Qualitative Treatment of Autonomous ODE

A differential equation is called autonomous if it has no independent term. Hence a first order general ode will look like
$$\dfrac{dy}{dt} = f(y) \phantom{csir net}\text{ [Independent of $t$]}.$$

Qualitative Treatment of Autonomous ODE
STEP I: Find the critical points $y_0$(solution of $f(y)=0$). These are horizontal solution of the given ode.

Since two solutions(called integral curve) can’t intersect, hence other solution will be in the region bounded by these horizontal solution.
STEP II: As we have partition the space with horizontal solution, now identify the region with $f(y)>0$ and $f(y)<0$.

As we know that
$$\begin{align*} \left [\dfrac{dy}{dt} = f(y) \right ] \begin{matrix} >0 &\Rightarrow y(t)\text{ is increasing}\\ <0 &\Rightarrow y(t)\text{ is decreasing} \end{matrix} \end{align*}$$
Hence the solution will look like

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Now we will apply these concept to a question of CSIR-NET.

Question: Let $y:\mathbb{R} \rightarrow \mathbb{R}$ satisfy the initial value problem
$$y'(t)=1-y^2,t\in \mathbb{R},~y(0)=0.$$ Then

1. $y(t_1)=1$ for some $t\in R$.
2. $y(t)>-1$ for all $t\in R$.
3. $y$ is strictly increasing in $\mathbb{R}$.
4. $y$ is increasing in $(0, 1)$ and decreasing in $(1,\infty)$.

Answer: Here $f(y)=1-y^2=(1-y)(1+y)$, so $y=\pm 1$ are two horizontal solution of the given ode
$$\begin{matrix} \text{Region 1}&y > 1 & f(y) < 0 & y(t) \text{ is decreasing}\\ \text{Region 2}&-1>y>1 & f(y) > 0 &y(t) \text{ is increasing}\\ \text{Region 3}&y <-1 & f(y) < 0 & y(t) \text{ is decreasing}\\ \end{matrix}$$
The required integral curve will pass through $(0,0)$, which lies in region 2, so the solution is increasing and bounded by $y \in (-1,1)$. Here 2nd and 4th option are correct.

Hence 2 and 4 are correct choices.

Problems on Permutation Group

Question: Let $H=\{e,(12)(34)\}$ and $K=\{e,(12)(34),(13)(24),(14)(23)\}$ be subgroup of $S_4$, where $e$ denotes the identify element of $S_4$. Then

1. $H$ and $K$ are normal subgroup of $S_4$
2. $H$ is normal in $K$ and $K$ are normal in $A_4$
3. $H$ is normal in $A_4$ but not in $S_4$
4. $H$ is normal in $S_4$ but $H$ is not

Answer: Recall

Conjugation preserve disjoint cyclic structure in permutation group, $i.e.$, for all $\sigma, \tau \in S_n$, the disjoint cyclic decomposition of $\tau$ and $\sigma \tau \sigma^{-1}$ will be the same.

Since $K$ contain all element with disjoint cyclic decompositon of type $(ab)(cd)$, hence it will be normal in $S_4$ and it will be normal in $A_4( \leq S_4)$.

$K$ is Abelian so all it’s subgroup will be normal. $H \trianglelefteq K$.

Hence 2 is correct option.

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Question: Let $S_n$ denote the symmetric group on $n$ symbols. The group $S_3 \oplus \mathbb{Z}/2\mathbb{Z}$ is isomorphic to which of the following groups?

1. $\mathbb{Z}/12\mathbb{Z}$
2. $\mathbb{Z}/6\mathbb{Z} \oplus \mathbb{Z}/6\mathbb{Z}$
3. $A_4$, the alternating group of order $12$
4. $D_6$, the dihedral group of order $12$

Answer: Note,

• The direct sum of a group will be Abelian iff all it’s constituent groups will be Abelian.
• The direct sum contains a normal subgroup isomorphic to each of it’s constituent group.

$S_3 \oplus \mathbb{Z}/2\mathbb{Z}$ is a non-abelian group of order $12$, hence it will be either $A_4$ or $D_6$.
Also $S_3 \oplus \mathbb{Z}/2\mathbb{Z}$ has a normal subgroup of order $2(\cong \mathbb{Z}_2)$, since $A_4$ does not contain a normal subgroup of order $2$, hence it has to be $D_6$.

Hence, 4 is correct choice.

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Question: Which of the following numbers can be orders of permutations $\sigma$ of $11$ symbols such that $\sigma$ does not fix any symbol?

$1.~18 \quad 2.~30 \quad 3.~15 \quad 4.~28$

Answer: The order of permutation is determined as follows

• The order of a $n-$cylce is $n$.
• The order any permutation is lcm of each lenght of individual cycle in the disjoint cyclic decomposition.

If a permutation does not fix any element of set, then it’s disjoint cyclic decomposition will contain all the elements of the set. So here we have to partition $11$, with cycle length($> 1$, as $1$ length cycle fixed the element inside it.) such that the lcm of it’s constituent will be given in the option.

Option 1: For example $2 + 9 = 11$ and $[2 , 9] = 11$, we can write a permutation consist of two disjoint cycle of lengths $2$ and $9$.
Option 2: Similarly, $5 + 6 = 11$ and $[5, 6]=30$.
Option 3: Here, $3+3+5 \neq 11$ and $[3,3,5]=15$.
Option 4: $4 + 7 = 11$ and $[4 , 7] = 28$.

Hence all are correct.

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Question: Let $\sigma = (1~2)(3~4~5)$ and $\tau = (1~2~3~4~5~6)$ be permutation in $S_6$, the group of permutations on six symbols. Which of the following statements are true?

1. The subgroup $\langle \sigma \rangle$ and $\langle \tau \rangle$ are isomorphic to each other.
2. $\langle \sigma \rangle$ and $\langle \tau \rangle$ are conjugate in $S_6$.
3. $\langle \sigma \rangle \cap \langle \tau \rangle$ is trivial group.
4. $\sigma$ and $\tau$ commute.

Answer: Since $|\sigma| = |\tau| = 6$, hence both will generate a group isomorphic to $\mathbb{Z}_6$.
Both $\sigma$ and $\tau$ has different disjoint cyclic structure, hence can’t be conjugate.
Also, $\langle \sigma \rangle \cap \langle \tau \rangle$ will be trivial.

Any two permutation will commute if and only if their disjoint cyclic decomposition has no elements in common.

Hence, $\sigma$ and $\tau$ will not-commute.

Hence 1 and 3 are correct choice.

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Question: Given the permutation $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 2 & 5 & 4 \end{pmatrix}$ the matrix $A$ is defined to be the one whose $i-$th column is the $\sigma(i)-$th column of the identity matrix $I$. Which of the following is correct?

$1.~A = A^{-2} \quad 2.~A = A^{-4} \quad 3.~A = A^{-5} \quad 4.~A = A^{-1}$

Answer: Note,

There is a natural isomorphism between $S_n$ and set of all permutation matrices of dimension $n \times n$, given by $f(\sigma) =$ the matrix whose $i-$th column is the $\sigma(i)-$th column of the identity matrix.

Then, $|\sigma| = |f(\sigma)|$

Also, $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 2 & 5 & 4 \end{pmatrix} = (132)(45)$,

Hence $|\sigma| =[2,3] = 6$, $i.e.$, $A^6 = I \Rightarrow A = A^{-5}$.

Hence 3 is correct choice.

AP-TS SET 2014 MATHEMATICS SOLUTIONS

AP-TS SET  2014 MATHEMATICS SOLUTIONS

PAPER-2 SOLUTIONS

Q 1

$S=\left\{ 1-\frac { { \left( -1 \right)  }^{ n } }{ n } \quad :\quad n=1,2,3,... \right\} \\ \quad =\left\{ 2,1-\frac { 1 }{ 2 } ,1+\frac { 1 }{ 3 } ,1-\frac { 1 }{ 4 } ,... \right\} \\ \alpha \quad =\quad inf\quad S\quad =\quad \frac { 1 }{ 2 } \\ \beta \quad =\quad sup\quad S\quad =\quad 2\\ \\ \therefore \quad \left( \alpha ,\beta  \right) \quad =\left( \frac { 1 }{ 2 } ,2 \right)$

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Q 2

$Let\quad { x }_{ n }\rightarrow l\\ { x }_{ n+1 }\quad =\frac { 1 }{ 2 } \left( { x }_{ n }+\frac { a }{ { x }_{ n } }  \right) ;\quad n\ge 1.\\ \Rightarrow \quad l=\frac { 1 }{ 2 } \left( l+\frac { a }{ l }  \right) \\ \Rightarrow { l }^{ 2 }=a\quad \Rightarrow l\quad =\sqrt { a } \\ By\quad AM-GM\quad inequality,\\ { x }_{ n+1 }=\frac { 1 }{ 2 } \left( { x }_{ n }+\frac { a }{ { x }_{ n } }  \right) \quad \ge \quad \sqrt { { x }_{ n }.\frac { a }{ { x }_{ n } }  } =\sqrt { a } \quad \forall n\\ and\quad { x }_{ n }\rightarrow \sqrt { a } \\ \therefore \quad { x }_{ n }\quad is\quad dicreasing.\\$

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Q 3

$f:{ R }\longrightarrow { R };f(x)=\begin{cases} 1 & \quad x\in { Q } \\ 0 & \quad x\in { Q^{ c } } \end{cases}\\ we\quad know\quad f\quad is\quad continuous\quad at\quad x\in { R }\\ \quad\quad\quad\quad\quad iff\quad { x }_{ n }\longrightarrow x\quad impies\quad that\quad f\left( { x }_{ n } \right) \longrightarrow f\left( x \right) .\\ let\quad x\in { Q },\quad then\quad \exists \quad ({ s }_{ n })\quad in\quad { Q }^{ c }\quad :\quad { s }_{ n }\longrightarrow x,\quad but\quad f\left( { s }_{ n } \right) =\quad 0\nrightarrow f\left( x \right) =1.\\ so\quad f\quad is\quad not\quad continuous\quad at\quad x\in { Q. }\\ similiarly\quad f\quad is\quad not\quad continuous\quad at\quad x\in { Q }^{ c }\quad as\quad \exists \quad { r }_{ n }\longrightarrow x\quad in\quad { Q }\\ but\quad f\left( { r }_{ n } \right) =1\nrightarrow f\left( x \right) =0.\\ threfore\quad set\quad of\quad continuity\quad points\quad is\quad empty.$

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Q 4

$Given f:\mathbb{R}\longrightarrow\mathbb{R} \quad is\quad monotonically\quad decreasing\quad\\ \\ f\left(x\right)\leq f\left(0\right)=a \quad \forall x\in\mathbb{R} \\ \therefore f\left((0,\infty\right))\subseteq (-\infty,a)\\$

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Q 5

$f:\left[ a,b \right] \longrightarrow \quad { R }.\quad Given\quad f\quad is\quad H\ddot { o } lder's\quad function\quad with\quad exponent\quad \delta \quad >0.\\ \quad \quad \left| f\left( x \right) -f\left( y \right)  \right| <{ \left| x-y \right|  }^{ 1+\delta  },\quad \forall \quad x,y\in \left[ a,b \right] \\ \Rightarrow \frac { \left| f\left( x \right) -f\left( y \right)  \right|  }{ \left| x-y \right|  } <\quad \left| x-y \right| ^{ \delta  }\\ \Rightarrow \quad \lim _{ y\rightarrow x }{ \frac { \left| f\left( x \right) -f\left( y \right)  \right|  }{ \left| x-y \right|  }  } <\quad \lim _{ y\rightarrow x }{ \left| x-y \right| ^{ \delta  } } \\ \Rightarrow \quad \left| f^{ ' }\left( x \right)  \right| =\quad 0\quad \forall x\in \left[ a,b \right] \\ \Leftrightarrow \quad f^{ ' }\left( x \right) =\quad 0\quad \quad \forall x\in \left[ a,b \right] \\ \Leftrightarrow \quad f\left( x \right) =\quad k\quad constant\quad \forall x\in \left[ a,b \right]$

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Q 6

$\left( \begin{matrix} 1, & -2, & 5 \end{matrix} \right) =\quad -6\left( \begin{matrix} 1, & 1, & 1 \end{matrix} \right) +3\left( \begin{matrix} 1, & 2, & 3 \end{matrix} \right) +2\left( \begin{matrix} 2, & -1, & 1 \end{matrix} \right) \\ \therefore \quad \left( \begin{matrix} \lambda _{ 1 }, & -\lambda _{ 2 }, & \lambda _{ 3 } \end{matrix} \right) =\left( \begin{matrix} -6 & 3 & 2 \end{matrix} \right).$

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Q 7

${ B }=\left\{ \alpha _{ 1 },\alpha _{ 2 },...,\alpha _{ n } \right\} \quad is\quad a\quad basis\quad for\quad { V }\quad a\quad vector\quad space.\\ i.e.\quad { B }\quad is\quad a\quad maximal\quad linearly\quad independent\quad set\quad in\quad { V }\\ { S }=\left\{ \beta _{ 1 },\beta _{ 2 },...,\beta _{ m } \right\} \quad is\quad a\quad linearly\quad independent\quad set\quad in\quad { V }.\\ \therefore \quad m\le n.\\ \\ Note:\quad { S }{ \subseteq B }\quad need\quad not\quad be\quad true.$

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Q 8

$T:{ { R } }^{ 3 }\longrightarrow { { R } }^{ 3 },\quad T(\begin{matrix} x, & y, & z \end{matrix})=(\begin{matrix} x+2y, & y-z & z+2z \end{matrix})\\ The\quad matrix\quad of\quad transformation\quad T\quad with\quad respect\quad to\quad standard\quad basis\quad is\\ \therefore \quad M=\begin{pmatrix} 1 & \quad 0 & \quad 1 \\ 2 & \quad 1 & \quad 0 \\ 0 & \quad -1 & \quad 2 \end{pmatrix}\\ by\quad row\quad operations\quad on\quad M\\ \begin{pmatrix} 1 & \quad 0 & \quad 1 \\ 2 & \quad 1 & \quad 0 \\ 0 & \quad -1 & \quad 2 \end{pmatrix}\xrightarrow { R_{ 2 }-2R_{ 1 } } \begin{pmatrix} 1 & \quad 0 & \quad \quad 1 \\ 0 & \quad 1 & \quad -2 \\ 0 & \quad -1 & \quad 2 \end{pmatrix}\xrightarrow { R_{ 3 }+R_{ 2 } } \begin{pmatrix} 1 & \quad \quad 0 & \quad \quad 1 \\ 0 & \quad 1 & \quad -2 \\ 0 & \quad 0 & \quad 0 \end{pmatrix}\\ \therefore \quad Rank\quad of\quad T\quad is\quad 2.$

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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}$
$= \begin{pmatrix}\alpha-\beta,&\beta-\gamma,&\gamma\end{pmatrix}$

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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}$

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Q 11

$Let\quad f(z)=zcosec(z)=\frac{z}{\sin(z)}.\\ $ 
Observe that f  is  analytic on

$\therefore \oint_Cf(z)dz=0\\ $

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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)\\ $
                          $=2\pi i*lim_{z\rightarrow 1}(z-1)f(z)\\$
                          $=2\pi i*lim_{z\rightarrow1}\frac{1}{z}=2\pi i$

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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.$

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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.

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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\}. $

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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.$

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Q 33

$P(X=5) =\quad P(X=5|Y=1)+P(X=5|Y=2)+P(X=5|Y=3) \\$
$\quad=\quad \frac{3}{20}+\frac{1}{5}+\frac{6}{20} \\$
$\quad = \quad \frac{13}{20}\\$


>$P(X=6) =\quad P(X=6|Y=1)+P(X=6|Y=2)+P(X=6|Y=3) \\$
$\quad=\quad \frac{2}{20}+\frac{1}{5}+\frac{1}{20} \\$
$\quad = \quad \frac{7}{20}\\$

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Q 34

$\varphi_X(t)\quad = \quad E(e^{itX})\\$
$=\int_{-\infty}^{\infty} e^{itx}.f_X(x) dx\\$
=$\int_{-\infty}^{\infty}e^{itx}.\frac{1}{\sqrt{2}\pi}e^{-\frac{x^2}{2}}dx\\$ 
=$\frac{1}{\sqrt{2}\pi}\int_{-\infty}^{\infty}e^{itx-\frac{x^2}{2}}dx\\$
=$\frac{1}{\sqrt{2}\pi}\int_{-\infty}^{\infty}e^{\frac{(it)^2}{2}}e^{-\frac{(x-it)^2}{2}}dx\\$
$=e^{-\frac{t^2}{2}}$

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Q 22 
Given  $y.f(x)=\tan{\left( \frac{\pi}{4}-\frac{1}{4}\cos^4{x}\right)} $; and 
$y\left(\frac{\pi}{2}\right)=1$
$\Rightarrow \quad y\left(\frac{\pi}{2}\right).f\left(\frac{\pi}{2}\right)=\tan{\left(\frac{\pi}{4}-\frac{1}{4}\cos^4{\frac{\pi}{2}}\right)}\\$

$\Rightarrow f\left(\frac{\pi}{2}\right)=\tan{\left(\frac{\pi}{4}-0\right)}=1.$

Problems on Euler's Theorem

Question: The number of elements in the set $\{m:1\leq m \leq 1000, m \text{ and } 1000 \text{ are relatively prime} \}$ is

$1.~100 \quad 2.~250 \quad 3.~300 \quad 4.~400$

Answer: The question asks for $\phi(1000)$.

And $\phi(n)$ can be calculated using following formula.
1. $\phi(p_1^{m_1}p_2^{m_2} \cdots p_r^{m_r}) = \phi(p_1^{m_1})\phi(p_2^{m_2}) \cdots \phi(p_r^{m_r})$.
2. $\phi(p^m) = \begin{cases}2^{m-1} &\text{ if } p= 2 \\ p^m - p^{m-1} &\text{ if } p \neq 2\end{cases}$
Or equivalently it can be calculated by single formula as follows
$$\phi(n) = n \prod_{p|n}\left(1-\frac{1}{p} \right)$$

$\phi(1000)=\phi(2^3)\cdot \phi(5^3) = 2^2(5^3-5^2)=400$.

Hence 4 is correct choice.

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Question: The last two digit of $7^{81}$ are

$1.~07 \quad 2.~17 \quad3.~37 \quad4.~47$

Answer: In this problem we have to use Euler’s Theorem

If $(a, n ) = 1$, then $a^{\phi(n)}$ = $1$ mod $n$.

Here $(7,100) = 1$ and $\phi(100) = \phi(2^2\cdot 5^2) = \phi(2^2)\phi(5^2)=2\cdot(5^2-5)=40$.
Thus $7^{40} = 1 \Rightarrow 7^{1+2\cdot 40} = 7$.

Hence 1 is correct option.

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Question: If $n$ is a positive integer such that the sum of all positive integers $a$ satisfy $1 \leq a \leq n$ and $\gcd(a,n) = 1$ is equal to $240n$, then the number of summands, namely, $\phi(n)$ is

$1.~120 \quad 2.~124 \quad 3.~240 \quad 4.~480$

Answer: The solution depends on the fact that relative prime number exist in pair.

Note that $(k,n) = (n-k,n)$. Hence $(k,n) =(n-k,n)=1$, $i.e.$, $k$ and $n-k$ are relatively prime together and their sum is equal to $n$. Hence, $$\sum_{\{1\leq a \leq n \text{ and } (a,n)=1\}}a = \frac{\phi(n)}{2} \times n.$$

Here,
$$\frac{\phi(n)}{2} \times n = 240\times n \Rightarrow \boxed{\phi(n) = 480 }.$$

Hence 4 is correct choice.

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Question: For a positive integer $m$, let $\phi(m)$ denote the number of integers $k$ such that $1 \leq k \leq m$ and $\gcd(k,m)=1$. Then which of the followong statements are necessarily true?

1. $\phi(n)$ divides $n$ for every positive integer $n$.
2. $n$ divides $\phi(a^n-1)$ for all positive integers $a$ and $n$.
3. $n$ divides $\phi(a^n-1)$ for all positive integers $a$ and $n$ such that $\gcd(a,n)=1$
4. $a$ divides $\phi(a^n-1)$ for all positive integers $a$ and $n$ such that $\gcd(a,n)=1$

Answer: More than one option can be true so we analyze each option separately.

Option 1: For any prime number $p$ we know that $\phi(p)=p-1 \nmid p$, hence false.

Option 2 and 3: For any positive integer $a, n$ we have
$$a^n - 1 = 0 \Rightarrow a^n = 1 \tag{$\!\!\!\!\mod (a^n - 1)$}$$
Also, note that $(a, a^n-1) = 1$, Using Euler’s theorem
$$a^{\phi(a^n - 1)} = 1 \tag{$\!\!\!\!\mod (a^n - 1)$}$$
Hence $n | \phi(a^n - 1)$ without any restriction on $a$ and $b$ regarding gcd.

Option 4: Take $n = 1$ then $(a,1)=1$. Here $a$ can’t divide $\phi(a - 1)<(a-1)$, hence false.

Hence 2 and 3 are correct options.

Finding Last Digit

Question: The unit digit of $2^{100}$ is

$1.~2 \quad 2.~4 \quad3.~6 \quad4.~8$

Answer: This problem can be solved by basic idea of periodicity. Calculating few terms mod $10$,

$2^1$	$2^2$	$2^3$	$2^4$	$2^5$	$2^6$
$2$	$4$	$8$	$6$	$2$	$4$

We can see that the periodicity is $4$.
Hence $2^{k + 4n}= 2^k$ for $k =\{1,2,3,4\}$.
Here $2^{100} = 2^{4+ 24\cdot 4} = 2^4 = 6$.

Hence 3 is correct answer.

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Question: The last digit of $38^{2011}$ is

$1.~6 \quad 2.~2 \quad 3.~4 \quad 4.~8$

Answer: First note that $38 = 8$ (mod $10$).
Compute the periodicity of $8$ mod $10$.

$8^1$	$8^2$	$8^3$	$8^4$	$8^5$
$8$	$4$	$2$	$6$	$8$

Again the period is $4$.
Hence $38^{2011} = 8^{2011}= 8^{3 + \text{ multiple of }4} = 8^3 = 2$.

Hence 2 is the correct choice.

Problems on Counting

Question: The number of surjective maps from a set of $4$ elements to a set of $3$ elements is

$1.~36 \quad 2.~64 \quad 3.~69 \quad 4.~81$

Answer: Let $A$ be a $4$ elements set and $B$ be a $3$ elements set. Here we need to find number of surjective maps between $A$ onto $B$. The idea is to partition the set $A$ into $3$ non-empty subset and assign each subset to a element of $B$.
Different partition of set $A$ into $3$ non-empty subset is equal to $\binom{4}{2}$, it can be represented by following figure(more generally it is given by Starling number of second kind)

Thus we got a total of $6$ partition. Each partition corresponds to $3!$ surjective maps. Hence total number of surjective maps $= 6 \times 3! = 36$.

Hence 1 is correct choice.

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Question: An ice cream shop sells ice cream in five possible falvours: Vanilla, Chocolate, Strawberry, Mango and Pineapple. How many combinations of three scoop cones are possible? [Note: The repeatation of flavours allowed but the order in which the flavours are chosen does not matter.]

$1.~10 \quad 2.~20 \quad 3.~35 \quad 4.~243$

Answer: Let’s count the possibilities one by one
1. All 3 cones are of same flavor, we have $5$ possiblities.
2. All 3 cones are of different flavor, we have $\binom{5}{3} = \frac{5!}{3!\cdot2!} = 10$ possibilities.
3. 2 are of same flavor and 1 of different flavor, then we have $5 \times 4 = 20$ possibilities.
Hence total choice is $5+10+20 = 35$.

Hence 3 is correct choice.

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Question: We are given a class consisting of $4$ boys and $4$ girls. A committee that consists of a President, a Vice-President and a Secretary is to be chosen among the $8$ students of the class. Let $a$ denote the number of ways of choosing the committee in such a way that the committee has at least one boy and at least one girl. Let $b$ deonote the number of ways of choosing the committee in such a way that the number of girls is greater than or equal to that of the boys. Then

$$\begin{align*}&1. ~a = 288 &2. ~a = 168 \\ &3.~b = 144 &4.~b = 192 \end{align*}$$

Answer: First we will choose a group of three people satisfying the condition and then multiply by $3!$, since the places are distinguishable.

Committee a: Choosing at least one boy and one girl = (Choosing 2 boy and 1 girl) + (Choosing 1 boy and 2 girl) = $2\binom{4}{2}\binom{4}{1}= 48$.
Hence no of ways choosing comittee a = $48 \times 6 = 288$.

Committee b: Choosing a committee in such a way that the number of girls is greater than or equal to that of the boys = (Choosing 2 girl and 1 boy)= $\binom{4}{2}\binom{4}{1}= 24$.
Hence no of ways choosing comittee a = $24 \times 6 = 144$.

Hence 1 and 3 are correct options.

AP-TS SET 2014 MATHEMATICAL SCIENCE OFFICIAL KEY

                                         AP-TS SET 2014 MATHEMATICAL SCIENCE OFFICIAL KEY