# ISI2014-DCG-33

150 views

Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties.

1. $f(0)=0$,
2. $f(1)=1$, and
3. $f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_2<1$.

Then the number of such functions is

1. $0$
2. $1$
3. $2$
4. $\infty$
in Calculus
recategorized
0
0
Now, it is fixed.
0
all series of polynomial functions? like x^2 , x^3 ?

## Related questions

1 vote
1
108 views
Let $f: \bigg( – \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg) \to \mathbb{R}$ be a continuous function, $f(x) \to +\infty$ as $x \to \dfrac{\pi^-}{2}$ and $f(x) \to – \infty$ as $x \to -\dfrac{\pi^+}{2}$. Which one of the following functions satisfies the above properties of $f(x)$? $\cos x$ $\tan x$ $\tan^{-1} x$ $\sin x$
Let $f(x) = \begin{cases}\mid \:x \mid +1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \mid \:x \mid -1, & \text{ if } x>0. \end{cases}$ Then $\underset{x \to a}{\lim} f(x)$ exists if $a=0$ for all $a \in R$ for all $a \neq 0$ only if $a=1$
Let $a_n=\bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1 – \frac{1}{\sqrt{n+1}} \bigg), \: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$
$\underset{x \to \infty}{\lim} \left( \frac{3x-1}{3x+1} \right) ^{4x}$ equals $1$ $0$ $e^{-8/3}$ $e^{4/9}$