Test Index

Functions Part 1

Section: Mathematics

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Question : 51 of 100

Marks: +1, -0

Let

f: R \rightarrow R

f(x)=2x-1

g: R-\{1\} \rightarrow R

g(x)=\frac{x-\frac{1}{2}}{x-1}

Then, the composition function

f(g(x))

[24 Feb 2021 Shift 1]

one-one but not onto
onto but not one-one
Neither one-one nor onto
Both one-one and onto

Solution:

Given,

f(x)=2x-1 ; f: R \rightarrow R

g(x)\;\;=\;\frac{x-\frac{1}{2}}{x-1} ; g: R-\{1\} \rightarrow R

f[g(x)]\;\;=2g(x)-1

\;\;=2 \times \left(\;\frac{x-\frac{1}{2}}{x-1}\right)-1=2 \times \left(\;\frac{2x-1}{2(x-1)}\right)-1

\;\;=\;\frac{2x-1}{x-1}-1=\;\frac{2x-1-x+1}{x-1}=\;\frac{x}{x-1}

\therefore\;\; f[g(x)]\;\;=1+\;\frac{1}{x-1}

Now, draw the graph of

1+\;\frac{1}{x-1}

,

because

Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.Hence, the required function is one-one into.

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