The function f is defined by \(f(x) = x\sqrt {9 - {x^2}}  + 2\arcsin \left( {\frac{x}{3}} \right)\).

(a)     Write down the largest possible domain, for each of the two terms of the function, f , and hence state the largest possible domain, D , for f .

(b)     Find the volume generated when the region bounded by the curve y = f(x) , the x-axis, the y-axis and the line x = 2.8 is rotated through \(2\pi \) radians about the x-axis.

(c)     Find \(f'(x)\) in simplified form.

(d)     Hence show that \(\int_{ - p}^p {\frac{{11 - 2{x^2}}}{{\sqrt {9 - {x^2}} }}} {\text{d}}x = 2p\sqrt {9 - {p^2}}  + 4\arcsin \left( {\frac{p}{3}} \right)\), where \(p \in D\) .

(e)     Find the value of p which maximises the value of the integral in (d).

(f)     (i)     Show that \(f''(x) = \frac{{x(2{x^2} - 25)}}{{{{(9 - {x^2})}^{\frac{3}{2}}}}}\).

  (ii)     Hence justify that f(x) has a point of inflexion at x = 0 , but not at \(x = \pm \sqrt {\frac{{25}}{2}} \) .

(a)     For \(x\sqrt {9 - {x^2}} \), \( - 3 \leqslant x \leqslant 3\) and for \(2\arcsin \left( {\frac{x}{3}} \right)\), \( - 3 \leqslant x \leqslant 3\)     A1

\( \Rightarrow D{\text{ is }} - 3 \leqslant x \leqslant 3\)     A1

[2 marks]

 

(b)     \(V = \pi \int_0^{2.8} {{{\left( {x\sqrt {9 - {x^2}}  = 2\arcsin \frac{x}{3}} \right)}^2}{\text{d}}x} \)     M1A1

= 181     A1

[3 marks]

 

(c)     \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(9 - {x^2})^{\frac{1}{2}}} - \frac{{{x^2}}}{{{{(9 - {x^2})}^{\frac{1}{2}}}}} + \frac{{\frac{2}{3}}}{{\sqrt {1 - \frac{{{x^2}}}{9}} }}\)     M1A1

\( = {(9 - {x^2})^{\frac{1}{2}}} - \frac{{{x^2}}}{{{{(9 - {x^2})}^{\frac{1}{2}}}}} + \frac{2}{{{{(9 - {x^2})}^{\frac{1}{2}}}}}\)     A1

\( = \frac{{9 - {x^2} - {x^2} + 2}}{{{{(9 - {x^2})}^{\frac{1}{2}}}}}\)     A1

\( = \frac{{11 - 2{x^2}}}{{\sqrt {9 - {x^2}} }}\)     A1

[5 marks]

 

(d)     \(\int_{ - p}^p {\frac{{11 - 2{x^2}}}{{\sqrt {9 - {x^2}} }}{\text{d}}x = \left[ {x\sqrt {9 - {x^2}} + 2\arcsin \frac{x}{3}} \right]_{ - p}^p} \)     M1

\( = p\sqrt {9 - {p^2}} + 2\arcsin \frac{p}{3} + p\sqrt {9 - {p^2}}  + 2\arcsin \frac{p}{3}\)     A1

\( = 2p\sqrt {9 - {p^2}} + 4\arcsin \left( {\frac{p}{3}} \right)\)     AG

[2 marks]

 

(e)     \(11 - 2{p^2} = 0\)     M1

\(p = 2.35\,\,\,\,\,\left( {\sqrt {\frac{{11}}{2}} } \right)\)     A1

Note: Award A0 for \(p = \pm 2.35\) .

 

[2 marks]

 

(f)     (i)     \(f''(x) = \frac{{{{(9 - {x^2})}^{\frac{1}{2}}}( - 4x) + x(11 - 2{x^2}){{(9 - {x^2})}^{ - \frac{1}{2}}}}}{{9 - {x^2}}}\)     M1A1

\( = \frac{{ - 4x(9 - {x^2}) + x(11 - 2{x^2})}}{{{{(9 - {x^2})}^{\frac{3}{2}}}}}\)     A1

\( = \frac{{ - 36x + 4{x^3} + 11x - 2{x^3}}}{{{{(9 - {x^2})}^{\frac{3}{2}}}}}\)     A1

\( = \frac{{x(2{x^2} - 25)}}{{{{(9 - {x^2})}^{\frac{3}{2}}}}}\)     AG

 

(ii)     EITHER

When \(0 < x < 3\), \(f''(x) < 0\). When \( - 3 < x < 0\), \(f''(x) > 0\).     A1

OR

\(f''(0) = 0\)     A1

THEN

Hence \(f''(x)\) changes sign through x = 0 , giving a point of inflexion.     R1

EITHER

\(x = \pm \sqrt {\frac{{25}}{2}} \) is outside the domain of f.     R1

OR

\(x = \pm \sqrt {\frac{{25}}{2}} \) is not a root of \(f''(x) = 0\) .     R1

[7 marks]

 

Total [21 marks]

It was disappointing to note that some candidates did not know the domain for arcsin. Most candidates knew what to do in (b) but sometimes the wrong answer was obtained due to the calculator being in the wrong mode. In (c), the differentiation was often disappointing with \(\arcsin \left( {\frac{x}{3}} \right)\) causing problems. In (f)(i), some candidates who failed to do (c) guessed the correct form of \(f'(x)\) (presumably from (d)) and then went on to find \(f''(x)\) correctly. In (f)(ii), the justification of a point of inflexion at x = 0 was sometimes incorrect – for example, some candidates showed simply that \(f'(x)\) is positive on either side of the origin which is not a valid reason.