Aspire Faculty ID #3665 · Topic: NIMCET 2012 · Just now
NIMCET 2012

The equation $(\cos p-1)x^2+(\cos p)x+\sin p=0$ where $x$ is a variable, has real roots. Then, the interval of $p$ is

Solution

For the quadratic equation to have real roots, its discriminant must be non-negative. 

Here, $a=\cos p-1,\qquad b=\cos p,\qquad c=\sin p$ 

Therefore, $\Delta=b^2-4ac$ 

$\Delta=\cos^2p-4(\cos p-1)\sin p$ 

$\Delta=\cos^2p+4(1-\cos p)\sin p$ 

For 0<p<0$\pi$, $\sin p >0$ and $1-\cos p>0$ 

Therefore, $\cos^2p+4(1-\cos p)\sin p>0$ 

Hence, the equation has two real roots for $p\in(0,\pi)$.
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