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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 interval
A.\[\left( {0,2\pi } \right)\]
B.\[\left( { - \pi ,0} \right)\]
C.\[\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)\]
D.\[\left[ {0,\pi } \right]\]

seo-qna
Last updated date: 13th Jun 2024
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Answer
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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.

Complete step-by-step answer:
 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.