GATE Overflow Test Series | Discrete Mathematics | Test 1 | Question: 22

Question

Consider the following function $f$ on $\left[-\frac{1}{2},3\right] \rightarrow [-1,1]$ such that $f(x) = \frac{2x-1}{4x+5}$. Which of the following statements is TRUE?

• A. $f$ is not one-one but is onto
• B. $f$ is neither one-one nor onto
• C. $f$ is bijective
• D. $f$ is one-one but not onto

Comments

In addition to the answer below: if you assign $f(x) = \frac{2x – 1}{4x + 5} = 1$ and try to find a suitable value for $x$, you will find that $x = -3$, which is not in the domain of $f$. Thus $1$ will never have a pre-image and the function is not onto.

Answer 1

To prove one-one. Let $f(x) = f(y).$

We get $\frac{2x-1}{4x+5} = \frac{2y-1}{4y+5}$

$\implies (2x-1)(4y+5) = (4x+5)(2y-1)$

$\implies 8xy +10x - 4y -5 = 8xy -4x +10y - 5$

$\implies 14x = 14y \implies x = y.$

So, $f$ is one-one.

For onto, we have to first find $x$ in terms of $f(x).$ Lets take $f(x) = y.$ So,
 $y = \frac{2x-1}{4x+5}$
 
 $\implies 2y = \frac{4x-2}{4x+5}$

 $\implies \frac{2y +1}{2y-1} = \frac{4x-2}{4x+5}$

 $\left[\because \frac{a}{b} = \frac{c}{d} \implies \frac{a+b}{a-b} = \frac{c+d}{c-d}\right]$
 
 $\implies \frac{2y +1}{2y-1} = \frac{8x+3}{-7}$
 
 $\implies 8x = \frac{-14y -7}{2y-1} - 3 $

 $\implies 8x = \frac{-14y -7-6y+3}{2y-1}  $

 $\implies 8x = \frac{-20y -4}{2y-1}  $
 
 $\implies x = \frac{-5y -1}{4y-2} = \frac{5y +1}{2-4y} $
 
 Now, our range is  $[-1,1]$ and so lets take any value in between that. So, we get $(5y+1)$ between $-4$ (putting $y=-1)$ and $6$ (putting $y=1)$ and $(2-4y)$ between $6$ and $2.$ Now, when $y = 0.5,$ we get $0$ in the denominator meaning we don't have an $x$ value in the range $\left[-\frac{1}{2},3\right].$ So, $f$ is not onto.

Comments

@samarpita

In the given function $f,$ there is NO pre-image of element $0.5.$

Assume, for some $x$ in the domain, we have $f(x) = 0.5,$ then we get $5=-2$ which is a contradiction.

Hence, the given function is Not Onto. 

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@Deepak Poonia but sir how the range [-½,3] is coming?

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In the given answer:

That's not Range of f.

That's the English word "range", Not the mathematical word " range".