Question

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

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

Answer 1

Answer: (C)

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

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

$6x^2+9x+2x+3=0$

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

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

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

Hence $co{t}^{-1}\left(\alpha \right)$ and $co{t}^{-1}\left(\alpha \right)$ are real, since $co{s}^{-1}\left(\alpha \right)\in [-1,1]$ and $cose{c}^{-1}\left(\alpha \right)\notin [-1,1]$.