मराठी

Let F : R → R F ( X ) = X 2 − 8 X 2 + 2 Then, F is (A) One-one but Not onto (B) One-one and onto (C) onto but Not One-one (D) Neither One-one Nor onto - Mathematics

Advertisements
Advertisements

प्रश्न

Let

\[f : R \to R\]
\[f\left( x \right) = \frac{x^2 - 8}{x^2 + 2}\]
Then,  f is

पर्याय

  • one-one but not onto

  • one-one and onto

  • onto but not one-one

  • neither one-one nor onto

MCQ

उत्तर

Injectivity:
Let x and y be two elements in the domain (R), such that

\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow \frac{x^2 - 8}{x^2 + 2} = \frac{y^2 - 8}{y^2 + 2}\] 
\[ \Rightarrow \left( x^2 - 8 \right)\left( y^2 + 2 \right) = \left( x^2 + 2 \right)\left( y^2 - 8 \right)\] 
\[ \Rightarrow x^2 y^2 + 2 x^2 - 8 y^2 - 16 = x^2 y^2 - 8 x^2 + 2 y^2 - 16\] 
\[ \Rightarrow 10 x^2 = 10 y^2 \] 
\[ \Rightarrow x^2 = y^2 \] 
\[ \Rightarrow x = \pm y \]

So, f is not one-one .

Surjectivity:

\[f\left( - 1 \right) = \frac{\left( - 1 \right)^2 - 8}{\left( - 1 \right)^2 + 2} = \frac{1 - 8}{1 + 2} = \frac{- 7}{3}\] 

\[ \text{ and} f\left( 1 \right) = \frac{\left( 1 \right)^2 - 8}{\left( 1 \right)^2 + 2} = \frac{1 - 8}{1 + 2} = \frac{- 7}{3}\]
\[\Rightarrow\] f is not onto.
The correct answer is (d).
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Functions - Exercise 2.6 [पृष्ठ ७७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 2 Functions
Exercise 2.6 | Q 23 | पृष्ठ ७७

व्हिडिओ ट्यूटोरियलVIEW ALL [5]

संबंधित प्रश्‍न

Check the injectivity and surjectivity of the following function:

f: → N given by f(x) = x3


Show that the modulus function f: → R given by f(x) = |x| is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x|  is − x if x is negative.


Give an example of a function which is not one-one but onto ?


Give an example of a function which is neither one-one nor onto ?


Classify the following function as injection, surjection or bijection : f : N → N given by f(x) = x2


Classify the following function as injection, surjection or bijection :  f : Z → Z given by f(x) = x2


Classify the following function as injection, surjection or bijection :

 f : Z → Z, defined by f(x) = x − 5 


Classify the following function as injection, surjection or bijection :

 f : R → R, defined by f(x) = sinx


Let A = [-1, 1]. Then, discuss whether the following function from A to itself is one-one, onto or bijective : `f (x) = x/2`


If f : R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.


Suppose f1 and f2 are non-zero one-one functions from R to R. Is `f_1 / f^2` necessarily one - one? Justify your answer. Here,`f_1/f_2 : R → R   is   given   by   (f_1/f_2) (x) = (f_1(x))/(f_2 (x))  for all  x in R .`


Find gof and fog when f : R → R and g : R → R is defined by  f(x) = 2x + x2 and  g(x) = x3


Let f : R → R and g : R → R be defined by f(x) = + 1 and (x) = x − 1. Show that fog = gof = IR.


Give examples of two functions f : N → Z and g : Z → Z, such that gof is injective but gis not injective.


If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.


State with reason whether the following functions have inverse :
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}


Consider f : R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1 `(x)= sqrt (x-4)` where R+ is the set of all non-negative real numbers.


If f : R → (−1, 1) defined by `f (x) = (10^x- 10^-x)/(10^x + 10 ^-x)` is invertible, find f−1.


 If f : R → R be defined by f(x) = x4, write f−1 (1).

If f : R → R is defined by f(x) = x2, find f−1 (−25).


Let f : R → Rg : R → R be two functions defined by f(x) = x2 + x + 1 and g(x) = 1 − x2. Write fog (−2).


If f(x) = x + 7 and g(x) = x − 7, x ∈ R, write fog (7).


The function \[f : [0, \infty ) \to \text {R given by } f\left( x \right) = \frac{x}{x + 1} is\]

 

 


The function

\[f : R \to R, f\left( x \right) = x^2\]
 

 Let
\[g\left( x \right) = 1 + x - \left[ x \right] \text{and} f\left( x \right) = \begin{cases}- 1, & x < 0 \\ 0, & x = 0, \\ 1, & x > 0\end{cases}\] where [x] denotes the greatest integer less than or equal to x. Then for all \[x, f \left( g \left( x \right) \right)\] is equal to


Mark the correct alternative in the following question:
Let f :  \[-\] \[\left\{ \frac{3}{5} \right\}\] \[\to\]  R be defined by f(x) = \[\frac{3x + 2}{5x - 3}\] Then,

 


Let A = ℝ − {3}, B = ℝ − {1}. Let f : A → B be defined by \[f\left( x \right) = \frac{x - 2}{x - 3}, \forall x \in A\] Show that f is bijective. Also, find
(i) x, if f−1(x) = 4
(ii) f−1(7)


The domain of the function f: R → R defined by f(x) = `sqrt(x^2 - 3x + 2)` is ______


Let A = R – {3}, B = R – {1}. Let f: A → B be defined by f(x) = `(x - 2)/(x - 3)` ∀ x ∈ A . Then show that f is bijective


Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:

f(x) = `x/2`


The mapping f : N → N is given by f(n) = 1 + n2, n ∈ N when N is the set of natural numbers is ____________.


A general election of Lok Sabha is a gigantic exercise. About 911 million people were eligible to vote and voter turnout was about 67%, the highest ever


Let I be the set of all citizens of India who were eligible to exercise their voting right in the general election held in 2019. A relation ‘R’ is defined on I as follows:

R = {(V1, V2) ∶ V1, V2 ∈ I and both use their voting right in the general election - 2019}

  • Mr. ’X’ and his wife ‘W’ both exercised their voting right in the general election-2019, Which of the following is true?

'If 'f' is a linear function satisfying f[x + f(x)] = x + f(x), then f(5) can be equal to:


Let a and b are two positive integers such that b ≠ 1. Let g(a, b) = Number of lattice points inside the quadrilateral formed by lines x = 0, y = 0, x = b and y = a. f(a, b) = `[a/b] + [(2a)/b] + ... + [((b - 1)a)/b]`, then the value of `[(g(101, 37))/(f(101, 37))]` is ______.

(Note P(x, y) is lattice point if x, y ∈ I)

(where [.] denotes greatest integer function)


If f : R `rightarrow` R is defined by `f(x) = (2x - 7)/4`, show that f(x) is one-one and onto.


The trigonometric equation tan–1x = 3tan–1 a has solution for ______.



The given function f : R → R is not ‘onto’ function. Give reason.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×