Question

If $\tan \theta$ and $\cot \theta$ are the roots of the equation $x^2+2x+1=0$ then the least value of $x^2+\tan \theta x+\cot \theta =0$ is

• A. $\frac{3}{4}$
• B. $\frac{5}{4}$
• C. $\frac{-5}{4}$
• D. $\frac{-3}{4}$

Answer 1

The correct option is C $\frac{-5}{4}$

$x^2+2x+1=0\Rightarrow {(x+1)}^2=0\Rightarrow x=-1\tan \theta and\cot \theta arerootsoftheequationsince\tan \theta =\cot \theta =-1sof\left(x\right)=x^2-x-1forleastvalueoff\left(x\right)f^{{}^{\prime }}\left(x\right)=0f^{\prime \prime }\left(x\right)>0f^{{}^{\prime }}\left(x\right)=2x-1=0\Rightarrow x=\frac{1}{2}{f}^{\prime \prime }\left(x\right)=2>0soleastvalueoff\left(x\right)={\left(\frac{1}{2}\right)}^2-\frac{1}{2}-1=\frac{-5}{4}sooptionCiscorrect$