Question

If $\left(\sin a\right)x^2+\left(\sin a\right)x+(1-\cos a)=0$, then the value of $a$ for which roots of this equation are real and distinct, is

• A. $\left(0,2{\tan }^{-1}\frac{1}{4}\right)$
• B. $\left(0,\frac{2\pi }{3}\right)$
• C. $(0,\pi )$
• D. $(0,2\pi )$

Answer 1

The correct option is A

$\left(0,2{\tan }^{-1}\frac{1}{4}\right)$

Explanation for the correct option.

Step 1: Find the determinant

Given that, $\left(\sin a\right)x^2+\left(\sin a\right)x+(1-\cos a)=0$.

We know that in equation $a{x}^2+bx+c=0$ roots are real & distinct when $D>0$.

And $D=b^2-4ac$

$\begin{array}{rcl}& \Rightarrow & {\sin }^{2}a-4\sin a(1-\cos a)>0\\ & \Rightarrow & (1-\cos a)(1+\cos a-4\sin a)>0\\ & \Rightarrow & (1+\cos a-4\sin a)>0\\ & \Rightarrow & 2{\cos }^2\frac{a}{2}-4\times 2\sin \frac{a}{2}\cos \frac{a}{2}>0\left[\because \sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\text{and}\cos \theta =2{\cos }^2\frac{\theta }{2}-1\right]\\ & \Rightarrow & 2{\cos }^2\frac{a}{2}\left[1-4\tan \frac{a}{2}\right]>0\end{array}$

Step 2: Find the value of $a$

As ${\cos }^2\frac{a}{2}>0$ so we only need to check another factor.

$$\begin{gathered} 1-4\tan \frac{a}{2}>0 \\ 4\tan \frac{a}{2}<1 \\ 0<a<2{\tan }^{-1}\frac{1}{4} \end{gathered}$$

Hence, option A is correct.