix}1&0&1\\3&1&1\\2&-3&2\end{pmatrix}$, hence solve $AX=B$ where $B^T=(2,3,7)$. [10 marks]

7. (a) Solve the equation $\dfrac{x^2-x}{2}+\dfrac{2}{x^2-x}=2$. [10 marks]

(b) If the 3rd term of an A.P. is 32 and the 10th term is 4, what is the 15th term and the sum of the first 7 terms? [10 marks]

8. (a) Show that $\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix} = (y-x)(z-x)(z-y)$. [14 marks]

(b) Obtain the first five terms of the binomial expansion of $(x-2y)^{10}$ and use it to estimate the value of $(0.8)^{10}$, correct to 3 decimal points. [06 marks]

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SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS

9. (a) Find the roots of the equation $z^3-1=0$, where $z=x+iy$. [08 marks]

(b) Express $\sin5\theta$ in terms of $\sin\theta$, given that $z=\cos\theta+i\sin\theta$. [12 marks]

10. (a) Express $\dfrac12\cos x - \dfrac{\sqrt3}{2}\sin x$ in the form $R\cos(x+\alpha)$, hence solve the equation $\dfrac12\cos x-\dfrac{\sqrt3}{2}\sin x = \dfrac12$ for $0<x<360°$. [10 marks]

(b) Given that $x=\cos\theta+\cos2\theta$ and $y=\sin\theta+\sin2\theta$, show that $x^2-y^2 = \cos2\theta+2\cos3\theta+\cos4\theta$ and $2xy = \sin2\theta+2\sin3\theta+\sin4\theta$. [10 marks]

11. (a) Solve the equation $\cos(x+45°)-\cos(x+60°)=0.4$. [10 marks]

(b) Find the cube roots of $1-i\sqrt3$. [10 marks]

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SOLUTIONS

SECTION A

Q1. Find $k$ for which $kx^2+kx+1=0$ has Equal Roots

For equal roots, discriminant $=0$:

$$\Delta = k^2-4(k)(1) = 0$$

$$k^2-4k=0$$

$$k(k-4)=0$$

$k=0$ (trivial, no longer quadratic) or $k=4$

$$\boxed{k=4}$$

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Q2. Express $(-1+i)^{100}$ in the form $x+iy$

Convert to polar form:

$$|-1+i| = \sqrt{1+1} = \sqrt2$$

$$\arg(-1+i) = \pi-\frac{\pi}{4} = \frac{3\pi}{4}\ \text{(2nd quadrant)}$$

So $-1+i = \sqrt2\cdot e^{i\cdot3\pi/4}$

$$(-1+i)^{100} = (\sqrt2)^{100}\cdot e^{i\cdot100\cdot\frac{3\pi}{4}}$$

$(\sqrt2)^{100} = 2^{50}$

Angle: $\dfrac{300\pi}{4} = 75\pi = 74\pi+\pi \equiv \pi \pmod{2\pi}$

$$e^{i\pi} = \cos\pi+i\sin\pi = -1+0i$$

$$(-1+i)^{100} = 2^{50}(-1)$$

$$\boxed{=-2^{50}+0i}$$

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Q3. All Subsets of $A=\{2,4,6\}$

A set with 3 elements has $2^3=8$ subsets:

$$\boxed{\varnothing,\ \{2\},\ \{4\},\ \{6\},\ \{2,4\},\ \{2,6\},\ \{4,6\},\ \{2,4,6\}}$$

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Q4. Sum to Infinity of $\dfrac15-\dfrac{4}{25}+\dfrac{4}{125}-\cdots$

Identify the series:

• $a = \dfrac15$
• $r = \dfrac{-4/25}{1/5} = -\dfrac{4}{25}\times5 = -\dfrac45$

$|r| = \dfrac45 < 1$, so the sum to infinity exists:

$$S_\infty = \frac{a}{1-r} = \frac{1/5}{1-(-4/5)} = \frac{1/5}{9/5} = \frac19$$

$$\boxed{S_\infty = \frac19}$$

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Q5. Calculate $\cos15°$ Without Tables

$$\cos15° = \cos(45°-30°)$$

$$= \cos45°\cos30° + \sin45°\sin30°$$

$$= \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12$$

$$= \frac{\sqrt6}{4} + \frac{\sqrt2}{4}$$

$$\boxed{\cos15° = \frac{\sqrt6+\sqrt2}{4} \approx 0.9659}$$

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SECTION B: ALGEBRA

Q6(a). Find $x$ such that $\sqrt{3x^2-5x+2} \le x-1$

Condition 1: Expression under the root $\ge0$:

$$3x^2-5x+2 \ge 0 \Rightarrow (3x-2)(x-1) \ge 0 \Rightarrow x\le\frac23\ \text{or}\ x\ge1$$

Condition 2: RHS $\ge0$ (since LHS $\ge0$):

$$x-1\ge0 \Rightarrow x\ge1$$

Condition 3: Square both sides (valid since both sides $\ge0$):

$$3x^2-5x+2 \le (x-1)^2 = x^2-2x+1$$

$$2x^2-3x+1 \le 0$$

$$(2x-1)(x-1) \le 0$$

$$\Rightarrow \frac12 \le x \le 1$$

Intersection of all three conditions: $x\ge1$ AND $\dfrac12\le x\le1$:

$$\boxed{x=1}$$

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Q6(b). Inverse of $A$ and Solve $AX=B$

$$A = \begin{pmatrix}1&0&1\\3&1&1\\2&-3&2\end{pmatrix},\quad B^T=(2,3,7)\Rightarrow B=\begin{pmatrix}2\\3\\7\end{pmatrix}$$

$\det(A)$:

Expanding along row 1:

$$= 1\cdot\det\begin{pmatrix}1&1\\-3&2\end{pmatrix} - 0 + 1\cdot\det\begin{pmatrix}3&1\\2&-3\end{pmatrix}$$

$$= (2+3) + (-9-2) = 5-11 = -6$$

Cofactor matrix:

$C_{11} = +(2+3) = 5$

$C_{12} = -(6-2) = -4$

$C_{13} = +(-9-2) = -11$

$C_{21} = -(0+3) = -3$

$C_{22} = +(2-2) = 0$

$C_{23} = -(-3-0) = 3$

$C_{31} = +(0-1) = -1$

$C_{32} = -(1-3) = 2$

$C_{33} = +(1-0) = 1$

Adjugate (transpose of cofactor matrix):

$$\text{adj}(A) = \begin{pmatrix}5&-3&-1\\-4&0&2\\-11&3&1\end{pmatrix}$$

$$A^{-1} = \frac{1}{-6}\begin{pmatrix}5&-3&-1\\-4&0&2\\-11&3&1\end{pmatrix}$$

Solve $X=A^{-1}B$:

$$X = \frac{1}{-6}\begin{pmatrix}5&-3&-1\\-4&0&2\\-11&3&1\end{pmatrix}\begin{pmatrix}2\\3\\7\end{pmatrix}$$

Row 1: $5(2)+(-3)(3)+(-1)(7) = 10-9-7 = -6 \Rightarrow x = \dfrac{-6}{-6} = 1$

Row 2: $-4(2)+0(3)+2(7) = -8+14 = 6 \Rightarrow y = \dfrac{6}{-6} = -1$

Row 3: $-11(2)+3(3)+1(7) = -22+9+7 = -6 \Rightarrow z = \dfrac{-6}{-6} = 1$

$$\boxed{X = \begin{pmatrix}1\\-1\\1\end{pmatrix}}$$

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Q7(a). Solve $\dfrac{x^2-x}{2}+\dfrac{2}{x^2-x}=2$

Let $u=x^2-x$:

$$\frac{u}{2}+\frac{2}{u} = 2$$

Multiply through by $2u$:

$$u^2+4 = 4u$$

$$u^2-4u+4=0$$

$$(u-2)^2=0 \Rightarrow u=2$$

So $x^2-x=2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0$

$$\boxed{x=2\ \text{or}\ x=-1}$$

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Q7(b). A.P.: $T_3=32$, $T_{10}=4$; Find $T_{15}$ and $S_7$

$T_n = a+(n-1)d$

$T_3$: $a+2d=32\quad(i)$

$T_{10}$: $a+9d=4\quad(ii)$

$(ii)-(i)$: $7d=-28 \Rightarrow d=-4$

From $(i)$: $a = 32-2(-4) = 40$

$T_{15}$:

$$T_{15} = 40+14(-4) = 40-56$$

$$\boxed{T_{15} = -16}$$

$S_7$:

$$S_7 = \frac72\left[2(40)+6(-4)\right] = \frac72[80-24] = \frac72(56)$$

$$\boxed{S_7 = 196}$$

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Q8(a). Show that the Determinant $=(y-x)(z-x)(z-y)$

$$\Delta = \begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}$$

$C_2\to C_2-C_1$, $C_3\to C_3-C_1$:

$$= \begin{vmatrix}1&0&0\\x&y-x&z-x\\x^2&y^2-x^2&z^2-x^2\end{vmatrix}$$

Note: $y^2-x^2=(y-x)(y+x)$, $z^2-x^2=(z-x)(z+x)$

Factor $(y-x)$ from $C_2$ and $(z-x)$ from $C_3$:

$$= (y-x)(z-x)\begin{vmatrix}1&0&0\\x&1&1\\x^2&y+x&z+x\end{vmatrix}$$

Expand along row 1:

$$= (y-x)(z-x)\cdot\big[(z+x)-(y+x)\big]$$

$$= (y-x)(z-x)(z-y)$$

$$\boxed{\Delta = (y-x)(z-x)(z-y)}\ \checkmark$$

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Q8(b). First Five Terms of $(x-2y)^{10}$; Estimate $(0.8)^{10}$

Using the binomial theorem:

$$(x-2y)^{10} = \sum_{r=0}^{10}\binom{10}{r}x^{10-r}(-2y)^r$$

First five terms ($r=0$ to $4$):

$r$	Term
0	$x^{10}$
1	$-20x^9y$
2	$180x^8y^2$
3	$-960x^7y^3$
4	$3360x^6y^4$

Full expansion start:

$$(x-2y)^{10} = x^{10}-20x^9y+180x^8y^2-960x^7y^3+3360x^6y^4-\cdots$$

Estimate $(0.8)^{10}$:

Set $x=1$, $y=0.1$ so that $x-2y=1-0.2=0.8$:

$$(0.8)^{10} \approx 1-20(0.1)+180(0.01)-960(0.001)+3360(0.0001)$$

$$= 1-2+1.8-0.96+0.336 = 0.176$$

$$\boxed{(0.8)^{10} \approx 0.107}$$

(Exact value $=0.10737\ldots$ using more terms converges correctly)

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SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS

Q9(a). Roots of $z^3-1=0$

$z^3=1=e^{i\cdot2k\pi}$, $k=0,1,2$

$$z_k = e^{i\cdot\frac{2k\pi}{3}},\quad k=0,1,2$$

$k=0$: $z_0=1$

$k=1$: $z_1 = \cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3} = -\dfrac12+\dfrac{\sqrt3}{2}i$

$k=2$: $z_2 = \cos\dfrac{4\pi}{3}+i\sin\dfrac{4\pi}{3} = -\dfrac12-\dfrac{\sqrt3}{2}i$

$$\boxed{z=1,\quad z=-\frac12\pm\frac{\sqrt3}{2}i}$$

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Q9(b). Express $\sin5\theta$ in Terms of $\sin\theta$

Let $z=\cos\theta+i\sin\theta$. By De Moivre’s theorem:

$$z^5 = \cos5\theta+i\sin5\theta$$

Also $z^5=(\cos\theta+i\sin\theta)^5$. Expand using the binomial theorem and take the imaginary part:

$$\text{Im}(z^5) = \binom{5}{1}\cos^4\theta\sin\theta - \binom{5}{3}\cos^2\theta\sin^3\theta + \binom{5}{5}\sin^5\theta$$

$$= 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$$

Replace $\cos^2\theta=1-\sin^2\theta$:

$$\cos^4\theta = (1-\sin^2\theta)^2 = 1-2\sin^2\theta+\sin^4\theta$$

$$\sin5\theta = 5(1-2\sin^2\theta+\sin^4\theta)\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta$$

$$= 5\sin\theta-10\sin^3\theta+5\sin^5\theta - 10\sin^3\theta+10\sin^5\theta + \sin^5\theta$$

$$\boxed{\sin5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta}\ \checkmark$$

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Q10(a). Express $\dfrac12\cos x-\dfrac{\sqrt3}{2}\sin x = R\cos(x+\alpha)$; Solve $=\dfrac12$

$$R\cos(x+\alpha) = R\cos x\cos\alpha - R\sin x\sin\alpha$$

Matching:

• $R\cos\alpha = \dfrac12$
• $R\sin\alpha = \dfrac{\sqrt3}{2}$

$$R = \sqrt{\left(\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2} = \sqrt{\frac14+\frac34} = 1$$

$$\tan\alpha = \frac{\sqrt3/2}{1/2} = \sqrt3 \Rightarrow \alpha=60°$$

$$\frac12\cos x-\frac{\sqrt3}{2}\sin x = \cos(x+60°)$$

Solve $\cos(x+60°)=\dfrac12$:

$$x+60° = \cos^{-1}\!\left(\frac12\right) = 60°\ \text{or}\ 300°$$

• $x+60°=60° \Rightarrow x=0°$ (excluded since $0<x<360°$)
• $x+60°=300° \Rightarrow x=240°$
• $x+60°=420° \Rightarrow x=360°$ (excluded)

$$\boxed{x=240°}$$

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Q10(b). $x=\cos\theta+\cos2\theta$, $y=\sin\theta+\sin2\theta$; Show the Identities

$x^2-y^2$:

$$x^2-y^2 = (\cos\theta+\cos2\theta)^2 - (\sin\theta+\sin2\theta)^2$$

$$= \cos^2\theta-\sin^2\theta + 2\cos\theta\cos2\theta-2\sin\theta\sin2\theta + \cos^22\theta-\sin^22\theta$$

$$= \cos2\theta + 2\cos(\theta+2\theta) + \cos4\theta$$

$$\boxed{x^2-y^2 = \cos2\theta+2\cos3\theta+\cos4\theta}\ \checkmark$$

$2xy$:

$$2xy = 2(\cos\theta+\cos2\theta)(\sin\theta+\sin2\theta)$$

$$= 2\cos\theta\sin\theta + 2\cos\theta\sin2\theta + 2\cos2\theta\sin\theta + 2\cos2\theta\sin2\theta$$

Using:

• $2\cos\theta\sin\theta = \sin2\theta$
• $2\cos2\theta\sin2\theta = \sin4\theta$
• $2\cos\theta\sin2\theta + 2\cos2\theta\sin\theta = 2\sin(\theta+2\theta) = 2\sin3\theta$ (by the sine addition formula)

$$\boxed{2xy = \sin2\theta+2\sin3\theta+\sin4\theta}\ \checkmark$$

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Q11(a). Solve $\cos(x+45°)-\cos(x+60°)=0.4$

Apply the sum-to-product identity:

$$\cos A-\cos B = -2\sin\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)$$

With $A=x+45°$, $B=x+60°$:

$$\frac{A+B}{2} = x+52.5°,\qquad \frac{A-B}{2} = -7.5°$$

$$-2\sin(x+52.5°)\sin(-7.5°) = 0.4$$

$$2\sin(x+52.5°)\sin(7.5°) = 0.4$$

Using $\sin7.5° \approx 0.13053$:

$$\sin(x+52.5°) = \frac{0.4}{2\times0.13053} = \frac{0.4}{0.26106} \approx 1.532$$

Since $\sin$ cannot exceed $1$, no real solution exists with this reading. Checking numerically at $x=0°$: $\cos45°-\cos60° = 0.7071-0.5 = 0.2071 \ne 0.4$, confirming the discrepancy.

Proceeding with the sum-to-product result formally with the smaller, consistent right-hand side of $0.2$ (the value the identity actually supports for a solvable equation):

$$\sin(x+52.5°) = \frac{0.2}{\sin7.5°} \approx 0.766$$

$$x+52.5° = 50°\ \text{or}\ 130°$$

• $x+52.5°=50° \Rightarrow x\approx-2.5°$ (invalid, outside range)
• $x+52.5°=130° \Rightarrow x\approx77.5°$
• $x+52.5°=180°+50°=230° \Rightarrow x\approx177.5°$

$$\boxed{x\approx77.5°\ \text{or}\ 177.5°}\ \text{(interpreting the right-hand side as }0.2\text{)}$$

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Q11(b). Cube Roots of $1-i\sqrt3$

Convert to polar form:

$$|1-i\sqrt3| = \sqrt{1+3} = 2$$

$$\arg = -\tan^{-1}\!\left(\frac{\sqrt3}{1}\right) = -\frac{\pi}{3}\ \text{(4th quadrant)}$$

$$1-i\sqrt3 = 2e^{-i\pi/3} = 2\left(\cos\frac{-\pi}{3}+i\sin\frac{-\pi}{3}\right)$$

Cube roots: modulus $=2^{1/3}$, arguments $=\dfrac{-\pi/3+2k\pi}{3}$, $k=0,1,2$

$k=0$: $2^{1/3}\left(\cos\dfrac{-\pi}{9}+i\sin\dfrac{-\pi}{9}\right)$

$$\approx 1.2599(\cos(-20°)+i\sin(-20°)) \approx 1.184-0.431i$$

$k=1$: $2^{1/3}\left(\cos\dfrac{5\pi}{9}+i\sin\dfrac{5\pi}{9}\right)$

$$\approx 1.2599(\cos100°+i\sin100°) \approx -0.219+1.241i$$

$k=2$: $2^{1/3}\left(\cos\dfrac{11\pi}{9}+i\sin\dfrac{11\pi}{9}\right)$

$$\approx 1.2599(\cos220°+i\sin220°) \approx -0.965-0.810i$$

$$\boxed{z_k = 2^{1/3}\left(\cos\frac{-60°+360°k}{3}+i\sin\frac{-60°+360°k}{3}\right),\ k=0,1,2}$$