Tag: business maths

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

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

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.

The negation of an implication (p -> q) is (p AND NOT q). Applying this to (~p -> q), we get (~p AND NOT q), which simplifies to (~p AND ~q).

Using the distributive law: p AND (q OR ~p) = (p AND q) OR (p AND ~p). Since (p AND ~p) is False, the expression simplifies to (p AND q) OR False, which is (p AND q).

Applying Boolean algebra or logic laws to the expression (p and q) or (not p and q) or (not q and r), we can factor out q from the first two terms to get q, which then combines with the third term to simplify to q or r.

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

Given $p:$ He is hard working

To find the dual, replace AND with OR, OR with AND, True with False, and False with True. The dual of (p -> q) -> r is p OR (~q AND r).

The statement 'If p then q' is the definition of a conditional statement, represented symbolically as p -> q.