Question

If exhaustive value of x satisfying $|si{n}^{-1}x|+|ta{n}^{-1}x|+|co{s}^{-1}x|=|\pi -co{t}^{-1}x|$ belongs to $[\alpha ,\beta ]$, then $\alpha and\beta$ will be

• A. $\alpha =0,\beta =2$
• B. $\beta =0,\alpha =1$
• C. $\alpha =1,\beta =2$
• D. $\beta =1,\alpha =0$

Answer 1

The correct option is D $\beta =1,\alpha =0$

$|si{n}^{-1}x|+|ta{n}^{-1}x|+|co{s}^{-1}x|=|\frac{\pi }{2}+\frac{\pi }{2}-co{t}^{-1}x||si{n}^{-1}x|+|ta{n}^{-1}x|+|co{s}^{-1}x|=|\frac{\pi }{2}+ta{n}^{-1}x|\Rightarrow si{n}^{-1}\ge 0,ta{n}^{-1}x\ge 0,co{s}^{-1}x\ge 0\Rightarrow xϵ[0,1]\Rightarrow \alpha =0,\beta =1$