Question

If $-2x^2+2x+\alpha <{\text{cosec}}^{-1}\left(\text{cosec 6}\right)+{\tan }^{-1}\left(\tan \right(-5\left)\right)\forall x\in R,$ then possible value(s) of $\alpha$ can be (where $\alpha \in Z$)

• A. $-3$
• B. $0$
• C. $-2$
• D. $1$

Answer 1

Answer: (A), (B), (C)

$-2$

Given:
$-2x^2+2x+\alpha <{\text{cosec}}^{-1}\left(\text{cosec 6}\right)+{\tan }^{-1}\left(\tan \right(-5\left)\right)$
we know that
${\text{cosec}}^{-1}\left(\text{cosec}x\right)=x-2\pi ;\frac{3\pi }{2}<x<2\pi$
${\tan }^{-1}\left(\tan x\right)=x+2\pi ;-2\pi <x<-\frac{3\pi }{2}$
Then inequality becomes
$-2x^2+2x+\alpha <6-2\pi -5+2\pi$
$\Rightarrow -2x^2+2x+\alpha -1<0$
$\because a{x}^2+bx+c<0\text{and}a<0\Rightarrow D<0$
$\Rightarrow 4+4\times 2(\alpha -1)<0$
$\Rightarrow \alpha <\frac{1}{2}$