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(b) Let function f : R → R be defined by f(x) = 2x + sinx for x ∈ R. Then f is
- a)one to one and onto
- b)one to one but NOT onto
- c)onto but NOT one to one
- d)neither one to one nor onto
Correct answer is option 'A'. Can you explain this answer?
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(b) Let function f : R R be defined by f(x) = 2x + sinx for x R. The...
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(b) Let function f : R R be defined by f(x) = 2x + sinx for x R. The...
Solution:
One-to-one:
To prove that f is one-to-one, let's assume that there exist two distinct elements a and b in R such that f(a) = f(b). Then we have:
2a sin(a) = 2b sin(b)
Dividing both sides by 2 and using the identity sin(a) = sin(b) if and only if a = b + 2nπ or a = π - b + 2nπ, we get:
a = b + 2nπ or a = π - b + 2nπ
where n is any integer. Since a and b are distinct, we cannot have a = b + 2nπ. Therefore, we must have a = π - b + 2nπ. Substituting this value of a in terms of b in f(a) = f(b), we get:
2(π - b + 2nπ) sin(π - b + 2nπ) = 2b sin(b)
Using the identity sin(π - x) = sin(x) and simplifying, we get:
2b sin(b) = 2b sin(b)
which is true. Therefore, we have shown that if f(a) = f(b), then a = π - b + 2nπ and hence a = b. Thus, f is one-to-one.
Onto:
To prove that f is onto, let y be any element in R. We need to show that there exists an element x in R such that f(x) = y. Let's consider the equation f(x) = y and try to solve for x. We have:
2x sin(x) = y
Dividing both sides by 2 and rearranging, we get:
x = arcsin(y/2)
Note that arcsin(y/2) is defined for all y in R since the range of the sine function is [-1, 1]. Therefore, we have shown that for any y in R, there exists an x in R such that f(x) = y. Thus, f is onto.
Conclusion:
Since f is both one-to-one and onto, we can conclude that f is a bijective function.
One-to-one:
To prove that f is one-to-one, let's assume that there exist two distinct elements a and b in R such that f(a) = f(b). Then we have:
2a sin(a) = 2b sin(b)
Dividing both sides by 2 and using the identity sin(a) = sin(b) if and only if a = b + 2nπ or a = π - b + 2nπ, we get:
a = b + 2nπ or a = π - b + 2nπ
where n is any integer. Since a and b are distinct, we cannot have a = b + 2nπ. Therefore, we must have a = π - b + 2nπ. Substituting this value of a in terms of b in f(a) = f(b), we get:
2(π - b + 2nπ) sin(π - b + 2nπ) = 2b sin(b)
Using the identity sin(π - x) = sin(x) and simplifying, we get:
2b sin(b) = 2b sin(b)
which is true. Therefore, we have shown that if f(a) = f(b), then a = π - b + 2nπ and hence a = b. Thus, f is one-to-one.
Onto:
To prove that f is onto, let y be any element in R. We need to show that there exists an element x in R such that f(x) = y. Let's consider the equation f(x) = y and try to solve for x. We have:
2x sin(x) = y
Dividing both sides by 2 and rearranging, we get:
x = arcsin(y/2)
Note that arcsin(y/2) is defined for all y in R since the range of the sine function is [-1, 1]. Therefore, we have shown that for any y in R, there exists an x in R such that f(x) = y. Thus, f is onto.
Conclusion:
Since f is both one-to-one and onto, we can conclude that f is a bijective function.
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Community Answer
(b) Let function f : R R be defined by f(x) = 2x + sinx for x R. The...
One one
2x1+sinx1=2x2+sinx2
2x1=2x2
X1=X2
hence prove this function is one one
onto
2x1+sinx€R
codamin €R
so,this function is onto
2x1+sinx1=2x2+sinx2
2x1=2x2
X1=X2
hence prove this function is one one
onto
2x1+sinx€R
codamin €R
so,this function is onto
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(b) Let function f : R R be defined by f(x) = 2x + sinx for x R. Then f isa)one to one and ontob)one to one but NOT ontoc)onto but NOT one to oned)neither one to one nor ontoCorrect answer is option 'A'. Can you explain this answer?
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(b) Let function f : R R be defined by f(x) = 2x + sinx for x R. Then f isa)one to one and ontob)one to one but NOT ontoc)onto but NOT one to oned)neither one to one nor ontoCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about (b) Let function f : R R be defined by f(x) = 2x + sinx for x R. Then f isa)one to one and ontob)one to one but NOT ontoc)onto but NOT one to oned)neither one to one nor ontoCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (b) Let function f : R R be defined by f(x) = 2x + sinx for x R. Then f isa)one to one and ontob)one to one but NOT ontoc)onto but NOT one to oned)neither one to one nor ontoCorrect answer is option 'A'. Can you explain this answer?.
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