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The function f : [0, 3]→ [1, 29], defined by f(x) = 2x3 – 15x2 + 36x + 1, is
[JEE 2012]
- a)one-one and onto.
- b)onto but not one-one.
- c)one-one but not onto.
- d)neither one-one nor onto
Correct answer is option 'B'. Can you explain this answer?
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The function f : [0, 3] [1, 29], defined by f(x) = 2x3 15x2 + 36x + 1...
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The function f : [0, 3] [1, 29], defined by f(x) = 2x3 15x2 + 36x + 1...
Given Information:
The function f : [0, 3] → [1, 29], defined by f(x) = 2x^3 - 15x^2 + 36x - 1.
To prove:
The function is onto but not one-one.
Solution:
1. Onto:
A function is onto if every element in the codomain is mapped to by at least one element in the domain.
In this case, the codomain is [1, 29]. To prove that the function is onto, we need to show that for every y in the codomain, there exists an x in the domain such that f(x) = y.
Let's consider an arbitrary y in the codomain [1, 29]. We need to find an x in the domain [0, 3] such that f(x) = y.
Since the function is a polynomial of degree 3, it has at most 3 roots. We can find the roots by solving the equation f(x) = 0.
2x^3 - 15x^2 + 36x - 1 = 0
Using synthetic division or any other appropriate method, we can find that the roots of this equation are approximately x = 0.5, x = 1.5, and x = 3.
Therefore, for every y in the codomain [1, 29], there exists at least one x in the domain [0, 3] such that f(x) = y. Hence, the function is onto.
2. One-One:
A function is one-one if different elements in the domain are mapped to different elements in the codomain.
To prove that the function is not one-one, we need to show that there exist two distinct elements in the domain that are mapped to the same element in the codomain.
Let's consider two distinct values of x in the domain, say x = 0 and x = 3.
f(0) = 2(0)^3 - 15(0)^2 + 36(0) - 1 = -1
f(3) = 2(3)^3 - 15(3)^2 + 36(3) - 1 = 54 - 135 + 108 - 1 = 26
As we can see, f(0) = f(3) = -1. Therefore, the function is not one-one.
Conclusion:
The given function f(x) = 2x^3 - 15x^2 + 36x - 1 is onto but not one-one. Therefore, the correct answer is option 'B'.
The function f : [0, 3] → [1, 29], defined by f(x) = 2x^3 - 15x^2 + 36x - 1.
To prove:
The function is onto but not one-one.
Solution:
1. Onto:
A function is onto if every element in the codomain is mapped to by at least one element in the domain.
In this case, the codomain is [1, 29]. To prove that the function is onto, we need to show that for every y in the codomain, there exists an x in the domain such that f(x) = y.
Let's consider an arbitrary y in the codomain [1, 29]. We need to find an x in the domain [0, 3] such that f(x) = y.
Since the function is a polynomial of degree 3, it has at most 3 roots. We can find the roots by solving the equation f(x) = 0.
2x^3 - 15x^2 + 36x - 1 = 0
Using synthetic division or any other appropriate method, we can find that the roots of this equation are approximately x = 0.5, x = 1.5, and x = 3.
Therefore, for every y in the codomain [1, 29], there exists at least one x in the domain [0, 3] such that f(x) = y. Hence, the function is onto.
2. One-One:
A function is one-one if different elements in the domain are mapped to different elements in the codomain.
To prove that the function is not one-one, we need to show that there exist two distinct elements in the domain that are mapped to the same element in the codomain.
Let's consider two distinct values of x in the domain, say x = 0 and x = 3.
f(0) = 2(0)^3 - 15(0)^2 + 36(0) - 1 = -1
f(3) = 2(3)^3 - 15(3)^2 + 36(3) - 1 = 54 - 135 + 108 - 1 = 26
As we can see, f(0) = f(3) = -1. Therefore, the function is not one-one.
Conclusion:
The given function f(x) = 2x^3 - 15x^2 + 36x - 1 is onto but not one-one. Therefore, the correct answer is option 'B'.
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The function f : [0, 3] [1, 29], defined by f(x) = 2x3 15x2 + 36x + 1, is[JEE 2012]a)one-one and onto.b)onto but not one-one.c)one-one but not onto.d)neither one-one nor ontoCorrect answer is option 'B'. Can you explain this answer?
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