Year: 2010
Paper: 1
Question Number: 4
Course: LFM Pure
Section: Integration
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Use the substitution $x=\dfrac{1}{t^{2}-1}\; $, where $t>1$,
to show that, for $ x>0$,
\[
\int \frac{1}{\sqrt{x\left(x+1\right) \; } \ }\; \d x
=2 \ln \left(\sqrt x+ \sqrt{x +1} \; \right)+c
\,.
\]
\textbf{Note:} You may use without proof the result
$\displaystyle \int \! \frac{1}{t^2-a^2} \, \d t = \frac{1}{2a} \ln \left| \frac{t-a}{t+a}\right| + \rm {constant}$.
The section of the curve
\[
y=\dfrac{1}{\sqrt{x}\; }-\dfrac{1}{\sqrt{x+1}\; }
\]
between $x=\frac{1}{8}$ and $x=\frac{9}{16}$ is rotated through $360^{o}$ about the $x$-axis. Show that the volume enclosed is $2\pi \ln \tfrac{5}{4}\,$. $\phantom{\dfrac AB}$\begin{align*}
&& x &= \frac{1}{t^2-1} \\
&& t &= \sqrt{\frac{x+1}{x}}\\
\Rightarrow && \frac{\d x}{\d t} &= \frac{-2t}{(t^2-1)^2} \\
\Rightarrow && I &= \int \frac{1}{\sqrt{x(x+1)}} \d x \\
&&&= \int \frac{1}{\sqrt{\frac1{t^2-1} \frac{t^2}{t^2-1}}} \cdot \frac{-2 t}{(t^2-1)^2} \d t \\
&&&= \int \frac{t^2-1}{t} \frac{-2t}{(t^2-1)^2} \d t \\
&&&= -\int \frac{2}{t^2-1} \d t \\
&&&= - \frac{2}{2 \cdot 1} \ln \left | \frac{t-1}{t+1} \right| +C \\
&&&= \ln \left | \frac{t+1}{t-1} \right| + C \\
&&&= \ln \left | \frac{\sqrt{\frac{x+1}{x}}+1}{\sqrt{\frac{x+1}{x}}-1} \right| + C \\
&&&= \ln \left | \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \right| + C \\
&&&= 2\ln \left | \sqrt{x+1}+\sqrt{x}\right| + C \\
&&&= 2\ln \left ( \sqrt{x+1}+\sqrt{x}\right) + C \\
\end{align*}
\begin{align*}
&& V&= \pi \int_{1/8}^{9/16} y^2 \d x \\
&&&= \pi \int_{1/8}^{9/16} \left ( \frac1x + \frac{1}{x+1} - \frac{2}{\sqrt{x(x+1)}}\right) \d x \\
&&&= \pi \left [ \ln x + \ln (x+1) - 4 \ln(\sqrt{x+1} + \sqrt{x}) \right]_{1/8}^{9/16} \\
&&&= \pi \left ( \ln \frac{9}{16} + \ln \frac{25}{16} - 4 \ln \left ( \frac54 + \frac34\right) \right) +\\
&&&\quad -\pi \left ( \ln \frac{1}{8} + \ln \frac{9}{8} - 4 \ln \left ( \frac1{2\sqrt{2}} + \frac3{2\sqrt{2}}\right) \right) \\
&&&= \pi \left ( 2 \ln 3 - 8 \ln 2 + 2 \ln 5 - 4\ln2 \right) - \pi \left ( -6 \ln 2 + 2\ln 3 - 2\ln 2\right) \\
&&&= \pi (2 \ln 5 - 4 \ln 2 ) \\
&&&= 2 \pi \ln \tfrac54
\end{align*}
This was another very popular question and, in spite of the challenges it posed, was answered well by many of the candidates. The majority of those who attempted the question knew how to perform an integration by substitution, and many were able to do so correctly in this challenging example. Almost every candidate was correctly able to determine dx/dt and substitute for x correctly in the integrand. The next step, simplifying the resulting expression, proved more challenging, and several candidates slipped up at this point (for example by forgetting a square root sign). Those who reached an integral of the form −2/(t2 − 1) dt generally used the given result to perform the integration, though some ignored the note and proceeded to use partial fractions; this did not gain them any extra credit. The majority either ignored the absolute value signs given in the note, quietly dropped them in the next step, or integrated their partial fraction expansion without the use of absolute value signs. While it is true, in this case, that they can (and should) be dropped, this did require some minimal justification. The simplification of the logarithm expression with t substituted with √((x + 1)/x) proved challenging. A good number of candidates wisely expanded the given right-hand side and showed that it gave the same as the left-hand side, while others floundered with the expression √(1 + 1/x). This question called for a deep understanding of simplification of fractions together with an equally sophisticated understanding of rationalising denominators, and this proved too hard for many candidates. Most candidates attempted the second part of the question, even if they had not completed the first part. A disturbing number of candidates had misremembered the formula for volume as 2π∫y² dx, and were therefore out by a factor of 2 throughout the rest of the question. They were only lightly penalised for this error. There was some noticeable difficulty experienced in expanding y², with many also making their lives somewhat harder by writing y as a single fraction before squaring. Even among the correct expansions, fewer attempts than expected noted that the integrand could be written as 1/x + 1/(x+1) − 2×(∗), where (∗) was the integrand from the first part of the question. Therefore many candidates ended up becoming stuck at this point. Of those candidates who reached the correct integrand, most were able to correctly substitute in the limits and get the signs correct (though this did require care, and not all managed it). The next step, simplifying the logarithms, left many struggling. While most were fairly comfortable with the rules for logs, far fewer were happy with simplifying as they went or cancelling common factors ("cross-cancelling") when multiplying fractions, leading some to work with fractions with large numerators and denominators. This approach frequently ended with arithmetical errors. Very few candidates applied the logarithm rules to logs of fractions, such as writing ln 9/16 = ln 9 − ln 16 = 2 ln 3 − 4 ln 2. This would have made the arithmetic far simpler and would have given them far more chance of reaching the correct answer.