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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.
Answer:

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.
Answer:

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.
Answer:

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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