Tag: functions and graphs

Questions Related to functions and graphs

If f is a constant function and f(100)=100  then f(2007)=_____

  1. 2007

  2. 100

  3. 0

  4. None of these


Correct Option: B
Explanation:

A constant function $ f(x) $ will have the result as a constant for any value of $ x $

So, if $ f(100) = 100 $
then $ f(2007 ) $ is also $ 100 $

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. $\displaystyle 2^{16}$


Correct Option: A
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.

On differentiating an identity function, we get?

  1. Signum function

  2. Sinc function

  3. Constant function

  4. None


Correct Option: C
Explanation:

Derivative of an Identity function gives a Constant function



$\dfrac{d(cx)}{dx} = c$ 
Where $c$ is a constant.

If $f,g,h$ are three functions from a set of positive real numbers into itself satisfying the condition,
$f(x) \cdot g(x)=h \sqrt{x^2 + y^2}$ such that $x,y \epsilon (0,\infty)$.then, $\dfrac{f(x)}{g(x)}$ is a?

  1. Constant function

  2. Identity function

  3. Zero function

  4. Signum function


Correct Option: A

A constant function is a periodic function.

  1. True

  2. False


Correct Option: A
Explanation:

$f(x)=f(x+a)$
A constant function is a period with any value of $'a'$ as a period

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. $1$


Correct Option: A

If $f\left( x \right)$ is a function satisfying  $f\left( x \right).f\left( {\frac{1}{x}} \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right)$ and $f\left( 4 \right) = 65$ then find $f\left( 6 \right)$

  1. $217$

  2. $215$

  3. $-216$

  4. $-217$


Correct Option: D

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. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$


Correct Option: C

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


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

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

  2. $f(\alpha)=10$

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

  4. none of these


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)