Tag: business maths

Questions Related to business maths

If f is even function and g is an odd function, then $f _og$ is ............function.

  1. Even

  2. Odd

  3. Neither even nor odd

  4. Either even


Correct Option: A
Explanation:

$fog$ function is an even function


Let $f\left(x \right)$ is a even function and $g \left( - x \right)$ is odd function.
So, $f\left( {g\left( { - x} \right)} \right) = f\left( { - g\left( x \right)} \right) = even$

State the whether given statement is true or false
If $f\left( x \right) = \dfrac{{x + 1}}{{x - 1}},$ then $f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = 0$

  1. True

  2. False


Correct Option: A
Explanation:

$f\left( x \right) = \dfrac{{x + 1}}{{x - 1}}$


$f\left( \dfrac 1 x \right) = \dfrac{{\dfrac1x + 1}}{{\dfrac1x- 1}}=\dfrac{1+x}{1-x}=-\dfrac{1+x}{x-1}$

Hence, $f(x)+f(\dfrac1x)=\dfrac{{x + 1}}{{x - 1}}-\dfrac{{x + 1}}{{x - 1}}=0$

The identity function on real numbers given by $f(x)=x$ is continuous at every real numbers.

  1. True

  2. False


Correct Option: A

The minimum value of $f\left( x \right) ={ x }^{ 2 }+2x+3 ,x\in R$ is equal to 

  1. $2$

  2. $3$

  3. $4$

  4. $1$


Correct Option: A
Explanation:

$f(x)=x^2+2x+3$


For minima: 


$f'(x)=0$

$f'(x)=2x+2=0$

$\implies x=-1$

Hence minimum value of $f(x)$ is at $x=-1$, i.e.

$f(-1)=(-1)^2-2+3=2$

Supposen(A) $= 3$ and $n ( B ) = 5 .$ find the number of elements in $A \times B$

  1. $15$

  2. $9$

  3. $25$

  4. $20$


Correct Option: A
Explanation:
Since $n(A)=3$ and $n(B)=5$
Number of elements in $A\times B=3\times 5=15$

If $f:\,\left( {3,6} \right) \to \left( {1,3} \right)$ is a function defined by $f\left( x \right) = x - \left[ {\frac{x}{3}} \right],\,then\,{f^{ - 1}}\left( x \right) = $

  1. $x-1$

  2. $x+1$

  3. $x$

  4. none of these


Correct Option: B

The tangents to the graph of the function  $y=f(x)$ at the point with abscissa $x=1$ forms an angle of $\pi/6$ and the point $x=2$ an angle of $\pi/3$ and at the point $x=3$ an angle of $\pi/4$. The value of 
$\displaystyle \int _{1}^{2}{f'(x)f''(x)dx}+\displaystyle \int _{2}^{3}{f''(x)dx}$

  1. $\dfrac{4\sqrt{3}-1}{3\sqrt{3}}$

  2. $\dfrac{3\sqrt{3}-1}{2}$

  3. $\dfrac{4-\sqrt{3}}{3}$

  4. $None\ of\ these$


Correct Option: A

The graph of the function $\cos x\cos x(x+2)-\cos^{2}(x+1)$ is  

  1. A straight line through $(0, -\sin^{2}1)$ with slope $2$.

  2. A straight line through $(0, 0)$

  3. A parabola with vertex $(1, -\sin^{2}1)$

  4. A straight line through $\left(\dfrac{\pi}{2},-\sin^{2}1\right)$ and parallel to the $x-axis$.


Correct Option: A

If $f(x)=\left | \sin x \right |$, then domain of $f$ for the existence of inverse is  

  1. $[0,\pi ]$

  2. $\left [ 0,\dfrac{\pi }{2} \right ]$

  3. $\left [ -\dfrac{\pi }{4},\dfrac{\pi }{4} \right ]$

  4. $\left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]$


Correct Option: B
Explanation:

We know that $-1 \leq \sin x \leq 1$ for $x \in \left [ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] $. 

For $| \sin x |$ to be invertible, the function has to be one-to-one. 
Thus, we need unique values of $x$ that give unique values of $f$ and vice versa.
For $x \in \left [0, \dfrac{\pi}{2} \right]$, $0 \leq \sin x \leq 1 \Rightarrow  0 \leq | \sin x \leq 1$
For $x \in \left [-\dfrac{\pi}{2},0 \right]$, $-1 \leq \sin x \leq 0 \Rightarrow  0 \leq | \sin x \leq 1$.
So, we have
$ \left [0, \dfrac{\pi}{2} \right] \rightarrow\left [0, 1 \right]$
$ \left [- \dfrac{\pi}{2},0 \right] \rightarrow\left [0, 1 \right]$
Since both the domains of $|\sin x|$ map to$\left [0, 1 \right]$, we consider only one of them for $x$ to be unique. 
Here, according to the options, the domain of $f$ must be$\left [0, \dfrac{\pi}{2} \right]$.

If $|z-1|+ |z+3| \le 8$, then the range of values of $|z-4|$ is

  1. $(0, 7)$

  2. $(1,8)$

  3. $[1,9]$

  4. $[2,5]$


Correct Option: C