- Show that, for ,
the largest value of
What is the smallest value?
- Find constants ,
such that, for all ,
- Hence, or otherwise, prove that
You should think about part (i) graphically, though it is not necessary to draw the graph: just set about it as if you were going to (starting point, ﬁnishing point, turning points, etc).
The equivalence sign in part (ii) indicates an equality that holds for all — you are not being asked to solve the equation for .
For part (iii), you need to know that inequalities can be integrated: this is ‘obvious’ if you think about integration in terms of area, though a formal proof requires a formal deﬁnition of integration and this is the sort of thing you would do in a ﬁrst university course in mathematical analysis.
Solution to problem 40
(i) Let Then
which is positive for . Therefore, increases in value from at to at .
(ii) Doing the differentiation gives
and multiplying by gives the following identity:
Equating coefficients of the different powers of on each side of the equivalence sign gives , , , , so , , and .
(iii) Using the results of parts (i) and (ii), we see that for
Inequalities can be integrated, so
from which the required result follows immediately.
Instead of equating coefficients in (), you could obtain equations for , , and by putting four carefully chosen values of into the equation. An obvious choice is , but (thinking ﬂexibly!) you could try to eliminate terms with factors of . Then it becomes more difficult to ﬁnd good choices.
You might wonder why, in part (ii), the term inside the derivative has only odd powers of . Would it not make it more general to include even powers as well? You could in fact include even powers but you would ﬁnd that their coefficients would be zero: all the other terms in the equation are even in , so the derivative has to be an even function which means that the function being differentiated must be odd.
Although the idea of this question is good, the ﬁnal result is a bit feeble. It only gives the value of the integral to an accuracy of about which is about . The actual value of the integral can be found fairly easily using the substitution and is so the inequalities can be used to give (rather bad) estimates for namely .