The equation $(\operatorname{cosp}-1) x^2+(\operatorname{cosp}) x+\sin \mathrm{p}=0$ in the variable $x$,…

The equation $(\operatorname{cosp}-1) x^2+(\operatorname{cosp}) x+\sin \mathrm{p}=0$ in the variable $x$, has real roots. Then p can take any value in the interval
  1. $(0,2 \pi)$
  2. $(-\pi, 0)$
  3. $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
  4. $(0, \pi)$

Solution

Given equation is $(\cos \mathrm{p}-1) x^2+(\cos \mathrm{p}) x+\sin \mathrm{p}=0$
Comparing with $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$, we get $a=\cos p-1, b=\cos p, c=\sin p$
It has real roots. $\therefore \quad b^2-4 a c \geq 0$ $\begin{aligned} & \Rightarrow \cos ^2 \mathrm{p}-4(\cos \mathrm{p}-1)(\sin \mathrm{p}) \geq 0 \\ & \Rightarrow \cos ^2 \mathrm{p}-4 \sin \mathrm{p} \cos \mathrm{p}+4 \sin \mathrm{p} \geq 0 \\ & \Rightarrow \cos ^2 \mathrm{p}-4 \sin \mathrm{p} \cos \mathrm{p}+4 \sin ^2 \mathrm{p} \\ & \quad+4 \sin \mathrm{p}-4 \sin ^2 \mathrm{p} \geq 0\end{aligned}$ $\begin{array}{ll} & \Rightarrow(\cos p-2 \sin p)^2+4 \sin p(1-\sin p) \geq 0 \\ \therefore \quad & (\cos p-2 \sin p) \text { is always positive } \\ \therefore \quad & 1-\sin p \geq 0 \text { for all values of } p, p \in(0, \pi)\end{array}$

Asked in: MHT CET 2024 (03 May Shift 2)

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