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Question: If roots of the equation \(2{x^{^2}} - 4x + 2\sin \theta - 1 = 0\) are of opposite sign\((where \the...

If roots of the equation 2x2−4x+2sin⁡θ−1=02{x^{^2}} - 4x + 2\sin \theta - 1 = 0 are of opposite sign(whereθ∈(0,π)),(where \theta \in (0,\pi )), then θ\theta belongs to
(A)(π6,5π6)(\dfrac{\pi }{6},\dfrac{{5\pi }}{6})
(B)(0,π6)∪(5π6,π)(0,\dfrac{\pi }{6}) \cup (\dfrac{{5\pi }}{6},\pi )
(C)(0,5π6)(0,\dfrac{{5\pi }}{6})
(D)(0,π)(0,\pi )

Explanation

Solution

Determinant of this quadratic equation should be greater than 0,roots of opposite sign mean if one is positive then second should be negative, so the product of both is negative.

Complete step-by-step answer:
2x2−4x+2sin⁡θ−1=02{x^2} - 4x + 2\sin \theta - 1 = 0
We are solving this question by using general equation, that is ax2+bx+c=0a{x^2} + bx + c = 0 if we suppose its root is $\alpha ,\beta $ and its of opposite sign that`s mean α.β<0\alpha .\beta < 0. In these type of question the value of α,β\alpha ,\beta low root get value of ca<0\dfrac{c}{a} < 0
comparing 2x2−4x+2sin⁡θ−1=02{x^2} - 4x + 2\sin \theta - 1 = 0 with general ax2+bx+c=0a{x^2} + bx + c = 0
a = 2 b = -4 c = 2sin⁡θ2\sin \theta

In this equation both roots are of opposite sing so we apply ca<0\dfrac{c}{a} < 0 (same equation) 2sin⁡θ−12<0\dfrac{{2\sin \theta - 1}}{2} < 0
2sin⁡θ−1<02\sin \theta - 1 < 0
Sin⁡θ<12\operatorname{Sin} \theta < \dfrac{1}{2}
D>0D > 0 [ where D = b2−4ac{b^2} - 4ac ]
Value of D will be 42−2.(2sin⁡θ−1)>0{4^2} - 2.(2\sin \theta - 1) > 0
= 16−4.2.(2sin⁡θ−1)>016 - 4.2.(2\sin \theta - 1) > 0
= 16−8(2sin⁡θ−1)>016 - 8(2\sin \theta - 1) > 0
Now open the brackets and multiply by 8
=16−16sin⁡θ+8>016 - 16\sin \theta + 8 > 0
Add the numbers which are 16+8
=24−16sin⁡θ>024 - 16\sin \theta > 0
compare the equation
= 24>16sin⁡θ24 > 16\sin \theta
= Sin⁡θ<2416\operatorname{Sin} \theta < \dfrac{{24}}{{16}}
This quantity is greater than 1. This is true for all values of θ\theta because the value of sin⁡θ\sin \theta is less than 1.
Sin⁡θ\operatorname{Sin} \theta should be less than 32=1.5\dfrac{3}{2} = 1.5
Sin⁡θ>0.5\operatorname{Sin} \theta > 0.5
θ\theta =(0,π6),(5π6,π)(0,\dfrac{\pi }{6}),(\dfrac{{5\pi }}{6},\pi )
So, the correct answer is “Option B”.

Note: Root of opposite sign means if one is positive then second should be negative, so the product of both must be negative. We can solve this question by solving other options. For roots of a given quadratic to be the opposite sign, the product of roots is negative. Sign of roots of a quadratic equation.
*Both roots are positive (If a and b are opposite in sign and a and c are same in sign)
*Both roots are negative (If a,b,c are all of same sign)
*Roots are of opposite sign (If a and c are of opposite sign)
*roots equal but opposite in sing (If b=0)