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Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P'(x) + P(x). Then,
- a)T is one -one but not on to .
- b)T is not one-one but onto.
- c)T is both one-one and onto.
- d)T is neither one-one nor onto
Correct answer is option 'C'. Can you explain this answer?
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Let P4 denote the real vector space of all polynomials with real coeff...
⇒ N(T) = {0}
⇒ T is one -one and onto.
Most Upvoted Answer
Let P4 denote the real vector space of all polynomials with real coeff...
⇒ N(T) = {0}
⇒ T is one -one and onto.
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Let P4 denote the real vector space of all polynomials with real coeff...
Explanation:
One-to-One:
- To show that the map T is one-to-one, we need to prove that if T[P(x)] = T[Q(x)], then P(x) = Q(x).
- Let P(x) = a4x^4 + a3x^3 + a2x^2 + a1x + a0 and Q(x) = b4x^4 + b3x^3 + b2x^2 + b1x + b0.
- T[P(x)] = P"(x) + P(x) + P(x) = 2a4 + 2a3x + 2a2x^2 + 2a1x^3 + 2a0x^4.
- T[Q(x)] = Q"(x) + Q(x) + Q(x) = 2b4 + 2b3x + 2b2x^2 + 2b1x^3 + 2b0x^4.
- Setting T[P(x)] = T[Q(x)], we get a4 = b4, a3 = b3, a2 = b2, a1 = b1, a0 = b0.
- Therefore, P(x) = Q(x), and T is one-to-one.
Onto:
- To show that the map T is onto, we need to prove that for every Q(x) in P4, there exists a P(x) such that T[P(x)] = Q(x).
- Let Q(x) = c4x^4 + c3x^3 + c2x^2 + c1x + c0.
- We need to find a P(x) such that P"(x) + P(x) + P(x) = Q(x).
- Solving this differential equation, we can find the expression for P(x) in terms of c4, c3, c2, c1, c0.
- Since we can express any Q(x) in terms of P(x), T is onto.
Therefore, the map T is both one-to-one and onto, making option 'C' the correct answer.
One-to-One:
- To show that the map T is one-to-one, we need to prove that if T[P(x)] = T[Q(x)], then P(x) = Q(x).
- Let P(x) = a4x^4 + a3x^3 + a2x^2 + a1x + a0 and Q(x) = b4x^4 + b3x^3 + b2x^2 + b1x + b0.
- T[P(x)] = P"(x) + P(x) + P(x) = 2a4 + 2a3x + 2a2x^2 + 2a1x^3 + 2a0x^4.
- T[Q(x)] = Q"(x) + Q(x) + Q(x) = 2b4 + 2b3x + 2b2x^2 + 2b1x^3 + 2b0x^4.
- Setting T[P(x)] = T[Q(x)], we get a4 = b4, a3 = b3, a2 = b2, a1 = b1, a0 = b0.
- Therefore, P(x) = Q(x), and T is one-to-one.
Onto:
- To show that the map T is onto, we need to prove that for every Q(x) in P4, there exists a P(x) such that T[P(x)] = Q(x).
- Let Q(x) = c4x^4 + c3x^3 + c2x^2 + c1x + c0.
- We need to find a P(x) such that P"(x) + P(x) + P(x) = Q(x).
- Solving this differential equation, we can find the expression for P(x) in terms of c4, c3, c2, c1, c0.
- Since we can express any Q(x) in terms of P(x), T is onto.
Therefore, the map T is both one-to-one and onto, making option 'C' the correct answer.
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Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer?
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Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer?.
Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
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Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P(x) + P(x). Then,a)T is one -one but not on to .b)T is not one-one but onto.c)T is both one-one and onto.d)T is neither one-one nor ontoCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Mathematics tests.
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