The maximum value of f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10) is?

2 Answers
Jul 25, 2018

f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10)

=((3sinx-10)-4cosx)((3sinx-10)+4cosx)

=(3sinx-10)^2-(4cosx)^2

=9sin^2x-60sinx+100-16cos^2x

=9sin^2x-60sinx+100-16+16sin^2x

=25sin^2x-60sinx+84

=(5sinx)^2-2*5sinx*6+6^2-6^2+84

=(5sinx-6)^2+48

f(x) will be maximum when (5sinx-6)^2 is maximum . It will be possible for sinx=-1

So

[f(x)]_"max"=(5(-1)-6)^2+48=169

Jul 25, 2018

Maximum is 169. Minimum is 50 (perhaps, nearly). This is graphical illustration, for Dilip's answer.

Explanation:

Let alpha = sin^(-1) ( 4/5 )..Then

f( x ) = 25 (sin ( x - alpha )-2)(sin (x + alpha) - 2)
See graph.
graph{(y - 25 (sin (x- 0.9273)-2)(sin (x+ 0.9273)-2))(y-169)(y-50)=0[-20 20 20 230]}
graph{(y - 25 (sin (x- 0.9273)-2)(sin (x+ 0.9273)-2))(y-169)=0[-1.75 -1.5 167 171]}