AP-TS SET 2014 MATHEMATICS SOLUTIONS
PAPER-2 SOLUTIONS
Q 1
$S=\left\{ 1-\frac { { \left( -1 \right) }^{ n } }{ n } \quad :\quad n=1,2,3,... \right\} \\ \quad =\left\{ 2,1-\frac { 1 }{ 2 } ,1+\frac { 1 }{ 3 } ,1-\frac { 1 }{ 4 } ,... \right\} \\ \alpha \quad =\quad inf\quad S\quad =\quad \frac { 1 }{ 2 } \\ \beta \quad =\quad sup\quad S\quad =\quad 2\\ \\ \therefore \quad \left( \alpha ,\beta \right) \quad =\left( \frac { 1 }{ 2 } ,2 \right)$
Q 2
$Let\quad { x }_{ n }\rightarrow l\\ { x }_{ n+1 }\quad =\frac { 1 }{ 2 } \left( { x }_{ n }+\frac { a }{ { x }_{ n } } \right) ;\quad n\ge 1.\\ \Rightarrow \quad l=\frac { 1 }{ 2 } \left( l+\frac { a }{ l } \right) \\ \Rightarrow { l }^{ 2 }=a\quad \Rightarrow l\quad =\sqrt { a } \\ By\quad AM-GM\quad inequality,\\ { x }_{ n+1 }=\frac { 1 }{ 2 } \left( { x }_{ n }+\frac { a }{ { x }_{ n } } \right) \quad \ge \quad \sqrt { { x }_{ n }.\frac { a }{ { x }_{ n } } } =\sqrt { a } \quad \forall n\\ and\quad { x }_{ n }\rightarrow \sqrt { a } \\ \therefore \quad { x }_{ n }\quad is\quad dicreasing.\\$
Q 3
$f:{ R }\longrightarrow { R };f(x)=\begin{cases} 1 & \quad x\in { Q } \\ 0 & \quad x\in { Q^{ c } } \end{cases}\\ we\quad know\quad f\quad is\quad continuous\quad at\quad x\in { R }\\ \quad\quad\quad\quad\quad iff\quad { x }_{ n }\longrightarrow x\quad impies\quad that\quad f\left( { x }_{ n } \right) \longrightarrow f\left( x \right) .\\ let\quad x\in { Q },\quad then\quad \exists \quad ({ s }_{ n })\quad in\quad { Q }^{ c }\quad :\quad { s }_{ n }\longrightarrow x,\quad but\quad f\left( { s }_{ n } \right) =\quad 0\nrightarrow f\left( x \right) =1.\\ so\quad f\quad is\quad not\quad continuous\quad at\quad x\in { Q. }\\ similiarly\quad f\quad is\quad not\quad continuous\quad at\quad x\in { Q }^{ c }\quad as\quad \exists \quad { r }_{ n }\longrightarrow x\quad in\quad { Q }\\ but\quad f\left( { r }_{ n } \right) =1\nrightarrow f\left( x \right) =0.\\ threfore\quad set\quad of\quad continuity\quad points\quad is\quad empty.$
Q 4
$Given f:\mathbb{R}\longrightarrow\mathbb{R} \quad is\quad monotonically\quad decreasing\quad\\ \\ f\left(x\right)\leq f\left(0\right)=a \quad \forall x\in\mathbb{R} \\ \therefore f\left((0,\infty\right))\subseteq (-\infty,a)\\$
Q 5
$f:\left[ a,b \right] \longrightarrow \quad { R }.\quad Given\quad f\quad is\quad H\ddot { o } lder's\quad function\quad with\quad exponent\quad \delta \quad >0.\\ \quad \quad \left| f\left( x \right) -f\left( y \right) \right| <{ \left| x-y \right| }^{ 1+\delta },\quad \forall \quad x,y\in \left[ a,b \right] \\ \Rightarrow \frac { \left| f\left( x \right) -f\left( y \right) \right| }{ \left| x-y \right| } <\quad \left| x-y \right| ^{ \delta }\\ \Rightarrow \quad \lim _{ y\rightarrow x }{ \frac { \left| f\left( x \right) -f\left( y \right) \right| }{ \left| x-y \right| } } <\quad \lim _{ y\rightarrow x }{ \left| x-y \right| ^{ \delta } } \\ \Rightarrow \quad \left| f^{ ' }\left( x \right) \right| =\quad 0\quad \forall x\in \left[ a,b \right] \\ \Leftrightarrow \quad f^{ ' }\left( x \right) =\quad 0\quad \quad \forall x\in \left[ a,b \right] \\ \Leftrightarrow \quad f\left( x \right) =\quad k\quad constant\quad \forall x\in \left[ a,b \right]$
Q 6
$\left( \begin{matrix} 1, & -2, & 5 \end{matrix} \right) =\quad -6\left( \begin{matrix} 1, & 1, & 1 \end{matrix} \right) +3\left( \begin{matrix} 1, & 2, & 3 \end{matrix} \right) +2\left( \begin{matrix} 2, & -1, & 1 \end{matrix} \right) \\ \therefore \quad \left( \begin{matrix} \lambda _{ 1 }, & -\lambda _{ 2 }, & \lambda _{ 3 } \end{matrix} \right) =\left( \begin{matrix} -6 & 3 & 2 \end{matrix} \right).$
Q 7
${ B }=\left\{ \alpha _{ 1 },\alpha _{ 2 },...,\alpha _{ n } \right\} \quad is\quad a\quad basis\quad for\quad { V }\quad a\quad vector\quad space.\\ i.e.\quad { B }\quad is\quad a\quad maximal\quad linearly\quad independent\quad set\quad in\quad { V }\\ { S }=\left\{ \beta _{ 1 },\beta _{ 2 },...,\beta _{ m } \right\} \quad is\quad a\quad linearly\quad independent\quad set\quad in\quad { V }.\\ \therefore \quad m\le n.\\ \\ Note:\quad { S }{ \subseteq B }\quad need\quad not\quad be\quad true.$
Q 8
$T:{ { R } }^{ 3 }\longrightarrow { { R } }^{ 3 },\quad T(\begin{matrix} x, & y, & z \end{matrix})=(\begin{matrix} x+2y, & y-z & z+2z \end{matrix})\\ The\quad matrix\quad of\quad transformation\quad T\quad with\quad respect\quad to\quad standard\quad basis\quad is\\ \therefore \quad M=\begin{pmatrix} 1 & \quad 0 & \quad 1 \\ 2 & \quad 1 & \quad 0 \\ 0 & \quad -1 & \quad 2 \end{pmatrix}\\ by\quad row\quad operations\quad on\quad M\\ \begin{pmatrix} 1 & \quad 0 & \quad 1 \\ 2 & \quad 1 & \quad 0 \\ 0 & \quad -1 & \quad 2 \end{pmatrix}\xrightarrow { R_{ 2 }-2R_{ 1 } } \begin{pmatrix} 1 & \quad 0 & \quad \quad 1 \\ 0 & \quad 1 & \quad -2 \\ 0 & \quad -1 & \quad 2 \end{pmatrix}\xrightarrow { R_{ 3 }+R_{ 2 } } \begin{pmatrix} 1 & \quad \quad 0 & \quad \quad 1 \\ 0 & \quad 1 & \quad -2 \\ 0 & \quad 0 & \quad 0 \end{pmatrix}\\ \therefore \quad Rank\quad of\quad T\quad is\quad 2.$
Q 9
$T:{ { R } }^{ 3 }\longrightarrow { { R } }^{ 3 },\quad T(\begin{matrix} x, & y, & z \end{matrix})=\begin{pmatrix} x+y+z,&y+z,&z\end{pmatrix}.\\ Let \quad T\begin{pmatrix}x,&y,&z\end{pmatrix}=\begin{pmatrix}\alpha&\beta&\gamma\end{pmatrix}\\ \Rightarrow\begin{pmatrix}x+y+z,&y+z,&z\end{pmatrix}=\begin{pmatrix}\alpha&\beta&\gamma\end{pmatrix}\\ \Rightarrow z=\gamma;\quad y+z=\beta \quad \Rightarrow y=\beta-\gamma;$
$\\x+y+z=\alpha \quad \Rightarrow x=\alpha-\beta.\\$
$ T^{-1} \begin{pmatrix}\alpha,&\beta,&\gamma\end{pmatrix}$
$= \begin{pmatrix}\alpha-\beta,&\beta-\gamma,&\gamma\end{pmatrix}$
Q 10
$3t^2+4t+5=a+b(1+t)+c(1+t^2)$
$ =(a+b+c)+bt+c+t^2;\\ $
$ \Rightarrow \quad a+b+c=5,\quad b=4, \quad c=3,$
$ \Rightarrow a=5-3-4=-2;\\$
$\therefore$$\begin{pmatrix}a,&b,&c\end{pmatrix}=\begin{pmatrix}-2,&4,&3\end{pmatrix}$
Q 11
$Let\quad f(z)=zcosec(z)=\frac{z}{\sin(z)}.\\ $
Observe that f is analytic on
$\therefore \oint_Cf(z)dz=0\\ $
Q 12
The path $ C: |z-1|=1/2,\quad f(z)=\frac{1}{z(z-1)}.\\$ f has simple polesat $z=0$ and $ z=1\\$
The $z=1$ lies inside the closed path C.
$\therefore \oint_Cf(z)dz=2\pi i*Res(f;z=1)\\ $
$=2\pi i*lim_{z\rightarrow 1}(z-1)f(z)\\$
$=2\pi i*lim_{z\rightarrow1}\frac{1}{z}=2\pi i$
Q 13
$f(z)=u+iv$ is an entire function with $u^2\leq v;\\$
$\Rightarrow f $ is bounded,
$\Rightarrow$ by Liovelli's theorem f is constant.
Therefore $ f(2014)=f(0)=1.$
Q 16
Given the order of Group o(G)=143=11$\times$13;
$let\quad n_{13}=\quad number\quad of\quad Syllow-3\quad subgroups\quad in\quad G\\$
by Syllow's theorem
$n_{13}\mid11\quad \Rightarrow n_{13}\in \{1,11\},\\$
$n_{13}\equiv1mod(13)\quad \Rightarrow n_{13}=1.\\$
also we know that unique is normal.
Therefore answer is 1.
Q 19
Given set $X=\{a,b,c,d,e\},A=\{a,c,d,e\}\subseteq X;\\$
topology on X is $T=\{\phi,X,\{a\},\{a,b\},\{a,b,e\},\{a,c,d\},\{a,b,c,d\}\}.\\$
Interior$(A^o)$ of A is the largest open in A.
Therefore $ A^o=\{a,c,d\}. $
Q 21
Given differential is $(x^2-y^2)dx+2xydy=0;\\$
$\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}=\frac{(\frac{y}{x})^2-1}{2(\frac{y}{x})}.\\$
$Let\quad y=zx\quad \Rightarrow \frac{dy}{dx}=z+x\frac{dz}{dx},\\$
$x\frac{dz}{dx}=\frac{z^2-1}{2z}-z=\frac{-1-z^2}{2z}\\$
$\Rightarrow \frac{1}{x}dx+\frac{2x}{1+z^2}dz=0\\$
$\Rightarrow\quad \log{x(z^2+1)}=C.\\$
$\Rightarrow\quad x(z^2+1)=C.\\$
Therefore the general solution is $ x^2+y^2=Cx.$
Q 33
$P(X=5) =\quad P(X=5|Y=1)+P(X=5|Y=2)+P(X=5|Y=3) \\$
$\quad=\quad \frac{3}{20}+\frac{1}{5}+\frac{6}{20} \\$
$\quad = \quad \frac{13}{20}\\$
>$P(X=6) =\quad P(X=6|Y=1)+P(X=6|Y=2)+P(X=6|Y=3) \\$
$\quad=\quad \frac{2}{20}+\frac{1}{5}+\frac{1}{20} \\$
$\quad = \quad \frac{7}{20}\\$
Q 34
$\varphi_X(t)\quad = \quad E(e^{itX})\\$
$=\int_{-\infty}^{\infty} e^{itx}.f_X(x) dx\\$
=$\int_{-\infty}^{\infty}e^{itx}.\frac{1}{\sqrt{2}\pi}e^{-\frac{x^2}{2}}dx\\$
=$\frac{1}{\sqrt{2}\pi}\int_{-\infty}^{\infty}e^{itx-\frac{x^2}{2}}dx\\$
=$\frac{1}{\sqrt{2}\pi}\int_{-\infty}^{\infty}e^{\frac{(it)^2}{2}}e^{-\frac{(x-it)^2}{2}}dx\\$
$=e^{-\frac{t^2}{2}}$
Q 22
Given $y.f(x)=\tan{\left( \frac{\pi}{4}-\frac{1}{4}\cos^4{x}\right)} $; and
$y\left(\frac{\pi}{2}\right)=1$
$\Rightarrow \quad y\left(\frac{\pi}{2}\right).f\left(\frac{\pi}{2}\right)=\tan{\left(\frac{\pi}{4}-\frac{1}{4}\cos^4{\frac{\pi}{2}}\right)}\\$
$\Rightarrow f\left(\frac{\pi}{2}\right)=\tan{\left(\frac{\pi}{4}-0\right)}=1.$
In Q 11 given function is not analytic.
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