Question

maxima and minima of y=3sinx+4cosx​

Answer 1

Answer:f(x)=3Sin(x)+4Cos(x)————(1)

Let 3=aCos(y)————(2)

and 4=aSin(y)————(3)

Now, (aCos(y))2+(aSin(y))2=32+42

=>a2(Cos2(y)+Sin2(y))=9+16

=>a2=25

=>a=5————(4)

=>3=5Cos(y)————(5)

and 4=5Sin(y)————(6)

And Tan(y)=43

y=Tan−1(43)————(7)

So equation (1) becomes,

f(x)=5Cos(y)Sin(x)+5Sin(y)Cos(x)

=5(Cos(y)Sin(x)+Sin(y)Cos(x))

=5Sin(x+y)

From (7) we have

f(x)=5Sin(x+Tan−1(43))————−(8)

As, −1≤Sin(x+Tan−1(43))≤1

=>−5≤Sin(x+Tan−1(43))≤5

So  max(f(x))=5

just change x with y

Explanation: