Question

Q.3 Find the minimum value of Function f(x) = (3 sin x - 4cosx - 10) X
(2 sinx + 4cosx - 10)​

Answer 1

Answer:

Given F(x)=3sinx+4cosx

Differentiate w.r.t x on both sides

Then f'(x)=3cosx-4sinx. (Since differentiation for sinx =cosx and differentiation for cosx=-sin x)

Again differentiate w.r.t x on both sides

Then f''(x)= -3sinx-4cosx=-(3sinx+4cosx). (If f''(x) >o it consists of minimum value and if f''(x)<0 maximum value exists)

So f''(x)=-F(x)

So I think

Its Minimum value does not exist

It consist of only max value

For all mini and max values f'(x)=0

So

3cosx-4sinx=0

So

3cosx=4sinx

Sinx/cosx=3/4

Tanx=3/4

So x=tan inverse 3/4

So max value of x is tan inverse 3/4