Tag: graphs of the form y=ax^2+bx+c

Questions Related to graphs of the form y=ax^2+bx+c

If $p = q$ then $px =$ ________

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

  1. $q$

  2. $qx$

  3. $q + x$

  4. $0$

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Correct Option: B
Explanation:

Given $p = q$
multiply both sides by $x$, we get
$px = qx$, which is required value.

Which of the following functions are identity functions?

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

  1. $f:R\rightarrow R, f(x) = x$

  2. $g : N \rightarrow Z, g(p)= 3$

  3. $h:z \rightarrow z, h(y)=y$

  4. $g:N\rightarrow N, g(z) =z$

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Correct Option: A,C,D
Explanation:

An identity function is a function that always returns the same value that was used as its argument.

Hence $f:R\rightarrow R,f(x)=x$ is an identity function

If ${ (x, 2), (4, y) }$ represents an identity function, then $( x, y)$ 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. (2, 4)

  2. (4, 2)

  3. (2, 2)

  4. (4, 4)

Show answer
Correct Option: A
Explanation:

Identity funtion returns the same value as of input.

Since input $x$ gives output $2$  
$\therefore$ $x=2$
Similarly input of $4$ gives output $y$
$\therefore y=4$
$\Rightarrow (x,y)=(2,4)$

Option (a) is correct.

Which of the following functions is/are constant ?

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

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

  2. $f(x)=x+\dfrac{1}{x}$

  3. $f(x)=7$

  4. $f(x)=6+x$

Show answer
Correct Option: C
Explanation:

$f(x)=7$ is constant function as its values do not depend on the variable $x$.

Its value is $7$ for any value of $x$.
Option $C$ is correct.

An identity function is a?

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

  1. Many to many function

  2. One to One function

  3. Many to one function

  4. None

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Correct Option: B
Explanation:

An identity function is of the form $f(x) = cx$


A one-to-one function is a function that preserves distinctness; it never maps distinct elements of its domain to the same element of its codomain.

In other words, every element of the function's codomain is the image of at most one element of its domain.

As $f(x) =cx$ has a different value at every value of $x$, it is a One to One function.

State whether the following statement is True or False.
The inverse of an identity function is the identity function itself.

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

  1. True

  2. False

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Correct Option: A
Explanation:

An identity function $f(x)$ is of the form $f(x) = cx$

Let inverse of $f(x)$ be $g(x)$
$\therefore g(x) = \dfrac{x}{c}$ (Identity Function)
Thus inverse of an identity function is the identity function itself.

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function such that for any irrational number $r,$ and any real number $x$ we have $f(x)=f(x+r)$. Then, $f$ 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. an identity function

  2. a constant function

  3. a zero function

  4. onto function

Show answer
Correct Option: B
Explanation:

A constant function is a function that has the same output value no matter what your input value is. Because of this, a constant function has the form y=k where k is a constant(a single value that doesn't change).

So in here our input is $x$ i f we change our input to say $x+r$
and still output does'nt change i.e.
$f(x)=f(x+r)$
it means it is a constant function,
example:
$ f(x)=5=5(x)^{0} $
$ f(x+r)=5(x+r)^{0}$
$ f(x+r)=5\times 1 $
$ f(x+r)=5=f(x) $

The graph of an Identity function 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. A straight line parallel to X axis

  2. A straight line parallel to Y axis

  3. A straight line passing through the origin

  4. None

Show answer
Correct Option: C
Explanation:

Identify function ,$f(x)=x$

$y=x$
So,the graph of an identity function is a straight line passing through the origin.

Let $f$ be a linear function for which $f (6)  - f (2) = 12$. The value of $f (12) - f(2)$ is equal the 

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

  1. $12$

  2. $18$

  3. $24$

  4. $30$

Show answer
Correct Option: D
Explanation:

Let $f(x)=ax+b$ [Since $f$ is given to be linear function] where $a$ and $b$ are constants.

According to the problem 
$f(6)-f(2)=12$
or, $6a-b-2a-b=12$
or, $4a=12$
or, $a=3$.
Now 
$f(12)-f(2)$
$=12a+b-2a-b$
$=10a=30$. [Using value of $a$]

The set values of $x$ for which function $f(x)=x\ln {x}-x+1$

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

  1. $\left( 1,\infty \right) $

  2. $\left( \cfrac { 1 }{ e } ,\infty \right) $

  3. $[e,\infty )$

  4. $\left( 0,1 \right) \cup \left( 1,\infty \right) $

Show answer
Correct Option: A
Explanation:
$f(x)=x \ln x -x +1$

we use the formula

$\log _{a}f(x)\Rightarrow f(x)>0$

$\Rightarrow x>0$

$\therefore x>0\Rightarrow (1,\infty )$