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

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

& \Rightarrow {{\left( x+1 \right)}^{2}}=0 \\\ & \Rightarrow x=-1 \\\ \end{aligned}$$ Here, we can see that there is only one value of $$'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{aligned} & \Rightarrow 2k-1=0 \\\ & \Rightarrow k=\dfrac{1}{2} \\\ \end{aligned}$$ Now we know that the least value is given as $$f\left( k \right)$$ By substituting the value of $$'k'$$ in equation (i) we get $$\begin{aligned} & \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{aligned}$$ Now, by taking the LCM and adding the terms in above equation we get $$\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{aligned} & \Rightarrow L=\dfrac{4\left( -1 \right)-{{\left( -1 \right)}^{2}}}{4} \\\ & \Rightarrow L=\dfrac{-4-1}{4}=\dfrac{-5}{4} \\\ \end{aligned}$$ Therefore the least value of $${{x}^{2}}+\tan \theta x+\cot \theta $$ is $$\dfrac{-5}{4}$$ So, option (c) is the correct answer.