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# Tutor profile: Abhinav V.

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Abhinav V.
Tutor for 7 years, Skill Development executive at ASAP - Govt of Kerala
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## Questions

### Subject:Trigonometry

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Question:

By giving a counter example show that the following statement is false “ If n is an even integer, then n is not prime".

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Abhinav V.

The given statement is in the form of if \$\$p\$\$ then \$\$q\$\$ We have to show that it is false. For we have to show that if \$\$p\$\$ then not \$\$q\$\$ \$\$2\$\$ is a counter example. \$\$2\$\$ is even which is also prime Thus we conclude that the given statement is false.

### Subject:Basic Math

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Question:

A course consists of \$\$15\$\$ papers which is divided into \$\$4\$\$ semesters. First , Second ,Third and Fourth semesters has \$\$5\$\$,\$\$4\$\$,\$\$3\$\$ and \$\$3\$\$ papers respectively from which a student joining this course has to select \$\$13\$\$ papers in all, selecting atleast \$\$3\$\$ papers from each semester. What is the possible number of ways in which a student can select the papers ?

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Abhinav V.

Number of available papers in first semester = \$\$5\$\$ Number of available papers in second semester = \$\$4\$\$ Number of available papers in third semester = \$\$3\$\$ Number of available papers in fourth semester = \$\$3\$\$ We have to select total \$\$13\$\$ papers for the course which should contain atleast \$\$3\$\$ papers in each semester \$\$3^{rd}\$\$ and \$\$4^{th} \$\$ semester has only \$\$3\$\$ papers each .So they are compulsory. Now we have to select courses for semesters \$\$1\$\$ and \$\$2\$\$. We have \$\$2\$\$ cases 1. \$\$4\$\$ from sem \$\$1\$\$, \$\$3\$\$ from sem \$\$2\$\$ 2. \$\$3\$\$ from sem \$\$1\$\$ , \$\$4\$\$ from sem \$\$2\$\$ So the total number of ways in which a student can select papers \$\$= (5C4*4C3*3C3*3C3) + (5C3*4C4*3C3*3C3)\$\$ \$\$= 20 + 10 = 30\$\$ ways

### Subject:Algebra

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Question:

Show that, \$\$\{ 6x :x ∈ Z \} = \{ 2x :x ∈ Z \} ∩ \{ 3x :x ∈ Z \} \$\$

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Abhinav V.

Let, \$\$A = \{ 6x :x ∈ Z \}\$\$ , \$\$B= \{ 2x :x ∈ Z \} \$\$ and \$\$C = \{ 3x :x ∈ Z \}\$\$ Let,\$\$ k ∈ A\$\$, then, \$\$k=6x\$\$ , \$\$x ∈ Z\$\$ So, \$\$k = 6x = 2* 3x\$\$ Let us call, \$\$3x = a\$\$ Since, \$\$x ∈ Z\$\$ , \$\$a ∈ Z\$\$ Therefore, \$\$k = 2a\$\$ , \$\$a∈ Z\$\$ So, \$\$k ∈ B\$\$ Also, \$\$k = 6x = 3* 2x\$\$ Let us call \$\$2x = b\$\$, \$\$b ∈ Z\$\$ \$\$6x = 3b\$\$ , \$\$b∈ Z\$\$ So,\$\$ k ∈ C\$\$, Since, \$\$k∈ B \$\$ and \$\$K ∈ C\$\$, \$\$K∈ B∩C\$\$ So, \$\$A ⊆ B∩C\$\$ …………………..\$\$(1)\$\$ Conversely let, \$\$K∈ B∩C\$\$ Then \$\$k=2x\$\$ for some \$\$x ∈ Z\$\$ and \$\$k= 3y\$\$ for some \$\$y ∈ Z\$\$, So \$\$2x = 3y\$\$ Since \$\$2\$\$ and \$\$3\$\$ are relatively prime \$\$2\$\$ must divide \$\$y\$\$ So \$\$y=2p\$\$ , \$\$p ∈ Z\$\$ Then, \$\$k=3y\$\$ \$\$K=3*2p\$\$ \$\$K=6p\$\$ , \$\$p∈ Z\$\$ \$\$K ∈ A\$\$ So, \$\$B∩C ⊆ A\$\$ …………………..\$\$(2)\$\$ From \$\$(1)\$\$ and \$\$(2)\$\$ \$\$A = B∩C\$\$ That is, \$\$\{ 6x :x ∈ Z \} = \{ 2x :x ∈ Z \} ∩ \{ 3x :x ∈ Z \} \$\$

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