Question

All the pairs $\left( x,y \right)$ that satisfy the inequality ${{2}^{\sqrt{{{\sin }^{2}}x-2\sin x+5}}}.\dfrac{1}{{{4}^{{{\sin }^{2}}y}}}\le 1$ also satisfy the equation
A. $\sin x=\left| \sin y \right|$
B. $\sin x=2\sin y$
C. $2\left| \sin x \right|=3\sin y$
D. $2\sin x=\sin y$

Answer 1

Hint: To solve this question, we should know the range of the function $\sin x$ and we should apply the concepts involved in inequalities. As ${{4}^{{{\sin }^{2}}y}}$ is positive for all values of y, we can multiply ${{4}^{{{\sin }^{2}}y}}$ on both the sides of inequality and the inequality doesn’t change. After that, we can write 4 as ${{2}^{2}}$ and apply the condition ${{a}^{x}}\le {{a}^{y}}\Rightarrow x\le y$. We know the range of $\sin x\in \left[ -1,1 \right]$ and the term the square root can be modified as $\sqrt{{{\left( \sin x-1 \right)}^{2}}+4}$ and by verifying the extreme values, we will get the required result.

Complete step-by-step answer:
We are given the inequality ${{2}^{\sqrt{{{\sin }^{2}}x-2\sin x+5}}}.\dfrac{1}{{{4}^{{{\sin }^{2}}y}}}\le 1$ and we are asked to find the relation between $\sin x$ and $\sin y$.
We know that ${{a}^{x}}>0\text{ }\forall a>0$.
Here a = 4 and ${{4}^{{{\sin }^{2}}y}}$. By multiplying by ${{4}^{{{\sin }^{2}}y}}$ on both sides, the inequality remains the same.
${{2}^{\sqrt{{{\sin }^{2}}x-2\sin x+5}}}\le {{4}^{{{\sin }^{2}}y}}$
We know that $4={{2}^{2}}$, we can write that
$\begin{align}
  & {{2}^{\sqrt{{{\sin }^{2}}x-2\sin x+5}}}\le {{\left( {{2}^{2}} \right)}^{{{\sin }^{2}}y}} \\
 & {{2}^{\sqrt{{{\sin }^{2}}x-2\sin x+5}}}\le {{2}^{2{{\sin }^{2}}y}} \\
\end{align}$
We have a property in the exponential powers which is
${{a}^{x}}\le {{a}^{y}}\Rightarrow x\le y$
Here $a=2,x=\sqrt{{{\sin }^{2}}x-2\sin x+5},y=2{{\sin }^{2}}y$
We can write that
$\sqrt{{{\sin }^{2}}x-2\sin x+5}\le 2{{\sin }^{2}}y$
We can write the 5 in the equation as 1 + 4, we get
$\sqrt{{{\sin }^{2}}x-2\sin x+1+4}\le 2{{\sin }^{2}}y$
We know the relation \[{{\left( \sin x-1 \right)}^{2}}={{\sin }^{2}}x-2\sin x+1\]. Using this in the above L.H.S, we get
$\sqrt{{{\left( \sin x-1 \right)}^{2}}+4}\le 2{{\sin }^{2}}y\to \left( 1 \right)$
Let us consider ${{\left( \sin x-1 \right)}^{2}}+4$.
We know the relation that $a+{{b}^{2}}\ge a$ and the value a is obtained when b = 0.
So, the value ${{\left( \sin x-1 \right)}^{2}}+4\ge 4$ and ${{\left( \sin x-1 \right)}^{2}}+4=4$ when $\sin x=1$
We also know that $\sin y\le 1$ and ${{\sin }^{2}}y\le 1$.
By multiplying with 2, we can write that $2{{\sin }^{2}}y\le 2$
By using the derived inequalities in equation – 1, we get
$\sqrt{\left( Value\ge 4 \right)}\le \left( Value\le 2 \right)$
By removing square root, we get
$\left( Value\ge 2 \right)\le \left( Value\le 2 \right)$
This inequality is only possible when both the values are equal and their value is equal to 2.
We can write that
$\begin{align}
  & {{\left( \sin x-1 \right)}^{2}}+4=4 \\
 & {{\left( \sin x-1 \right)}^{2}}=0 \\
 & \sin x=1 \\
\end{align}$
$\begin{align}
  & 2{{\sin }^{2}}y=2 \\
 & {{\sin }^{2}}y=1 \\
 & \left| \sin y \right|=1 \\
\end{align}$
From these two relations, we get
$\sin x=\left| \sin y \right|=1$
$\therefore \sin x=\left| \sin y \right|=1$. The answer is option-A.

Note: The key to solve this question is to note that the two expressions on the L.H.S and R.H.S are at their extremes. The smaller term is at the smallest value and the larger term is at the smallest value possible. Generally in these kinds of inequalities in trigonometric functions, the answer will be at the either of the extreme values in the range.