
The equation \[{{\tan }^{4}}x-2{{\sec }^{2}}x+a=0\] will have at least one solution if:
A. \[1\le a\le 4\]
B. \[a\ge 2\]
C. \[a\le 3\]
D. none of these
Answer
612.6k+ views
Hint:In the above question we will substitute the value of \[{{\sec }^{2}}x\] in terms of \[{{\tan }^{2}}x\], i.e. we know the formula-\[{{\sec }^{2}}x=1+{{\tan }^{2}}x\]. Also, we will use the property that the square of any term (real number) is always greater than or equal to zero.
Complete step-by-step answer:
We have been given the equation \[{{\tan }^{4}}x-2{{\sec }^{2}}x+a=0\].
So by substituting \[{{\sec }^{2}}x=1+{{\tan }^{2}}x\] in the above equation, we get as the following:
\[{{\tan }^{4}}x-2\left( 1+{{\tan }^{2}}x \right)+a=0\]
On further solving the above equation, we get as the following:
\[{{\tan }^{4}}x-2{{\tan }^{2}}x-2+a=0\]
Now we will make the above equation a perfect square as shown below.
\[{{\left( {{\tan }^{2}}x \right)}^{2}}-2\left( 1 \right)\left( {{\tan }^{2}}x \right)+1-1-2+a=0\]
On simplification of the equation above, we get as the following:
\[{{\left( {{\tan }^{2}}x-1 \right)}^{2}}-3+a=0\]
As we happen to know that the square term is somehow always greater than or equal to zero, we will eventually use it here to find the condition upon ‘a’ for at least one solution of the above equation.
\[\begin{align}
& {{\left( {{\tan }^{2}}x-1 \right)}^{2}}=3-a \\
& {{\left( {{\tan }^{2}}x-1 \right)}^{2}}\ge 0 \\
& 3-a\ge 0 \\
& a\le 3 \\
\end{align}\]
Here, we get the answer of the above equation. We get the values as a is smaller than and equal to 3.
The equation \[{{\tan }^{4}}x-2{{\sec }^{2}}x+a=0\] will have at least one solution if \[a\le 3\].
Therefore, the correct option of the given question is option C.
Note: Just be careful while doing calculation as there is a chance that you might make a silly mistake and end up getting a wrong answer.Remember the property that the square term is always greater than equal to zero as it will help you a lot in many questions.Also, remember the other trigonometric identities and formula which helps you a lot in these types of questions.
Complete step-by-step answer:
We have been given the equation \[{{\tan }^{4}}x-2{{\sec }^{2}}x+a=0\].
So by substituting \[{{\sec }^{2}}x=1+{{\tan }^{2}}x\] in the above equation, we get as the following:
\[{{\tan }^{4}}x-2\left( 1+{{\tan }^{2}}x \right)+a=0\]
On further solving the above equation, we get as the following:
\[{{\tan }^{4}}x-2{{\tan }^{2}}x-2+a=0\]
Now we will make the above equation a perfect square as shown below.
\[{{\left( {{\tan }^{2}}x \right)}^{2}}-2\left( 1 \right)\left( {{\tan }^{2}}x \right)+1-1-2+a=0\]
On simplification of the equation above, we get as the following:
\[{{\left( {{\tan }^{2}}x-1 \right)}^{2}}-3+a=0\]
As we happen to know that the square term is somehow always greater than or equal to zero, we will eventually use it here to find the condition upon ‘a’ for at least one solution of the above equation.
\[\begin{align}
& {{\left( {{\tan }^{2}}x-1 \right)}^{2}}=3-a \\
& {{\left( {{\tan }^{2}}x-1 \right)}^{2}}\ge 0 \\
& 3-a\ge 0 \\
& a\le 3 \\
\end{align}\]
Here, we get the answer of the above equation. We get the values as a is smaller than and equal to 3.
The equation \[{{\tan }^{4}}x-2{{\sec }^{2}}x+a=0\] will have at least one solution if \[a\le 3\].
Therefore, the correct option of the given question is option C.
Note: Just be careful while doing calculation as there is a chance that you might make a silly mistake and end up getting a wrong answer.Remember the property that the square term is always greater than equal to zero as it will help you a lot in many questions.Also, remember the other trigonometric identities and formula which helps you a lot in these types of questions.
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