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Problem 36:  Root counting ( ) 1999 Paper III

Let f(x) = (1 + x2)ex k, where k is a given constant. Show that f(x) 0 and sketch the graph of f(x) in the three cases k < 0,   0 < k < 1 and k > 1.

Hence, or otherwise, show that the equation

(1 + x2)ex k = 0

has exactly one real root if k > 0 and no real roots if k 0.

Determine the number of real roots of the equation (ex 1) ktan1x = 0

in the different cases that arise according to the value of the constant k.

Note: If y = tan1x,  then dy dx = 1 1 + x2.


In good STEP style, this question has two related parts. In this case, the first part not only gives you guidance for the second part, but also provides a result that is useful for the second part.

In part (ii) of the original question you were asked to determine the number of real roots in the cases 0 < k 2π  and  2π < k < 1. I thought that you would like to work out all the different cases for yourself — but it makes the question considerably harder, especially if you think about the special values of k as well as the ranges of k.

For part (i) you need to know that xex 0 as x . This is just a special case of the result that exponentials go to zero faster than any power of x.

For part (ii), it is not really necessary to do all the sketches, but I think you should (though you perhaps wouldn’t under examination conditions) because it gives you a complete understanding of the way the function depends on the parameter k. For the graphs, you may want to consider the sign of f(0); this will give you an important clue. Remember that, by definition, 1 2π < tan1x < 1 2π.

Solution to problem 36

(i) Differentiating gives f(x) = 2xex + (1 + x2)ex = (1 + x)2ex which is non-negative because the square is non-negative and the exponential is positive.

There is one stationary point, at x = 1, which is a point of inflection, since the gradient on either side is positive.

Also, f(x) k as x , f(0) = 1 k and f(x) as x +. The graph is essentially exponential, with a hiccup at x = 1. The differences between the three cases are the positions of the horizontal asymptote (x ) and the place where the graph cuts the y axis (above or below the x axis?).

Since the graph is increasing and 0 < f(x) < , the given equation has one real root if k > 0 and no real roots if k 0.


(ii) First we collect up information required to sketch the graph, as in part (i). We have

f(0) = 0,    f(x) 1 2kπ 1 as x ,    f(x) as x ,    f(x) = ex k(1 + x2)1 (which we know from the first part can only be zero only if k > 0),    f(0) = 1 k.

The important values of k seem to be 0, 2π and 1 so we will have to consider four cases.



As you see, the equation (ex 1) ktan1x = 0 has one root if k 1 2π and two roots otherwise. The case k = 1 can be thought of as having one root at x = 0 or as having two roots, both at x = 0 (it is a double root).


I think I made a bit of a meal of this; got carried away with the sketches. But it was reassuring (compulsive, I found) to consider every case. My explanation is pretty compact, so I think you’ll have to work it through on your own to convince yourself of the details.