Expressing in different form and finding a maximum? See picture below (7 marks)

Question

1502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-11T03:24:08

enter image source here

(a)

$sqrt15sin(2x)+sqrt5cos(2x)$

$=sqrt20(sqrt15/sqrt20sin(2x)+sqrt5/sqrt20cos(2x))$

[Taking $sqrt15/sqrt20=cosalpha and sqrt5/sqrt20=sinalpha$ ]

$=sqrt20(cosalphasin(2x)+sinalphacos(2x))$

$=2sqrt5sin(2x+alpha)$

Where $alpha=tan^-1(sqrt5/sqrt15)=tan^-1(1/sqrt3)=pi/6$

(b)

(i)
$f(x)=2/(5+sqrt15sin(2x)+sqrtcos(2x))$

$=>f(x)=2/(5+2sqrt5sin(2x+alpha))$

The value of $f(x)$ will be maximum when $sin(2x+alpha)$ is minimum i.e. $sin(2x+alpha)=-1$

So

$f(x)_"max"=2/(5-2sqrt5)$

$=>f(x)_"max"=(2(5+2sqrt5))/((5-2sqrt5)(5+2sqrt5))$

$=>f(x)_"max"=(10+4sqrt5)/(25-20)=2+4/5sqrt5$

(ii)Here $f(x)$ is maximum when

$sin(2x+alpha)=-1$

$=>sin(2x+pi/6)=sin((3pi)/2)$

$=>2x+pi/6=(3pi)/2$

$=>2x=(3pi)/2-pi/6=(4pi)/3$

$=>x=(2pi)/3$