Question

Find the range of values of $t$ for which $2\sin t = \dfrac{{5{x^2} - 2x + 1}}{{3{x^2} - 2x - 1}}$, $t \in [\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}]$

Answer 1

Hint: We need to make the value of the fraction equal to a constant $y$ and make $\sin t = y$ and then get a new equation and solve it, find the discriminant value. Now by further simplifying, we will get the required answer.

Complete step by step answer:
The equation given in the question is;
$2\sin t = \dfrac{{5{x^2} - 2x + 1}}{{3{x^2} - 2x - 1}}$ and $t \in [\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}]$
Now, the first step would be equating $\sin t = y$. After this, by substituting the value of $y$, we get:
$2y = \dfrac{{5{x^2} - 2x + 1}}{{3{x^2} - 2x - 1}}$
By multiplying the denominator with $2y$ , we get:
$2y(3{x^2} - 2x - 1) = 5{x^2} - 2x + 1$
By gathering all the $x$ terms on one side and equating it to zero, we get:
$6{x^2}y - 4xy - 2y = 5{x^2} - 2x + 1$
$\Rightarrow {x^2}(6y - 1) + x( - 4y + 2) - (1 + 2y) = 0$

Now, for this equation to have real roots, the discriminant should be equal to or greater than zero. In a quadratic equation $a{x^2} + bx + c = 0$ , the discriminant is given by the formula;
$D = {b^2} - 4ac$
Now substituting the values of a, b and c in the equation, we get:
${( - 4y + 2)^2} + 4(6y - 5)(1 + 2y) \geqslant 0$
By simplification we get:
$16{y^2} + 4 - 16y + 4(6y - 5 + 12{y^2} - 10y) \geqslant 0$
By taking out 4 as common, we get:
$4{y^2} + 1 - 4y + 6y - 5 + 12{y^2} - 10y \geqslant 0$
$\Rightarrow 16{y^2} - 8y - 4 \geqslant 0$
$\Rightarrow 4{y^2} - 2y - 1 \geqslant 0$

To solve the above quadratic equation, we need to use discriminant to find the roots. The discriminant of the above equation is:
$D = {2^2} + 4 \times 4 \times 1$
$\Rightarrow D = 4 + 16 = 20$
Now the roots of the equation are:
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using the above formula, we get the roots are:
$\dfrac{{2 \pm \sqrt {20} }}{8} = \dfrac{{2 \pm 2\sqrt 5 }}{8} \\
\Rightarrow \dfrac{{2 \pm \sqrt {20} }}{8} = \dfrac{{1 \pm \sqrt 5 }}{4}$

Now the range of $y$ is:
$\Rightarrow y \leqslant \dfrac{{1 - \sqrt 5 }}{4}$ and $y \geqslant \dfrac{{1 + \sqrt 5 }}{4}$
Now we know that $\sin t = y$. After substituting the value of y, we get:
$ - 1 \leqslant \sin t \leqslant \dfrac{{1 - \sqrt 5 }}{4}$ and $1 \geqslant \sin t \geqslant \dfrac{{1 + \sqrt 5 }}{4}$
Applying ${\sin ^{ - 1}}$ on both the equations, we get:
${\sin ^{ - 1}}( - 1) \leqslant {\sin ^{ - 1}}(\sin t) \leqslant {\sin ^{ - 1}}(\dfrac{{1 - \sqrt 5 }}{4})$ $\Rightarrow {\sin ^{ - 1}}(1) \geqslant {\sin ^{ - 1}}(\sin t) \geqslant {\sin ^{ - 1}}(\dfrac{{1 + \sqrt 5 }}{4})$
After solving the equations, we get:
$\dfrac{{ - \pi }}{2} \leqslant t \leqslant \dfrac{{ - \pi }}{{10}}$ and $\dfrac{{3\pi }}{{10}} \leqslant t \leqslant \dfrac{\pi }{2}$
Therefore, we can say that:
$\therefore t \in [\dfrac{{ - \pi }}{2},\dfrac{{ - \pi }}{{10}}] \cup [\dfrac{{3\pi }}{{10}},\dfrac{\pi }{2}]$

Therefore, the correct option is A.

Note: For any equation to have real and distinct roots, the discriminant should be greater than zero, to have real and equal roots, the discriminant should be equal to zero. To have imaginary roots, the discriminant should be less than zero. Here since they mentioned the range of $t$ as $t \in [\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}]$, the answer is $t \in [\dfrac{{ - \pi }}{2},\dfrac{{ - \pi }}{{10}}] \cup [\dfrac{{3\pi }}{{10}},\dfrac{\pi }{2}]$ . The general answer to this question would be:
$t \in [2n\pi - \dfrac{\pi }{2},2n\pi - \dfrac{\pi }{{10}}] \cup [2n\pi + \dfrac{{3\pi }}{{10}},2n\pi + \dfrac{\pi }{2}]$..........(the above question is $n = 0$ )