Question

Find the real value of x for which ${{27}^{\cos 2x}}{{.81}^{\sin 2x}}$ is minimum. Also find this minimum value.

Answer 1

Hint:Here, we can use derivatives to find the minimum value of the given function. First, we can equate the derivative of the function to 0 to get the critical points and then we can check whether it is a point of minima.

Complete step-by-step answer:
The maxima and minima of a function known collectively as extrema are the largest and the smallest value of the function, either within a given range or on the entire domain.
In order to find the maxima or minima of a function, first of all we find the derivative of the function. After that, we equate the derivative to 0 to get certain values of the independent variable. Then, we find the double derivative of the function and check its value at those values known as critical points.
If the double derivative is positive at a critical point, we say that it is a point of minima and if it is negative then it is a point of maxima.
Now, the function given here is ${{27}^{\cos 2x}}{{.81}^{\sin 2x}}$.
We can also write it as:
${{3}^{3\cos 2x}}{{.3}^{4\sin 2x}}={{3}^{\left( 3\cos 2x+4\sin 2x \right)}}$
We know that the function ${{a}^{x}}$ is always increasing in its domain wherever ‘a’ is an integer. So, ${{3}^{\left( 3\cos 2x+4\sin 2x \right)}}$ is always increasing.
In order to find its minimum value, we have to find the minimum value of $3\cos 2x+4\sin 2x$.
So,
$\begin{aligned} & \dfrac{d\left( 3\cos 2x+4\sin 2x \right)}{dx} \\\ & =-3\left( \sin 2x \right)\times 2+4\cos 2x\times 2 \\\ & =-6\sin 2x+8\cos 2x \\\ \end{aligned}$
On equating it to 0, we get:
$\begin{aligned} & -6\sin 2x+8\cos 2x=0 \\\ & \Rightarrow \dfrac{\sin 2x}{\cos 2x}=\dfrac{-8}{-6} \\\ & \Rightarrow \tan 2x=\dfrac{4}{3} \\\ \end{aligned}$
So, we have the value of 2x as:
$2x={{53}^{0}},\pi +{{53}^{0}},2\pi +{{53}^{0}},......$
Now again differentiating it we get:
$\begin{aligned} & \dfrac{d\left( -6\sin 2x+8\cos 2x \right)}{dx} \\\ & =-6\left( \cos 2x \right)\times 2+8\left( -\sin 2x \right)\times 2 \\\ & =-12\cos 2x-16\sin 2x \\\ \end{aligned}$
When, $2x={{53}^{0}}$, $\dfrac{d\left( -6\sin 2x+8\cos 2x \right)}{dx}<0$, which means that this point corresponds to maxima.
When $2x=\pi +{{53}^{0}}$, $\dfrac{d\left( -6\sin 2x+8\cos 2x \right)}{dx}>0$, which corresponds to minima.
Therefore, we get minimum value of $3\cos 2x+4\sin 2x$ at $2x=\pi +{{53}^{0}}$, or $x=\dfrac{\left( \pi +{{53}^{0}} \right)}{2}$.
And, the minimum value is:
$\begin{aligned} & =3\cos \left( \pi +{{53}^{0}} \right)+4\sin \left( \pi +{{53}^{0}} \right) \\\ & =3\times \left( \dfrac{-3}{5} \right)+4\times \left( \dfrac{-4}{5} \right) \\\ & =\dfrac{-9-16}{5}=\dfrac{-25}{5}=-5 \\\ \end{aligned}$
Hence, the minimum value of the given function is = ${{3}^{-5}}=\dfrac{1}{243}$.

Note: Students should note here that $\tan 2x$ repeats its value after each ${{180}^{0}}$ and hence its value at ${{53}^{0}}$ will be same as its value at $\pi +{{53}^{0}}$.To know that which point is maxima and which point is minima we have to double derivative the function. If the sign of $f''(x)$ is positive then it is a point of minima , if the sign is negative then it is a point of maxima.