FANDOM


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 $

Q14编辑

除了特别提示,社区内容遵循CC-BY-SA 授权许可。