- A$R$
- ✓$(-1,1)$
- C$R-\{0\}$
- D$[-1,1]$
If $x > 0,\left| x \right| = x \Rightarrow f\left( x \right) = \frac{x}{{1 + x}}$
which is not defined for $x=-1$
If $x < 0,\left| x \right| = - x \Rightarrow f\left( x \right) = \frac{x}{{1 - x}}$
which is not defined for $x=1$
Thus $f\left( x \right)$ defined for all value of $R$ except $1$ and $-1$
Hence, range $=(-1,1)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
- stir the liquid in $J_1$ and transfer $10\,ml$ from $J_1$ into $J_2$
- stir the liquid in $J_2$ and transfer $10\, ml$ from $J_2$ into $J_3$
- stir the liquid in $J_3$ and transfer $10 \,ml$ from $J_3$ into $J_1$.
After performing the operation four times, let $x, y, z$ be the amounts of $X, Y, Z$ respectively, in $J_1$. Then,
$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$
then value of $\lambda^{2}$ is equal to $.....$
