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...

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 H and K are normal subgroup of S_4 H is normal in K and K are normal in A_4 H is normal in A_4 but not in S_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...

AP-TS SET 2014 MATHEMATICS SOLUTIONS

AP-TS SET  2014 MATHEMATICS SOLUTIONS PAPER-2 SOLUTIONS Q 1 $S=\left\{ 1-\frac { { \left( -1 \right)  }^{ n } }{ n } \quad :\quad n=1,2,3,... \right\} \\ \quad =\left\{ 2,1-\frac { 1 }{ 2 } ,1+\frac { 1 }{ 3 } ,1-\frac { 1 }{ 4 } ,... \right\} \\ \alpha \quad =\quad inf\quad S\quad =\quad \frac { 1 }{ 2 } \\ \beta \quad =\quad sup\quad S\quad =\quad 2\\ \\ \therefore \quad \left( \alpha ,\beta  \right) \quad =\left( \frac { 1 }{ 2 } ,2 \right)$ Q 2 $Let\quad { x }_{ n }\rightarrow l\\ { x }_{ n+1 }\quad...

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...

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. 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)....

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},...

AP-TS SET 2014 MATHEMATICAL SCIENCE OFFICIAL KEY

                                         AP-TS SET 2014 MATHEMATICAL SCIENCE OFFICIAL KEY  ...