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# Maximum value of the expression $2\sin x+4\cos x+3$ isA. $2\sqrt{5}+3$B. $2\sqrt{5}-3$C. $\sqrt{5}+3$D. None of these.

Last updated date: 20th Mar 2023
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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.

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 $\sqrt{20}$.