The equation $ (\cos p-1) x^{2}+\cos p x+\sin p=0 $ in $ x $ has real roots. Then the set of values of $ p $ is

$ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The equation

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

x

has real roots. Then the set of values of

p

View answer and explanation

Correct option: D

[0, \pi]

Option D (

[0, \pi]

) is correct because the discriminant of the quadratic equation must be non-negative for real roots, which leads to the condition that

\sin p

must be non-negative, restricting p to the interval

[0, \pi]

Practice All Chapter MCQs →

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

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

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)