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

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Answer 1 · 2017-04-11T03:24:08

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(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#