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Question: The equation \((\cos p - 1){x^2} + \cos px + \sin p = 0\),in the variable \(x\) has real roots. Th...

The equation (cos⁡p−1)x2+cos⁡px+sin⁡p=0(\cos p - 1){x^2} + \cos px + \sin p = 0,in the variable xx has real roots.
Then pp can take any value in the interval
A.(0,2π)(0,2\pi )
B.(−π,0)( - \pi ,0)
C.(−π2,π2)(\dfrac{{ - \pi }}{2},\dfrac{\pi }{2})
D.(0,π)(0,\pi )

Explanation

Solution

Hint : Use the Quadratic equation formula, since the given equation is a quadratic equation. Real roots means that Discriminant should be greater than zero. Check for the range of trigonometric values, their signs and then put the value.

Complete step-by-step answer :
We are given with the equation (cos⁡p−1)x2+cos⁡px+sin⁡p=0(\cos p - 1){x^2} + \cos px + \sin p = 0.Since, the highest degree of xx is DD so it is a Quadratic Equation.
According to the question, it is saying that the equation gives real roots which mean that the Discriminant should be greater than zero.
Or, Discriminant⩾0 \geqslant 0
From the discriminant formula we know that Discriminant (DD)=b2−4ac = {b^2} - 4ac
So, b2−4ac⩾0{b^2} - 4ac \geqslant 0
Equating (cos⁡p−1)x2+cos⁡px+sin⁡p=0(\cos p - 1){x^2} + \cos px + \sin p = 0 with the basic quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 , we get:

a=(cos⁡p−1) b=cos⁡p c=sin⁡p   a = (\cos p - 1) \\\ b = \cos p \\\ c = \sin p \;

Putting these values in Discriminant formula, we get:

b2−4ac⩾0 (cos⁡p)2−4(cos⁡p−1)sin⁡p⩾0   {b^2} - 4ac \geqslant 0 \\\ {(\cos p)^2} - 4(\cos p - 1)\sin p \geqslant 0 \;

On Further solving:
(cos⁡p)2⩾4(cos⁡p−1)sin⁡p{(\cos p)^2} \geqslant 4(\cos p - 1)\sin p ……….(i)
We know that for real roots the range of cos⁡p\cos p and sin⁡p\sin p should be as follows:

0⩽cos⁡p⩽1 sin⁡p⩾0   0 \leqslant \cos p \leqslant 1 \\\ \sin p \geqslant 0 \;

From (i), we are getting that if sin⁡p\sin p is greater than zero, then cos⁡p\cos p will also become greater than zero.
Now, let’s check that in which quadrant sin⁡p\sin p will be greater than zero.
Since, we know that for the first and second quadrant sin⁡p\sin p will be greater than zero.
So, the range will be (0,π)(0,\pi ) which is the first and second quadrant but, the values are greater than or equal to so it will be a closed interval.
Therefore, it becomes [0,π][0,\pi ] .
Hence, the equation (cos⁡p−1)x2+cos⁡px+sin⁡p=0(\cos p - 1){x^2} + \cos px + \sin p = 0,in the variable xx has real roots.
Then pp can take any value in the interval [0,π][0,\pi ] .
Therefore, the correct option is Option D that is [0,π]\left[ {0,\pi } \right]
So, the correct answer is “OPTION D”.

Note : There can be a possibility of error in checking for the range of the trigonometric values. Use of Quadratic Formula is must, without this the solution would become lengthy. Always check for the closed and open intervals.