Question

The area is bounded by $x+x_1,y=y_1$ and $y=-(x+1{)}^2$. Where $x_1,y_1$ are the values of $x,y$ satisfying the equation $si{n}^{-1}x+si{n}^{-1}y=-\pi$ will be (nearer to origin)

• A. $1/3$
• B. $3/2$
• C. $1$
• D. $2/3$

Answer 1

Answer: (D)

$2/3$

( Wrong question, $x=x$, instead of $x+x_1$ )

Given that:

$x=x_1$

$y=y_1$

$y=-(x+1{)}^2$

$\text{\&}{x}_1,y_1\text{satisfy}{\sin }^{-1}x+{\sin }^{-1}y=z$

We know:

$-\frac{z}{2}\le {\sin }^{-1}x\le \frac{z}{2}$

$\frac{-z}{2}\le {\sin }^{-1}y\le \frac{z}{2}$

$\therefore -z\le {\sin }^{-1}x+{\sin }^{-1}y\le z$

$\text{So, for}$

${\sin }^{-1}x+{\sin }^{-1}y=-z$

$\Rightarrow {\sin }^{-1}x=-\frac{\pi }{2}\text{\&}{\sin }^{-1}y=-z/2$

$x=-1\text{and}y=-1$

$\text{area required}=1-|{\int }_1^0-(x+1{)}^{2}dx\mid$

$=1-|{\int }_1^0(x^2+1+2x)dx|$

$=1-|\frac{-{(x^3+x+x^2)|}_{-1}^0}{3}$

$=1-|(\frac{-1}{3}-1+1)|$

$=1-\frac{1}{3}$

$=2/3sq\text{. Units}$