JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 4)
Let f : R → R be defined as f (x) = 2x – 1 and g : R - {1} → R be defined as g(x) =
$${{x - {1 \over 2}} \over {x - 1}}$$.
Then the composition function f(g(x)) is :
one-one but not onto
onto but not one-one
both one-one and onto
neither one-one nor onto
Explanation
Given, f(x) = 2x $$-$$ 1; f : R $$\to$$ R
$$g(x) = {{x - 1/2} \over {x - 1}};g:R - \{ 1) \to R$$
$$f[g(x)] = 2g(x) - 1$$
$$ = 2 \times \left( {{{x - {1 \over 2}} \over {x - 1}}} \right) - 1 = 2 \times \left( {{{2x - 1} \over {2(x - 1)}}} \right) - 1$$
$$ = {{2x - 1} \over {x - 1}} - 1 = {{2x - 1 - x + 1} \over {x - 1}} = {x \over {x - 1}}$$
$$\therefore$$ $$f[g(x)] = 1 + {1 \over {x - 1}}$$
Now, draw the graph of $$1 + {1 \over {x - 1'}}$$
_24th_February_Morning_Shift_en_4_1.png)
$$\because$$ 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.
$$g(x) = {{x - 1/2} \over {x - 1}};g:R - \{ 1) \to R$$
$$f[g(x)] = 2g(x) - 1$$
$$ = 2 \times \left( {{{x - {1 \over 2}} \over {x - 1}}} \right) - 1 = 2 \times \left( {{{2x - 1} \over {2(x - 1)}}} \right) - 1$$
$$ = {{2x - 1} \over {x - 1}} - 1 = {{2x - 1 - x + 1} \over {x - 1}} = {x \over {x - 1}}$$
$$\therefore$$ $$f[g(x)] = 1 + {1 \over {x - 1}}$$
Now, draw the graph of $$1 + {1 \over {x - 1'}}$$
$$\because$$ 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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