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Suppose f(x) is a continuous function such that $0.4\leq f(x) \leq 0.6 for 0 \leq x\leq 1.$. Which of the following is always true?

a) f(0.5)=0.5

b) There exists x between 0 and 1 such that f(x)=0.8x

c) There exists x between 0 and  0.5 such that f(x)=x

d) f(0.5)>0.5

e) None of the above statements are always true

How do we evaluate such ques?
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Why answer is B here ? Can anyone prove it ?

+1 vote

(A) f(0.5)=0.5, we cannot say here f(x) value always true Because we need to know f(x) value between

0.4≤ f(x) ≤ 0.6, and here we are getting f(x) value when x=0.5

(C)Here we know f(x) value between 0 to 0.5. But when f(x)=0.6 , x value may be ≥1

(D) Here also we cannot predict f(x) value when 0.4≤ f(x) ≤ 0.6

f(0.5)>0.5 is an inequality. So, we cannot get any exact value of x

Now for (B) Here we can see the f(x) value 0.4≤ f(x) ≤ 0.6 when x  between 0 to 1

for eg: f(0.5)=0.4, where x value is 0.5

f(0.6)=0.48,where x value is 0.6

f(0.7)=0.56 , where x value is 0.7

here we are only concern about f(x) is between 0.4 and 0.6.

so, here value of x always between 0 ≤ x ≤ 1 when 0.4≤ f(x) ≤ 0.6

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I could not understand the explanation for C option...could u please clarify it again ?
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if f(x)=0.6, then what could be x value? That is simple question here
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Okay since function is not defined we cannot be sure about its value.. Right?
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exactly
+2
Why answer is B here ? Can anyone prove it ? I don't think it is proved in your answer !
+1 vote

The straight line is function $y = 0.8*x$.

Now no matter how weird the function is, provided that it is continuous it has to cross the line $y = 0.8*x$ at least once. The point where it crosses aforementioned line is the point where $f(x) = 0.8x$.

Proof:

If we want $f(x) = 0.8*x$ we want $g(x) = f(x) - 0.8*x = 0$

To prove that $g(x)$ has a root in $[0,1]$, all we need to do is prove that there is $a$ and $b$, such that $0 \leq a, b \leq 1$, $g(a) < 0$ and $g(b) > 0$

Let's see.
at $x = 0$
the minimum g can take is $0.4 - 0 = 0.4 > 0$
the maximum g can take is $0.6 - 0 = 0.6 > 0$
so at 0 g is positive

at $x = 1$
the minimum g can take is $0.4 - 0.8 = -0.4 < 0$
the maximum g can take is $0.6 - 0.8 = -0.2 < 0$
so at 1 g is negative

thus by intermediate value theorem there is c, s.t. $0 < c < 1, g(c) = f(c) - 0.8*c = 0 \Rightarrow f(c) = 0.8*c$.

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