Question

The only value of \[x\] for which \[{2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}\] holds, is
A. \[\dfrac{{5\pi }}{4}\]
B. \[\dfrac{{3\pi }}{4}\]
C. \[\dfrac{\pi }{2}\]
D. All values of \[x\].

Answer 1

Hint: Use Arithmetic mean\[(AM) \ge \] Geometric mean\[(GM)\]between\[{2^{\sin x}}\]and\[{2^{\cos x}}\]. Here we have to find the minimum value of\[{2^{\sin x}} + {2^{\cos x}}\]. We know that, Arithmetic means\[ \ge \]Geometric means or, \[AM \ge GM\]. The maximum value can reach up to infinity. Now, using the formulas above, it is possible to determine the minimum and maximum values of various trigonometric functions. But keep in mind that we must first infer as much of the equation as we can.




Complete step by step solution:Given the equation,
\[{2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}\]
For any two values, say\[a\]and\[b\],
\[AM = \dfrac{{a + b}}{2}\]
And \[GM = \sqrt {ab} \]
Considering
\[a = {2^{\sin x}}\]and\[b = {2^{\cos x}}\]
We get:
\[AM = \dfrac{{{2^{\sin x}} + {2^{\cos x}}}}{2}\]
Since\[AM \ge GM\],
We have
\[\dfrac{1}{2}\left( {{2^{\sin x}} + {2^{\cos x}}} \right) \ge \sqrt {{2^{\sin x}} \cdot {2^{\cos x}}} \]
After solving, it becomes:
\[ \Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge 2 \cdot {2^{\dfrac{{\sin x + \cos x}}{2}}}\]
Simplify:
\[ \Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \dfrac{{\sin x + \cos x}}{2}}}\]
And, we know that;
\[\sin x + \cos x \ge - \sqrt 2 \].
Therefore,
\[{2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}\]
For:
\[x = \dfrac{{5\pi }}{4}\]
The expression\[M = A + \left| B \right|\]is the function's maximum value. This maximum value is reached whenever sin x or cos x are equal to 1. The expression \[m = A - \left| B \right|\]is the function's minimal value. When there is\[cosx = 1\]or\[sinx = 1\], this minimum is reached.
Hence, \[{2^{\sin x}} + {2^{\cos x}}\]is always greater than or equal to
\[{2^{\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right)}}\]
That means, the minimum value of\[{2^{\sin x}} + {2^{\cos x}}\] is
\[2\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right)\]
Hence, the only value of \[x\] for which\[{2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}\]holds, is
\[x = \dfrac{{5\pi }}{4}\]



Option ‘A’ is correct

Note: In trigonometry, questions involving minima and maxima it is a must try to use the technique of\[AM \ge GM\]sometimes. Do not always try to solve the question only through trigonometric equations and functions. You do not have to learn this formula, just observe here that if the equation is of type\[a{\sin ^2}\phi + b{\cos ^2}\phi a{\sin ^2}\phi + b{\cos ^2}\phi \], no matter what, the maximum value is the larger of values \[(a,b)\]and minimum value is smaller of values\[(a,b)\].