I.S.I. 2015 subjective solution (Problem 3)


Teacher: You may use the notion of derivative of a function which is \(\displaystyle {\lim_ {y to x} \frac{f(x) – f(y)} {x-y} } \) . Though it is not given that the function is differentiable, we can use this as a motivating factor to ‘see’ what is happening.

Student: Okay. So in case the function is differentiable then is the derivative of the function in question is +2 or -2?

Teacher: That is right but is it possible that function has derivative +2 in some interval and -2 interval in some other interval?

Student: It is possible, for example f(x) = 2x if x > 0 and -2x other wise. But I don’t think that a function like this will meet the conditions of this problem for all sets of points. I think here we are dealing with a ‘constant slope’ function.

Teacher: How can you prove that?

Student: Let me try. Suppose a be an arbitrary point in the domain. Now we will consider two other points, a+t and a-k in domain such that t and k are positive quantities. We consider two cases: Case 1 – constant slope and Case 2 – changing slope

Without loss of generality let us assume that f(a+t) – f(a) = 2t.

Case 1: Constant Slope

If f(a) – f(a-k) = 2k then we see that

f(a+t) – f(a) + f(a) – f(a-k) = 2t + 2k or f(a+t) – f(a-k) = 2{(a+t) – (a-k)} hence it is consistent with the problem’s statement.

Case 2: Non constant slope

However if f(a) – f(a-k) = -2k then

f(a+t) – f(a-k) = 2t – 2k …(i)

But we know that |f(a+t) – f(a-k)| = 2|(a+t) – (a-k)| = 2(t+k). … (ii)

Combining (i) and (ii) we conclude that either

2t – 2k = 2t + 2k => 4k = 0 => k = 0


2k – 2t = 2t + 2k => 4t = 0 => t = 0

But both t and k are positive quantities hence we arrive at a contradiction by assuming +2 and -2 slopes for two different pairs of points.

I think this proves that slope of function concerned is constant. So we can now say that it is differentiable and f'(x) = 2 or -2

Teacher: Very nice! Now this problem can be easily completed by solving the differential equation f’(x) = 2 or -2.

f(x) must be 2x + c or -2x + c where c is an arbitrary constant.

16 Replies to “I.S.I. 2015 subjective solution (Problem 3)”

  1. I have doubt! How can we assume that, the function is differentiable at any point! It is not given in the question! Can’t we give this kind of arguments, like-
    {f(x_1)-f(x_2)}/{x_1 – x_2} is the slope of the line joining of the points (x_1, f(x_1) ) and (x_2, f(x_2) ) which is constant i.e. if we fix a point (x_1, f(x_1) ) on the graph of the function and move the other point (x_2, f(x_2) ), then the slope will be always constant. Hence, the function is a linear function. Now the slope is +or -2. Hence f(x)=2x+c or -2x+c.
    Is this argument correct?

    1. Debjit, you are right. However in the argument that we used, if you see carefully, we have not used the fact that f is differentiable. We used it as a motivating factor to ‘see’ inside the problem. Rest of the solution rests on simple arithmetic manipulation. I have edited the solution to make the point clear

    2. Also Debjit, concerning your solution: the fine point of this problem is that the ‘modulus’ of the slope is 2 (and not the slope). Hence we may have two different slopes for two different pairs of point. Trick is to show that this not the case. You may read the revised solution presented above.

      1. Yes, “‘modulus’ of the slope is 2 (and not the slope)”. So, we can consider two cases- (i) slope is +2 and (ii) slope is -2. That is why two different general forms of the function is coming! 🙂

        1. True Debjit. But how do we know that a single function does not have both the slopes in two separate intervals? How do we know that it has either exactly one slope through out?

        2. Also it is easy to see that if one says : oh then the function has either +2 or -2 slope then it must be linear etc.; it is too easy be an I.S.I problem. Most students will miss the part that one has to establish that a single function satisfying the given criterion cannot have both the slopes for pairs of points in its domain

      2. Dear Sir,
        I thought it obvious that the slope is constant throughout any interval because in that case the graph of a function will be like a mountain and the line joining two points in intervals having different slopes cannot be equal to either +2 or -2 . So I didnt prove that part and merely wrote that slope is constant . How many marks will I be awarded for this question


        1. Biswarup,

          The hunch is that it will be like a mountain… and that is a right guess. But you have to mathematically establish that.

          I am not sure how much marks will be deducted for missing that part. Hopefully you will get some marks.

      3. Dear Sir,

        This was the thought in my mind at that moment of examination ( but unfortunately I didnt think it important to write this in the answer sheet) :

        Let us assume that the slope is not constant in a particular interval [a,b]

        Let us WLOG take the largest possible subinterval [a,t] such that the slope of f on [a,t] is 2.
        Then let us take the largest possible subinterval [t,t1] such that the slope of f on [t,t1] is -2.
        Let us take 2 take two points a1 and b1 on the intervals [a,t] and [t,t1] respectively.

        Then f(a1)=f(t)-2(t-a1) and f(b1)=f(t)-2(b1-t)

        Then f(b1)-f(a1)=2(2t-a1-b1) and it easily follows that slope is neither +2 or -2

      4. Dear Sir ,

        What do think the cutoff is going to be this time around. Thank you very much if you could kindly reply

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