TS ICET 2019 Question Paper Shift-1 (23rd May)

If p and q are statements, then "$$\sim p \Rightarrow p \wedge q$$"

let solve  that through the navigation of  "$$\sim p \Rightarrow p \wedge q$$"

p→(p∨∼q)
−(p→(p∨∼q))
∵ − (p → q) = p∧ ∼ q
= p∧ ∼ ( p∨ ∼ q)= p∧(∼ q∧ ∼ q) = (p∧∼p)∧(p∧q)=t∧ ( p ∧ q) =t  Answer 

From the above table, the truth value of the expression (∼p⇒p)∧(p⇒∼p) is always F. Hence it's a contradiction. so the p is always true and q is false  Answer

$$A^1 = X - A$$, then $$\left(A \cup \left(B^1 \cap C^1\right)\right) = $$

• Let S{x|x∈A−(B∩C)}S{x|x∈A−(B∩C)} Let Q{y|y∈(A−B)∪(A−C)}Q{y|y∈(A−B)∪(A−C)}
• All x∈Ax∈A and x∉B,Cx∉B,C.
• All y∈Ay∈A and y∉B,Cy∉B,C.
• Because all xx fit the definition of QQ then we say S⊆QS⊆Q
• Because all yy fit the definition of SS then we say Q⊆SQ⊆S
• Since S⊆QS⊆Q and Q⊆SQ⊆S then S=Q⟹A−(B∩C)=(A−B)∪(A−C)S=Q⟹A−(B∩C)=(A−B)∪(A−C) 

But then quickly realized it was wrong because the xx from set SS must meet the following criteria:

$$\left(A \cup \left(B^1 \cap C^1\right)\right) $$  Answer

NOTE- 

Whether the empty relation is reflexive or not depends on the set on which you are defining this relation -- you can define the empty relation on any set XX.

1- The statement "RR is reflexive" says: for each x∈Xx∈X, we have (x,x)∈R(x,x)∈R. This is vacuously true if X=∅X=∅, and it is false if XX is nonempty.

2- The statement "RR is symmetric" says: if (x,y)∈R(x,y)∈R then (y,x)∈R(y,x)∈R. This is vacuously true, since (x,y)∉R(x,y)∉R for all x,y∈Xx,y∈X.

3- The statement "RR is transitive" says: if (x,y)∈R(x,y)∈R and (y,z)∈R(y,z)∈R then (x,z)∈R(x,z)∈R. Similarly to the above, this is vacuously true.

4- To summarize, RR is an equivalence relation if and only if it is defined on the empty set. It fails to be reflexive if it is defined on a nonempty set.

If X has m elements and Y has n elements, the number if onto functions are,

$$n^m-\left(\begin{array}{c}n\\ 1\end{array}\right)(n-1)^{m} + \left(\begin{array}{c}n\\ 2\end{array}\right)(n-2^{m}) + ................+(-1)^{n-1}\left(\begin{array}{c}n\\ n-1\end{array}\right)1^{m}$$

here m= 5 and n= 4 put the value and we get 

620 Answer 

we know the line fourmula  y = mx + c

we have two line equation 

ax - yb = 0 and bx - ay = 0 

then we get 

$$y_{1}$$ = $$m_{1}$$x + $$ c_{1}$$          and         $$y_{2}$$ = $$m_{2}$$x + $$ c_{2}$$ 

put the value of x and y  

we get

-b = a$$m_{1}$$ + $$ c_{1}$$                  and           -a = b $$m_{2}$$  +   $$ c_{2}$$ 

where   $$ c_{1}$$ and   $$ c_{2}$$ are constant  = 0 then 

$$m_{1}$$ = -b\a                                  and            $$m_{2}$$ = -a\b 

then we know that the angle is 

$$\tan\theta$$ = |$$\frac {(m_{1}-m_{2})}{(1 +m_{1}m_{2})}$$ | 

put the value of $$m_{1}$$ and $$m_{2}$$ we get 

$$\tan\theta$$ = $$\frac {(a^{2}-b^{2})}{2ab}$$  Answer 

The general line through two points (a,b)(a,b) and (c,d)(c,d) is

(c − a) (y−b) = (d −b) (x − a) 

In our case, that is 

(-5 - 13)(y + 3) = 12 (x − 13) =   -18(y +2 ) = 12 (x−13)

which we can write

12x + 18y = -120 

2x + 18y = -20  Answer 

Tan(480) =Tan(360+120) = Tan 120     $$\because$$ tan(360 + $$\theta$$) = tan$$\theta$$

Tan120 = Tan(90+30) = -Cot 30

Tan30 = 1/√3, so, Cot 30 = √3

So, Tan (480) = - cot 30 = -√3     Answer 

$$\cos \frac{2 \pi}{6} + \sin \frac{5 \pi}{6} =$$

we can  write this 

cos60 + sin150 =                     equation 1

and sin150 = sin( 90 + 60) 

                = sin90co60 + cos90sin60

put sin90 = 1  and cos60 = 1\2 

we get 

sin150 = 1\2 and cos 60  = 1\2            put value in equation 1 

cos60 + sin150 =  1\2 + 1\2 = 1            Answer      

 we have given that 

   $$6 \cos \theta - 7 \sin \theta = 0 $$              equation 1  and

   $$(7 \cos 2\theta + 6 \sin 2\theta)^2 =$$        equation 2 

then solving equation 1 we get 

$$6 \cos \theta - 7 \sin \theta = 0 $$

$$ \tan \theta$$ = 6\7 

now solving equation 2 

$$(7 \cos 2\theta + 6 \sin 2\theta)^2 =$$    here we know that 

$$ \cos 2\theta$$ = $$\cos\theta^{2}$$ - $$\sin\theta^{2}$$  and   $$ \sin 2\theta$$ = 2$$\sin\theta$$ $$\cos\theta$$

put the value in equation 2 we get 

=  (7($$\cos\theta^{2}$$ - $$\sin\theta^{2}$$) + 6 ( 2$$ \sin \theta$$ $$\cos\theta$$))^2  

on solving ge get 

= 7\6$$\times$$ $$tan\theta$$ $$7^{2}$$

= 49 answer