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# The equation $\left( {\cos p - 1} \right){x^2} + \cos px + \sin p = 0$, in the variable $x$ has real roots. Then $p$ can take any value in the intervalA.$\left( {0,2\pi } \right)$B.$\left( { - \pi ,0} \right)$C.$\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$D.$\left[ {0,\pi } \right]$

Last updated date: 13th Jun 2024
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Hint: Here, we will use the formula of discriminant is calculated ${b^2} - 4ac$ of the standard form of quadratic equation is $a{x^2} + bx + c$. Then we will compare the given expression with the standard form of the quadratic equation to find the value of $a$, $b$ and $c$. Then we will substitute the above value of $a$, $b$ and $c$ in the formula of discriminant. Since we know that when $y$ is always positive, the discriminant $D$ is less than 0 and then simplifies to find the required interval.

We are given that the equation $\left( {\cos p - 1} \right){x^2} + \cos px + \sin p = 0$, in the variable $x$ has real roots.
We know that any quadratic equation has real roots when its discriminant is non-negative.
We know that the discriminant is calculated using the formula ${b^2} - 4ac$ of the standard form of quadratic equation is $a{x^2} + bx + c$.
Comparing the given expression with the standard form of quadratic equation to find the value of $a$, $b$ and $c$, we get
$a = \cos p - 1$
$b = \cos p$
$c = \sin p$
Substituting the above value of $a$, $b$ and $c$ in the formula of discriminant, we get
$\Rightarrow {\cos ^2}p - 4 \cdot \sin p \cdot \left( {\cos p - 1} \right) \\ \Rightarrow {\cos ^2}p - 4\sin p\left( {\cos p - 1} \right) \\$
Since we know that when is always non-negative, the discriminant $D$ is greater than equal to 0.
$\Rightarrow {\cos ^2}p - 4\sin p\left( {\cos p - 1} \right) \geqslant 0$
So we have observed that ${\cos ^2}x \geqslant 0$,$4\sin p\left( {\cos p - 1} \right) \geqslant 0$ $\forall x \in \mathbb{R}$.
Since the discriminant is always non-negative $\sin p \geqslant 0$ and we know that $\sin x$ is positive only in the 1st and 2nd quadrant.
Therefore, the interval of $p$ is $\left[ {0,\pi } \right]$.

Hence, option D is correct.

Note: A quadratic is a type of problem that deals with a variable multiplied by itself and an operation known as squaring. One should know that in an equation $a{x^2} + bx + c$, the sum of roots of the equation is equal $- a$ and product is equal to $b$. We need to know the product of a negative number and a positive number is a negative. Substitute the values properly and avoid calculation mistakes.