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# Solve the following inequality.${{2}^{\dfrac{1}{{{\cos }^{2}}x}}}\sqrt{{{y}^{2}}-y+\dfrac{1}{2}}\le 1$

Last updated date: 10th Aug 2024
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Hint: For solving the given inequality, we need to divide both sides of the inequality by ${{2}^{{{\sec }^{2}}x}}$ to get $\sqrt{{{y}^{2}}-y+\dfrac{1}{2}}\le \dfrac{1}{{{2}^{{{\sec }^{2}}x}}}$. Then, we need to obtain the range of each side of the inequality. The range of the LHS can be obtained by using the completing the square method, and that of RHS can be obtained by using the range for the function ${{\sec }^{2}}x$ which is $\left[ 1,\infty \right)$. From these ranges, we can determine the values common to both the ranges which can be solved for both the sides to obtain the required values of $x$ and $y$.

Complete step by step solution:
The inequality given in the above question is
${{2}^{\dfrac{1}{{{\cos }^{2}}x}}}\sqrt{{{y}^{2}}-y+\dfrac{1}{2}}\le 1$
Now, we know that
\begin{align} & \Rightarrow \sec x=\dfrac{1}{\cos x} \\ & \Rightarrow {{\sec }^{2}}x=\dfrac{1}{{{\cos }^{2}}x} \\ \end{align}
Putting this in the above inequality, we get
$\Rightarrow {{2}^{{{\sec }^{2}}x}}\sqrt{{{y}^{2}}-y+\dfrac{1}{2}}\le 1$
Now, we know that the exponential function is always positive. So we have ${{2}^{{{\sec }^{2}}x}}>0$. So we can divide the above inequality by ${{2}^{{{\sec }^{2}}x}}$ to get
$\Rightarrow \sqrt{{{y}^{2}}-y+\dfrac{1}{2}}\le \dfrac{1}{{{2}^{{{\sec }^{2}}x}}}.......(i)$
Let us simplify the LHS of the above inequality, which is given by
$\Rightarrow LHS=\sqrt{{{y}^{2}}-y+\dfrac{1}{2}}$
We apply the completing the square method inside the square root to get
\begin{align} & \Rightarrow LHS=\sqrt{{{y}^{2}}-2y\left( \dfrac{1}{2} \right)+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{1}{2}} \\ & \Rightarrow LHS=\sqrt{{{y}^{2}}-2y\left( \dfrac{1}{2} \right)+{{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{1}{2}-\dfrac{1}{4}} \\ & \Rightarrow LHS=\sqrt{{{y}^{2}}-2y\left( \dfrac{1}{2} \right)+{{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{1}{2}-\dfrac{1}{4}} \\ & \Rightarrow LHS=\sqrt{{{y}^{2}}-2y\left( \dfrac{1}{2} \right)+{{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{1}{4}}........(ii) \\ \end{align}
Now, we know that ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$.
Putting $a=y$ and $b=\dfrac{1}{2}$, we get
$\Rightarrow {{a}^{2}}-2y\left( \dfrac{1}{2} \right)+{{\left( \dfrac{1}{2} \right)}^{2}}={{\left( y-\dfrac{1}{2} \right)}^{2}}$
Substituting this in (iii) we get
$\Rightarrow LHS=\sqrt{{{\left( y-\dfrac{1}{2} \right)}^{2}}+\dfrac{1}{4}}$
Now, we know that the range of the square function is $\left[ 0,\infty \right)$. So the range of the function ${{\left( y-\dfrac{1}{2} \right)}^{2}}$ will also be $\left[ 0,\infty \right)$. This means that the range of the function ${{\left( y-\dfrac{1}{2} \right)}^{2}}+\dfrac{1}{4}$ is $\left[ 0+\dfrac{1}{4},\infty \right)=\left[ \dfrac{1}{4},\infty \right)$. This in turn gives us the range of the function $\sqrt{{{\left( y-\dfrac{1}{2} \right)}^{2}}+\dfrac{1}{4}}$ as $\left[ \sqrt{\dfrac{1}{4}},\infty \right)=\left[ \dfrac{1}{2},\infty \right)$.
This means that
$LHS\ge \dfrac{1}{2}........(iii)$
Now, we simplify the RHS of the inequality (i) which is
$\Rightarrow RHS=\dfrac{1}{{{2}^{{{\sec }^{2}}x}}}$
We know that the range of the function ${{\sec }^{2}}x$ is $\left[ 1,\infty \right)$. So the range of the function ${{2}^{{{\sec }^{2}}x}}$ is $\left[ {{2}^{1}},\infty \right)=\left[ 2,\infty \right)$. This means that the range of the function $\dfrac{1}{{{2}^{{{\sec }^{2}}x}}}$ becomes $\left( 0,\dfrac{1}{2} \right]$.
This means that
$0\le RHS\le \dfrac{1}{2}........(iv)$
From (iii) and (iv) the only value common between the ranges of the LHS and the RHS is $\dfrac{1}{2}$. So the inequality (i) $\sqrt{{{y}^{2}}-y+\dfrac{1}{2}}\le \dfrac{1}{{{2}^{{{\sec }^{2}}x}}}$ is defined only for the “equal to” sign, and not for the “less than” sign, where each of the two sides is equal to $\dfrac{1}{2}$. So the given inequality can be reduced to
$\Rightarrow \sqrt{{{y}^{2}}-y+\dfrac{1}{2}}=\dfrac{1}{{{2}^{{{\sec }^{2}}x}}}=\dfrac{1}{2}$
On solving the above equation, we get $y=\dfrac{1}{2}$ and $x=n\pi$, where $n\in Z$.
Hence, the solution of the given inequality is $x=n\pi$ and $y=\dfrac{1}{2}$.

Note: While dividing an inequality, it is important to check its sign. This is because the sign of an inequality is reversed when divided or multiplied by a negative number. We could divide the given inequality by ${{2}^{{{\sec }^{2}}x}}$ since it is positive for each value of $x$.