Tag: business maths

Questions Related to business maths

if $\begin{bmatrix}2 &1 \ 7 &4 \end{bmatrix}$A$\begin{bmatrix}-3 &2 \ 5 &-3 \end{bmatrix}=\begin{bmatrix}1 &0 \ 0&1 \end{bmatrix}$, then matrix A equals

  1. $\begin{bmatrix}7 &5 \ -11 &-8 \end{bmatrix}$

  2. $\begin{bmatrix}2 & 1 \ 5 & 3 \end{bmatrix}$

  3. $\begin{bmatrix}7 & 34 \ 1 & 5 \end{bmatrix}$

  4. $\begin{bmatrix}5 & 13 \ 3 & 8 \end{bmatrix}$


Correct Option: A
Explanation:

$\begin{bmatrix}2 &1 \ 7 &4 \end{bmatrix}$A$\begin{bmatrix}-3 &2 \ 5 &-3 \end{bmatrix}=\begin{bmatrix}1 &0 \ 0&1 \end{bmatrix}$

$P=\begin{bmatrix}2 &1 \ 7 &4 \end{bmatrix},Q=\begin{bmatrix}-3 &2 \ 5 &-3 \end{bmatrix}, R=\begin{bmatrix}1 & 0 \ 0 & 1 \end{bmatrix}$

$PAQ = R \Rightarrow  A = P^{-1}RQ^{-1}$

$\Rightarrow A=P^{-1}Q^{-1}=(QP)^{-1}$

$QP=\begin{bmatrix}8 &5 \ -11 &-7 \end{bmatrix}$

$\therefore A=(QP)^{-1}=\begin{bmatrix}7 &5 \ -11 &-8 \end{bmatrix}$ 

Hence, option A.

Lets $A=\begin{bmatrix} 0&5 \-5 & 0\end{bmatrix}$ be a skew symmetric matrix and $I + A$ is non singular, then the matrix $B = (I - A)(I + A)^{-1}$ is

  1. an Orthogonal Matrix

  2. an Idempotent Matrix

  3. a Nilpotent Matrix

  4. Data Insufficient


Correct Option: A
Explanation:

$B=(I-A)(I+A)^{-1}$

$B^{T}=[(I+A)^{-1}]^{T}(I-A)^{T}$
$\Rightarrow B^{T}=[(I+A)^{T}]^{-1}(I-A)^{T}$
$\Rightarrow B^{T}=(I-A)^{-1}(I+A)$             since $A^{T}=-A$
$B^{-1}=(I+A)(I-A)^{-1}$
In this case commutativity holds, so,
$B^{T}=B^{-1}\Rightarrow B\text{ is Orthogonal Matrix}$

Consider the following statements :
$S _1$ : If $f(x)$ and $g(x)$ both are discontinuous function and $f(x) + g(x)$ is continuous, then $f(x) - g(x)$ is discontinuous.

$S _2$ : If a tangent to the standard ellipse $\displaystyle \frac{x^2}{a^2}+\displaystyle \frac{y^2}{b^2} = 1$ intersects the principal axis at A and  B, then least value of $AB$ is $(a+b)$.

$S _3$ : If $A$ and $B$ are two matrices such that $AB = O$, where $O$ is null matrix, then at least one of the matrices $A$ and $B$ must be a null matrix.

$S _4$ : If $a,b,c \epsilon R$ and $D$ is a perfect square of a rational number, then both roots of the quadratic equation $a{ x }^{ 2 }+bx+c=0$ are rational.

State, in order, whether ${ S } _{ 1 },{ S } _{ 2 },{ S } _{ 3 }$ or $ { S } _{ 4 }$ are true or false.
  1. FFTT

  2. TTFF

  3. FFFT

  4. TTTF


Correct Option: B
Explanation:

S1 : If $f(x)$ and $g(x)$ both are discontinuous function and $f(x) + g(x)$ is continuous, then $f(x) - g(x)$ is discontinuous.
True
eg: $f(x) = [x]$ and $g(x) = [1-x]$ for non integral values of x, where [.] is a greatest integer function
$f(x)+g(x) = [x]+[1-x] = 0$ is continuous and
$f(x)-g(x)$ is discontinuous.
S2 : If a tangent to the standard ellipse $\displaystyle\frac { x^{ 2 } }{ a^{ 2 } } +\displaystyle\frac { y ^2}{b^2  } =1$  intersects the principal axis at $A$ and  $B$, then least value of $AB$ is $(a+b)$
True
Tanget at $\left( a\cos { \theta , } b\sin { \theta  }  \right)$ is $\displaystyle\frac { x\cos { \theta  }  }{ a } +\displaystyle\frac { y\sin { \theta  }  }{ b } =1$ which intersects axis at $A=(\displaystyle\frac { a }{ \cos { \theta  }  } ,0)$ and $B=(0,\displaystyle\frac { b }{ \sin { \theta  }  })$
$AB^{2} = \displaystyle\frac { a^{ 2 } }{ \cos ^{ 2 }{ \theta  }  } +\displaystyle\frac { b^{ 2 } }{ \sin ^{ 2 }{ \theta  }  } $
AB is minimum at $\theta =\tan ^{ -1 }{ \left(\displaystyle \frac { \sqrt { b }  }{ \sqrt { a }  }  \right)  } $ and minimum value is $a+b$

S3 : If $A$ and $B$ are two matrices such that $AB=O$, where $O$ is null matrix, then at least one of the matrices $A$ and $B$ must be a null matrix.
False
product of two non zero  matrices can be a null matrix.
S4 : If $a,b,c R$ and $D$ is a perfect square of a rational number, then both roots of the quadratic equation $ax^{2}+bx+c=0$ are rational.
False
eg $\sqrt { 3 } x^{ 2 }+\sqrt { 28 } x+\sqrt { 3 } =0$
where $D$ is a perfect square and roots are not rational.


Negation of $(\sim p\rightarrow q)$ is ________________.

  1. $\sim { p }{ \wedge }\sim q$

  2. $\sim \left( p\vee q \right) \vee \left( p\vee \left( \sim p \right) \right) $

  3. $\sim \left( p\vee q \right) \wedge \left( p\vee \left( \sim p \right) \right) $

  4. $\left( \sim p\vee q \right) \wedge \left( p\vee \sim q \right) $


Correct Option: A

$p \wedge ( q \vee \sim p ) =?$

  1. $p \vee q$

  2. $p \wedge q$

  3. $p \rightarrow  q$

  4. none of these


Correct Option: B

$( p \wedge q ) \vee ( \sim p \wedge q ) \vee ( \sim q \wedge r ) =? $

  1. $q \vee r$

  2. $q \wedge r$

  3. $q \rightarrow r$

  4. none of these


Correct Option: B

$p$: He is hard working.
$q$: He is intelligent.
Then $ \sim q\Rightarrow\sim p$, represents

  1. If he is hard working, then he is not intelligent.

  2. If he is not hard working, then he is intelligent.

  3. If he is not intelligent, then he is not had working.

  4. If he is not intelligent, then he is hard working.


Correct Option: C
Explanation:

p:she is hardworking
q:she is intelligent

~p:she is not hardworking
~q:she is not intelligent

~q=>~p 
means She is not intelligent implies she is not hardworking
Hence, Option C

$p:$ He is hard working.
$q:$ He will win.
The symbolic form of "If he will not win then he is not hard working", is

  1. $ p\Rightarrow q$

  2. $ (\sim p)\Rightarrow (\sim q)$

  3. $ (\sim q)\Rightarrow (\sim p)$

  4. $ (\sim q)\Rightarrow p$


Correct Option: C
Explanation:

Given $p:$ He is hard working

and $q:$ He will win
we get $\sim p:$ He is not hard working

and $\sim q:$ He will not win
Now the given statement in the question is "If he will not win then he is not hard working" which means 
"If he will not win then he is not hard working"
For this conditional statement, the symbolic form is $\left( \sim q \right) \Rightarrow \left( \sim p \right) $

Dual of $( p \rightarrow q ) \rightarrow r$ is _________________.

  1. $p\vee (\sim  q\wedge r)$

  2. $p\vee q\wedge r$

  3. $p\vee (\sim  q\wedge \sim r)$

  4. $\sim p\vee (\sim q\wedge r)$


Correct Option: A

P: he studies hard, q: he will get good marks. The symbolic form of " If he studies hard then he will get good marks "is_____

  1. $\sim q\Rightarrow p$

  2. $p\Rightarrow q$

  3. $\sim p\vee q$

  4. $p\Leftrightarrow q$


Correct Option: B