Answer :

A function is bijective iff it is one-one and onto.


Option A. f (x) = x3


Let f(x1) = f(x2)


x13= x23


x1= x2


f is one one


Let f(x) = y, y Z


y = x3


x = y1/3 but y1/3 Z


f is not onto


Thus, f is not bijective.


Option B. f (x) = x + 2


Let f(x1) = f(x2)


x1+2 = x2+2


x1= x2


f is one one


Let f(x) = y, y Z


y = x + 2


x = y – 2


for each y Z there exists x Z (domain) such that f(x) = y.


f is onto


Thus, f is bijective.


Option C. f (x) = 2x + 1


Let f(x1) = f(x2)


2x1+1 = 2x2+1


x1= x2


f is one one


Let f(x) = y, y Z


y = 2x + 1


y - 1 = 2x



We observe that if we put y=0, then .


Thus, y = 0 Z does not have pre image in Z (domain)


f is not onto.


Thus, f is not bijective.


Option D. f (x) = x2 + 1


let f(x1) = f(x2)


x12 + 1 = x22 + 1


x12 = x22


x1 = ± x2


x1= x2 and x1= - x2


For e.g., f(-1) = |-1| = 1 and f(1) = |1| = 1


f is not one-one.


Since, f is not one one it cannot be bijective.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Fill in theMathematics - Exemplar

Let f : [2, ∞) <sMathematics - Exemplar

Let f : N Mathematics - Exemplar

Fill in theMathematics - Exemplar

Let f :R →<Mathematics - Exemplar

Let f : [0, 1] <sMathematics - Exemplar

Which of the follMathematics - Exemplar