Question

Which of the following is valid If $p={{\sin }^{2}}x+{{\cos }^{4}}x$, then
A. $\dfrac{3}{4}\le p\le 1$
B. $\dfrac{3}{16}\le p\le \dfrac{1}{4}$
C. $\dfrac{1}{4}\le p\le 1$
D. None of these

Answer 1

Hint: In this problem we will be using the trigonometric identity i.e. ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. First, we will expand the term ${{\cos }^{4}}x$ as ${{\cos }^{2}}x.{{\cos }^{2}}x$. Now we will substitute the value of ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ from the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in $p$ and we will find the value of $p$. Again, we will find the value of $p$ by substituting ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in $p$. From these two values we will find the range of the $p$.

Complete step-by-step solution
Given that, $p={{\sin }^{2}}x+{{\cos }^{4}}x$
We will be substituting ${{\cos }^{4}}x={{\cos }^{2}}x.{{\cos }^{2}}x$ in the above equation, then
$p={{\sin }^{2}}x+{{\cos }^{2}}x.{{\cos }^{2}}x$
We have the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, from this identity we will be substituting ${{\cos }^{2}}x=\left( 1-{{\sin }^{2}}x \right)$ in $p$, then
$\begin{align}
  & p={{\sin }^{2}}x+{{\cos }^{2}}x\left( 1-{{\sin }^{2}}x \right) \\
 & ={{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x.{{\cos }^{2}}x
\end{align}$
Substituting ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in the above equation, then
$p=1-{{\sin }^{2}}x.{{\cos }^{2}}x$
From the above equation, we can say that for any value of $x$ the value of $p$is less than or equal to $1$. Mathematically
$p\le 1....\left( \text{i} \right)$
Again, $p={{\sin }^{2}}x+{{\cos }^{4}}x$
Now we are going to substituting ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in the above equation, then
$p=1-{{\cos }^{2}}x+{{\cos }^{4}}x$
Now we will rearrange the above terms as below,
$p={{\left( {{\cos }^{2}}x \right)}^{2}}-2.\dfrac{1}{2}{{\cos }^{2}}x+1$
Now, we will performing method of completing square by adding and subtracting ${{\left( \dfrac{1}{2} \right)}^{2}}$ in the above equation, then
\[\begin{align}
  & p={{\left( {{\cos }^{2}}x \right)}^{2}}-2.\dfrac{1}{2}{{\cos }^{2}}x+1+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}} \\
 & ={{\left( {{\cos }^{2}}x \right)}^{2}}-2.\dfrac{1}{2}{{\cos }^{2}}x+{{\left( \dfrac{1}{2} \right)}^{2}}+1-\dfrac{1}{4}
\end{align}\]
We know that ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$, then
$p={{\left( {{\cos }^{2}}x-\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}$
From the above equation we can say that, the value of $p$ is greater than or equal to $\dfrac{3}{4}$. Mathematically
$p\ge \dfrac{3}{4}....\left( \text{ii} \right)$
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$, we have
$\dfrac{3}{4}\le p\le 1$

Note: The problem is completely based on the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we have some other trigonometric identities, they are
$\Rightarrow {{\sec }^{2}}x-{{\tan }^{2}}x=1$
$\Rightarrow {{\csc }^{2}}x-{{\cot }^{2}}x=1$
While solving this problem student may do mistakes by writing ${{\cos }^{4}}x$ as ${{\cos }^{2}}x+{{\cos }^{2}}x$ and writing $p$ as
$\begin{align}
  & p={{\sin }^{2}}x+{{\cos }^{2}}x+{{\cos }^{2}}x \\
 & =1+{{\cos }^{2}}x \\
\end{align}$
Which is not correct.