Question

Consider the function defined implicitly by the equation $y^2-2y{e}^{si{n}^{-1}x}+x^2-1+\left[x\right]+e^{2si{n}^{-1}x}=0$ (Where, [x] denotes the greatest integer function). Then the area of the region bonded by the curve and the line $x=-1$ is

• A. $(\frac{\pi }{2}-1)$ sq unit
• B. $(\frac{\pi }{2}+1)$ sq unit
• C. $(\pi +1)$ sq unit
• D. $(\pi -1)$ sq unit

Answer 1

The correct option is C $(\pi +1)$ sq unit

$y^2-2y{e}^{sin-1x}+x^2-1+\left[x\right]+e^{2si{n}^{-1}x}=0$ $x\in [-1,1]$
$y^2-2y{e}^{si{n}^{-1}x}+e^{2si{n}^{-1}x}+x^2-1+\left[x\right]=0$
$(y-e^{si{n}^{-1}x}{)}^2=1-x^2-[x]$
$y=e^{si{n}^{-1}x}\pm \sqrt{1-x^2\left[x\right]}$
Area bounded between them $={\int }_{-1}^1[(e^{si{n}^{-1}x}+\sqrt{1-x^2-\left[x\right]})-(e^{si{n}^{-1}x}-\sqrt{1-x^2-\left[x\right]}dx)]$
$=2{\int }_{-1}^1\sqrt{1-x^2-\left[x\right]}dx$
$=2{\int }_{-1}^o\sqrt{1-x^2+1}dx+2{\int }_0^1\sqrt{1-x^2-0}dx$
$=2{\int }_{-1}^0\sqrt{2-x^2}dx+2{\int }_0^1\sqrt{1-x^2}dx$
$=2{\left[\frac{1}{2}(x\sqrt{2-x^2}+2si{n}^{-1}\left(\frac{x}{\sqrt{2}}\right))\right]}_{-1}^0+2{\left[\frac{1}{2}(x\sqrt{1-x^2}+si{n}^{-1}x)\right]}_0^1$
$=1\sqrt{2-1}-2si{n}^{-1}\left(\frac{-1}{\sqrt{2}}\right)+si{n}^{-1}1-si{n}^{-1}0$
$=1-2\left(\frac{-\pi }{4}\right)+\frac{\pi }{2}$
$=1+\frac{1}{2}+\frac{\pi }{2}$
$=(\pi +1)sq.unts$