PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

If the function\[f : R \to \text{A given by} f\left( x \right) = \frac{x^2}{x^2 + 1}\] is a surjection, then A =

If a function\[f : [2, \infty )\text{ to B defined by f}\left( x \right) = x^2 - 4x + 5\] is a bijection, then B =

The function

\[f : R \to R\] defined by\[f\left( x \right) = \left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)\]

(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto

The  function f : [-1/2, 1/2, 1/2] → [-π /2,π/2], defined by f (x) = `sin^-1` (3x - `4x^3`), is

Let

\[f : R \to R\]  be a function defined by

Let

\[f : R - \left\{ n \right\} \to R\]

Let

\[f : R \to R\] is defined by

\[f\left( x \right) = \frac{e^{x^2} - e^{- x^2}}{e^{x^2 + e^{- x^2}}} is\]

The function

A function f from the set of natural numbers to the set of integers defined by

\[f\left( n \right)\begin{cases}\frac{n - 1}{2}, & \text{when n is odd} \\ - \frac{n}{2}, & \text{when n is even}\end{cases}\]

Which of the following functions from

\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\}\]

\[f : Z \to Z\]  be given by

 ` f (x) = {(x/2, ", if  x is even" ) ,(0 , ", if  x  is  odd "):}`

Then,  f is

The function \[f : R \to R\] defined by

\[f\left( x \right) = 6^x + 6^{|x|}\] is 

Let  \[f\left( x \right) = x^2 and g\left( x \right) = 2^x\] Then, the solution set of the equation

If \[f : R \to R is given by f\left( x \right) = 3x - 5, then f^{- 1} \left( x \right)\] 

If \[g \left( f \left( x \right) \right) = \left| \sin x \right| \text{and} f \left( g \left( x \right) \right) = \left( \sin \sqrt{x} \right)^2 , \text{then}\]

The inverse of the function

\[f : R \to \left\{ x \in R : x < 1 \right\}\] given by

\[f\left( x \right) = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] is 

Let

 \[A = \left\{ x \in R : x \geq 1 \right\}\] The inverse of the function, 

\[f : A \to A\] given by

\[f\left( x \right) = 2^{x \left( x - 1 \right)} , is\]

Let

\[A = \left\{ x \in R : x \leq 1 \right\} and f : A \to A\] be defined as

\[f\left( x \right) = x \left( 2 - x \right)\] Then,

\[f^{- 1} \left( x \right)\] is

Let  \[f\left( x \right) = \frac{1}{1 - x} . \text{Then}, \left\{ f o \left( fof \right) \right\} \left( x \right)\]