NIMCET 2012 — Mathematics PYQ

From the given equation $(\cos p-1)x^2+(\cos p)x+\sin p=0$:

• $a=\cos p-1$

• $b=\cos p$

• $c=\sin p$

2. Set up the Discriminant ($D$):

3. Analyze the expression:

Expanding the inequality:

To find the valid interval for $p$, we test the provided options to see which range consistently satisfies the condition $D\ge 0$.

• For $p\in (0,\pi )$ (Option d):

  In this interval, $\sin p$ is always positive (> 0).

  We know that $\cos p-1$ is always less than or equal to 0 (since the max value of $\cos p$ is $1$).

  If $p\in (0,\pi )$, then \cos p < 1, making $(\cos p-1)$ negative.

  Looking back at $D=b^2-4ac$:

  • $b^2=(\cos p{)}^2$ is always $\ge 0$.

  • $-4(a)(c)=-4(\text{negative})(\text{positive})=\text{positive}$.

  Since we are adding a positive value to a non-negative value ($b^2$), the sum $D$ will always be greater than 0.

Conclusion:

The condition for real roots is satisfied when $p$ is in the interval $(0,\pi )$.

Correct Option:

(d) $(0,\pi )$