Functions Ques 26
Find the range of values of $t$ for which
$ 2 \sin t=\frac{1-2 x+5 x^{2}}{3 x^{2}-2 x-1}, t \in[-\frac{\pi}{2}, \frac{\pi}{2}] \text {. } $
(2005, 2M)
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Answer:
Correct Answer: 26.$t \in[-\frac{\pi}{2}, \frac{\pi}{10} ]\cup [\frac{3 \pi}{10}, \frac{\pi}{2}]$
Solution:
Formula:
- Given, $2 \sin t=\frac{1-2 x+5 x^{2}}{3 x^{2}-2 x-1}, t \in[-\frac{\pi}{2}, \frac{\pi}{2}]$
Put $\quad 2 \sin t=y \Rightarrow-2 \leq y \leq 2$
$\therefore \quad y=\frac{1-2 x+5 x^{2}}{3 x^{2}-2 x-1}$
$\Rightarrow \quad(3 y-5) x^{2}-2 x(y-1)-(y+1)=0$
Since, $x \in R-{1,-1 / 3}$
$ \begin{aligned} & {\left[\text { as, } 3 x^{2}-2 x-1 \neq 0 \Rightarrow(x-1)(x+1 / 3) \neq 0\right]} \\ & \therefore \quad D \geq 0 \\ & \Rightarrow \quad 4(y-1)^{2}+4(3 y-5)(y+1) \geq 0 \\ & \Rightarrow \quad y^{2}-y-1 \geq 0 \\ & \Rightarrow \quad (y-\frac{1}{2})^{2}-\frac{5}{4} \geq 0 \\ & \Rightarrow \quad (y-\frac{1}{2}-\frac{\sqrt{5}}{2} )\quad( y-\frac{1}{2}+\frac{\sqrt{5}}{2} )\geq 0 \\ & \Rightarrow \quad y \leq \frac{1-\sqrt{5}}{2} \\ & \text { or } \quad y \geq \frac{1+\sqrt{5}}{2} \\ & \Rightarrow \quad 2 \sin t \leq \frac{1-\sqrt{5}}{2} \\ & \text { or } \quad 2 \sin t \geq \frac{1+\sqrt{5}}{2} \\ & \Rightarrow \quad \sin t \leq \sin (-\frac{\pi}{10}) \\ & \text { or } \quad \sin t \geq \sin (\frac{3 \pi}{10}) \\ & \Rightarrow \quad t \leq-\frac{\pi}{10} \quad \text { or } \quad t \geq \frac{3 \pi}{10} \end{aligned} $
Hence, range of $t$ is $[-\frac{\pi}{2},-\frac{\pi}{10} ]\cup [\frac{3 \pi}{10}, \frac{\pi}{2}]$.
BETA
