मराठी

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.

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प्रश्न

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.

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर
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उत्तर

This statement is True.

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पाठ 1: Relations And Functions - Solved Examples [पृष्ठ १०]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 1 Relations And Functions
Solved Examples | Q 33 | पृष्ठ १०

व्हिडिओ ट्यूटोरियलVIEW ALL [5]

संबंधित प्रश्‍न

Check the injectivity and surjectivity of the following function:

f : N → N given by f(x) = x3


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

(Hint: Consider f(x) = x and g(x) =|x|)


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


Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = 5x3 + 4


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


Set of ordered pair of a function ? If so, examine whether the mapping is injective or surjective :{(a, b) : a is a person, b is an ancestor of a} 


Find gof and fog when f : R → R and g : R → R is  defined by  f(x) = 8x3 and  g(x) = x1/3.


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.


Let f : R → R and g : R → R be defined by f(x) = x2 and g(x) = x + 1. Show that fog ≠ gof.


Let f(x) = x2 + x + 1 and g(x) = sin x. Show that fog ≠ gof.


   if `f (x) = sqrt(1-x)` and g(x) = `log_e` x are two real functions, then describe functions fog and gof.


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A function f : R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).


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Let f : R → R be the function defined by f(x) = 4x − 3 for all x ∈ R Then write f .   [NCERT EXEMPLAR]


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Let\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\} = \text{B and C} = \left\{ x \in R : x \geq 0 \right\} and\]\[S = \left\{ \left( x, y \right) \in A \times B : x^2 + y^2 = 1 \right\} \text{and } S_0 = \left\{ \left( x, y \right) \in A \times C : x^2 + y^2 = 1 \right\}\]

Then,



Let f be an injective map with domain {x, y, z} and range {1, 2, 3}, such that exactly one of the following statements is correct and the remaining are false.

\[f\left( x \right) = 1, f\left( y \right) \neq 1, f\left( z \right) \neq 2 .\]

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\[A = \left\{ x : - 1 \leq x \leq 1 \right\} \text{and} f : A \to \text{A such that f}\left( x \right) = x|x|\]

 


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{(a, b): a is a person, b is an ancestor of a}


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