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Suppose that $f(x)$ 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?

1. $f(0.5) = 0.5$.
2. There exists $x$ between $0$ and $1$ such that $f(x) = 0.8x$.
3. There exists $x$ between $0$ and $0.5$ such that $f(x) = x$.
4. $f(0.5) > 0.5$.
5. None of the above statements are always true.
asked in Calculus | 532 views
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Why answer is B here ? Can anyone prove it ?

This is a repeating question on continuity. Let me solve it a non-standard way -- which should be useful in GATE.

From the question $f$ is a function mapping the set of real (or rational) numbers between [0,1] to [0.4,0.6]. So, clearly the co-domain here is smaller than the domain set. The function is not given as onto and so, there is no requirement that all elements in co-domain set be mapped to by the domain set. We are half done now. Lets see the options:

A. $f(0.5) = 0.5$. False, as we can have $f(0.5) = 0.4$, continuity does not imply anything other than all points being mapped being continuous.

C. Again false, we can have $f(x) = 0.6$ for all $x$.

D. False, same reason as for A.

Only B option left- which needs to be proved as correct now since we also have E option. We know that for a function all elements in domain set must have a mapping. All these can map to either 1 or more elements but at least one element must be there in the range set. i.e., $f(x) = y$ is true for some $y$ which is in $[0.4, 0.6]$. In the minimal case this is a single element say $c$. Now for $x = 1/0.8$, option B is true. In the other case, say the minimal value of $f(x) = a$ and the maximum value be $f(x) = b$. Now,

as per Intermediate Value theorem (see: https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch7.pdf),  all points between $a$ and $b$ are also in the range set as $f$ is continuous. Now, we need to consider $x$ in the range $[0.5, 0.75]$ as then only $f(x)$ can be $0.8 x$ and be in $[0.4, 0.6]$. In our case we have

$f(x_1) = a, f(x_2) = b$. Lets assume $a != 0.8x_1$ and $b!=0.8x_2$. Now, for all other points in $[0.5, 0.75]$, $f(x)$ must be between $a$ and $b$ and all points between $a$ and $b$ must be mapped by some $x$.

Moreover, for $x =0.5$, $f(x) \geq 0.4$ and for $x=0.75$, $f(x) \leq 0.6$. So, if we plot $g(x) = 0.8x$, this line should cross $f(x)$ at some point between $0.5$ and $0.75$ because at $x= 0.5$, $f(x)$ must be above or equal to the line $0.8x$ (shown below) and for $x = 0.75$ it must be below or equal which means an intersection must be there. This shows there exist some $x$ between $0.5$ and $0.75$ for which $f(x) = 0.8x$ a stronger case than option B. So, B option is true. Now please try for $f(x) = 0.9x$ and see if it is true.

edited
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@Arjun Sir why u took range in between $[0.5,0.75]$, means here told x can be in between $[0,1]$.
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I did not take that range-- but that is the range within which 0.8x can be within 0.4 and 0.6 and hence the intersection point 'x' must be within 0.5-0.75.
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y=mx ; where m>=0.6

then 100% this straight line and our function f(x) will intersect each other at some point b/w [0,1]

so if y=mx ; m>=0.6 then there exists x b/w 0 and 1 such that y=mx.

tell me if i'm making some wrong claim.

Another approach...

It may be possible that... -> +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

edited by
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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. $0 < c < 1, g(c) = f(c) - 0.8*c = 0 \Rightarrow f(c) = 0.8*c$.

edited

(A) f(0.5)=0.5, we cannot say here f(x) value always trueBecause 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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you haven't proved the existence of such an $x$.
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then whats the answer sir ? @ Arjun
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option B is correct only, I have given answer now.
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@srestha  ma,am you had not provided any reason to eliminate e, you are choosing b by eliminating a,c,d.

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+1 vote