Question

Let [a] denote the greatest integer less than or equal to a. The set of all values of x satisfying the inequality-
$|2x^2-4x-7|<[1+\frac{1}{2}{\left(\frac{\cos \theta }{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}\right)}^2],-\frac{\pi }{2}\le \theta <\frac{\pi }{2}$ is

• A. (-1, 3)
• B. $(1-\sqrt{5},1+\sqrt{5})$
• C. $(1-\sqrt{5},-1)\cup (3,1+\sqrt{5})$
• D. none of these

Answer 1

The correct option is C $(1-\sqrt{5},-1)\cup (3,1+\sqrt{5})$

$f\left(\theta \right)=$ $[1+\frac{1}{2}{\left(\frac{\cos \theta }{\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}\right]}^2],-\frac{\pi }{2}\le \theta <\frac{\pi }{2}$

$=[1+\frac{1}{2}\frac{{\cos }^2\theta }{{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}^2}]$

$=[1+\frac{1}{2}\left(\frac{1-{\sin }^2\theta }{1+\sin \theta }\right)]$

$=[1+\frac{1}{2}(1-\sin \theta \left)\right]$

$=\left[\frac{3-\sin \theta }{2}\right]$

$\text{i.e.}1<f\left(\theta \right)\le 2,(-\frac{\pi }{2}\le 0<\frac{\pi }{2})$

Now, For the inequality to be defined

$|2x^2-4x-7|<1$

$-1<2x^2-4x-7<1$

$2x^2-4x-6>0-\left(i\right)$

$2x^2-4x-8<0-\left(ii\right)$

On solving above eq",

we get $x\in \left(\right(-\infty ,-1)\cup (3,\infty \left)\right)\cap (1-\sqrt{5},1+\sqrt{5})$

Hence, $x\in (1-\sqrt{5},-1)\cup (3,1+\sqrt{5})$.