Question

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

• A. $\text{both}{\cos }^{-1}\alpha \text{and}{\cos }^{-1}\beta \text{are}\text{real}$
• B. $\text{both}\cos \text{e}{\text{c}}^{-1}\alpha \text{and}\cos \text{e}{\text{c}}^{-1}\beta \text{are}\text{real}$
• C. $\text{both}{\cot }^{-1}\alpha \text{and}{\cot }^{-1}\beta \text{are}\text{real}$
• D. none of these

Answer 1

Answer: (B), (C)

$\text{both}\cos \text{e}{\text{c}}^{-1}\alpha \text{and}\cos \text{e}{\text{c}}^{-1}\beta \text{are}\text{real}$
$\text{both}{\cot }^{-1}\alpha \text{and}{\cot }^{-1}\beta \text{are}\text{real}$

Solution : -

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

$6x^2+11x+3=0$

$\Rightarrow (2x+3)(3x+1)=0$

$\Rightarrow x=\frac{-3}{2}$ & $x=-1/3$

Let $\alpha =\frac{-3}{2}$ and $\beta =\frac{-1}{3}$

Now $co{s}^{-1}\alpha =co{s}^{-1}\left(\frac{-3}{2}\right)$ which is not defined

since domain of $co{s}^{-1}\alpha$ is $[-1,1]$

$si{n}^{-1}\beta =si{n}^{-1}\left(\frac{-1}{3}\right)$ exists i.e, real

$cose{c}^{-1}\alpha$ and $cose{c}^{-1}\beta$ are real, since value lies in

domain, and similarly for $co{t}^{-1}\alpha$ and $co{t}^{-1}\beta$

$\therefore$ option B and C are correct