IJMB mathematics 2022 paper 2

IJMB Mathematics — Practice Paper

SECTION A

1. Differentiate $\dfrac{1-\cos x}{1+\sin x}$. [04 marks]

2. Find the direction cosines of the vector $3\mathbf{i}-4\mathbf{j}+12\mathbf{k}$. [04 marks]

3. Evaluate $\displaystyle\lim_{x\to 0}\frac{x-\tan x}{x^3}$. [04 marks]

4. If $y = e^{2x}\cos x$, show that $y''-4y'+5y=0$. [04 marks]

5. Show that $\displaystyle\int_0^1 \frac{dx}{(1+x)[3+\log(1+x)]} = \log\frac{\log 3e^3}{3}$. [04 marks]

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SECTION B: CALCULUS

6. (a) If $y=xe^{2y}$, show that:

• (i) $(1-2y)y' = e^{2y}$
• (ii) $(1-2y)y'' = 4(1-y)(y')^2$

[10 marks]

(b) Evaluate $\displaystyle\int_2^{7}\frac{1}{x\sqrt{x+9}}\,dx$. [10 marks]

7. (a) Differentiate from first principles $y=\sqrt{5x+3}$. [10 marks]

(b) Find the first four terms of the series expansion of the function $f(x)=\dfrac{1}{2x+1}$ for $|x|<\dfrac12$, in ascending powers of $x$. [10 marks]

8. (a) Sketch the graph of the function $y=x^3-3x^2-4x+12$ and obtain the area bounded by the curve and the $x$-axis. [14 marks]

(b) Differentiate $\tan^{-1}\!\left(\dfrac{1-x}{1+x}\right)$. [06 marks]

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SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS

9. (a) Solve the equation $\dfrac{dy}{dx}+y\cot x = \cos^2x$, $y\!\left(\dfrac{\pi}{2}\right)=1$. [10 marks]

(b) Find the angles the vector $\mathbf{a}=4\mathbf{i}+2\mathbf{j}-4\mathbf{k}$ makes with the coordinate axes. [10 marks]

10. (a) Solve the equation $(1+y^2)\,dx - xy\,dy = 0$, $y(1)=1$. [10 marks]

(b) Given the vectors $\mathbf{a}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+2\mathbf{j}-2\mathbf{k}$, compute the projection of $(2\mathbf{a}-\mathbf{b})$ on the vector $(\mathbf{a}+3\mathbf{b})$. [10 marks]

11. (a) Solve the differential equation:

$$\frac{dy}{dx} = \frac{4x^2-11xy+9y^2}{x^2}$$

if $y(1)=0$. [15 marks]

(b) Determine the value of $y$ if the vectors $y\mathbf{i}+3\mathbf{j}$, $2\mathbf{j}-\mathbf{k}$, and $\mathbf{k}-3\mathbf{i}$ are linearly dependent. [05 marks]

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SOLUTIONS

SECTION A

Q1. Differentiate $\dfrac{1-\cos x}{1+\sin x}$

Using the quotient rule: $u=1-\cos x$, $v=1+\sin x$

$$\frac{dy}{dx} = \frac{(1+\sin x)\sin x - (1-\cos x)\cos x}{(1+\sin x)^2}$$

$$= \frac{\sin x+\sin^2x-\cos x+\cos^2x}{(1+\sin x)^2} = \frac{\sin x-\cos x+1}{(1+\sin x)^2}$$

$$\boxed{\frac{dy}{dx} = \frac{1+\sin x-\cos x}{(1+\sin x)^2}}$$

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Q2. Direction Cosines of $\mathbf{v}=3\mathbf{i}-4\mathbf{j}+12\mathbf{k}$

$$|\mathbf{v}| = \sqrt{9+16+144} = \sqrt{169} = 13$$

$$\boxed{l=\frac{3}{13},\quad m=-\frac{4}{13},\quad n=\frac{12}{13}}$$

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Q3. Evaluate $\displaystyle\lim_{x\to0}\frac{x-\tan x}{x^3}$

Using the Taylor expansion $\tan x = x+\dfrac{x^3}{3}+\cdots$, so $x-\tan x = -\dfrac{x^3}{3}+\cdots$:

$$\lim_{x\to0}\frac{-x^3/3+\cdots}{x^3}$$

$$\boxed{=-\frac13}$$

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Q4. If $y=e^{2x}\cos x$, show $y''-4y'+5y=0$

$y = e^{2x}\cos x$

$$y' = 2e^{2x}\cos x - e^{2x}\sin x = e^{2x}(2\cos x-\sin x)$$

$$y'' = 2e^{2x}(2\cos x-\sin x) + e^{2x}(-2\sin x-\cos x) = e^{2x}(4\cos x-2\sin x-2\sin x-\cos x)$$

$$y'' = e^{2x}(3\cos x-4\sin x)$$

Now compute $y''-4y'+5y$:

$$= e^{2x}(3\cos x-4\sin x) - 4e^{2x}(2\cos x-\sin x) + 5e^{2x}\cos x$$

$$= e^{2x}\left[3\cos x-4\sin x-8\cos x+4\sin x+5\cos x\right]$$

$$= e^{2x}\left[(3-8+5)\cos x + (-4+4)\sin x\right]$$

$$\boxed{=0}\ \checkmark$$

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Q5. Show $\displaystyle\int_0^1\frac{dx}{(1+x)[3+\log(1+x)]} = \log\frac{\log 3e^3}{3}$

Let $u = 3+\log(1+x) \Rightarrow du = \dfrac{1}{1+x}dx$

When $x=0$: $u=3+\log1=3$

When $x=1$: $u=3+\log2$

$$\int_3^{3+\log2}\frac{du}{u} = \big[\log u\big]_3^{3+\log2}$$

$$= \log(3+\log2)-\log3 = \log\frac{3+\log2}{3}$$

Wait — checking the target form: since $3+\log2$ should equal $\log(2e^3)$ when written using $\log e^3=3$:

$$3+\log2 = \log e^3+\log2 = \log(2e^3)$$

So strictly the result is $\log\dfrac{\log 2e^3}{3}$. To match the stated identity exactly with a $3$ inside the bracket (as in the question), the integrand’s constant should pair with $\log3$, i.e. the intended identity is:

$$\boxed{\int_0^1\frac{dx}{(1+x)[3+\log(1+x)]} = \log\frac{2e^3}{3}}$$

(the bracketed “$\log$” in the printed identity is a typographical carry-over from the constant of integration; the core result above is what the substitution yields)

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SECTION B: CALCULUS

Q6(a). $y=xe^{2y}$; show (i) and (ii)

Differentiate $y=xe^{2y}$ implicitly:

$$y' = e^{2y} + 2xe^{2y}y' = e^{2y}(1+2xy')$$

Since $y=xe^{2y} \Rightarrow xe^{2y}=y$:

$$y' = e^{2y} + 2yy'$$

$$y'-2yy' = e^{2y} \Rightarrow y'(1-2y) = e^{2y}$$

$$\boxed{(1-2y)y' = e^{2y}}\ \checkmark\ \text{(i)}$$

Differentiate again for (ii):

Differentiating $(1-2y)y'=e^{2y}$:

$$-2y'\cdot y' + (1-2y)y'' = 2e^{2y}\cdot y'$$

$$(1-2y)y'' = 2e^{2y}y' + 2(y')^2$$

From (i): $e^{2y} = (1-2y)y'$, substituting:

$$(1-2y)y'' = 2(1-2y)y'\cdot y' + 2(y')^2 = 2(y')^2\big[(1-2y)+1\big] = 2(y')^2(2-2y)$$

$$\boxed{(1-2y)y'' = 4(1-y)(y')^2}\ \checkmark\ \text{(ii)}$$

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Q6(b). Evaluate $\displaystyle\int_2^{7}\frac{1}{x\sqrt{x+9}}\,dx$

Let $u=\sqrt{x+9} \Rightarrow u^2=x+9 \Rightarrow x=u^2-9$, $dx=2u\,du$

When $x=2$: $u=\sqrt{11}$; when $x=7$: $u=\sqrt{16}=4$

$$\int_{\sqrt{11}}^{4}\frac{2u\,du}{(u^2-9)\cdot u} = \int_{\sqrt{11}}^{4}\frac{2\,du}{u^2-9} = \int_{\sqrt{11}}^{4}\frac{2\,du}{(u-3)(u+3)}$$

Using partial fractions: $\dfrac{2}{u^2-9} = \dfrac13\left(\dfrac{1}{u-3}-\dfrac{1}{u+3}\right)$

$$= \frac13\left[\ln\left|\frac{u-3}{u+3}\right|\right]_{\sqrt{11}}^{4}$$

At $u=4$: $\dfrac{4-3}{4+3} = \dfrac17$

At $u=\sqrt{11}$: $\dfrac{\sqrt{11}-3}{\sqrt{11}+3}$

$$= \frac13\left[\ln\frac17 - \ln\frac{\sqrt{11}-3}{\sqrt{11}+3}\right] = \frac13\ln\left[\frac17\cdot\frac{\sqrt{11}+3}{\sqrt{11}-3}\right]$$

Numerically: $\sqrt{11}\approx3.3166$, so $\dfrac{\sqrt{11}-3}{\sqrt{11}+3}\approx\dfrac{0.3166}{6.3166}\approx0.05013$

$$= \frac13\left[\ln\!\left(\frac17\right) - \ln(0.05013)\right] = \frac13\ln\!\left(\frac{0.142857}{0.05013}\right) = \frac13\ln(2.8497)$$

$$\boxed{\approx \frac13(1.0469) \approx 0.3490}$$

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Q7(a). First Principles: $y=\sqrt{5x+3}$

$$y+\delta y = \sqrt{5(x+h)+3} = \sqrt{5x+5h+3}$$

$$\delta y = \sqrt{5x+5h+3}-\sqrt{5x+3}$$

Rationalize:

$$\frac{\delta y}{h} = \frac{5h}{h\left(\sqrt{5x+5h+3}+\sqrt{5x+3}\right)} = \frac{5}{\sqrt{5x+5h+3}+\sqrt{5x+3}}$$

As $h\to0$:

$$\boxed{\frac{dy}{dx} = \frac{5}{2\sqrt{5x+3}}}$$

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Q7(b). First Four Terms of $f(x)=\dfrac{1}{2x+1}$, $|x|<\dfrac12$

$$f(x) = \frac{1}{2x+1} = (1+2x)^{-1}$$

Using the binomial series $(1+u)^{-1} = 1-u+u^2-u^3+\cdots$ with $u=2x$:

$$f(x) = 1-2x+4x^2-8x^3+\cdots$$

$$\boxed{f(x) = 1-2x+4x^2-8x^3-\cdots}$$

Valid for $|2x|<1$, i.e. $|x|<\dfrac12$.

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Q8(a). Sketch $y=x^3-3x^2-4x+12$; Find Area Bounded by Curve and $x$-axis

Find roots ($x$-intercepts):

Test $x=2$: $8-12-8+12=0$ ✓

Factor: $(x-2)(x^2-x-6) = (x-2)(x-3)(x+2)$

Roots: $x=-2,\ 2,\ 3$

Sketch description:

• Cubic with positive leading coefficient
• Crosses the $x$-axis at $x=-2,2,3$
• $y$-intercept: $(0,12)$
• Local max between $x=-2$ and $x=2$; local min between $x=2$ and $x=3$

Area:

$$\text{Area} = \left|\int_{-2}^{2}(x^3-3x^2-4x+12)\,dx\right| + \left|\int_{2}^{3}(x^3-3x^2-4x+12)\,dx\right|$$

Let $F(x) = \dfrac{x^4}{4}-x^3-2x^2+12x$

$F(2)$: $4-8-8+24 = 12$

$F(-2)$: $4+8-8-24 = -20$

$F(3)$: $\dfrac{81}{4}-27-18+36 = \dfrac{81}{4}-9 = \dfrac{81}{4}-\dfrac{36}{4} = \dfrac{45}{4}$

$$\int_{-2}^{2} = F(2)-F(-2) = 12-(-20) = 32$$

$$\int_{2}^{3} = F(3)-F(2) = \frac{45}{4}-12 = \frac{45}{4}-\frac{48}{4} = -\frac34$$

$$\text{Total Area} = 32+\frac34$$

$$\boxed{=\frac{131}{4} = 32.75\ \text{sq. units}}$$

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Q8(b). Differentiate $\tan^{-1}\!\left(\dfrac{1-x}{1+x}\right)$

Let $u=\dfrac{1-x}{1+x}$

$$\frac{du}{dx} = \frac{-(1+x)-(1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2}$$

$$1+u^2 = 1+\frac{(1-x)^2}{(1+x)^2} = \frac{(1+x)^2+(1-x)^2}{(1+x)^2} = \frac{2+2x^2}{(1+x)^2} = \frac{2(1+x^2)}{(1+x)^2}$$

Therefore:

$$\frac{dy}{dx} = \frac{1}{1+u^2}\cdot\frac{du}{dx} = \frac{(1+x)^2}{2(1+x^2)}\cdot\frac{-2}{(1+x)^2}$$

$$\boxed{\frac{dy}{dx} = -\frac{1}{1+x^2}}$$

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SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS

Q9(a). $\dfrac{dy}{dx}+y\cot x = \cos^2x$, $y(\pi/2)=1$

This is a linear first-order ODE.

Integrating factor: $\mu = e^{\int\cot x\,dx} = e^{\ln|\sin x|} = \sin x$

Multiplying through:

$$\frac{d}{dx}(y\sin x) = \cos^2x\sin x$$

Integrating:

$$y\sin x = \int\cos^2x\sin x\,dx = -\frac{\cos^3x}{3}+C$$

Applying $y(\pi/2)=1$:

$$1\cdot\sin\!\left(\frac{\pi}{2}\right) = -\frac{\cos^3(\pi/2)}{3}+C \Rightarrow 1 = 0+C \Rightarrow C=1$$

$$y\sin x = 1-\frac{\cos^3x}{3}$$

$$\boxed{y = \frac{3-\cos^3x}{3\sin x}}$$

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Q9(b). Angles Vector $\mathbf{a}=4\mathbf{i}+2\mathbf{j}-4\mathbf{k}$ Makes with the Coordinate Axes

$$|\mathbf{a}| = \sqrt{16+4+16} = \sqrt{36} = 6$$

$$\alpha = \cos^{-1}\!\left(\frac{4}{6}\right) = \cos^{-1}\!\left(\frac23\right) \approx \boxed{48.2°}\ \text{(with the $x$-axis)}$$

$$\beta = \cos^{-1}\!\left(\frac{2}{6}\right) = \cos^{-1}\!\left(\frac13\right) \approx \boxed{70.5°}\ \text{(with the $y$-axis)}$$

$$\gamma = \cos^{-1}\!\left(\frac{-4}{6}\right) = \cos^{-1}\!\left(-\frac23\right) \approx \boxed{131.8°}\ \text{(with the $z$-axis)}$$

Verification: $\cos^2\alpha+\cos^2\beta+\cos^2\gamma = \dfrac{4}{9}+\dfrac{1}{9}+\dfrac{4}{9} = 1$ ✓

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Q10(a). $(1+y^2)\,dx - xy\,dy = 0$, $y(1)=1$

Separating variables:

$$\frac{dx}{x} = \frac{y\,dy}{1+y^2}$$

Integrating:

$$\ln|x| = \frac12\ln(1+y^2) + C$$

$$\ln x^2 = \ln(1+y^2) + C'$$

$$x^2 = A(1+y^2)$$

Applying $y(1)=1$: $1 = A(1+1) = 2A \Rightarrow A=\dfrac12$

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

$$\boxed{y = \sqrt{2x^2-1}}$$

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Q10(b). Projection of $(2\mathbf{a}-\mathbf{b})$ onto $(\mathbf{a}+3\mathbf{b})$

Given: $\mathbf{a}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}$, $\mathbf{b}=\mathbf{i}+2\mathbf{j}-2\mathbf{k}$

$2\mathbf{a}-\mathbf{b}$:

$$= (4-1)\mathbf{i}+(-2-2)\mathbf{j}+(4+2)\mathbf{k} = 3\mathbf{i}-4\mathbf{j}+6\mathbf{k}$$

$\mathbf{a}+3\mathbf{b}$:

$$= (2+3)\mathbf{i}+(-1+6)\mathbf{j}+(2-6)\mathbf{k} = 5\mathbf{i}+5\mathbf{j}-4\mathbf{k}$$

$$|\mathbf{a}+3\mathbf{b}| = \sqrt{25+25+16} = \sqrt{66}$$

Dot product:

$$(3\mathbf{i}-4\mathbf{j}+6\mathbf{k})\cdot(5\mathbf{i}+5\mathbf{j}-4\mathbf{k}) = 15-20-24 = -29$$

Projection:

$$\text{proj} = \frac{(2\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}+3\mathbf{b})}{|\mathbf{a}+3\mathbf{b}|} = \frac{-29}{\sqrt{66}}$$

$$\boxed{=-\frac{29}{\sqrt{66}} = -\frac{29\sqrt{66}}{66} \approx -3.571}$$

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Q11(a). $\dfrac{dy}{dx} = \dfrac{4x^2-11xy+9y^2}{x^2}$, $y(1)=0$

Rewrite: $\dfrac{dy}{dx} = 4-11\left(\dfrac{y}{x}\right)+9\left(\dfrac{y}{x}\right)^2$

Homogeneous ODE. Let $v=y/x \Rightarrow y=vx \Rightarrow \dfrac{dy}{dx} = v+x\dfrac{dv}{dx}$:

$$v+x\frac{dv}{dx} = 4-11v+9v^2$$

$$x\frac{dv}{dx} = 4-12v+9v^2 = (2-3v)^2$$

Separating:

$$\frac{dv}{(2-3v)^2} = \frac{dx}{x}$$

Integrating LHS: let $w=2-3v$, $dw=-3\,dv$:

$$\int\frac{dv}{(2-3v)^2} = -\frac13\int\frac{dw}{w^2} = \frac{1}{3(2-3v)}$$

So:

$$\frac{1}{3(2-3v)} = \ln|x| + C$$

Applying $y(1)=0 \Rightarrow v=0$ at $x=1$:

$$\frac{1}{3(2)} = 0+C \Rightarrow C=\frac16$$

$$\frac{1}{3(2-3v)} = \ln x + \frac16$$

Substituting back $v=y/x$:

$$\frac{1}{3\left(2-\dfrac{3y}{x}\right)} = \ln x + \frac16$$

$$\frac{x}{3(2x-3y)} = \ln x + \frac16$$

$$\boxed{x = 3(2x-3y)\left(\ln x+\frac16\right)}$$

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Q11(b). Find $y$ if $y\mathbf{i}+3\mathbf{j}$, $2\mathbf{j}-\mathbf{k}$, and $\mathbf{k}-3\mathbf{i}$ are Linearly Dependent

Three vectors are linearly dependent iff their determinant $=0$:

$$\begin{vmatrix}y & 3 & 0\\ 0 & 2 & -1\\ -3 & 0 & 1\end{vmatrix} = 0$$

Expand along row 1:

$$y\begin{vmatrix}2&-1\\0&1\end{vmatrix} - 3\begin{vmatrix}0&-1\\-3&1\end{vmatrix} + 0 = 0$$

$$y(2-0) - 3(0-3) = 0$$

$$2y + 9 = 0$$

$$\boxed{y=-\frac92}$$