Question

If $1$ lies between the roots of the equation $3x^2-3\sin \alpha x-2{\cos }^2\alpha =0$, then $\alpha$ lies in the interval

• A. $(0,\frac{\pi }{2})$
• B. $(\frac{\pi }{12},\frac{\pi }{2})$
• C. $(\frac{\pi }{6},\frac{5\pi }{6})$
• D. $(\frac{\pi }{6},\frac{\pi }{2})\cup (\frac{\pi }{2},\frac{5\pi }{6})$

Answer 1

Answer: (D)

$(\frac{\pi }{6},\frac{\pi }{2})\cup (\frac{\pi }{2},\frac{5\pi }{6})$

If a,b are the roots, then

$f\left(x\right)=3(x-a)(x-b)$

$f\left(1\right)=3(1-a)(1-b)=$ -ve as $a<1<b$

on $f\left(1\right)=3-3sin\alpha -2(1-si{n}^2\alpha )<0$

or $2si{n}^2\alpha -3sin\alpha +1<0$

$(2sin\alpha -1)(sin\alpha -1)<0$

$2(sin\alpha -\frac{1}{2})(sin\alpha -1)<0$

$\therefore \frac{1}{2}<sin\alpha <1.$ Now $sin{30}^{\circ }=sin{150}^{\circ }=\frac{1}{2}$

$\therefore \alpha$ lies between $\frac{\pi }{6}to\frac{5\pi }{6}$ excluding $\frac{\pi }{2}$ as $sin\alpha$ is not equal to 1 and hence (d) is correct.