Tag: business maths

Questions Related to business maths

If $\displaystyle f(x)=27x^{3}+\frac{1}{x^{3}}$ and $\alpha,\beta$ are the roots of $\displaystyle 3x+\frac{1}{x}=2$ is

business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

  1. $f(\alpha)=f(\beta)$

  2. $f(\alpha)=10$

  3. $f(\beta)=-10$

  4. none of these

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Correct Option: A,C
Explanation:
$3x+\dfrac{1}{x}=2$
Cubing the above equation, we get
$ \left(3x+\dfrac{1}{x}\right)^3=2^3$
$\therefore 27x^3+\dfrac{1}{x^3}+3(3x)\left(\dfrac{1}{x}\right)\left(3x+\dfrac{1}{x}\right)=8$

$\therefore 27x^3+\dfrac{1}{x^3}+9\left(3x+\dfrac{1}{x}\right)=8$

$\therefore 27x^3+\dfrac{1}{x^3}+9(2)=8$

$\therefore 27x^3+\dfrac{1}{x^3}=-10$

$\alpha$ and $\beta$ are roots of above equation.
$\therefore 27\alpha^3+\dfrac{1}{\alpha^3}=-10$ ...(1)
and $27\beta^3+\dfrac{1}{\beta^3}=-10$         ...(2)

$f(x)=27x^3+\dfrac{1}{x^3}$

$f(\alpha)=27\alpha^3+\dfrac{1}{\alpha^3}$
$\implies f(\alpha) =-10$   ...(from 1)

Similarly, $f(\beta)=27\beta^3+\dfrac{1}{\beta^3}$
$\implies f(\beta)=-10$   ...(from 2)

$\therefore f(\alpha)=f(\beta)=-10$


So, the correct options are option (A) and (C)

If a function satisfies $(x-y)f(x+y)-(x+y)f(x-y)=2(x^{2}y-y^{3}),\forall x,y\in R$ and $ f(1)=2,$ then

business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

  1. $f(x)$ must be polynomial function

  2. $f(3)=12$

  3. $f(0)=0$

  4. $f(x)$ may not be differentiable

Show answer
Correct Option: A,B,C
Explanation:

$(x-y)f(x+y)-(x+y)f(x-y)$ $=2y((x-y)(x+y))$


Let $x-y=u, x+y=v$

$uf(v)-vf(u)=uv(v-u)$

$\displaystyle \frac{f(v)}{v}-\frac{f(u)}{u}=v-u$

$\Rightarrow  \displaystyle \left ( \frac{f(v)}{v}-v \right )=\left ( \frac{f(u)}{u}-u \right )=constant$

Let $\displaystyle \frac{f(x)}{x}-x=\lambda $

$\Rightarrow f(x)=(\lambda x+x^{2})$

$f(1)=2$

$\lambda +1=2\Rightarrow \lambda =1$

$f(x)=x^{2}+x$

If $g(x)$ is a polynomial satisfying $g(x) g(y) = g(x) + g(y) + g(xy) - 2$ for all real $x$ and $y$ and $g(2) = 5$ then $g(3)$ is equal  to -

business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

  1. $10$

  2. $24$

  3. $21$

  4. none of these

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Correct Option: A
Explanation:

$g(x) g(y) = g (x) + g (y) + g (xy) - 2$


Substitute $x = 2$ & $y = 1$

$g (2) g (1) = g (2) + g (1) + g (2) - 2$

$\Rightarrow 4g (1) = 8 $

$\Rightarrow g (1) = 2$

$g (x) g(y) = g (x) + g (y) + g(xy)-2$, now substitute $y\, =\, \displaystyle \frac{1}{x}$

Now $g (x)\, g \left (\displaystyle \frac{1}{x} \right )\, =\, g (x)\, +\, g \left (\displaystyle \frac{1}{x} \right )$

$g(x) = 1\, \pm\, x^n$

$\therefore\, 5\, =\, 1\, \pm\, 2^n\, (\because\, g (2)\, =\, 5)$

$2^n=4$

So, $n = 2$

Now $g (3) = 1$ + $3^2=10$

Write a rational function $f$ that has vertical asymptote at $x=4$, a horizontal asymptote at $y=5$ and a zero at $x=-7$.

business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

  1. $f(x)=\dfrac{5(x-7)}{(x-4)}$

  2. $f(x)=\dfrac{5(x+7)}{(x-4)}$

  3. $f(x)=\dfrac{5(x-7)}{(x+4)}$

  4. $f(x)=\dfrac{(x+7)}{(x+4)}$

Show answer
Correct Option: B
Explanation:

Since the rational function $f$ has the vertical asymptote at $x=4$, then the denominator of $f$ contains the term $(x-4)$.

Thus function $f(x)$ is of the form $f=\dfrac{g(x)}{x-4}$.
Since the horizontal asymptote exists $y=5$, the numerator $g(x)$ of $f(x)$ has to be of the same degree as the denominator with a leading coefficient equal to $5$.
Also $g(x)$ must contain the term $(x+7)$ since $f$ has zero at $x=-7$.
Hence, $f(x)=\dfrac{5(x+7)}{(x-4)}$

A large mixing tank currently contains $200$ gallons of water into which $10$ pounds of sugar have been mixed. A tap will open pouring $20$ gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of $2$ pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after $14$ minutes. Then 

business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

  1. the concentration is greater than at the beginning?

  2. the concentration lesser than at the beginning?

  3. the concentration equal to the concentration at the beginning?

  4. None of the above

Show answer
Correct Option: A
Explanation:

Let $t$ be the number of minutes since the tap opened. 

Since the water increases at $20$ gallons per minute, and the sugar increases at $2$ pound per minute, these are constant rates of change. 

This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. 

We can write an equation independently for each:

$\text{Water}: \: W(t)=200+20t$ in gallons

$\text{Sugar}: \: S(t)=10+2t$ in pounds​​

The concentration $C$ will be the ratio of pounds of sugar to gallons of water.

$C(t)=\dfrac{10+2t}{200+20t}$​​

The concentration after $14$ minutes is given by evaluating $C(t)$ at $t=14$

$\therefore C(14)=\dfrac{10+28}{200+280}=\dfrac{19}{240}$

This means the concentration is $19$ pounds of sugar to $240$ gallons of water.

At the beginning, the concentration is $C(0)=\dfrac{10+0}{200+0}=\dfrac{1}{20}$.

Since $\dfrac{19}{240}\approx 0.08 > \dfrac{1}{20}=0.05$, the concentration is greater after $14$ minutes than at the beginning.

The given table shows the number of cars manufactured in four different colours on a particular day. Study it carefully and answer the question.

 Colour    Number of cars manufactured
 Vento  Creta
 Red  65
 White  54
 Black  66
 Sliver 37

What was the total number of black cars manufactured?

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  1. $240$

  2. $206$

  3. $205$

  4. $159$

Show answer
Correct Option: B
Explanation:

The number of Black cars manufactured 


$=$ no. of black $Vento$ +no. of black $Creta$ +no. of black $WagonR$ 

$=66+52+88 =206$          

" the relation S is defined on $N\times N\left ( a,b \right )S\left ( c,d \right )\Leftrightarrow bc\left ( a+d \right )=ad$ in an equivalance " that statement is ?

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. True

  2. False

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Correct Option: A

If a = b then ax = ...........

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $b + x$

  2. bx

  3. b - x

  4. $b\, \div\, x$

Show answer
Correct Option: B
Explanation:

Given,

a = b
Multiplying both sides by x.
ax = bx

If  x + y = 3 and xy = 2, then the value of $\displaystyle x^{3}-y^{3}$ is equal to 

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. 6

  2. 7

  3. 8

  4. 0

Show answer
Correct Option: B
Explanation:

Formula used:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
               $=(\sqrt{(x+y)^2-4xy})[(x+y)^2-xy]$
               $=(\sqrt{(3)^2-4(2)})[(3)^2-2]$
               $=(\sqrt 1)(7)=7$
Option B is correct.

What is the solution of $x\le 4,y\ge 0$ and $x\le -4,y\le 0$ ?

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $x\ge -4,y\le 0$

  2. $x\le 4,y\ge 0$

  3. $x\le -4,y=0$

  4. $x\ge -4,y=0$

Show answer
Correct Option: C
Explanation:

$x \le 4$ and $x \le -4$

$\Rightarrow x \le -4$
Also,
$y \ge 0$ and $y \le 0$
$\Rightarrow y = 0$
Hence the solutione is $x \le -4, \; y = 0$.