Question

The equation $(\cos p-1)x^2+\cos px+\sin p=0$, in the variable x has real roots. Then p can take any value in the interval

• A. $(0,2\pi )$
• B. $(-\pi ,0)$
• C. $(-\pi /2,\pi /2)$
• D. $(0,\pi )$

Answer 1

The correct option is D

$(0,\pi )$

Explanation for the correct option:

For a quadratic equation to possess real roots, the discriminant should be non-negative

$$\begin{gathered} b^2-4ac\ge 0 \\ \Rightarrow {\cos }^{2}p-4\sin p(\mathrm{co}sp-1)\ge 0 \end{gathered}$$

We can observe that ${\cos }^{2}p\ge 0,\cos p-1\le 0\forall x\in R$

So for discriminant to be always non-negative $\sin p\ge 0$

and we know $\sin p$positive in 1st and 2nd quadrant,

Hence interval of p is$(0,\pi )$.

Hence, the correct answer is option(D).