Tag: business maths

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Questions Related to business maths

Multiple choice business maths functions and graphs graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $f\left( {g\left( { - x} \right)} \right) = f\left( { - g\left( x \right)} \right) = even$

Multiple choice business maths functions and graphs graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $f(x)+f(\dfrac1x)=\dfrac{{x + 1}}{{x - 1}}-\dfrac{{x + 1}}{{x - 1}}=0$

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given f(x) = x - [x/3] for x in (3, 6). For x in (3, 6), x/3 is in (1, 2), so [x/3] = 1. Thus f(x) = x - 1. Solving y = x - 1 for x gives x = y + 1, so f^-1(x) = x + 1.

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

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. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given integral can be evaluated by substitution. Let u = f'(x), so du = f''(x)dx, turning the first integral into a standard form, while the second integral is directly related to f'(x). Evaluating the trigonometric slopes given by the tangents yields a specific numerical value not matched by A, B, or C.

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

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 $x-axis$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Expanding and simplifying the given expression using trigonometric identities reveals that the function is a constant with respect to x or simplifies to a line parallel to the x-axis passing through the specified point.

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

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. $\left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$\left [0, \dfrac{\pi}{2} \right]$.

Multiple choice business maths limits and continuity of a function graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

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. $[2,5]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation |z-1| + |z+3| = 8 represents an ellipse with foci at 1 and -3. The distance between foci is 4. The major axis 2a = 8, so a = 4. The center is at -1. The vertices are at -1 +/- 4, i.e., 3 and -5. The range of |z-4| is the distance from 4 to the ellipse. The minimum distance is |3-4| = 1 and the maximum is |-5-4| = 9.