Nanyang Junior College H2 Math papers, with video solutions

The motion of a particle is described by the differential equation \(\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{t}^{2}}}-4\frac{\mathrm{d}x}{\mathrm{d}t}+4x={{\mathrm{e}}^{2t}}\), where \(x\) is the displacement of the particle in metres along the \(x\)- axis with respect to origin \(O\) at time \(t\) seconds. By using the substitution \(x=u{{\mathrm{e}}^{2t}}\), show that this differential equation may be reduced to the form \(\frac{{{\mathrm{d}}^{2}}u}{\mathrm{d}{{t}^{2}}}=a\), where \(a\) is a constant to be determined.

Hence, find the displacement of the particle at time \(t\) if the particle moves off from the initial point \(x=1\) with a velocity of \(1\)m/s.[7]

After some time, the particle is returned to its initial point and set into a new motion along the \(x\)- axis. The motion of the particle is described by the differential equation \(\frac{\mathrm{d}v}{\mathrm{d}t}=2{{\mathrm{e}}^{-v}}\), where \(v\) is the velocity of the particle in metres per second at time \(t\) seconds after it starts on this new motion. If the particle is at rest just before it begins this new motion, find an expression for \(t\) in terms of \(v\).

Show that \(2x=3+{{\mathrm{e}}^{v}}(v-1)\), where \(x\) is the horizontal displacement of the particle in metres.[6]

1. 1. Show that $\tan 2\theta =\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ using the addition formula from the List of Formulae (MF26).

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   2. By using the substitution $x=\tan \theta $, or otherwise, find the exact value of  $\displaystyle\int\nolimits_{0}^{1}{{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)}\,\mathrm{d}x$.

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2. Find the exact value of $\displaystyle\int\nolimits_{-1}^{1}{\frac{2+\left| x \right|}{2+x}\mathrm{d}x}$.

   [4]

The diagram below shows a city map of two towns, $A$ and $B$ separated by a river. A bridge is to be built between the two towns, which are on opposite sides of a straight river of uniform width $r$km, and the two towns are $p$ km apart measured along the riverbank. Town $A$ is $1$km from the riverbank, and Town $B$ is $b$km away from riverbank.

A bridge is to be built perpendicular to the riverbank at a distance of $x$km from Town $B$, measured along the riverbank, allowing traffic to flow between the two towns.

Find the distance $x$, in terms of $b$ and $p$, such that the distance of travel between Town $A$ and Town $B$ can be minimised if $b>1$. (It is not necessary to verify that the distance is minimum.)

A spherical tank with negligible thickness and internal radius \(a\)cm contains water. At time \(t\)s, the water surface is at a height \(x\)cm above the lowest point of the tank and the volume of water in the tank, \(V\)cm\(^{3}\), is given by \(V=\frac{1}{3}\pi {{x}^{2}}\left( 3a-x \right)\). Water flows from the tank, through an outlet at its lowest point, at a rate \(\pi k\sqrt{x}\)cm\(^{3}\) s\(^{-1}\), where \(k\) is a positive constant.

1. Show that \(\left( 2ax-{{x}^{2}} \right)\frac{\mathrm{d}x}{\mathrm{d}t}=-k\sqrt{x}\).[2]
2. Find the general solution for \(t\) in terms of \(x\),\(a\) and \(k\).[3]
3. Find the ratio \({{T}_{1}}:{{T}_{2}}\), where \({{T}_{1}}\) is the time taken to empty the tank when initially it is completely full, and \({{T}_{2}}\) is the time taken to empty the tank when initially it is half full.[4]

Do not use a calculator in answering this question.

1. Explain why the equation ${{z}^{3}}+a{{z}^{2}}+az+7=0$ cannot have more than two non-real roots, where $a$ is a real constant.

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2. Given that $z=-7$ is a root of the equation in (i), find the other roots, leaving your answers in the form $r{{e}^{\mathrm{i}\theta }}$, where $r>0$ and $-\pi <\theta \le \pi $ .

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3. Hence, solve the equation $\mathrm{i}{{z}^{3}}+8{{z}^{2}}-8\mathrm{i}z-7=0$, leaving your answers in the form $r{{e}^{\mathrm{i}\theta }}$, where $r>0$ and $-\pi <\theta \le \pi $ .

   [2]