# Maximum value of the expression \[2\sin x+4\cos x+3\] is

A. \[2\sqrt{5}+3\]

B. \[2\sqrt{5}-3\]

C. \[\sqrt{5}+3\]

D. None of these.

Answer

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Hint: Use trigonometric R-method to find the maxima of a given expression. Find the formula of the R-method and compare it with the expression given. Then solve it to get the maximum value.

“Complete step-by-step answer:”

We have been given the expression, \[2\sin x+4\cos x+3\].

So, let us put, \[f\left( x \right)=2\sin x+4\cos x+3\].

The maximum value of \[a\sin x+b\cos x\]is equal to \[\sqrt{{{a}^{2}}+{{b}^{2}}}\].

This equation \[\left( a\sin x+b\cos x \right)\] is similar to the expression, \[2\sin x+4\cos x+3\].

Let us use the Trigonometric R-method to solve this expression.

The R-method is used to find the extrema (maxima and minimum) of combinations of trigonometric function.

Let us consider, \[y=A\sin x+b\sin x\].

Thus by using the R-formula, let us express y as, \[y=a\sin x+b\cos x=R\sin \left( x+\theta \right)\].

For maximum value of y, \[\sin \left( x+\theta \right)=1\].

\[\therefore \]Maximum value of, \[y=R\left( 1 \right)=R\].

We know, \[\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b\].

We need to find the values of R.

\[a\sin x+b\cos x=\left( R\cos \theta \right)\sin x+\left( R\sin \theta \right)\cos x\]

By comparing we can see that, \[a=R\cos \theta \] and \[b=R\sin \theta \].

\[\begin{align}

& \therefore \dfrac{b}{a}=\dfrac{R\sin \theta }{R\cos \theta }=\tan \theta \\

& \therefore \tan \theta =\dfrac{b}{a} \\

& \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right) \\

\end{align}\]

\[a=R\cos \theta \]

\[\begin{align}

& R=\dfrac{a}{\cos \theta }=\dfrac{a}{\cos {{\tan }^{-1}}\left( \dfrac{b}{a} \right)}=\dfrac{a}{\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}} \\

& R=\sqrt{{{a}^{2}}+{{b}^{2}}} \\

& \therefore a\sin x+b\cos x=\sqrt{{{a}^{2}}+{{b}^{2}}} \\

\end{align}\]

So, value of a = 2 and b = 4.

\[\begin{align}

& f\left( x \right)=2\sin x+4\cos x+3 \\

& f\left( x \right)=\sqrt{{{a}^{2}}+{{b}^{2}}}+3 \\

& f\left( x \right)=\sqrt{{{2}^{2}}+{{4}^{2}}}+3 \\

& \therefore f\left( x \right)=\sqrt{20}+3=\sqrt{2\times 2\times 5}+3 \\

& f\left( x \right)=2\sqrt{5}+3 \\

\end{align}\]

Therefore, the maximum value of f (x) becomes, \[\sqrt{20}+3=2\sqrt{5}+3\].

Hence, option (a) is the correct answer.

Note: Maxima and Minima are important concepts in trigonometry. Here we did the proof for finding maximum value. You can simply apply the expression in the given function f (x) as they are similar.

If \[f\left( x \right)=2\sin x+4\cos x\], then \[f\left( x \right)=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{2}^{2}}+{{4}^{2}}}=\sqrt{20}\].

The maximum value of f (x) is \[\sqrt{20}\].

“Complete step-by-step answer:”

We have been given the expression, \[2\sin x+4\cos x+3\].

So, let us put, \[f\left( x \right)=2\sin x+4\cos x+3\].

The maximum value of \[a\sin x+b\cos x\]is equal to \[\sqrt{{{a}^{2}}+{{b}^{2}}}\].

This equation \[\left( a\sin x+b\cos x \right)\] is similar to the expression, \[2\sin x+4\cos x+3\].

Let us use the Trigonometric R-method to solve this expression.

The R-method is used to find the extrema (maxima and minimum) of combinations of trigonometric function.

Let us consider, \[y=A\sin x+b\sin x\].

Thus by using the R-formula, let us express y as, \[y=a\sin x+b\cos x=R\sin \left( x+\theta \right)\].

For maximum value of y, \[\sin \left( x+\theta \right)=1\].

\[\therefore \]Maximum value of, \[y=R\left( 1 \right)=R\].

We know, \[\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b\].

We need to find the values of R.

\[a\sin x+b\cos x=\left( R\cos \theta \right)\sin x+\left( R\sin \theta \right)\cos x\]

By comparing we can see that, \[a=R\cos \theta \] and \[b=R\sin \theta \].

\[\begin{align}

& \therefore \dfrac{b}{a}=\dfrac{R\sin \theta }{R\cos \theta }=\tan \theta \\

& \therefore \tan \theta =\dfrac{b}{a} \\

& \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right) \\

\end{align}\]

\[a=R\cos \theta \]

\[\begin{align}

& R=\dfrac{a}{\cos \theta }=\dfrac{a}{\cos {{\tan }^{-1}}\left( \dfrac{b}{a} \right)}=\dfrac{a}{\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}} \\

& R=\sqrt{{{a}^{2}}+{{b}^{2}}} \\

& \therefore a\sin x+b\cos x=\sqrt{{{a}^{2}}+{{b}^{2}}} \\

\end{align}\]

So, value of a = 2 and b = 4.

\[\begin{align}

& f\left( x \right)=2\sin x+4\cos x+3 \\

& f\left( x \right)=\sqrt{{{a}^{2}}+{{b}^{2}}}+3 \\

& f\left( x \right)=\sqrt{{{2}^{2}}+{{4}^{2}}}+3 \\

& \therefore f\left( x \right)=\sqrt{20}+3=\sqrt{2\times 2\times 5}+3 \\

& f\left( x \right)=2\sqrt{5}+3 \\

\end{align}\]

Therefore, the maximum value of f (x) becomes, \[\sqrt{20}+3=2\sqrt{5}+3\].

Hence, option (a) is the correct answer.

Note: Maxima and Minima are important concepts in trigonometry. Here we did the proof for finding maximum value. You can simply apply the expression in the given function f (x) as they are similar.

If \[f\left( x \right)=2\sin x+4\cos x\], then \[f\left( x \right)=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{2}^{2}}+{{4}^{2}}}=\sqrt{20}\].

The maximum value of f (x) is \[\sqrt{20}\].

Last updated date: 01st Oct 2023

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