Question

find the maximum and minimum value of f(x)= sinx+root 3 cosx

Answer 1

Given,

$\longrightarrow\sf{f(x)=\sin x+\sqrt3\,\cos x}$

Dividing and multiplying 2 on RHS,

$\longrightarrow\sf{f(x)=2\left[\dfrac{\sin x+\sqrt3\,\cos x}{2}\right]}$

$\longrightarrow\sf{f(x)=2\left[\dfrac{1}{2}\,\sin x+\dfrac{\sqrt3}{2}\,\cos x\right]}$

$\longrightarrow\sf{f(x)=2\left[\cos60^o\sin x+\sin60^o\cos x\right]}$

$\longrightarrow\sf{f(x)=2\left[\sin x\cos60^o+\cos x\sin60^o\right]}$

Since $\sf{\sin A\cos B+\cos A\sin B=\sin(A+B),}$

$\longrightarrow\sf{f(x)=2\sin(x+60^o)}$

Well, since $\sf{\sin x\in[-1,\ 1],}$

$\longrightarrow\sf{\sin(x+60^o)\in[-1,\ 1]}$

$\longrightarrow\sf{2\sin(x+60^o)\in[-2,\ 2]}$

That is,

$\longrightarrow\sf{f(x)\in[-2,\ 2]}$

Hence,

$\longrightarrow\sf{\underline{\underline{\min(f(x))=-2}}}$

$\longrightarrow\sf{\underline{\underline{\max(f(x))=2}}}$