299 个页面

## Q1编辑

$\frac{d}{dx}(sin2x)$
$=\lim_{\Delta x\to 0} \frac{2sin\frac{[2(x+\Delta x)-2x]}{2}cos\frac{[2(x+\Delta x)+2x]}{2}}{\Delta x}$
$=\lim_{\Delta x\to 0} 2\frac{sin\Delta x}{\Delta x}cos(2x+2\Delta x)$
$=2\cdot(1)\cdot cos[2x+2\cdot(0)]$
$=2cos2x$

## Q2编辑

$(1+ax)^n$
$=1+nax+\frac{n(n-1)}{2}a^2x^2+...$

The coefficients of $x$ and $x^2$ are 20 and 180 respectively. So we have

$\begin{cases} na=-20 & \\ \frac{n(n-1)}{2}a^2=180 \end{cases}$

On solving, we have

$\begin{cases} a=-2 \\ n=10 \end{cases}$

## Q3编辑

Let $P(n)$ be the proposition

"$1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{4n+1}{3n+1}$

for all positive integers $n$."

For $n=1$,

L.S.$=1+\frac{1}{1\times4}=\frac{5}{4}$

R.S.$=\frac{4\cdot(1)+1}{3\cdot(1)+1}=\frac{5}{4}$

L.S.=R.S. $\Rightarrow P(1)$ is true.

Next, suppose $P(n)$ is true for some positive integers $k$,

i.e. "$1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{4k+1}{3k+1}$"

Then for $n=k+1$,

L.S.$=1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{[3(k+1)-2][3(k+1)+1]}$

$=\frac{4k+1}{3k+1}+\frac{1}{(3k+1)(3k+4)}$
$=\frac{4k+1}{3k+1}+\frac{1}{(3k+1)(3k+4)}$
$=\frac{(4k+1)(3k+4)+1}{(3k+1)(3k+4)}$
$=\frac{12k^2+19k+5}{(3k+1)(3k+4)}$
$=\frac{(3k+1)(4k+5)}{(3k+1)(3k+4)}$
$=\frac{4k+5}{3k+4}$

R.S.$=\frac{4(k+1)+1}{3(k+1)+1}$

$=\frac{4k+5}{3k+4}$

L.S.=R.S.$\Rightarrow P(k+1)$ is true.

By the principle of mathematical induction, $P(n)$ is true for all positive integers $n$.

## Q4编辑

a)

$\frac{dy}{dx}=e^x-1$
$\int \frac{dy}{dx} dx=\int e^x-1 dx$
$y=e^x-x+C$

At $(1,e)$,

$(e)=e^{(1)}-(1)+C \Rightarrow C=1$

Therefore $y=e^x-x+1$ is the required equation.

b) At $x=0$,

$y=e^{(0)}-(0)+1=2$
$\frac{dy}{dx}|_{x=0}=e^{(0)}-1=0$

So the slope of the required tangent has a slope of 0 and an y-intercept of 2. Therefore $y=2$ is the required tangent.

## Q5编辑

a)

Since the sign of $f'(x)$ changes from $+$ to $-$ as $x$ passes through 0 in the positive direction, a maximum point occurs at $x=0$.

Since the sign of $f''(x)$ changes as $x$ passes through 1 and -1, inflexion points occur at $x=1$ and $x=-1$.

At $x=0$,

$f(0)=\frac{3-3\cdot(0)^2}{3+(0)^2}=1$

At $x=1$,

$f(1)=\frac{3-3\cdot(1)^2}{3+(1)^2}=0$

At $x=-1$,

$f(-1)=\frac{3-3\cdot(-1)^2}{3+(-1)^2}=0$

So a maximum point occurs at $(0,1)$ and two inflexion points occur at $(1,0)$ and $(-1,0)$.

b)

Since $x^2\geq 0$ for any real number $x$, $x^2+3\geq 3$ for any real number $x$. $f(x)$ has no vertical asymptote.

Since $f(x)=\frac{3-3x^2}{3+3x^2}=\frac{-3(x^2+3)+12}{3+x^2}=-3+\frac{12}{3+x^2}$,

$\lim_{x\to \pm\infty} f(x)=\lim_{x\to \pm\infty} -3+\frac{12}{3+x^2}=-3+(0)=-3$.

So $y=-3$ is a horizontal asymptote of $f(x)$.

No oblique asymptote.

c)

To be drawn

## Q6编辑

a)

Required area

$=\int_0^4 -\frac{x^2}{2}+2x+4-4 dx+\int_4^5 4-(-\frac{x^2}{2}+2x+4-4) dx$
$=-\frac{x^3}{6}|_0^4+x^2|_0^4+\frac{x^3}{6}|_4^5-x^2|_4^5$
$=-\frac{64}{6}+16+\frac{125}{6}-\frac{64}{6}-25+16$
$=-\frac{13}{2}$ sq. units

b)

Required volume

$=|\pi \int_0^4 (-\frac{x^2}{2}+2x+4-4)^2 dx+\pi \int_4^5 [4-(-\frac{x^2}{2}+2x+4)]^2 dx|$
$=|\pi \int_0^4 (-\frac{x^2}{2}+2x)^2 dx+\pi \int_4^5 (\frac{x^2}{2}-2x)]^2 dx|$
$=|\pi \int_0^5 (-\frac{x^2}{2}+2x)^2 dx|$
$=|\pi \int_0^5 \frac{x^4}{4}-2x^3+4x^2 dx|$
$=|\pi (\frac{x^5}{20}-\frac{x^4}{2}+\frac{x^3}{4})|_0^5|$
$=\frac{123\pi}{12}$ cu. units

## Q7编辑

a)

$tanx$
$=\frac{sinx}{cosx}$
$=\frac{sinx}{cosx}\times \frac{2cosx}{2cosx}$
$=\frac{2sinxcosx}{2cos^2x}$
$=\frac{sin2x}{1+cos2x}$

b)

$tany$
$=\frac{sin2y}{1+cos2y}$ (By (a))
$=\frac{sin2y}{1+cos2y}\times \frac{sec2y}{sec2y}$
$=\frac{tan2y}{sec2y+1}$
$=\frac{\frac{sin4y}{1+cos4y}}{sec2y+1}$ (By (a))
$=\frac{sin4y}{(1+cos4y)(sec2y+1)}\times \frac{sec4y}{sec4y}$
$=\frac{tan4y}{(sec4y+1)(sec2y+1)}$
$=\frac{\frac{sin8y}{1+cos8y}}{(sec4y+1)(sec2y+1)}$ (By (a))
$=\frac{sin8y}{(1+cos8y)(sec4y+1)(sec2y+1)}\times \frac{cos2ycos4y}{cos2ycos4y}$
$=\frac{sin8ycos2ycos4y}{(1+cos8y)(1+cos4y)(1+cos2y)}$

## Q8编辑

$M=\begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 1 \\ k & 0 & 0 \end{bmatrix}, k \ne 0$

a)

$detM=$
$1 \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} - k \begin{vmatrix} 0 & 1 \\ k & 0 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ k & 0 \end{vmatrix}$
$=k^2$
$M^{-1}$
$=\frac{1}{detM}{\begin{bmatrix} 0 & k & -k \\ 0 & 0 & k^2 \\ k & -1 & 1 \end{bmatrix}}^T$
$=\frac{1}{(k^2)}\begin{bmatrix} 0 & 0 & k \\ k & 0 & -1 \\ -k & k^2 & 1 \end{bmatrix}$

b)

$M\begin{bmatrix} x \\ 1 \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}$
$\begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 1 \\ k & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}$
$\begin{cases} x+k=2 ......(1) & \\ 1+z=2 ......(2) & \\ kx=1 ......(3) \end{cases}$

On solving (1) and (3), we have $k=1$.

## Q9编辑

a)

$\left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 2 & 1-2a & 2-b & a+4 \\ 3 & 1-3a & 3-ab & 4 \end{array}\right]\sim \left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 0 & 1 & -b & a \\ 0 & 1 & -ab & -2 \end{array}\right]$

We require the system to have infinitely solutions, i.e.

$\begin{cases} a=-2 ......(1) & \\ -b=-ab ......(2) & \end{cases}$

From (2), $a=1$ or $b=0$, but since we have already required $a=-2$, we reject the solution $a=1$.

Therefore, for infinitely many solutions we are left with

$\begin{cases} a=-2 & \\ b=0 & \end{cases}$

b)
Since $(E)$ has infinitely many solutions, $a=-2$ and $b=0$.

$\left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 2 & 1-2a & 2-b & a+4 \\ 3 & 1-3a & 3-ab & 4 \end{array}\right]\sim \left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 0 & 1 & -b & a \\ 0 & 1 & -ab & -2 \end{array}\right]$(From (a))
$\sim \left[\begin{array}{rrr|r} 1 & -(-2) & 1 & 2 \\ 0 & 1 & -(0) & (-2) \\ 0 & 1 & -(-2)(0) & -2 \end{array}\right]\sim \left[\begin{array}{rrr|r} 1 & 2 & 1 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 1 & 0 & -2 \end{array}\right]$
$\sim \left[\begin{array}{rrr|r} 1 & 2 & 1 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right]\Rightarrow \begin{cases} z=t & \\ y=-2-t & \\ x=2-t-2(-2-t)=6+t & \end{cases}$ for any real number t.

## Q10编辑

a)

$\overrightarrow{ON}$
$=\frac{k}{k+1}\overrightarrow{OA}+\frac{1}{k+1}\overrightarrow{OB}$
$=\frac{k}{k+1}(2\hat{\imath})+\frac{1}{k+1}(\hat{\imath}+2\hat{\jmath})$
$=\frac{2k+1}{k+1}\hat{\imath}+\frac{2}{k+1}\hat{\jmath}$

b)

$\overrightarrow{MB}$
$=\overrightarrow{OB}-\overrightarrow{OM}$
$=\overrightarrow{OB}-\frac{1}{2}\overrightarrow{OA}$
$=(\hat{\imath}+2\hat{\jmath})-[\frac{1}{2}(2\hat{\imath})]$
$=2\hat{\jmath}$
$\overrightarrow{OA}\cdot\overrightarrow{MB}=0$ since $\hat{\imath}\perp\hat{\jmath}$
$\Rightarrow \angle PMA=90^{\circ}$
$\angle ANO$
$=180^{\circ}-\angle PMA$ (opp. $\angle$, cyclic quad.)
$=180^{\circ}-(90^{\circ})=90^{\circ}$
$\Rightarrow \overrightarrow{ON}\perp\overrightarrow{AB}\Rightarrow \overrightarrow{ON}\cdot\overrightarrow{AB}=0$
$\overrightarrow{AB}$
$=\overrightarrow{OB}-\overrightarrow{OA}$
$=(\hat{\imath}+2\hat{\jmath})-(2\hat{\imath})$
$=-\hat{\imath}+2\hat{\jmath}$
$\overrightarrow{ON}\cdot\overrightarrow{AB}=0$
$(\frac{2k+1}{k+1}\hat{\imath}+\frac{2}{k+1}\hat{\jmath})\cdot(-\hat{\imath}+2\hat{\jmath})=0$
$-\frac{2k+1}{k+1}+\frac{4}{k+1}=0$
$3-2k=0$
$k=\frac{3}{2}$

## Q11编辑

a)

$\frac{d}{d\theta}[\ln(sec\theta+tan\theta)]$
$=\frac{1}{sec\theta+tan\theta}sec\theta tan\theta+sec^2\theta$
$=\frac{sec\theta+tan\theta}{sec\theta+tan\theta}sec\theta$
$=sec\theta$

Therefore $\int sec\theta d\theta=\int\frac{d}{d\theta}\ln(sec\theta+tan\theta)d\theta+C=\ln(sec\theta+tan\theta)+C$

b) i)

$\int \frac{du}{\sqrt{u^2-1}}$

Let $u=sec\theta$ for $0\le\theta\le\frac{\pi}{2}$, so that $du=sec\theta tan\theta d\theta$.

$\int \frac{du}{\sqrt{u^2-1}}$
$=\int \frac{sec\theta tan\theta d\theta}{\sqrt{sec^2\theta-1}}$
$=\int \frac{sec\theta tan\theta d\theta}{\sqrt{tan^2\theta}}$
$=\int sec\theta$
$=\ln(sec\theta+tan\theta)+C$ (By (a))
$=\ln(u+\sqrt{u^2-1})+C, u\ge 1$

Diagram to be inserted here

ii)

$\int_0^1 \frac{2x}{\sqrt{4x^2+x^4+3}} dx$
$=\int_0^1 \frac{2x}{\sqrt{x^4+4x^2+4-1}} dx$
$=\int_0^1 \frac{d(x^2+2)}{\sqrt{(x^2+2)^2-1}}$
$=\ln[(x^2+2)+\sqrt{(x^2+2)^2-1}]|_0^1$ (By (b))
$=\ln[\frac{((1)^2+2)+\sqrt{((1)^2+2)^2-1}}{((0)^2+2)+\sqrt{((0)^2+2)^2-1}}]$
$=\ln(\frac{3+2\sqrt{2}}{2+\sqrt{3}})$
$=\ln(\frac{3+2\sqrt{2}}{2+\sqrt{3}}\frac{2-sqrt{3}}{2-sqrt{3}})$
$=\ln[\frac{(3+2\sqrt{2})(2-sqrt{3})}{4-3}]$
$=\ln(4\sqrt{2}-3\sqrt{3}-2\sqrt{6}+6)$

c)

$t=tan\phi$
$1=sec^2\phi\frac{d\phi}{dt}$
$\frac{d\phi}{dt}=cos^2\phi$
$\frac{d\phi}{dt}=(\frac{1}{\sqrt{1+t^2}})^2=\frac{1}{1+t^2}$

So $d\phi=\frac{dt}{1+t^2}$

$\int_0^{\frac{\pi}{4}} \frac{tan\phi}{\sqrt{1+2cos^2\phi}} d\phi$
$=\int_0^1 \frac{t}{\sqrt{1+\frac{2}{1+t^2}}} (\frac{dt}{1+t^2})$
$=\int_0^1 \frac{t dt}{\sqrt{(1+t^2)^2+2(1+t^2)}}$
$=\int_0^1 \frac{t dt}{\sqrt{(1+t^2)^2+2(1+t^2)+1-1}}$
$=\int_0^1 \frac{t dt}{\sqrt{[(1+t^2)+1]^2-1}}$
$=\frac{1}{2}\int_0^1 \frac{d(t^2+2)}{\sqrt{(t^2+2)^2-1}}$
$=\frac{1}{2}\ln(4\sqrt{2}-3\sqrt{3}-2\sqrt{6}+6)$ (By (b)(ii))

## Q12编辑

a)

i)

$T$
$=\frac{x}{7}+\frac{\sqrt{(40-x)^2+30^2}}{1.4}$
$=\frac{x+5\sqrt{(40-x)^2+900}}{7}$

ii)

$\frac{dT}{dx}=\frac{1+5\cdot\frac{1}{2}[(40-x)^2+900]^{-\frac{1}{2}}[2\cdot(40-x)\cdot(-1)]}{7}$

When $T$ is minimum, $\frac{dT}{dx}=0$. So,

$(0)=1+5\cdot\frac{1}{2}[(40-x)^2+900]^{-\frac{1}{2}}[2\cdot(40-x)\cdot(-1)]$
$0=1-5[(40-x)^2+900]^{-\frac{1}{2}}(40-x)]$
$1=5[(40-x)^2+900]^{-\frac{1}{2}}(40-x)]$
$(40-x)^2+900=[5(40-x)]^2$
$[(40-x)^2+900]=25(40-x)^2$
$24(40-x)^2-900=0$
$(40-x)^2-\frac{900}{24}=0$
$2x^2-160x+3125=0$
$x=\frac{160}{2\cdot2}\pm\frac{\sqrt{(-160)^2-4\cdot(2)\cdot(3125)}}{2\cdot2}$
$x=40\pm\frac{75}{2}=40\pm5\frac{3}{2}$

First derivative test hand-drawn table here

So, $x=40+5\sqrt{\frac{3}{2}}$ for minimum $T$.

$QB$
$=\sqrt{[40-(40+5\sqrt{\frac{3}{2}})]^2+30^2}$ (Pyth. thm.)
$=\sqrt{\frac{1875}{2}}$
$=25\sqrt{\frac{3}{2}}\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{25\sqrt{6}}{2}$ m

b) i)

## Q13编辑

a)

$M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}, tr(M)\equiv a+d.$
$BAB^{-1}=\begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}$

i)

$MN=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} e & f \\ g & h \end{bmatrix}=\begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}$
$NM=\begin{bmatrix} e & f \\ g & h \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} ae+fc & eb+fd \\ ga+ch & gb+hd \end{bmatrix}$
$tr(MN)=ae+bg+cf+dh$
$tr(NM)=ae+fc+gb+hd=tr(MN)$

ii)

$tr(BAB^{-1})$
$=tr(AB^{-1}B)$ (By (a)(i))
$=tr(AI)$
$=tr(A)$

Since $tr(BAB^{-1})=1+3=4$, $tr(A)=4.$

iii)

$det(BAB^{-1})$
$=detBdetAdet(B^{-1})$
$=detBdet(B^{-1})detA$
$=det(BB^{-1})detA$
$=det(I)detA$
$=(1)\cdot detA$
$=detA$

Since $det(BAB^{-1})=1\times3-0\times0=3$, $detA=3.$

b)

$C=\begin{bmatrix} p & q \\ r & s \end{bmatrix}$
$C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda_1\begin{bmatrix} x \\ y \end{bmatrix}$ and $C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda_2\begin{bmatrix} x \\ y \end{bmatrix}$ for $\begin{bmatrix} x \\ y \end{bmatrix}\ne0$, $\lambda_1\ne\lambda_2$.

i)

$C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda_i\begin{bmatrix} x \\ y \end{bmatrix}$ for $i=1, 2$
$(C-\lambda_i)\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$
$\begin{bmatrix} p-lambda_i & q \\ r & s-lambda_i \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$

As we have required $\begin{bmatrix} x \\ y \end{bmatrix}$ to be non-trivial, i.e. $\begin{bmatrix} x \\ y \end{bmatrix}\ne0$, we must have

$\begin{vmatrix} p-\lambda_1 & q \\ r & s-\lambda_1 \end{vmatrix}=\begin{vmatrix} p-\lambda_2 & q \\ r & s-\lambda_2 \end{vmatrix}=0$.

ii)

$\begin{vmatrix} p-\lambda_i & q \\ r & s-\lambda_i \end{vmatrix}=0$ for $i=1, 2$
$(p-\lambda_i)(s-\lambda_i)-qr=0$
$ps-p\lambda_i-s\lambda_i+\lambda_i^2-qr=0$
$\lambda_i^2-(p+s)\lambda_i+(ps-qr)=0$

Notice that $detC=ps-qr$ and $trC=p+s$.

So, $\lambda_i^2-trC\lambda_i+detC=0$ for $i=1, 2$.

$\lambda_1$ and $\lambda_2$ are therefore the roots of the equation $\lambda^2-trC\lambda+detC=0$.

c)

$\lambda^2-trC\lambda+detC=0$ (By (b)(ii))
$\lambda^2-(4)\lambda+(3)=0$ (By (a)(ii) and (iii))
$(\lambda-1)(\lambda-3)=0$
$\lambda=1$ or $3$