Question

If \[\tan \theta \] and \[\cot \theta \] are the roots of the equation \[{{x}^{2}}+2x+1=0\] then the least value of
\[{{x}^{2}}+\tan \theta x+\cot \theta \] is
(a) \[\dfrac{3}{4}\]
(b) \[\dfrac{5}{4}\]
(c) \[\dfrac{-5}{4}\]
(d) \[\dfrac{-3}{4}\]

Answer 1

Hint: We solve this problem first by finding the value of \[\tan \theta \] and \[\cot \theta \] using the condition that \[\tan \theta \] and \[\cot \theta \] are the roots of the equation \[{{x}^{2}}+2x+1=0\] then the least value of \[f\left( x \right)\] is given as \[f\left( k \right)\] such that \[{f}'\left( k \right)=0\]. That means we differentiate the given polynomial and make it zero to find \['k'\] then the least value is given as \[f\left( k \right)\]

Complete step by step answer:
We are given with quadratic equation that is
\[\Rightarrow {{x}^{2}}+2x+1=0\]
We know that the formula of square of sum of numbers that is
\[\Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
By using this formula to given equation we get
\[\begin{align}
  & \Rightarrow {{\left( x+1 \right)}^{2}}=0 \\
 & \Rightarrow x=-1 \\
\end{align}\]
Here, we can see that there is only one value of \['x'\]
We are given that \[\tan \theta \] and \[\cot \theta \] are the roots of the equation \[{{x}^{2}}+2x+1=0\]
We know that if a quadratic equation has only one root that means both the roots are equal.
By using this condition we can write that
\[\Rightarrow \tan \theta =\cot \theta =-1\]
We are asked to find the least value of \[{{x}^{2}}+\tan \theta x+\cot \theta \]
Let us assume that this polynomial as
\[\Rightarrow f\left( x \right)={{x}^{2}}+\tan \theta x+\cot \theta \]
Now, by substituting the value of \[\tan \theta \] and \[\cot \theta \] in above function we get
\[\Rightarrow f\left( x \right)={{x}^{2}}-x-1.........equation(i)\]
We know that the least value of \[f\left( x \right)\] is given as \[f\left( k \right)\] such that \[{f}'\left( k \right)=0\]
Now, by differentiating on both sides in the above equation we get
\[\Rightarrow {f}'\left( x \right)=2x-1\]
Now let us find the value of \['k'\] such that \[{f}'\left( k \right)=0\] that is
\[\begin{align}
  & \Rightarrow 2k-1=0 \\
 & \Rightarrow k=\dfrac{1}{2} \\
\end{align}\]
Now we know that the least value is given as \[f\left( k \right)\]
By substituting the value of \['k'\] in equation (i) we get
\[\begin{align}
  & \Rightarrow f\left( k \right)={{\left( \dfrac{1}{2} \right)}^{2}}-\left( \dfrac{1}{2} \right)-1 \\
 & \Rightarrow f\left( k \right)=\dfrac{1}{4}-\dfrac{1}{2}-1 \\
\end{align}\]
Now, by taking the LCM and adding the terms in above equation we get
\[\Rightarrow f\left( k \right)=\dfrac{1-2-4}{4}=\dfrac{-5}{4}\]
Therefore the least value of \[{{x}^{2}}+\tan \theta x+\cot \theta \] is \[\dfrac{-5}{4}\]

So, the correct answer is “Option c”.

Note: We can find the least value of a function directly.
The least value of a quadratic function \[a{{x}^{2}}+bx+c\] is given as
\[L=\dfrac{4ac-{{b}^{2}}}{4a}\]
We have the function given as
\[\Rightarrow f\left( x \right)={{x}^{2}}+\tan \theta x+\cot \theta \]
By using the above formula we get the least value as
\[\Rightarrow L=\dfrac{4\left( 1 \right)\left( \cot \theta \right)-{{\left( \tan \theta \right)}^{2}}}{4\left( 1 \right)}\]
Now, by substituting the value of \[\tan \theta \] and \[\cot \theta \] in above function we get
\[\begin{align}
  & \Rightarrow L=\dfrac{4\left( -1 \right)-{{\left( -1 \right)}^{2}}}{4} \\
 & \Rightarrow L=\dfrac{-4-1}{4}=\dfrac{-5}{4} \\
\end{align}\]
Therefore the least value of \[{{x}^{2}}+\tan \theta x+\cot \theta \] is \[\dfrac{-5}{4}\]
So, option (c) is the correct answer.