For any two real numbers, an operation * defined by a * b  = 1 + ab is

• neither commutative nor associative

• commutative but not associative

• both commutative and associative

• associative but not commutative

Let f : N $\to$ N defined by f(n) = $$\begin{gathered} \left\{\frac{\mathrm{n} +1}{2}, \mathrm{if} \mathrm{n} \mathrm{is} \mathrm{odd} \\ \frac{\mathrm{n}}{2}, \mathrm{if} \mathrm{n} \mathrm{is} \mathrm{even}\right\} \end{gathered}$$, then f is

• onto but not one-one

• one-one and onto

• neither one-one nor onto

• one-one but not onto

Suppose f(x) = (x + 1)² for x $\ge$ - 1. If g(x) is a function whose graph is the reflection of the graph of f(x) in the line y = x, then g(x) is equal to

• $\frac{1}{{\left(\mathrm{x} + 1\right)}^2}\mathrm{x} > - 1$

• $- \sqrt{\mathrm{x}} - 1$

• $\sqrt{\mathrm{x}} + 1$

• $\sqrt{\mathrm{x}} - 1$

Let * be a binary operation defined on R by a * b = $\frac{\mathrm{a} + \mathrm{b}}{4}, \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}$, then the operation * is

• commutative and associative

• commutative but not associative

• associative but not commutative

• neither associative nor commutative

305.

Let f : $\mathrm{R} \to \mathrm{R}$ be defined by f(x) = x⁴, then

• f may be one-one and onto

• f is neither one-one nor onto

• f is one-one and onto

• f is one-one but not onto

Binary operation * on R - {- 1} defined by $\mathrm{a} * \mathrm{b} = \frac{\mathrm{a}}{\mathrm{b} + 1}$ is

• * is neither associative not commutative

• * is associative but not commutative

• * is commutative but not commutative

• * is associative and commutative

A function f from the set of natural numbers to integers defined by f(n) = $$\begin{gathered} \left\{\frac{\mathrm{n} - 1}{2}, \mathrm{when} \mathrm{n} \mathrm{is} \mathrm{odd} \\ - \frac{\mathrm{n}}{2}, \mathrm{when} \mathrm{n} \mathrm{is} \mathrm{even}\right\} \end{gathered}$$, is

• one - one but not onto

• onto but not one - one

• one - one and onto both

• neither one - one nor onto

If g(x) = x² + x - 2 and gof(x) = 2x² - 5x+ 2, then f(x) is equal to

• 2x - 3

• 2x + 3

• 2x² + 3x + 1

• 2x² - 3x - 1

Inverse of the function f(x) = $\frac{{\mathrm{e}}^{\mathrm{x}} - {\mathrm{e}}^{- \mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}} + {\mathrm{e}}^{- \mathrm{x}}} + 2$ is

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x} - 2}{\mathrm{x} - 1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x} - 1}{3 - \mathrm{x}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x}}{2 - \mathrm{x}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• ${\log }_{\mathrm{e}}{\left(\frac{\mathrm{x} - 1}{\mathrm{x} + 1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

If f(x) = $\frac{\mathrm{x} - 1}{\mathrm{x}}, \mathrm{x} \ne 0, 0 \in \mathrm{R}$ and g(u) = u² + 1, $\mathrm{u} \in \mathrm{R}$ then g[f(1)] and f[g(- 1)] is equal to

• 1, 1/2

• - 1, 1/2

• 0, - 1

• None of these