Tag: functions and graphs

Quizzes (5) Questions (65)

Questions Related to functions and graphs

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

The number of elements of an identity function defined on a set containing four elements is______

  1. $\displaystyle 2^{2}$
  2. $\displaystyle 2^{4}$
  3. $\displaystyle 2^{8}$
  4. $\displaystyle 2^{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If an element is related to itself, it is called an identity function. That is $ f(x) = x $

So, if  the set has $ 4 $ elements, then the function will also have $ 4 = 2^2 $ elements.

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

Let $f(-2, 2)\rightarrow(-2, 2)$ be a continuous function given $f(x)=f{(x}^{2})$. Given $f(0)=\dfrac{1}{2}$ then the $4f(\dfrac{1}{2})$

  1. $4$
  2. $2$
  3. $-2$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given f(x) = f(x^2), by repeated substitution we have f(x) = f(x^4) = f(x^8) and so on. For any x in (-1, 1), x^(2^n) approaches 0 as n goes to infinity. Since f is continuous and f(0) = 1/2, f(x) must be equal to 1/2 for all x in (-1, 1). Thus, f(1/2) = 1/2, and 4 * f(1/2) = 4 * (1/2) = 2.

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

Let $f\left( x \right) = p{x^2} + qx - \left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right),\,\left( {p,q,a,b,c \in R} \right)(a,b,c$ are distinct). If both roots of $f(x)=0$ are non-real, then 

  1. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$
  2. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$
  3. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$
  4. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression (a-b)^2 + (b-c)^2 + (c-a)^2 is always positive for distinct a, b, c. If the roots of px^2 + qx - K = 0 are non-real, the discriminant q^2 + 4pK < 0. The options involve complex algebraic manipulations of these coefficients.

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

  1. $2f(4) =3 f(6)$
  2. $14f(1) = f(3)$
  3. $ 9f(3) = 2f(5)$
  4. $f(10) = f(11)$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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. 

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

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

  1. $f(\alpha)=f(\beta)$
  2. $f(\alpha)=10$
  3. $f(\beta)=-10$

  4. none of these
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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)