JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The equation

\left( {\cos p - 1} \right){x^2} + \cos p\;x + \sin p = 0,

in variable

x

, has real roots. Then,

p

belongs to the interval

View answer and explanation

Correct option: D

\left( {0,\pi } \right)

Option D (

\left( {0,\pi } \right)

) is correct because for the quadratic equation to have real roots, the discriminant must be greater than or equal to zero, leading to the condition

\sin p \ge 0

, which is satisfied in the interval

\left( {0,\pi } \right)

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{x^2} + 4ax + 2 > 0

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