2020 IJMB mathematics paper 1

SECTION A

1. Differentiate $y = \sec\left(e^{x^{1/3}}\right)$ [04 marks]

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

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

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

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

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SECTION B: CALCULUS

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

(b) If $y = e^{-2px}\sin 3qx$, show that $y'' + 4py' + (4p^2+9q^2)y = 0$. [10 marks]

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

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

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

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

(i) $\displaystyle\int_1^3 x^2\ln x\,dx$

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

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SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS

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

(b) The vertices of $\triangle PQR$ are represented by the vectors $2\mathbf{i}+\mathbf{j}+3\mathbf{k}$, $\mathbf{i}-2\mathbf{j}+\mathbf{k}$, and $3\mathbf{i}+\mathbf{j}-\mathbf{k}$, 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 $2\mathbf{a}-3\mathbf{b}$, given that $\mathbf{a}=2\mathbf{i}+\mathbf{j}-\mathbf{k}$ and $\mathbf{b}=\mathbf{i}-\mathbf{j}+2\mathbf{k}$. Hence obtain the angle between it and vector $\mathbf{a}$. [10 marks]

(b) Solve the equation $y\dfrac{dy}{dx} = \sec^2 x\, e^{3\tan x - 8y^2}$. [10 marks]

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

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

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SOLUTIONS

SECTION A

Q1. Differentiate $y = \sec\left(e^{x^{1/3}}\right)$

Let $u = e^{x^{1/3}}$, so $y=\sec(u)$

$$\frac{dy}{du} = \sec(u)\tan(u)$$

$\dfrac{du}{dx}$: Let $v = x^{1/3}$, $\dfrac{dv}{dx} = \dfrac13 x^{-2/3}$

$$\frac{du}{dx} = e^{x^{1/3}}\cdot\frac{1}{3x^{2/3}}$$

By the chain rule:

$$\boxed{\frac{dy}{dx} = \frac{e^{x^{1/3}}\sec\!\left(e^{x^{1/3}}\right)\tan\!\left(e^{x^{1/3}}\right)}{3x^{2/3}}}$$

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

Factor numerator:

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

Therefore:

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

Taking the limit:

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

$$\boxed{=3}$$

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

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

Let $\mathbf{b}=\mathbf{i}-2\mathbf{j}+0\mathbf{k}=(1,-2,0)$

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

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

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

Separating variables:

$$\frac{dy}{1-y} = \cos x\,dx$$

Integrating both sides:

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

$$\ln|1-y| = -\sin x - C$$

$$1-y = Ae^{-\sin x}$$

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

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

Substitution: Let $u = 3+2\tan x$, $du = 2\sec^2x\,dx$

Limits:

• $x=0$: $u=3$
• $x=\pi/3$: $\tan(\pi/3)=\sqrt3$, $u=3+2\sqrt3$

$$\int_3^{3+2\sqrt3}\frac{du}{u} = \ln(3+2\sqrt3)-\ln3$$

$$\boxed{=\ln\!\left(\frac{3+2\sqrt3}{3}\right)}$$

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SECTION B: CALCULUS

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

Implicit differentiation:

$$3x^2+3y^2\frac{dy}{dx}+3p+3q\frac{dy}{dx}=0$$

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

At $(2,1)$:

$$\frac{dy}{dx}\bigg|_{(2,1)} = \frac{-(4+p)}{1+q}$$

Second derivative — differentiating implicitly:

$$\frac{d^2y}{dx^2} = \frac{-2x(y^2+q) + 2y(x^2+p)\frac{dy}{dx}}{(y^2+q)^2}$$

Substituting at $(2,1)$ with $\dfrac{dy}{dx}=\dfrac{-(4+p)}{1+q}$:

$$\frac{d^2y}{dx^2}\bigg|_{(2,1)} = \frac{-2(1+q) + 2\cdot\dfrac{-(4+p)^2}{1+q}}{(1+q)^2} = \frac{-2(1+q)^2-2(4+p)^2}{(1+q)^3}$$

$$\boxed{\frac{d^2y}{dx^2}\bigg|_{(2,1)} = \frac{-2\left[(1+q)^2+(4+p)^2\right]}{(1+q)^3}}$$

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Q6(b). Show that $y''+4py'+(4p^2+9q^2)y=0$ for $y=e^{-2px}\sin3qx$

First derivative:

$$y' = -2pe^{-2px}\sin3qx + 3qe^{-2px}\cos3qx = e^{-2px}(-2p\sin3qx+3q\cos3qx)$$

Second derivative:

$$y'' = -2pe^{-2px}(-2p\sin3qx+3q\cos3qx) + e^{-2px}(-6pq\cos3qx-9q^2\sin3qx)$$

$$y'' = e^{-2px}\left[(4p^2-9q^2)\sin3qx - 12pq\cos3qx\right]$$

Compute $y''+4py'$:

$$y''+4py' = e^{-2px}\left[(4p^2-9q^2)\sin3qx-12pq\cos3qx\right] + 4pe^{-2px}(-2p\sin3qx+3q\cos3qx)$$

$$= e^{-2px}\left[(4p^2-9q^2-8p^2)\sin3qx + (-12pq+12pq)\cos3qx\right] = e^{-2px}\left[-(4p^2+9q^2)\sin3qx\right]$$

Adding $(4p^2+9q^2)y$:

$$= e^{-2px}\sin3qx\left[-(4p^2+9q^2)+(4p^2+9q^2)\right]$$

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

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

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

$f(x)=\sin^2 2x$, $f(x+h)=\sin^2(2x+2h)$

Using $\sin^2A-\sin^2B = \sin(A+B)\sin(A-B)$ with $A=2x+2h$, $B=2x$:

$$f(x+h)-f(x) = \sin(4x+2h)\sin(2h)$$

Therefore:

$$\frac{dy}{dx} = \lim_{h\to0}\frac{\sin(4x+2h)\sin(2h)}{h} = \lim_{h\to0}\sin(4x+2h)\cdot2\cdot\frac{\sin(2h)}{2h}$$

As $h\to0$: $\dfrac{\sin2h}{2h}\to1$, $\sin(4x+2h)\to\sin4x$

$$\boxed{\frac{dy}{dx} = 2\sin4x}$$

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

Partial fractions:

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

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

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

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

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

Therefore:

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

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Q8(a). Show that $(1-x^2)y''-xy'-m^2y=0$ for $y=\sinh(m\sin^{-1}x)$

Let $u=m\sin^{-1}x$

$$y' = \cosh(m\sin^{-1}x)\cdot\frac{m}{\sqrt{1-x^2}} = \frac{m\cosh u}{\sqrt{1-x^2}}$$

Rearranging: $\sqrt{1-x^2}\,y' = m\cosh u$

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

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

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

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

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

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

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Q8(b)(i). $\displaystyle\int_1^3 x^2\ln x\,dx$ (Integration by Parts)

Let $u=\ln x \Rightarrow du=\dfrac1x dx$; let $dv=x^2dx \Rightarrow v=\dfrac{x^3}{3}$

$$\int x^2\ln x\,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C$$

Evaluating from 1 to 3:

At $x=3$: $\dfrac{27}{3}\ln3 - \dfrac{27}{9} = 9\ln3 - 3$

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

$$= 9\ln3-3-\left(-\frac19\right)$$

$$\boxed{= 9\ln3 - \frac{26}{9}}$$

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

Let $u=\sin^{-1}x \Rightarrow du=\dfrac{1}{\sqrt{1-x^2}}dx$; let $dv=dx \Rightarrow v=x$

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

Evaluating from 0 to 1:

At $x=1$: $1\cdot\dfrac{\pi}{2}+0 = \dfrac{\pi}{2}$

At $x=0$: $0 + 1 = 1$

$$\boxed{=\frac{\pi}{2}-1}$$

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SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS

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

Rearranging:

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

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

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

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

Separating variables:

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

Integrating:

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

$$\ln\left[x(1+v^2)\right] = C \Rightarrow x(1+v^2) = K$$

Substituting back $v=y/x$:

$$x\left(1+\frac{y^2}{x^2}\right) = K \Rightarrow x + \frac{y^2}{x} = K$$

$$\boxed{x^2+y^2 = Kx}$$

(a family of circles)

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

Position vectors:

• $P = 2\mathbf{i}+\mathbf{j}+3\mathbf{k} = (2,1,3)$
• $Q = \mathbf{i}-2\mathbf{j}+\mathbf{k} = (1,-2,1)$
• $R = 3\mathbf{i}+\mathbf{j}-\mathbf{k} = (3,1,-1)$

Side vectors:

$$PQ = Q-P = (-1,-3,-2),\quad |PQ| = \sqrt{1+9+4} = \sqrt{14}$$

$$QR = R-Q = (2,3,-2),\quad |QR| = \sqrt{4+9+4} = \sqrt{17}$$

$$PR = R-P = (1,0,-4),\quad |PR| = \sqrt{1+0+16} = \sqrt{17}$$

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

$$PQ\cdot PR = (-1)(1)+(-3)(0)+(-2)(-4) = -1+0+8 = 7$$

$$\boxed{\cos P = \frac{7}{\sqrt{14}\cdot\sqrt{17}} = \frac{7}{\sqrt{238}}}$$

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

$QP=(1,3,2)$, $QR=(2,3,-2)$

$$QP\cdot QR = 2+9-4 = 7$$

$$\boxed{\cos Q = \frac{7}{\sqrt{14}\cdot\sqrt{17}} = \frac{7}{\sqrt{238}}}$$

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

$RP=(-1,0,4)$, $RQ=(-2,-3,2)$

$$RP\cdot RQ = 2+0+8 = 10$$

$$\boxed{\cos R = \frac{10}{\sqrt{17}\cdot\sqrt{17}} = \frac{10}{17}}$$

Area of $\triangle PQR$:

$$PQ\times PR = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&-3&-2\\1&0&-4\end{vmatrix}$$

$$= \mathbf{i}[(-3)(-4)-(-2)(0)] - \mathbf{j}[(-1)(-4)-(-2)(1)] + \mathbf{k}[(-1)(0)-(-3)(1)]$$

$$= \mathbf{i}(12) - \mathbf{j}(6) + \mathbf{k}(3) = (12,-6,3)$$

$$|PQ\times PR| = \sqrt{144+36+9} = \sqrt{189} = 3\sqrt{21}$$

$$\boxed{\text{Area} = \frac12\times3\sqrt{21} = \frac{3\sqrt{21}}{2}\ \text{sq. units}}$$

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

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

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

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

$$2\mathbf{a}-3\mathbf{b} = (1,5,-8)$$

$$|2\mathbf{a}-3\mathbf{b}| = \sqrt{1+25+64} = \sqrt{90} = 3\sqrt{10}$$

Unit vector:

$$\boxed{\hat{u} = \frac{1}{3\sqrt{10}}(\mathbf{i}+5\mathbf{j}-8\mathbf{k})}$$

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

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

$$\hat{u}\cdot\mathbf{a} = \frac{(1)(2)+(5)(1)+(-8)(-1)}{3\sqrt{10}} = \frac{2+5+8}{3\sqrt{10}} = \frac{15}{3\sqrt{10}} = \frac{5}{\sqrt{10}}$$

$$\cos\theta = \frac{\hat{u}\cdot\mathbf{a}}{|\mathbf{a}|} = \frac{5/\sqrt{10}}{\sqrt6} = \frac{5}{\sqrt{60}} = \frac{5}{2\sqrt{15}} = \frac{\sqrt{15}}{6}$$

$$\boxed{\theta = \cos^{-1}\left(\frac{\sqrt{15}}{6}\right)}$$

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Q10(b). Solve $y\dfrac{dy}{dx} = \sec^2x\cdot e^{3\tan x - 8y^2}$

Rewriting:

$$y\frac{dy}{dx} = \sec^2x\cdot e^{3\tan x}\cdot e^{-8y^2}$$

Separating variables:

$$y\,e^{8y^2}\,dy = \sec^2x\cdot e^{3\tan x}\,dx$$

Integrating left side: let $u=8y^2$, $du=16y\,dy$

$$\int y\,e^{8y^2}dy = \frac{e^{8y^2}}{16}$$

Integrating right side: let $v=3\tan x$, $dv=3\sec^2x\,dx$

$$\int\sec^2x\cdot e^{3\tan x}dx = \frac{e^{3\tan x}}{3}$$

Therefore:

$$\frac{e^{8y^2}}{16} = \frac{e^{3\tan x}}{3} + C$$

$$\boxed{e^{8y^2} = \frac{16}{3}e^{3\tan x} + K}$$

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

Observe the left side:

$$\frac{d}{dx}(y\sin x) = y'\sin x + y\cos x$$

which is exactly the left-hand side of the equation. So:

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

Integrating both sides:

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

$$\boxed{y = \frac{C - \tfrac12\cos2x}{\sin x}}$$

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Q11(b). Projection of $2\mathbf{a}-\mathbf{b}$ on $\mathbf{a}+2\mathbf{b}$

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

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

$$2\mathbf{a}=(4,-2,6),\quad \mathbf{b}=(1,2,-1)$$

$$2\mathbf{a}-\mathbf{b} = (3,-4,7)$$

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

$$\mathbf{a}=(2,-1,3),\quad 2\mathbf{b}=(2,4,-2)$$

$$\mathbf{a}+2\mathbf{b} = (4,3,1)$$

$$|\mathbf{a}+2\mathbf{b}| = \sqrt{16+9+1} = \sqrt{26}$$

$(2\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}+2\mathbf{b})$:

$$= (3)(4)+(-4)(3)+(7)(1) = 12-12+7 = 7$$

$$\boxed{\text{Projection} = \frac{7}{\sqrt{26}} = \frac{7\sqrt{26}}{26}}$$