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Let $g$ be a function defined on the interval $[0,2]$ and $x \le g(x) \le x^2-x+1$ for $0 \le x \le 2$. Then

$(A)$ $g$ must be a polynomial.

$(B)$ $g$ must be continuous at $x = 1$.

$(C)$ $g$ must be continuous at $x = 0$ and $x = 2$.

$(D)$ $g$ must be a continuous function.

Since $\lim_{x\to 1} x = 1$ and $\lim_{x\to 1} x^2 - x + 1 = 1$. So by sandwich rule of functions $g$ is continuous at $x = 1$. So $B$ option is right.

Since $\lim_{x\to 0} x = 0$ and $\lim_{x\to 0} x^2 - x + 1 = 1$. So I am not able to conclude anything about continuity at $x = 0$.

Further I draw the graph of $x$ and $x^2 - x + 1$ and conclude that we can define a function satisfy given condition is possible but i did not define this function. So Plz help me to check out other options.

Deepak Gupta
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  • Well if you can rule out continuity at any point, then quite obviously it is not a polynomial... – Macavity Oct 23 '17 at 14:30
  • B is correct as you said. You cannot be sure for the others(limited information). So the others are incorrect, you can find functions that don't satisfy A,C or D – mxaxc Oct 23 '17 at 14:34

2 Answers2

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You have already proved (B) is correct. For the other ones, consider the function $$g(x)=\begin{cases} 1 & \text{if } x=0\\ x &\text{if } x\in (0,2) \\ 3 & \text{if } x= 2\end{cases}$$

Surb
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  • Thanks sir . I think we can define more function satisfying the given conditions. – Deepak Gupta Oct 23 '17 at 14:48
  • @DeepakGupta There are indeed infinitely many functions with these properties, already in the example above, you could have taken any choice of $g(0)\in(0,1)$ and $g(2)\in (2,3)$ to derive the same conclusions. – Surb Oct 23 '17 at 15:44
  • Okk sir. I think also same. Thanks again. – Deepak Gupta Oct 23 '17 at 15:51
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None of the four properties changes if you pass to $h(x) := g(x)-x$. Then your condition is $$0 \leq h(x) \leq (x-1)^2 \text{ for } x \in [0,2].$$

Hence obviously (B) holds while all other things do not even make sense.

MooS
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