Question

The graph of the curve $y=\frac{(x^2+1)}{(x^2-1)}$ is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

(i) Since only even powers of $x$ occurs in the equation, the curve is symmetrical about the $y-$ axis
(ii) When $x=0,y=-1$ so that the curve cuts the $y-$axis at $(0,-1).$ But when $y=0,x$ is imaginary and hence the curve does not cut the $x-$axis.
(iii) Writing the equation as $x^2=\frac{(y+1)}{(y-1)},$ we find that $x$ is imaginary when $-1<y<1.$ Hence no part of the curve lies between the lines $y=+1$ and $y=-1.$
(iv)As $x\to -1-h$ (i.e., from the left) or $x\to 1+h,h\to 0^{+}$ (i.e., from the right) $y\to +\infty .$
and )As $x\to -1+h$ (i.e., from the right) or $x\to 1-h,h\to 0^{+}$ (i.e., from the left) $y\to -\infty .$
$\therefore$ The part of the curve between the lines $x=-1$ and $x=+1$ lies below the line $y=-1$ and approches the lines $x=-1$ and $x=1$ asymptotically.
But when $x>1,y$ is +ve and as $x\to 1+0y\to +\infty .$ Also as $x\to +\infty ,y\to +1+0.$
Hence for $x>1$ the part of the curve lies above the line $y=1$ and to the right of the line $x=1$ and approaches both of them asymptotically.
The part of the curve to the left of the line $x=-1$ and above the line $y=1$ is symmetrical with the part just considered.
It is easily seen that the curve attains a maximum value $-1$ when $x=0$ so that $(0,-1)$ is a maximum point.
The general shape of the curve is as shown in figure: