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

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

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

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