If both $\displaystyle :\alpha, \beta $ and $\displaystyle :6x^{2}+11x+3= 0$ are real roots of the equation then
- both $\displaystyle \:\cos ^{-1}\alpha $ and $\displaystyle \:\cos ^{-1}\beta $ are real
- both $\displaystyle cosec ^{-1}\alpha $ and $\displaystyle \cos ^{-1}\beta $ are real
- both $\displaystyle \:\cot ^{-1}\alpha $ and $\displaystyle \:\cot ^{-1}\beta $ are real
-
none of these
For cos^-1(x) or cot^-1(x) to be real, the input must be in the domain [-1, 1] for cos^-1 and any real number for cot^-1. The roots of 6x^2 + 11x + 3 = 0 are found via quadratic formula: (-11 +/- sqrt(121 - 72))/12 = (-11 +/- 7)/12. Roots are -4/12 = -1/3 and -18/12 = -3/2. Since -3/2 is outside [-1, 1], cos^-1(-3/2) is not real. However, cot^-1(x) is defined for all real x.
Using the quadratic formula to solve 6x^2 + 11x + 3 = 0, the discriminant is 121 - 72 = 49, yielding roots of -1/3 and -3/2. The inverse cosine function is only defined for arguments between -1 and 1, so cos^(-1)(-3/2) is not real; cosec^(-1) is undefined for arguments between -1 and 1, making option B invalid. The inverse cotangent function is defined for all real numbers, so both cot^(-1)(-1/3) and cot^(-1)(-3/2) are real.