Question

What is the maximum value of \[f\left( x \right) = 2 \sin x + \cos 2x\], \[0 \le x \le \dfrac{\pi }{2}\] occurs at \[x\] ?
A. 0
B. \[\dfrac{\pi }{6}\]
C. \[\dfrac{\pi }{2}\]
D. None of these

Answer 1

Hint: First, differentiate the given function with respect to \[x\]. Then simplify the equation by substituting \[f'\left( x \right) = 0\] and get the critical points. Again, differentiate the first differential equation with respect to \[x\]. After that, substitute the critical points in the second derivative and check the maxima and minima of the given function.

Formula used:
1. \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
2. \[\dfrac{d}{{dx}}\left( {\cos nx} \right) = - n \sin nx\]
3. \[\dfrac{d}{{dx}}\left( {\sin nx} \right) = n \cos nx\]
4. \[\sin2x = 2\sin x \cos x\]

Complete step by step solution:
The given function is \[f\left( x \right) = 2 \sin x + \cos 2x\], where \[0 \le x \le \dfrac{\pi }{2}\].
Let’s differentiate the above function with respect to \[x\].
Apply the formulas \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\] and \[\dfrac{d}{{dx}}\left( {\cos nx} \right) = - n \sin nx\].
\[f'\left( x \right) = 2 \cos x – 2\sin 2x\] \[.....\left( 1 \right)\]

To find the critical points equate the first derivative to zero.
\[f'\left( x \right) = 0\]
\[ \Rightarrow 2 \cos x – 2\sin 2x = 0\]
Divide both sides by 2.
\[ \Rightarrow \cos x - \sin 2x = 0\]
Apply the trigonometric identity \[\sin2x = 2\sin x \cos x\].
\[\cos x – 2\sin x \cos x = 0\]
Factor out the common term.
\[\cos x\left( {1 – 2\sin x} \right) = 0\]
\[ \Rightarrow \cos x = 0\] or \[1 – 2\sin x = 0\]
\[ \Rightarrow \cos x = 0\] or \[\sin x = \dfrac{1}{2}\]
\[ \Rightarrow x = \dfrac{\pi }{2}\] or \[x = \dfrac{\pi }{6}\]
Thus, the critical points of the function are \[x = \dfrac{\pi }{6}\] and \[x = \dfrac{\pi }{2}\].

Differentiate the equation \[\left( 1 \right)\] with respect to \[x\].
Apply the formula \[\dfrac{d}{{dx}}\left( {\sin nx} \right) = n \cos nx\].
\[f''\left( x \right) = - 2 \sin x – 4\cos 2x\] \[.....\left( 2 \right)\]

Now to find the maximum and minimum points of the function, substitute the critical points in the equation \[\left( 2 \right)\].
For \[x = \dfrac{\pi }{2}\]:
\[f''\left( {\dfrac{\pi }{2}} \right) = - 2 \sin\left( {\dfrac{\pi }{2}} \right) – 4\cos 2\left( {\dfrac{\pi }{2}} \right)\]
\[ \Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = - 2 \left( 1 \right) - 4\left( { - 1} \right)\] \[....\left[ {\sin\left( {\dfrac{\pi }{2}} \right) = 1 ,\cos \left( \pi \right) = - 1} \right]\]
\[ \Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = - 2 + 4\]
\[ \Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = 2 \]
Since \[f''\left( {\dfrac{\pi }{2}} \right) > 0\]. So, it is the minima of the given function.

For \[x = \dfrac{\pi }{6}\]:
\[f''\left( {\dfrac{\pi }{6}} \right) = - 2 \sin\left( {\dfrac{\pi }{6}} \right) – 4\cos 2\left( {\dfrac{\pi }{6}} \right)\]
\[ \Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 2 \sin\left( {\dfrac{\pi }{6}} \right) – 4\cos \left( {\dfrac{\pi }{3}} \right)\]
\[ \Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 2 \left( {\dfrac{1}{2}} \right) - 4\left( {\dfrac{1}{2}} \right)\] \[....\left[ {Since \sin\left( {\dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2} } \right]\]
\[ \Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 1 - 2\]
\[ \Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 3 \]
Since \[f''\left( {\dfrac{\pi }{6}} \right) < 0\]. So, it is the maxima of the given function.
Thus, the maximum point of the given function is \[x = \dfrac{\pi }{6}\].
Hence the correct option is B.

Note: Students often get confused about the maximum and minimum point of a function.
A function \[f\left( x \right)\] has a maximum value at \[x = a\], if \[f'\left( a \right) = 0\], and \[f''\left( a \right) < 0\].
A function \[f\left( x \right)\] has a minimum value at \[x = a\], if \[f'\left( a \right) = 0\], and \[f''\left( a \right) > 0\].