Question

If $\alpha ,\beta$ are roots of the equation $6x^2+11x+3=0$ then

• A. Both ${\cos }^{-1}\alpha$ and ${\cos }^{-1}\beta$ are real
• B. Both ${\text{cosec}}^{-1}\alpha$ and ${\text{cosec}}^{-1}\beta$ are real
• C. Both ${\cot }^{-1}\alpha$ and ${\cot }^{-1}\beta$ are real
• D. None of the above

Answer 1

The correct option is C

Both ${\cot }^{-1}\alpha$ and ${\cot }^{-1}\beta$ are real

$6x^2+11x+3=0\Rightarrow x=\frac{-1}{3},\frac{-3}{2}$
$-1<\frac{-1}{3}<1$
$\frac{-3}{2}<-1$
$\therefore co{s}^{-1}\left(\frac{1}{3}\right)$ exists but $co{s}^{-1}\left(\frac{-3}{2}\right)$ does not;
$cose{c}^{-1}\left(\frac{-3}{2}\right)$ exists but $cose{c}^{-1}\left(\frac{-1}{3}\right)$ does not;
$co{t}^{-1}\left(\frac{-1}{3}\right)$ and $co{t}^{-1}\left(\frac{-3}{2}\right)$ exist