Question

The maximum value of $f\left( x \right) = 2\sin x + \cos 2x$ , $0 \leqslant x \leqslant \dfrac{\pi }{2}$ occurs at x is
$\left( 1 \right){\text{ }}0$
$\left( 2 \right){\text{ }}\dfrac{\pi }{6}$
$\left( 3 \right){\text{ }}\dfrac{\pi }{2}$
$\left( 4 \right)$ None of these

Answer 1

Hint : First differentiate the given function. Then let the first derivative equal to zero and find the critical numbers. Then find the second derivative of the given function and put those critical numbers in the second derivative. If the second derivative of the function is less than zero then the function is maximum at that value of x.

Complete step-by-step answer :
The value of the function at a maximum point is called the maximum value of the function.
It is given that $f\left( x \right) = 2\sin x + \cos 2x$ .On differentiating it with respect to x we get
$\Rightarrow f'\left( x \right) = 2\cos x - 2\sin 2x$ ---------- (i)
Now let $f'\left( x \right) = 0$ to find critical numbers. So,
$\Rightarrow 2\cos x - 2\sin 2x = 0$
We know that $\sin 2x = 2\sin x\cos x$ .Therefore by putting this value in the above equation we get
$\Rightarrow 2\cos x - 2\left( {2\sin x\cos x} \right) = 0$
On multiplying we get
$\Rightarrow 2\cos x - 4\sin x\cos x = 0$
By taking $2\cos x$ common from the above equation we get
$\Rightarrow 2\cos x\left( {1 - 2\sin x} \right) = 0$
From here we have $2\cos x = 0$ and $1 - 2\sin x = 0$ .On further solving the equation $2\cos x = 0$ we get
$\Rightarrow \cos x = 0$
We know that $\cos \dfrac{\pi }{2} = 0$ .Therefore,
$\Rightarrow \cos x = \cos \dfrac{\pi }{2}$
From the above equation we get the value of $x = \dfrac{\pi }{2}$
On further solving the equation $1 - 2\sin x = 0$ we get
$\Rightarrow 2\sin x = 1$
$\Rightarrow \sin x = \dfrac{1}{2}$
We know that $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ .Therefore the above equation becomes
$\Rightarrow \sin x = \sin \dfrac{\pi }{6}$
sin on both sides will cancel out each other and we get
$x = \dfrac{\pi }{6}$
Hence, the required critical values are $x = \dfrac{\pi }{2}$ and $x = \dfrac{\pi }{6}$
Now to find the second derivative of the function, differentiate equation (i) with respect to x
$f''\left( x \right) = - 2\sin x - 4\cos 2x$
Now apply here the values of x that we find
At $x = \dfrac{\pi }{2}$ ,
$\Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = - 2\sin \dfrac{\pi }{2} - 4\cos 2\left( {\dfrac{\pi }{2}} \right)$
$\Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = - 2\sin \dfrac{\pi }{2} - 4\cos \pi$
The value of $\sin \dfrac{\pi }{2}$ is $1$ and the value of $\cos \pi$ is $- 1$ .Therefore,
$\Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = - 2 - 4\left( { - 1} \right)$
$\Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = - 2 + 4$
$\Rightarrow f''\left( {\dfrac{\pi }{2}} \right) = 2$ which is $\geqslant 0$ that is function is minimum
At $x = \dfrac{\pi }{6}$ ,
$\Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 2\sin \dfrac{\pi }{6} - 4\cos 2\left( {\dfrac{\pi }{6}} \right)$
$\Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 2\sin \dfrac{\pi }{6} - 4\cos \dfrac{\pi }{3}$
Value of $\sin \dfrac{\pi }{6}$ is $\dfrac{1}{2}$ and the value of $\cos \dfrac{\pi }{3}$ is $\dfrac{1}{2}$ .Therefore,
$\Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 2\left( {\dfrac{1}{2}} \right) - 4\left( {\dfrac{1}{2}} \right)$
On further solving we get
$\Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 1 - 2$
$\Rightarrow f''\left( {\dfrac{\pi }{6}} \right) = - 3$ which is $\leqslant 0$ that is here the function is maximum
Thus, the function is maximum at $x = \dfrac{\pi }{6}$ .
Hence, the correct option is $\left( 2 \right){\text{ }}\dfrac{\pi }{6}$
So, the correct answer is “Option 2”.

Note : Remember that the function f(x) is maximum when f’’(x) is less than zero and the function f(x) is minimum when f’’(x) is greater than zero. Remember the formula of Sin2x that we used in the solution. Keep in mind the trigonometric values at every angle or radian because it is important and is used in various questions.