## Question

The function *f *is defined by \(f(x) = \left\{ \begin{array}{r}{e^{ – x^3}}( – {x^3} + 2{x^2} + x),x \le 1\\ax + b,x > 1\end{array} \right.\), where \(a\) and \(b\) are constants.

Find the exact values of \(a\) and \(b\) if \(f\) is continuous and differentiable at \(x = 1\).

(i) Use Rolle’s theorem, applied to \(f\), to prove that \(2{x^4} – 4{x^3} – 5{x^2} + 4x + 1 = 0\) has a root in the interval \(\left] { – 1,1} \right[\).

(ii) Hence prove that \(2{x^4} – 4{x^3} – 5{x^2} + 4x + 1 = 0\) has at least two roots in the interval \(\left] { – 1,1} \right[\).

**Answer/Explanation**

## Markscheme

\(\mathop {{\text{lim}}}\limits_{x \to {1^ – }} {{\text{e}}^{ – {x^2}}}\left( { – {x^3} + 2{x^2} + x} \right) = \mathop {{\text{lim}}}\limits_{x \to {1^ + }} (ax + b)\) \(( = a + b)\) *M1*

\(2{{\text{e}}^{ – 1}} = a + b\) **A1**

differentiability: attempt to differentiate **both **expressions *M1*

\(f'(x) = – 2x{{\text{e}}^{ – {x^2}}}\left( { – {x^3} + 2{x^2} + x} \right) + {{\text{e}}^{ – {x^2}}}\left( { – 3{x^2} + 4x + 1} \right)\) \((x < 1)\) **A1**

(or \(f'(x) = {{\text{e}}^{ – {x^2}}}\left( {2{x^4} – 4{x^3} – 5{x^2} + 4x + 1} \right)\))

\(f'(x) = a\) \((x > 1)\) *A1*

substitute \(x = 1\) in **both **expressions and equate

\( – 2{{\text{e}}^{ – 1}} = a\) *A1*

substitute value of \(a\) and find \(b = 4{{\text{e}}^{ – 1}}\) *M1A1*

*[8 marks]*

(i) \(f'(x) = {{\text{e}}^{ – {x^2}}}\left( {2{x^4} – 4{x^3} – 5{x^2} + 4x + 1} \right)\) (for \(x \leqslant 1\)) *M1*

\(f(1) = f( – 1)\) **M1**

Rolle’s theorem statement *(A1)*

by Rolle’s Theorem, \(f'(x)\) has a zero in \(\left] { – 1,1} \right[\) *R1*

hence quartic equation has a root in \(\left] { – 1,1} \right[\) *AG*

(ii) let \(g(x) = 2{x^4} – 4{x^3} – 5{x^2} + 4x + 1\).

\(g( – 1) = g(1) < 0\) and \(g(0) > 0\) *M1*

as \(g\) is a polynomial function it is continuous in \(\left[ { – 1,0} \right]\) and \(\left[ {0,{\text{ 1}}} \right]\). *R1*

(or \(g\) is a polynomial function continuous in any interval of real numbers)

then the graph of \(g\) must cross the *x*-axis at least once in \(\left] { – 1,0} \right[\) *R1*

and at least once in \(\left] {0,1} \right[\).

*[7 marks]*

## Examiners report

[N/A]

[N/A]

## Question

In this question you may assume that \(\arctan x\) is continuous and differentiable for \(x \in \mathbb{R}\).

Consider the infinite geometric series

\[1 – {x^2} + {x^4} – {x^6} + \ldots \;\;\;\left| x \right| < 1.\]

Show that the sum of the series is \(\frac{1}{{1 + {x^2}}}\).

Hence show that an expansion of \(\arctan x\) is \(\arctan x = x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots \)

\(f\) is a continuous function defined on \([a,{\text{ }}b]\) and differentiable on \(]a,{\text{ }}b[\) with \(f'(x) > 0\) on \(]a,{\text{ }}b[\).

Use the mean value theorem to prove that for any \(x,{\text{ }}y \in [a,{\text{ }}b]\), if \(y > x\) then \(f(y) > f(x)\).

(i) Given \(g(x) = x – \arctan x\), prove that \(g'(x) > 0\), for \(x > 0\).

(ii) Use the result from part (c) to prove that \(\arctan x < x\), for \(x > 0\).

Use the result from part (c) to prove that \(\arctan x > x – \frac{{{x^3}}}{3}\), for \(x > 0\).

Hence show that \(\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}\).

**Answer/Explanation**

## Markscheme

\(r = – {x^2},\;\;\;S = \frac{1}{{1 + {x^2}}}\) *A1AG*

*[1 mark]*

\(\frac{1}{{1 + {x^2}}} = 1 – {x^2} + {x^4} – {x^6} + \ldots \)

**EITHER**

\(\int {\frac{1}{{1 + {x^2}}}{\text{d}}x} = \int {1 – {x^2} + {x^4} – {x^6} + \ldots } {\text{d}}x\) *M1*

\(\arctan x = c + x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots \) *A1*

**Note: **Do not penalize the absence of *\(c\) *at this stage.

when \(x = 0\) we have \(\arctan 0 = c\) hence \(c = 0\) *M1A1*

**OR**

\(\int_0^x {\frac{1}{{1 + {t^2}}}{\text{d}}t = } \int_0^x {1 – {t^2} + {t^4}} – {t^6} + \ldots {\text{d}}t\) *M1A1A1*

**Note: **Allow *\(x\) *as the variable as well as the limit.

** M1 **for knowing to integrate,

**for each of the limits.**

*A1*\([\arctan t]_0^x = \left[ {t – \frac{{{t^3}}}{3} + \frac{{{t^5}}}{5} – \frac{{{t^7}}}{7} + \ldots } \right]_0^x\) *A1*

hence \(\arctan x = x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots \) *AG*

*[4 marks]*

applying the \(MVT\) to the function \(f\) on the interval \([x,{\text{ }}y]\) *M1*

\(\frac{{f(y) – f(x)}}{{y – x}} = f'(c)\;\;\;({\text{for some }}c \in ]x,{\text{ }}y[)\) *A1*

\(\frac{{f(y) – f(x)}}{{y – x}} > 0\;\;\;({\text{as }}f'(c) > 0)\) *R1*

\(f(y) – f(x) > 0{\text{ as }}y > x\) *R1*

\( \Rightarrow f(y) > f(x)\) *AG*

**Note: **If they use *\(x\) *rather than \(c\) they should be awarded ** M1A0R0**, but could get the next

**.**

*R1***[4 marks]**

(i) \(g(x) = x – \arctan x \Rightarrow g'(x) = 1 – \frac{1}{{1 + {x^2}}}\) *A1*

this is greater than zero because \(\frac{1}{{1 + {x^2}}} < 1\) *R1*

so \(g'(x) > 0\) *AG*

(ii) (\(g\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on \(]0,{\text{ }}b[\) with \(g'(x) > 0\) on \(]0,{\text{ }}b[\) for all \(b \in \mathbb{R}\))

(If \(x \in [0,{\text{ }}b]\) then) from part (c) \(g(x) > g(0)\) *M1*

\(x – \arctan x > 0 \Rightarrow \arctan x < x\) *M1*

(as *\(b\) *can take any positive value it is true for all \(x > 0\)) *AG*

*[4 marks]*

let \(h(x) = \arctan x – \left( {x – \frac{{{x^3}}}{3}} \right)\) *M1*

(\(h\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on \(]0,{\text{ }}b[\) with \(h'(x) > 0\) on \(]0,{\text{ }}b[\))

\(h'(x) = \frac{1}{{1 + {x^2}}} – (1 – {x^2})\) *A1*

\( = \frac{{1 – (1 – {x^2})(1 + {x^2})}}{{1 + {x^2}}} = \frac{{{x^4}}}{{1 + {x^2}}}\) *M1A1*

\(h'(x) > 0\) hence \(({\text{for }}x \in [0,{\text{ }}b]){\text{ }}h(x) > h(0)( = 0)\) *R1*

\( \Rightarrow \arctan x > x – \frac{{{x^3}}}{3}\) *AG*

**Note: **Allow correct working with \(h(x) = x – \frac{{{x^3}}}{3} – \arctan x\).

**[5 marks]**

use of \(x – \frac{{{x^3}}}{3} < \arctan x < x\) *M1*

choice of \(x = \frac{1}{{\sqrt 3 }}\) *A1*

\(\frac{1}{{\sqrt 3 }} – \frac{1}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\) *M1*

\(\frac{8}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\) *A1*

**Note: **Award final ** A1 **for a correct inequality with a single fraction on each side that leads to the final answer.

\(\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}\) *AG*

*[4 marks]*

*Total [22 marks]*

## Examiners report

Most candidates picked up this mark for realizing the common ratio was \( – {x^2}\).

Quite a few candidates did not recognize the importance of ‘hence’ in this question, losing a lot of time by trying to work out the terms from first principles.

Of those who integrated the formula from part (a) only a handful remembered to include the ‘\( + c\)’ term, and to verify that this must be equal to zero.

Most candidates were able to achieve some marks on this question. The most commonly lost mark was through not stating that the inequality was unchanged when multiplying by \(y – x{\text{ as }}y > x\).

The first part of this question proved to be very straightforward for the majority of candidates.

In (ii) very few realized that they had to replace the lower variable in the formula from part (c) by zero.

Candidates found this part difficult, failing to spot which function was required.

Many candidates, even those who did not successfully complete (d) (ii) or (e), realized that these parts gave them the necessary inequality.