Show that the function $f: R \rightarrow R$ defined as $f(x)=x^{2},$ is neither one-one nor onto.
Show that the function $f: R \rightarrow R$ defined as $f(x)=x^{2},$ is neither one-one nor onto.
since $f(-1)=1=f(1), \,f$ is not oneone. Also, the element $-2$ in the co-domain $R$ is not image of any element $x$ in the domain $R$ (Why ?). Therefore $f$ is not onto.
Similar Questions
Consider the function $f (x) = x^3 - 8x^2 + 20x -13$
Number of positive integers $x$ for which $f (x)$ is a prime number, is
Consider the function $f (x) = x^3 - 8x^2 + 20x -13$
Number of positive integers $x$ for which $f (x)$ is a prime number, is
If $f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x,x \ne 0$ and $S = \left\{ {x \in R:f\left( x \right) = f\left( { - x} \right)} \right\}$;then $S :$
If $f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x,x \ne 0$ and $S = \left\{ {x \in R:f\left( x \right) = f\left( { - x} \right)} \right\}$;then $S :$
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The range of the function,
$\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)$
$-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ is :
The range of the function,
$\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)$
$-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ is :
- [JEE MAIN 2021]
Let $f\left( n \right) = \left[ {\frac{1}{3} + \frac{{3n}}{{100}}} \right]n$ , where $[n]$ denotes the greatest integer less than or equal to $n$. Then $\sum\limits_{n = 1}^{56} {f\left( n \right)} $ is equal to
Let $f\left( n \right) = \left[ {\frac{1}{3} + \frac{{3n}}{{100}}} \right]n$ , where $[n]$ denotes the greatest integer less than or equal to $n$. Then $\sum\limits_{n = 1}^{56} {f\left( n \right)} $ is equal to
- [JEE MAIN 2014]
Solve $|x\,-\,2| + |x\,-\,1| = x\,-\,3$
Solve $|x\,-\,2| + |x\,-\,1| = x\,-\,3$