Question

If \[\alpha ,\beta (\alpha < \beta )\] are the roots of the equation \[6{x^2} + 11x + 3 = 0\] then which of the following is real?
A) \[{\cos ^{ - 1}}\alpha \]
B) \[{\sin ^{ - 1}}\beta \]
C) \[\cos e{c^{ - 1}}\alpha \]
D) Both \[{\cot ^{ - 1}}\alpha ,{\cot ^{ - 1}}\beta \]

Answer 1

Hint: At first, we will find out the roots of the given quadratic equation. We will apply Sreedhar Acharya’s formula.
Let us consider, the general equation of quadratic equation is \[a{x^2} + bx + c = 0\]
The roots are, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
After finding the roots we will substitute the roots in the given options and try to find out which one is real.

Complete step-by-step answer:
It is given that, \[\alpha ,\beta (\alpha < \beta )\] are the roots of the equation\[6{x^2} + 11x + 3 = 0\].
As an initial step, we will find the values of \[\alpha ,\beta \] by Sreedhar Acharya’s formula.
Let us consider, the general equation of quadratic equation is \[a{x^2} + bx + c = 0\]
The roots are, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Now let us compare the given equation \[6{x^2} + 11x + 3 = 0\] with the general form of quadratic equation, then we get,
\[a = 6,b = 11,c = 3\]
So, the roots are found to be,
\[x = \dfrac{{ - 11 \pm \sqrt {{{11}^2} - 4 \times 6 \times 3} }}{{2 \times 6}}\]
Here let us choose,
\[\alpha = \dfrac{{ - 11 + \sqrt {{{11}^2} - 4 \times 6 \times 3} }}{{2 \times 6}}\] and \[\beta = \dfrac{{ - 11 - \sqrt {{{11}^2} - 4 \times 6 \times 3} }}{{2 \times 6}}\]
Let us solve the values of\[\alpha ,\beta \], then we get,
\[\alpha = \dfrac{{ - 11 + 7}}{{12}}\] and \[\beta = \dfrac{{ - 11 - 7}}{{12}}\]
Simplifying the fraction again we get,
\[\alpha = \dfrac{{ - 1}}{3}\] and \[\beta = \dfrac{{ - 3}}{2}\]
Now let us consider the option
Substitute the values of \[\alpha = \dfrac{{ - 1}}{3}\] and \[\beta = \dfrac{{ - 3}}{2}\] in the given options we get,
\[{\cos ^{ - 1}}(\dfrac{{ - 1}}{3}) = {1.9^ \circ }\]
\[{\sin ^{ - 1}}(\dfrac{{ - 3}}{2})\] not real. Since, the value of the sin function lies between -1 to 1.
\[\cos e{c^{ - 1}}(\dfrac{{ - 1}}{3})\] has real value.
\[{\cot ^{ - 1}}(\dfrac{{ - 1}}{3}) = {143^ \circ }\] and \[{\cot ^{ - 1}}(\dfrac{{ - 3}}{2}) = {146^ \circ }\]
Hence,
\[{\cos ^{ - 1}}\alpha \], \[\cos e{c^{ - 1}}\alpha \], both \[{\cot ^{ - 1}}\alpha ,{\cot ^{ - 1}}\beta \] have real values.

Hence options (A),(C) and (D) are correct.

Note:
We can also find the roots of the given equation by middle term factor method.
\[6{x^2} + 11x + 3 = 0\]
We will rewrite \[11\] in terms of the factors of \[6 \times 3 = 18\]
So, we have the following form of equation,
\[6{x^2} + 9x + 2x + 3 = 0\]
By simplifying we get,
\[3x(2x + 3) + 1(2x + 3) = 0\]
On simplifying again we get,
\[(3x + 1)(2x + 3) = 0\]
The roots are \[\alpha = \dfrac{{ - 1}}{3}\] and \[\beta = \dfrac{{ - 3}}{2}\].