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

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 $\beta = \dfrac{{ - 3}}{2}$.