Question

If $1$ lies between the roots of the equation $16x^2-8x\sin \theta -15{\cos }^2\theta =0$, then

• A. $-\frac{1}{5}<\sin \theta <\frac{1}{5}$
• B. $\frac{1}{5}<\sin \theta <\frac{1}{3}$
• C. $-\frac{1}{5}<\sin \theta <0$
• D. $-\frac{1}{3}<\sin \theta <-\frac{1}{5}$

Answer 1

Answer: (B)

$\frac{1}{5}<\sin \theta <\frac{1}{3}$

$1$ lies between the roots of the equation $16x^2-8x\sin \theta -15{\cos }^{2}x=0$ and coefficient of $x^2>0$.
$\therefore f\left(1\right)<0$ and $D>0$ $\Rightarrow 16-8\sin \theta -15{\cos }^2<0$ $\Rightarrow 16-8\sin \theta -15(1-{\sin }^2)<0$ $\Rightarrow 1-8\sin \theta +15{\sin }^2\theta <0$ $\Rightarrow 15{\sin }^2\theta -8\sin \theta +1<0$
$\Rightarrow (3\sin \theta -1)(5\sin \theta -1)<0$ $\Rightarrow \frac{1}{5}<\sin \theta <\frac{1}{3}$.