(a)     Show that the solution of the differential equation

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \cos x{\cos ^2}y{\text{,}}$$

given that $y = \frac{\pi }{4}{\text{ when }}x = \pi {\text{, is }}y = \arctan (1 + \sin x){\text{.}}$

(b)     Determine the value of the constant a for which the following limit exists

$$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\arctan (1 + \sin x) - a}}{{{{\left( {x - \frac{\pi }{2}} \right)}^2}}}$$

and evaluate that limit.

(a)     this separable equation has general solution

$\int {{{\sec }^2}y{\text{d}}y = \int {\cos x{\text{d}}x} }$     (M1)(A1)

$\tan y = \sin x + c$     A1

the condition gives

$\tan \frac{\pi }{4} = \sin \pi  + c \Rightarrow c = 1$     M1

the solution is $\tan y = 1 + \sin x$     A1

$y = \arctan (1 + \sin x)$     AG

[5 marks]

(b)     the limit cannot exist unless $a = \arctan \left( {1 + \sin \frac{\pi }{2}} \right) = \arctan 2$     R1A1

in that case the limit can be evaluated using l’Hopital’s rule (twice) limit is

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\left( {\arctan (1 + \sin x)} \right)}^\prime }}}{{2\left( {x - \frac{\pi }{2}} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y'}}{{2\left( {x - \frac{\pi }{2}} \right)}}$     M1A1

where y is the solution of the differential equation

the numerator has zero limit (from the factor $\cos x$ in the differential equation)     R1

so required limit is

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{y''}}{2}$     M1A1

finally,

$y'' = - \sin x{\cos ^2}y - 2\cos x\cos y\sin y \times y'(x)$     M1A1

since $\cos y\left( {\frac{\pi }{2}} \right) = \frac{1}{{\sqrt 5 }}$     A1

$y'' = - \frac{1}{5}{\text{ at }}x = \frac{\pi }{2}$     A1

the required limit is $- \frac{1}{{10}}$     A1

[12 marks]

Total [17 marks]

Many candidates successfully obtained the displayed solution of the differential equation in part(a). Few complete solutions to part(b) were seen which used the result in part(a). The problem can, however, be solved by direct differentiation although this is algebraically more complicated. Some successful solutions using this method were seen.