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Let \(f: R \to R \), be defined as \(f(x) = x^4\). Choose the correct answer.

\begin{array}{1 1}(A)\;f\;is\;one-one\;onto. & (B)\;f\;is\;many-one\;onto.\\(C) \;f\;is\;one-one\;but\;not\;onto. & (D)\;f\;is\;neither\;one-one\;nor\; onto.\end{array}

1 Answer

  • A function $f: X \rightarrow Y$ is one to one or injective function if $\forall x_1, x_2 \in X, f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
  • A function$ f : X \rightarrow Y$ is said to be onto or surjective, if for every element say $(y)$ of $ Y$ there exists an element say $(x)$ in $X$ so that $y$ is the image of $x$ under $ f$, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f: R \to R \,be\, defined\, as\,f(x) = x^4$
Step1: Injective or One-One function:
For an injective or one-one function, Let $f(x_1) = f(x_2)$ for any $x_1,x_2 \in R$
$\Rightarrow\: x_1^4=x_2^4$
$\Rightarrow\:x_1=\pm x_2$
For example $2,-2 \in R $ and $f(-2)=f(2)=2^4=16$ but $-2\neq2$
$\therefore $f: R \to R \,be\, defined\, as\,f(x) = x^4$is not one-one
Step 2: Surjective or On-to function:
Consider an element $-2 \in R$.
But there is no real number $x$ such that $f(x)=x^4=-2$
$therefore\: f: R \to R \,\, defined\, as\,f(x) = x^4$is not onto.
$\therefore\:f: R \to R \,be\, defined\, as\,f(x) = x^4$is neither one one not onto
Solution:'D' option is correct


answered Mar 19, 2013 by thagee.vedartham
edited Dec 26, 2013 by rvidyagovindarajan_1

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