Mathematical Reasoning Test

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Mathematical Reasoning Test - 33

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Question 1
1 / -0

While simplifying $$\sqrt { \frac { 1-cosx }{ 1+cosx } } $$, two students got the following two answers A & B.
A) cosec x - cot x (B) $$\frac { 1 }{ cosecx+cotx } $$ What can you say about answers ?

A
Both A & B are wrong

B
Both A & B are right

C
A is right B is wrong

D
B is right A is wrong

Solution
Question 2
1 / -0

Which of the following proportional is a tautology?

A
$$\sim (p \rightarrow q) \vee (p \sim \wedge q)$$

B
$$(p \rightarrow p) \vee (p \sim \wedge q)$$

C
$$(p \rightarrow q) \vee (p \sim \wedge q)$$

D
$$(p \rightarrow q) \wedge (p \wedge \sim q)$$

Solution
Question 3
1 / -0

The logically equivalent proportion of $$p \leftrightarrow q$$ is

A
$$(p \wedge q) \vee (p \vee q)$$

B
$$(p \rightarrow q) \wedge (p \rightarrow q)$$

C
$$(p \rightarrow q) \vee (p \rightarrow q)$$

D
$$(p \wedge q) \rightarrow (p \vee q)$$

Solution
Question 4
1 / -0

If p, q, r are statements with truth values false, true and false respectively. then truth value of $$(\sim pv \sim q) v r$$ is

A
True

B
False

C
False, if r is true

D
False, if q is false

Solution
Question 5
1 / -0

The converse of the contra positive of $$
\sim p \rightarrow q
$$ is ........(1)

A
$$

q \rightarrow p

$$

B
$$

\sim p \rightarrow p

$$

C
$$

p \rightarrow \sim q

$$

D
$$

\sim q \rightarrow \sim p

$$
Question 6
1 / -0

The contrapositive of $$( p \wedge q ) \Rightarrow r$$ is

A
$$\sim r \Rightarrow ( p \vee q )$$

B
$$r \Rightarrow ( p \vee q )$$

C
$$- r \Rightarrow ( - p \vee - q )$$

D
$$p \Rightarrow ( q \vee r)$$

Solution
Question 7
1 / -0

Consider the statement :$$"P\left( n \right) :{ n }^{ 2 }-n+41$$ is prime." Then which one of the following is true?

A
P(3) is false but P(3) is true.

B
P(3)is false but P(5) is true.

C
Both P(3) and P(5) are true.

D
Both P(3) and P(5) are false.

Solution
Question 8
1 / -0

Two pairs of statement are:
p: If a quadrilateral is a rectangle, then its opposite sides are equal.
q: If opposite sides of a quadrilateral are equal, then the quadrilateral is a rectangle.
The combined statement of these pairs using If and only if is:

A
A quadrilateral is a rectangle if and only if its all sides are equal.

B
A quadrilateral is a rectangle if and only if its opposite sides are equal.

C
A quadrilateral is a square if and only if its opposite sides are equal.

D
A quadrilateral is not a rectangle if and only if its opposite sides are equal.

Solution
Question 9
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Let A and B denote the statements
A:$$\cos { \alpha } +\cos { \beta } +\cos { \gamma } =0$$ & B:$$\sin { \alpha } +\sin { \beta } +\sin { \gamma } =0$$
If $$\cos { (\beta -\gamma ) } +\cos { (\gamma -\alpha ) } +\cos { (\alpha -\beta ) } =-\frac { 3 }{ 2 } $$ then

A
A is true and Bis false

B
A is false and B is true

C
Both A and B are true

D
Both A and B are false

Solution

Given ,

$$A:\cos\alpha+\cos\beta+\cos\gamma=0$$..........(1)

$$A:\sin\alpha+\sin\beta+\sin\gamma=0$$..........(2)

Now,

on squaring both sides of equation (1) and (2)

we get,

$$(\cos\alpha+\cos\beta+\cos\gamma)^2=0$$........(3)
and

$$(\sin\alpha+\sin\beta+\sin\gamma)^2=0$$..........(4)
on adding equation (3) and (4) we get,

$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2(\cos\alpha\cos\beta+\cos\beta\cos\gamma+\cos\gamma\cos\alpha)+$$
$$\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2(\sin\alpha\sin\beta+\sin\beta\sin\gamma+\sin\gamma\\sin\alpha)=0$$

$$\therefore (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$

$$\Rightarrow (\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)+(\cos^2\gamma+\sin^2\gamma)+2[(\cos\alpha\cos\beta+\sin\alpha\sin\beta)+$$
$$(\cos\beta\cos\gamma+\sin\beta\sin\gamma)+(\cos\gamma\cos\alpha+\sin\gamma\sin\alpha)]=0$$

$$\Rightarrow 1+1+1+2[\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)]=0$$

$$\because$$ we know that $$\cos(\alpha-\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$
Also, $$\cos^2x+\sin^2x=1$$

$$\Rightarrow 3+2[\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)]=0$$

$$\Rightarrow 2[\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)]=-3$$

$$\Rightarrow \cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=\dfrac{-3}{2}$$

Thus, A and B both statement are true.
Question 10
1 / -0

If $$x = 5$$ and $$y = - 2$$ then $$x - 2 y = 9 .$$ The contrapositive of this statement is

A
If $$x - 2 y = 9$$ then $$x = 5$$ and $$y = - 2$$

B
If $$x - 2 y \neq 9$$ then $$x \neq 5$$ and $$y \neq - 2$$

C
If $$x - 2 y \neq 9$$ then $$x \neq 5$$ or $$y \neq - 2$$

D
If $$x - 2 y \neq 9$$ then either $$x \neq 5$$ or $$y = - 2$$