Tag: introduction to sets

Questions Related to introduction to sets

If $f(x)$ is a polynomial function satisfying the condition $f(x) \times f\left(\dfrac{1}{x}\right)=f(x)+f\left(\dfrac{1}{x}\right)$ and $f(2)=9$ 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. $2f(4) =3 f(6)$

  2. $14f(1) = f(3)$

  3. $ 9f(3) = 2f(5)$

  4. $f(10) = f(11)$

Show answer
Correct Option: B,C
Explanation:

The polynomial which satisfies $f(x)f(1/x)=f(x)+f(1/x)$ is $ \pm x^n+1$ (standard result)
Given that $f(2) = 9 \ \Rightarrow \pm 2^n + 1 = 9 \ \Rightarrow 2^n = 8 $
(-ve sign not possible here)
$ \Rightarrow n=3$
Hence the function is $ f(x)=x^3+1$
$ \Rightarrow f(1) = 2, \; f(3)=28 , \; f(5)=126$
$ f(4) = 65, \; f(6) = 217$
Using these, we see only option B and C are correct. 

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

Show answer
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

Show answer
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.