Show that the function $f : R \rightarrow R$ given by $f(x) = x^{3}$ is injective.
(N/A) To check if the function $f(x) = x^{3}$ is injective (one-one),we assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in R$.
$f(x_1) = f(x_2) \Rightarrow x_1^{3} = x_2^{3}$.
Taking the cube root on both sides,we get $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in R$,the function $f$ is injective.
$f(x_1) = f(x_2) \Rightarrow x_1^{3} = x_2^{3}$.
Taking the cube root on both sides,we get $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in R$,the function $f$ is injective.
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The function $f: C \rightarrow C$ defined by $f(x) = \frac{ax + b}{cx + d}$ for $x \in C$,where $ad - bc \neq 0$,reduces to a constant function if:
If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x+4 & \text{for } x < -4 \\ 3x+2 & \text{for } -4 \leq x < 4 \\ x-4 & \text{for } x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is:
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| List-$I$ | List-$II$ |
| $(A)$ $f(-5) + f(-4)$ | $(i)$ $14$ |
| $(B)$ $f(|f(-8)|)$ | $(ii)$ $4$ |
| $(C)$ $f(f(-7) + f(3))$ | $(iii)$ $-11$ |
| $(D)$ $f(f(f(f(0)))) + 1$ | $(iv)$ $-1$ |
| $(v)$ $1$ | |
| $(vi)$ $0$ |
DifficultAP EAMCET 2006
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MediumAIEEE 2012
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