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Question: If \(f\) and \(g\) are one- one functions from \(R \to R\), then, A. \[f + g\] is one-one B. \(f...

If ff and gg are one- one functions from R→RR \to R, then,
A. f+gf + g is one-one
B. fgfg is one-one
C. fogfog is one-one
D. none of these$$$$

Explanation

Solution

Since, ff and gg are one- one functions from R→RR \to R, then if x1≠x2{x_1} \ne {x_2}, then f(x1)≠f(x2)f\left( {{x_1}} \right) \ne f\left( {{x_2}} \right) and g(x1)≠g(x2)g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right). Let f(x)=−xf\left( x \right) = - x and g(x)=xg\left( x \right) = x and then determine whether f+gf + g and fgfg are one- one functions by taking examples. Similarly, prove if fogfog is one-one using the definition of one-one function.

Complete step-by-step answer:
We are given that ff and gg are one- one functions from R→RR \to R
That is, if x1≠x2{x_1} \ne {x_2}, then f(x1)≠f(x2)f\left( {{x_1}} \right) \ne f\left( {{x_2}} \right) and g(x1)≠g(x2)g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)
Let us first see if f+gf + g is one-one or not.
Let f(x)=−xf\left( x \right) = - x and g(x)=xg\left( x \right) = x
Then, f(x1)=f(x2)f\left( {{x_1}} \right) = f\left( {{x_2}} \right) if and only if x1=x2{x_1} = {x_2}
Similarly, g(x1)=g(x2)g\left( {{x_1}} \right) = g\left( {{x_2}} \right) if and only if x1=x2{x_1} = {x_2}
That is ff and gg are one-one functions.
Then, f+g(x)=−x+x=0f + g\left( x \right) = - x + x = 0
For example, f+g(−2)=f+g(1)=0f + g\left( { - 2} \right) = f + g\left( 1 \right) = 0 but −2≠1 - 2 \ne 1
Thus, f+gf + g is not none-one.

Similarly, f.g(x)=−x(x)=−x2f.g\left( x \right) = - x\left( x \right) = - {x^2}
Let x1=2,x2=−2{x_1} = 2,{x_2} = - 2, then,
f.g(2)=−(2)2=−4f.g\left( 2 \right) = - {\left( 2 \right)^2} = - 4 and f.g(−2)=−(−2)2=−4f.g\left( { - 2} \right) = - {\left( { - 2} \right)^2} = - 4
Here, fg(2)=fg(−2)=−4fg\left( 2 \right) = fg\left( { - 2} \right) = - 4 but 2≠−22 \ne - 2
Therefore, fgfg is not one-one.

Now, if x1≠x2{x_1} \ne {x_2}, then g(x1)≠g(x2)g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right) which is a given condition.
And if g(x1)≠g(x2)g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right) then f(g(x1))≠f(g(x2))f\left( {g\left( {{x_1}} \right)} \right) \ne f\left( {g\left( {{x_2}} \right)} \right) because ff is a one-one function.
Hence, fogfog is one-one.
Thus, option C is correct.

Note: If a function is one-one, then each element of the domain has a unique image. Not two different elements can map to the same element. Examples should be taken carefully such that it helps to check if the function is one-one or not.