2019 IJMB mathematics paper 1

SECTION A

1. Differentiate $y = \operatorname{cosec}(e^{\sqrt{x}})$ [04 marks]

2. Evaluate $\displaystyle\lim_{x\to 1}\left(\frac{1-x^3}{1-x}\right)$ [04 marks]

3. Show that the vectors $\mathbf{i}-\mathbf{j}-\mathbf{k}$ and $\mathbf{i}+2\mathbf{j}-\mathbf{k}$ are perpendicular [04 marks]

4. Solve the equation $\dfrac{dy}{dx} = \sin x(1+y)$. [04 marks]

5. Evaluate $\displaystyle\int_0^{\pi/4} \frac{3\sec^2 x}{4+3\tan x}\,dx$. [04 marks]

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SECTION B: CALCULUS

6. (a) Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ at $(1,1)$ if $x^3+y^3+3gx+3fy+c=0$ [10 marks]

(b) If $y = e^{-ax}\cos 2bx$, show that $y'' + 2ay' + (a^2+4b^2)y = 0$. [10 marks]

7. (a) Differentiate from first principles $y = \cos^2 3x$. [10 marks]

(b) Find $\displaystyle\int \frac{dx}{(x-1)(x-2)(x+3)}$. [10 marks]

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8. (a) If $y = \cosh(k\cos^{-1}x)$, show that $(1-x^2)y'' - xy' - k^2y = 0$. [06 marks]

(b) Using integration by parts, evaluate the following:

(i) $\displaystyle\int_1^2 x^2\log x\,dx$

(ii) $\displaystyle\int_0^1 \cos^{-1}x\,dx$ [14 marks]

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SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS

9. (a) Solve the equation $(x^2+y^2)dx - 3xy\,dy = 0$. [12 marks]

(b) The vertices of $\triangle PQR$ are represented by the vectors $\mathbf{i}+3\mathbf{j}+\mathbf{k}$, $3\mathbf{i}-\mathbf{j}+\mathbf{k}$, and $\mathbf{i}+\mathbf{j}-3\mathbf{k}$, respectively. Calculate the cosine of the angles of $\triangle PQR$ and obtain its area. [08 marks]

10. (a) Calculate the unit vector in the direction of $3\mathbf{a}+2\mathbf{b}$, given that $\mathbf{a}=\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k}$. Hence obtain the angle between it and vector $\mathbf{b}$. [10 marks]

(b) Solve the equation $y\dfrac{dy}{dx} = \sec^2 x\, e^{2\tan x - 10y^2}$. [10 marks]

11. (a) Solve the equation $\cos x\dfrac{dy}{dx} - y\sin x = \cos^2 x - 1$. [10 marks]

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

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SOLUTIONS

SECTION A

Q1. Differentiate $y = \operatorname{cosec}(e^{\sqrt{x}})$

Let $u = e^{\sqrt{x}}$, so $y = \operatorname{cosec}(u)$

$\dfrac{dy}{du} = -\operatorname{cosec}(u)\cot(u)$

$\dfrac{du}{dx}$: Let $v = \sqrt{x} = x^{1/2}$

$\dfrac{dv}{dx} = \dfrac{1}{2\sqrt{x}}$

$\dfrac{du}{dx} = e^{\sqrt{x}}\cdot\dfrac{1}{2\sqrt{x}}$

By chain rule:

$$\frac{dy}{dx} = -\operatorname{cosec}(e^{\sqrt{x}})\cot(e^{\sqrt{x}})\cdot\frac{e^{\sqrt{x}}}{2\sqrt{x}}$$

$$\boxed{\frac{dy}{dx} = \frac{-e^{\sqrt{x}}\operatorname{cosec}(e^{\sqrt{x}})\cot(e^{\sqrt{x}})}{2\sqrt{x}}}$$

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Q2. Evaluate $\displaystyle\lim_{x\to1}\left[\frac{1-x^3}{1-x}\right]$

Factor numerator:

$$1-x^3 = (1-x)(1+x+x^2)$$

Therefore:

$$\frac{1-x^3}{1-x} = \frac{(1-x)(1+x+x^2)}{1-x} = 1+x+x^2$$

Taking the limit:

$$\lim_{x\to1}(1+x+x^2) = 1+1+1$$

$$\boxed{=3}$$

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Q3. Show that $\mathbf{i}-\mathbf{j}-\mathbf{k}$ and $\mathbf{i}+2\mathbf{j}-\mathbf{k}$ are Perpendicular

Two vectors are perpendicular if their dot product = 0.

Let $\mathbf{a} = \mathbf{i}-\mathbf{j}-\mathbf{k} = (1,-1,-1)$

Let $\mathbf{b} = \mathbf{i}+2\mathbf{j}-\mathbf{k} = (1,2,-1)$

$$\mathbf{a}\cdot\mathbf{b} = (1)(1)+(-1)(2)+(-1)(-1) = 1-2+1 = \mathbf{0}\ \checkmark$$

Since $\mathbf{a}\cdot\mathbf{b}=0$, the vectors are perpendicular. $\blacksquare$

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Q4. Solve $\dfrac{dy}{dx} = \sin x(1+y)$

Separating variables:

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

Integrating both sides:

$$\int\frac{dy}{1+y} = \int\sin x\,dx$$

$$\ln|1+y| = -\cos x + C$$

$$|1+y| = e^{-\cos x + C} = Ae^{-\cos x}$$

$$\boxed{1+y = Ae^{-\cos x}}$$

or $y = Ae^{-\cos x} - 1$

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Q5. Evaluate $\displaystyle\int_0^{\pi/4}\left[\frac{3\sec^2 x}{4+3\tan x}\right]dx$

Substitution: Let $u = 4+3\tan x$

$\dfrac{du}{dx} = 3\sec^2 x \Rightarrow du = 3\sec^2 x\,dx$

Limits:

• $x=0$: $u = 4+3(0) = 4$
• $x=\pi/4$: $u = 4+3(1) = 7$

Integral becomes:

$$\int_4^7\frac{du}{u} = \big[\ln|u|\big]_4^7 = \ln7-\ln4$$

$$\boxed{=\ln\frac{7}{4}}$$

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SECTION B: CALCULUS

Q6(a). Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ at $(1,1)$ for $x^3+y^3+3gx+3fy+c=0$

Implicit differentiation:

$$3x^2 + 3y^2\frac{dy}{dx} + 3g + 3f\frac{dy}{dx} = 0$$

$$\frac{dy}{dx}(3y^2+3f) = -3x^2-3g$$

$$\frac{dy}{dx} = \frac{-(x^2+g)}{y^2+f}$$

At $(1,1)$:

$$\frac{dy}{dx}\bigg|_{(1,1)} = \frac{-(1+g)}{1+f}$$

Second derivative — differentiating $\dfrac{dy}{dx}$ implicitly:

Let $p = \dfrac{dy}{dx} = \dfrac{-(x^2+g)}{y^2+f}$

Using the quotient rule:

$$\frac{d^2y}{dx^2} = \frac{-2x(y^2+f) - (-(x^2+g))\left(2y\frac{dy}{dx}\right)}{(y^2+f)^2} = \frac{-2x(y^2+f) + 2y(x^2+g)\frac{dy}{dx}}{(y^2+f)^2}$$

Substituting $\dfrac{dy}{dx} = \dfrac{-(x^2+g)}{y^2+f}$ at $(1,1)$:

$$\frac{d^2y}{dx^2}\bigg|_{(1,1)} = \frac{-2(1+f) + 2\cdot\frac{-(1+g)^2}{(1+f)}}{(1+f)^2} = \frac{-2(1+f)^2 - 2(1+g)^2}{(1+f)^3}$$

$$\boxed{\frac{d^2y}{dx^2}\bigg|_{(1,1)} = \frac{-2\left[(1+f)^2+(1+g)^2\right]}{(1+f)^3}}$$

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Q6(b). Show that $y''+2ay'+(a^2+4b^2)y=0$ for $y=e^{-ax}\cos 2bx$

First derivative:

$$y' = -ae^{-ax}\cos2bx - 2be^{-ax}\sin2bx = e^{-ax}(-a\cos2bx - 2b\sin2bx)$$

Second derivative:

$$y'' = -ae^{-ax}(-a\cos2bx-2b\sin2bx) + e^{-ax}(2ab\sin2bx - 4b^2\cos2bx)$$

$$y'' = e^{-ax}\big[a^2\cos2bx + 2ab\sin2bx + 2ab\sin2bx - 4b^2\cos2bx\big]$$

$$y'' = e^{-ax}\big[(a^2-4b^2)\cos2bx + 4ab\sin2bx\big]$$

Now compute $y''+2ay'+(a^2+4b^2)y$:

$$y''+2ay' = e^{-ax}\big[(a^2-4b^2)\cos2bx + 4ab\sin2bx\big] + 2a\cdot e^{-ax}\big[-a\cos2bx - 2b\sin2bx\big]$$

$$= e^{-ax}\big[(a^2-4b^2-2a^2)\cos2bx\big] = e^{-ax}\big[-(a^2+4b^2)\cos2bx\big]$$

Adding $(a^2+4b^2)y$:

$$= e^{-ax}\big[-(a^2+4b^2)\cos2bx\big] + (a^2+4b^2)e^{-ax}\cos2bx = e^{-ax}\cos2bx\big[-(a^2+4b^2)+(a^2+4b^2)\big]$$

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

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Q7(a). Differentiate $y=\cos^2 3x$ from First Principles

Definition: $\dfrac{dy}{dx} = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$f(x) = \cos^2 3x$, $\quad f(x+h) = \cos^2(3x+3h)$

$f(x+h)-f(x)$:

$$\cos^2(3x+3h) - \cos^2 3x$$

Using the identity $\cos^2 A - \cos^2 B = -\sin(A+B)\sin(A-B)$:

$$= -\sin(6x+3h)\sin(3h)$$

Therefore:

$$\frac{dy}{dx} = \lim_{h\to0}\frac{-\sin(6x+3h)\sin(3h)}{h} = \lim_{h\to0}-\sin(6x+3h)\cdot3\cdot\frac{\sin(3h)}{3h}$$

As $h\to0$: $\dfrac{\sin(3h)}{3h}\to1$ and $\sin(6x+3h)\to\sin(6x)$

$$\boxed{\frac{dy}{dx} = -3\sin6x}$$

(Note: this equals $-2\cdot3\sin3x\cos3x = -3\sin6x$ using the double-angle identity) ✓

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Q7(b). Find $\displaystyle\int\frac{dx}{(x-1)(x-2)(x+3)}$

Partial fractions:

$$\frac{1}{(x-1)(x-2)(x+3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+3}$$

$$1 = A(x-2)(x+3) + B(x-1)(x+3) + C(x-1)(x-2)$$

$x=1$: $1 = A(-1)(4) \Rightarrow A = -\dfrac14$

$x=2$: $1 = B(1)(5) \Rightarrow B = \dfrac15$

$x=-3$: $1 = C(-4)(-5) \Rightarrow C = \dfrac{1}{20}$

Therefore:

$$\int\frac{dx}{(x-1)(x-2)(x+3)} = \int\left[-\frac14\cdot\frac{1}{x-1}+\frac15\cdot\frac{1}{x-2}+\frac{1}{20}\cdot\frac{1}{x+3}\right]dx$$

$$\boxed{= -\frac14\ln|x-1| + \frac15\ln|x-2| + \frac{1}{20}\ln|x+3| + C}$$

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Q8(a). Show that $(1-x^2)y''-xy'-k^2y=0$ for $y=\cosh(k\cos^{-1}x)$

Let $u = k\cos^{-1}x$

$$y' = \sinh(k\cos^{-1}x)\cdot k\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{k\sinh u}{\sqrt{1-x^2}}$$

Rearranging: $\sqrt{1-x^2}\cdot y' = -k\sinh u$

Squaring: $(1-x^2)(y')^2 = k^2\sinh^2 u = k^2(\cosh^2 u - 1) = k^2(y^2-1)$

Differentiating both sides w.r.t. $x$:

$$-2x(y')^2 + (1-x^2)\cdot2y'y'' = 2k^2yy'$$

Dividing by $2y'$ ($y'\neq0$):

$$-xy' + (1-x^2)y'' = k^2y$$

$$\boxed{(1-x^2)y'' - xy' - k^2y = 0}\ \checkmark\ \blacksquare$$

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Q8(b)(i). $\displaystyle\int_1^2 x^2\log x\,dx$ (Integration by Parts)

Let $u=\log x \Rightarrow du = \dfrac1x dx$

Let $dv = x^2 dx \Rightarrow v = \dfrac{x^3}{3}$

$$\int x^2\log x\,dx = \frac{x^3}{3}\log x - \int\frac{x^3}{3}\cdot\frac1x\,dx = \frac{x^3}{3}\log x - \frac13\int x^2\,dx$$

$$= \frac{x^3}{3}\log x - \frac{x^3}{9} + C$$

Evaluating from 1 to 2:

$$\left[\frac{x^3}{3}\log x - \frac{x^3}{9}\right]_1^2$$

At $x=2$: $\dfrac83\log2 - \dfrac89$

At $x=1$: $\dfrac13\log1 - \dfrac19 = 0 - \dfrac19 = -\dfrac19$

$$= \frac{8\log2}{3} - \frac89 + \frac19$$

$$\boxed{= \frac{8\log2}{3} - \frac79}$$

(Using log base 10; if natural log: $= \dfrac{8\ln2}{3} - \dfrac79$)

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Q8(b)(ii). $\displaystyle\int_0^1\cos^{-1}x\,dx$ (Integration by Parts)

Let $u=\cos^{-1}x \Rightarrow du = -\dfrac{1}{\sqrt{1-x^2}}dx$

Let $dv = dx \Rightarrow v = x$

$$\int\cos^{-1}x\,dx = x\cos^{-1}x - \int x\cdot\left(\frac{-1}{\sqrt{1-x^2}}\right)dx = x\cos^{-1}x + \int\frac{x}{\sqrt{1-x^2}}dx$$

For the remaining integral: let $w = 1-x^2$, $dw=-2x\,dx$

$$\int\frac{x}{\sqrt{1-x^2}}dx = -\sqrt{1-x^2}$$

Therefore:

$$\int\cos^{-1}x\,dx = x\cos^{-1}x - \sqrt{1-x^2} + C$$

Evaluating from 0 to 1:

At $x=1$: $1\cdot\cos^{-1}(1) - \sqrt0 = 1\cdot0 - 0 = 0$

At $x=0$: $0\cdot\cos^{-1}(0) - \sqrt1 = 0-1 = -1$

$$\boxed{=0-(-1)=1}$$

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SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS

Q9(a). Solve $(x^2+y^2)dx - 3xy\,dy=0$

Rearranging:

$$\frac{dy}{dx} = \frac{x^2+y^2}{3xy}$$

This is homogeneous. Let $y=vx \Rightarrow \dfrac{dy}{dx} = v+x\dfrac{dv}{dx}$

$$\frac{dy}{dx} = \frac{x^2+v^2x^2}{3x\cdot vx} = \frac{1+v^2}{3v}$$

Substituting:

$$v + x\frac{dv}{dx} = \frac{1+v^2}{3v}$$

$$x\frac{dv}{dx} = \frac{1+v^2}{3v} - v = \frac{1+v^2-3v^2}{3v} = \frac{1-2v^2}{3v}$$

Separating variables:

$$\frac{3v\,dv}{1-2v^2} = \frac{dx}{x}$$

Integrating: Left side: let $w=1-2v^2$, $dw=-4v\,dv$

$$\int\frac{3v}{1-2v^2}dv = -\frac34\int\frac{dw}{w} = -\frac34\ln|1-2v^2|$$

$$\int\frac{dx}{x} = \ln|x|$$

Therefore:

$$-\frac34\ln|1-2v^2| = \ln|x| + C$$

$$\ln|1-2v^2|^{-3/4} = \ln|x| + C$$

Substituting back $v=y/x$:

$$\left(1-\frac{2y^2}{x^2}\right)^{-3/4} = Ax$$

$$\boxed{\left(\frac{x^2-2y^2}{x^2}\right)^{3/4} = \frac{K}{x}}$$

or equivalently: $(x^2-2y^2)^3 = Kx^2$ (after simplification)

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Q9(b). Triangle PQR — Cosines of Angles and Area

Position vectors:

• $P = \mathbf{i}+3\mathbf{j}+\mathbf{k} = (1,3,1)$
• $Q = 3\mathbf{i}-\mathbf{j}+\mathbf{k} = (3,-1,1)$
• $R = \mathbf{i}+\mathbf{j}-3\mathbf{k} = (1,1,-3)$

Side vectors:

$$PQ = Q-P = (2,-4,0),\quad |PQ| = \sqrt{4+16+0} = \sqrt{20} = 2\sqrt5$$

$$QR = R-Q = (-2,2,-4),\quad |QR| = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt6$$

$$PR = R-P = (0,-2,-4),\quad |PR| = \sqrt{0+4+16} = \sqrt{20} = 2\sqrt5$$

Cosine of angle at $P$ (between $PQ$ and $PR$):

$$\cos P = \frac{PQ\cdot PR}{|PQ||PR|}$$

$$PQ\cdot PR = (2)(0)+(-4)(-2)+(0)(-4) = 0+8+0 = 8$$

$$\cos P = \frac{8}{2\sqrt5\cdot2\sqrt5} = \frac{8}{20} = \boxed{\frac25}$$

Cosine of angle at $Q$ (between $QP$ and $QR$):

$QP = -PQ = (-2,4,0)$, $\quad QR = (-2,2,-4)$

$$QP\cdot QR = (-2)(-2)+(4)(2)+(0)(-4) = 4+8+0 = 12$$

$$\cos Q = \frac{12}{2\sqrt5\cdot2\sqrt6} = \frac{12}{4\sqrt{30}} = \boxed{\frac{3}{\sqrt{30}}}$$

Cosine of angle at $R$ (between $RP$ and $RQ$):

$RP = -PR = (0,2,4)$, $\quad RQ = -QR = (2,-2,4)$

$$RP\cdot RQ = (0)(2)+(2)(-2)+(4)(4) = 0-4+16 = 12$$

$$\cos R = \frac{12}{2\sqrt5\cdot2\sqrt6} = \frac{12}{4\sqrt{30}} = \boxed{\frac{3}{\sqrt{30}}}$$

Area of $\triangle PQR$:

Using the cross product: Area $= \dfrac12|PQ\times PR|$

$$PQ\times PR = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -4 & 0\\ 0 & -2 & -4\end{vmatrix}$$

$$= \mathbf{i}[(-4)(-4)-(0)(-2)] - \mathbf{j}[(2)(-4)-(0)(0)] + \mathbf{k}[(2)(-2)-(-4)(0)]$$

$$= \mathbf{i}(16) - \mathbf{j}(-8) + \mathbf{k}(-4) = (16,8,-4)$$

$$|PQ\times PR| = \sqrt{256+64+16} = \sqrt{336} = 4\sqrt{21}$$

$$\boxed{\text{Area} = \frac12\times4\sqrt{21} = 2\sqrt{21}\ \text{sq. units}}$$

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Q10(a). Unit Vector in Direction of $3\mathbf{a}+2\mathbf{b}$ and Angle with $\mathbf{b}$

Given: $\mathbf{a}=(1,-2,1)$, $\mathbf{b}=(1,2,-1)$

$3\mathbf{a}+2\mathbf{b}$:

$$3\mathbf{a} = (3,-6,3),\quad 2\mathbf{b} = (2,4,-2)$$

$$3\mathbf{a}+2\mathbf{b} = (5,-2,1)$$

$$|3\mathbf{a}+2\mathbf{b}| = \sqrt{25+4+1} = \sqrt{30}$$

Unit vector:

$$\boxed{\hat{u} = \frac{1}{\sqrt{30}}(5\mathbf{i}-2\mathbf{j}+\mathbf{k})}$$

Angle between $\hat u$ and $\mathbf{b}$:

$$|\mathbf{b}| = \sqrt{1+4+1} = \sqrt6$$

$$\hat{u}\cdot\mathbf{b} = \frac{5}{\sqrt{30}}(1) + \frac{-2}{\sqrt{30}}(2) + \frac{1}{\sqrt{30}}(-1) = \frac{5-4-1}{\sqrt{30}} = \frac{0}{\sqrt{30}} = 0$$

$$\boxed{\theta = 90°}$$

The unit vector $3\mathbf{a}+2\mathbf{b}$ is perpendicular to $\mathbf{b}$.

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Q10(b). Solve $y\dfrac{dy}{dx} = \sec^2x\cdot e^{2\tan x-10y^2}$

Rewriting:

$$y\frac{dy}{dx} = \sec^2x\cdot e^{2\tan x}\cdot e^{-10y^2}$$

Separating variables:

$$y\cdot e^{10y^2}\,dy = \sec^2x\cdot e^{2\tan x}\,dx$$

Integrating left side: Let $u=10y^2$, $du=20y\,dy \Rightarrow y\,dy = \dfrac{du}{20}$

$$\int y\,e^{10y^2}dy = \frac{1}{20}\int e^u\,du = \frac{e^{10y^2}}{20}$$

Integrating right side: Let $v=2\tan x$, $dv=2\sec^2x\,dx \Rightarrow \sec^2x\,dx = \dfrac{dv}{2}$

$$\int\sec^2x\cdot e^{2\tan x}dx = \frac12\int e^v\,dv = \frac{e^{2\tan x}}{2}$$

Therefore:

$$\frac{e^{10y^2}}{20} = \frac{e^{2\tan x}}{2} + C$$

$$\boxed{e^{10y^2} = 10e^{2\tan x} + K}$$

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Q11(a). Solve $\cos x\dfrac{dy}{dx} - y\sin x = \cos^2x-1$

Rewriting in standard form:

$$\frac{dy}{dx} - y\tan x = \cos x - \sec x$$

This is a first-order linear ODE: $\dfrac{dy}{dx}+P(x)y=Q(x)$

where $P(x)=-\tan x$, $Q(x)=\cos x-\sec x$

Integrating factor:

$$\mu = e^{\int-\tan x\,dx} = e^{\ln|\cos x|} = \cos x$$

Multiplying through by $\cos x$:

$$\frac{d}{dx}[y\cos x] = \cos x(\cos x-\sec x) = \cos^2x - 1 = -\sin^2x$$

Integrating:

$$y\cos x = \int -\sin^2x\,dx = \int -\frac{1-\cos2x}{2}\,dx = -\frac{x}{2} + \frac{\sin2x}{4} + C$$

$$\boxed{y\cos x = -\frac{x}{2} + \frac{\sin2x}{4} + C}$$

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Q11(b). Projection of $3\mathbf{a}+\mathbf{b}$ on $\mathbf{a}-3\mathbf{b}$

Given: $\mathbf{a}=(1,-3,1)$, $\mathbf{b}=(2,-1,1)$

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

$$3\mathbf{a} = (3,-9,3),\quad \mathbf{b}=(2,-1,1)$$

$$3\mathbf{a}+\mathbf{b} = (5,-10,4)$$

Compute $\mathbf{a}-3\mathbf{b}$:

$$\mathbf{a} = (1,-3,1),\quad 3\mathbf{b}=(6,-3,3)$$

$$\mathbf{a}-3\mathbf{b} = (-5,0,-2)$$

$$|\mathbf{a}-3\mathbf{b}| = \sqrt{25+0+4} = \sqrt{29}$$

Projection formula:

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

$(3\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}-3\mathbf{b})$:

$$= (5)(-5)+(-10)(0)+(4)(-2) = -25+0-8 = -33$$

$$\boxed{\text{Projection} = \frac{-33}{\sqrt{29}} = \frac{-33\sqrt{29}}{29}}$$