Problem 36:  Root counting ($✓$ $✓$ $✓$) 1999 Paper III

(i)
Let $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)=\left(1+{x}^{2}\right){e}^{x}-k\phantom{\rule{0.3em}{0ex}}$, where $k$ is a given constant. Show that ${\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}}^{\prime }\left(x\right)\ge 0$ and sketch the graph of $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)\phantom{\rule{0.3em}{0ex}}$ in the three cases $k<0\phantom{\rule{0.3em}{0ex}}$,   $0 and $k>1\phantom{\rule{0.3em}{0ex}}$.

Hence, or otherwise, show that the equation

$\left(1+{x}^{2}\right){\mathrm{e}}^{x}-k=0\phantom{\rule{2.77695pt}{0ex}}$

has exactly one real root if $k>0$ and no real roots if $k\le 0\phantom{\rule{0.3em}{0ex}}$.

(ii)
Determine the number of real roots of the equation $\left({\mathrm{e}}^{x}-1\right)-k{tan}^{-1}x=0$

in the different cases that arise according to the value of the constant $k\phantom{\rule{0.3em}{0ex}}$.

Note: If $y={tan}^{-1}x$,  then $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{1+{x}^{2}}\phantom{\rule{0.3em}{0ex}}$.

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  and  $2∕\pi . 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 $k\phantom{\rule{0.3em}{0ex}}$.

For part (i) you need to know that $x{\mathrm{e}}^{x}\to 0$ as $x\to -\infty \phantom{\rule{0.3em}{0ex}}$. This is just a special case of the result that exponentials go to zero faster than any power of $x\phantom{\rule{0.3em}{0ex}}$.

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 $k\phantom{\rule{0.3em}{0ex}}$. For the graphs, you may want to consider the sign of ${\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}}^{\prime }\left(0\right)\phantom{\rule{0.3em}{0ex}}$; this will give you an important clue. Remember that, by deﬁnition, $-\frac{1}{2}\pi <{tan}^{-1}x<\frac{1}{2}\pi \phantom{\rule{0.3em}{0ex}}$.

Solution to problem 36

(i) Differentiating gives ${\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}}^{\prime }\left(x\right)=2x{\mathrm{e}}^{x}+\left(1+{x}^{2}\right){e}^{x}={\left(1+x\right)}^{2}{\mathrm{e}}^{x}$ which is non-negative because the square is non-negative and the exponential is positive.

There is one stationary point, at $x=-1\phantom{\rule{0.3em}{0ex}}$, which is a point of inﬂection, since the gradient on either side is positive.

Also, $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)\to -k$ as $x\to -\infty \phantom{\rule{0.3em}{0ex}}$, $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(0\right)=1-k$ and $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)\to \infty$ as $x\to +\infty \phantom{\rule{0.3em}{0ex}}$. The graph is essentially exponential, with a hiccup at $x=-1\phantom{\rule{0.3em}{0ex}}$. The differences between the three cases are the positions of the horizontal asymptote ($x\to -\infty$) and the place where the graph cuts the $y$ axis (above or below the $x$ axis?).

Since the graph is increasing and $0<\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)<\infty$, the given equation has one real root if $k>0$ and no real roots if $k\le 0$.

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

$\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(0\right)=0$,    $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)\to \frac{1}{2}k\pi -1$ as $x\to -\infty \phantom{\rule{0.3em}{0ex}}$,    $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)\to \infty$ as $x\to \infty \phantom{\rule{0.3em}{0ex}}$,    ${\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}}^{\prime }\left(x\right)={e}^{x}-k{\left(1+{x}^{2}\right)}^{-1}\phantom{\rule{0.3em}{0ex}}$ (which we know from the ﬁrst part can only be zero only if $k>0\phantom{\rule{0.3em}{0ex}}$),    ${\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}}^{\prime }\left(0\right)=1-k\phantom{\rule{0.3em}{0ex}}$.

The important values of $k$ seem to be $0\phantom{\rule{0.3em}{0ex}}$, $2∕\pi$ and $1\phantom{\rule{0.3em}{0ex}}$ so we will have to consider four cases.

As you see, the equation $\left({\mathrm{e}}^{x}-1\right)-k{tan}^{-1}x=0$ has one root if $k\le \frac{1}{2}\pi$ 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).

Post-mortem

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.