मराठी

Find Gof And Fog When F : R → R And G : R → R Is Defined By F(X) = X2 + 8 And G(X) = 3x3 + 1

Advertisements
Advertisements

प्रश्न

Find gof and fog when f : R → R and g : R → R is defined by  f(x) = x2 + 8 and g(x) = 3x3 + 1 .

Advertisements

उत्तर

Given, f : R → R and g : R → R
So, gof R → R  and fog : R → R

f(x) = x2 + 8  and g(x) = 3x3 + 1

(gof) (x)

g (x))

g x)

3 2+)3 1

(fog) (x)

f g ))

f 3x3 )

3x3+)2 8

9x6 6x318

=9x6+6x3+9

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Functions - Exercise 2.2 [पृष्ठ ४६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 2 Functions
Exercise 2.2 | Q 1.3 | पृष्ठ ४६

व्हिडिओ ट्यूटोरियलVIEW ALL [5]

संबंधित प्रश्‍न

Show that the function f in `A=R-{2/3} ` defined as `f(x)=(4x+3)/(6x-4)` is one-one and onto hence find f-1


Check the injectivity and surjectivity of the following function:

f : Z → Z given by f(x) = x2


Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = `((x - 2)/(x - 3))`. Is f one-one and onto? Justify your answer.


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


Classify the following function as injection, surjection or bijection :

f : R → R, defined by f(x) = `x/(x^2 +1)`


Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.


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


Find fog and gof  if : f(x)= x + 1, g (x) = 2x + 3 .


If f(x) = |x|, prove that fof = f.


If f(x) = 2x + 5 and g(x) = x2 + 1 be two real functions, then describe each of the following functions:
(1) fog
(2) gof
(3) fof
(4) f2
Also, show that fof ≠ f2


Let  f  be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.


State with reason whether the following functions have inverse:

h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}


Consider the function f : R→  [-9 , ∞ ]given by f(x) = 5x2 + 6x - 9. Prove that f is invertible with -1 (y) = `(sqrt(54 + 5y) -3)/5`             [CBSE 2015]


Write the total number of one-one functions from set A = {1, 2, 3, 4} to set B = {abc}.


If f : C → C is defined by f(x) = (x − 2)3, write f−1 (−1).


Let A = {1, 2, 3, 4} and B = {ab} be two sets. Write the total number of onto functions from A to B.


If the mapping f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, then write fog. [NCERT EXEMPLAR]


Let 

\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\} = B\] Then, the mapping\[f : A \to \text{B given by} f\left( x \right) = x\left| x \right|\] is 

 


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


Let

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


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

 


Mark the correct alternative in the following question:
Let A = {1, 2, ... , n} and B = {a, b}. Then the number of subjections from A into B is


Let f, g: R → R be two functions defined as f(x) = |x| + x and g(x) = x – x ∀ x ∈ R. Then, find f o g and g o f


Let N be the set of natural numbers and the function f: N → N be defined by f(n) = 2n + 3 ∀ n ∈ N. Then f is ______.


Set A has 3 elements and the set B has 4 elements. Then the number of injective mappings that can be defined from A to B is ______.


Let A be a finite set. Then, each injective function from A into itself is not surjective.


For sets A, B and C, let f: A → B, g: B → C be functions such that g o f is surjective. Then g is surjective.


Let f: `[2, oo)` → R be the function defined by f(x) = x2 – 4x + 5, then the range of f is ______.


If f(x) = (4 – (x – 7)3}, then f–1(x) = ______.


Let f : [0, ∞) → [0, 2] be defined by `"f" ("x") = (2"x")/(1 + "x"),` then f is ____________.


Range of `"f"("x") = sqrt((1 - "cos x") sqrt ((1 - "cos x")sqrt ((1 - "cos x")....infty))`


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: R → R given by f(x) =(3 − x3)1/3, find f0f(x)


A function f: x → y is/are called onto (or surjective) if x under f.


Let the function f: R → R be defined by f(x) = 4x – 1, ∀ x ∈ R then 'f' is


Let [x] denote the greatest integer ≤ x, where x ∈ R. If the domain of the real valued function f(x) = `sqrt((|[x]| - 2)/(|[x]| - 3)` is (–∞, a) ∪ [b, c) ∪ [4, ∞), a < b < c, then the value of a + b + c is ______.


Let f(x) be a polynomial function of degree 6 such that `d/dx (f(x))` = (x – 1)3 (x – 3)2, then

Assertion (A): f(x) has a minimum at x = 1.

Reason (R): When `d/dx (f(x)) < 0, ∀  x ∈ (a - h, a)` and `d/dx (f(x)) > 0, ∀  x ∈ (a, a + h)`; where 'h' is an infinitesimally small positive quantity, then f(x) has a minimum at x = a, provided f(x) is continuous at x = a.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×