GATE CSE
First time here? Checkout the FAQ!
x
0 votes
136 views

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 by Veteran (29.1k points)   | 136 views

2 Answers

+3 votes

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 $x =0.5$, $f(x) \geq 0.4$ aand 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.

answered by Veteran (281k points)  
@Arjun Sir why u took range in between $[0.5,0.75]$, means here told x can be in between $[0,1]$.
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.
0 votes

(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

So, answer will be (B)

answered by Veteran (52.4k points)  
you haven't proved the existence of such an $x$.
then whats the answer sir ? @ Arjun
option B is correct only, I have given answer now.


Top Users Mar 2017
  1. rude

    4272 Points

  2. sh!va

    2994 Points

  3. Rahul Jain25

    2804 Points

  4. Kapil

    2608 Points

  5. Debashish Deka

    2244 Points

  6. 2018

    1414 Points

  7. Vignesh Sekar

    1338 Points

  8. Akriti sood

    1246 Points

  9. Bikram

    1246 Points

  10. Sanjay Sharma

    1016 Points

Monthly Topper: Rs. 500 gift card

21,452 questions
26,771 answers
60,972 comments
22,985 users