Commerce (English Medium) Class 12 - CBSE Question Bank Solutions

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

If the function

\[f : R \to R\]  be such that

\[f\left( x \right) = x - \left[ x \right]\] where [x] denotes the greatest integer less than or equal to x, then \[f^{- 1} \left( x \right)\]

If  \[F : [1, \infty ) \to [2, \infty )\] is given by

\[f\left( x \right) = x + \frac{1}{x}, then f^{- 1} \left( x \right)\]

Let  \[f\left( x \right) = \frac{\alpha x}{x + 1}, x \neq - 1\] Then, for what value of α is \[f \left( f\left( x \right) \right) = x?\]

The distinct linear functions that map [−1, 1] onto [0, 2] are

Let

\[f : [2, \infty ) \to X\] be defined by

\[f\left( x \right) = 4x - x^2\] Then, f is invertible if X =

If  \[f : R \to \left( - 1, 1 \right)\] is defined by

\[f\left( x \right) = \frac{- x|x|}{1 + x^2}, \text{ then } f^{- 1} \left( x \right)\] equals

Let [x] denote the greatest integer less than or equal to x. If \[f\left( x \right) = \sin^{- 1} x, g\left( x \right) = \left[ x^2 \right]\text{  and } h\left( x \right) = 2x, \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}\]

If  \[g\left( x \right) = x^2 + x - 2\text{ and} \frac{1}{2} gof\left( x \right) = 2 x^2 - 5x + 2\] is equal to