Mathematics Exam  >  Mathematics Questions  >  Let V be the vector space of all polynomial f... Start Learning for Free
Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1, f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis is
  • a)
    (1, - 1, 1)
  • b)
    (-1 . - 1, 0)
  • c)
    (0 , - 1. 2)
  • d)
    ( 1, 0, 1)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let V be the vector space of all polynomial functions of degree ≤ 2...
∵ 1 + X + X2 = 1.1 + (-1 ).(x + 1) + 1. (x + 1 )2 = 1 .f1 + (-1 )f2 + 1 .f3 
∴ Required co-ordinates are(1, -1, 1).
View all questions of this test
Most Upvoted Answer
Let V be the vector space of all polynomial functions of degree ≤ 2...
To determine the dimension of V, we need to find the number of linearly independent vectors in V.

Since V is the vector space of all polynomial functions of degree less than or equal to 5, we can write any polynomial function in V as:

f(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + a₅x⁵

where a₀, a₁, a₂, a₃, a₄, and a₅ are constants.

To find the dimension of V, we need to find the maximum number of linearly independent vectors that can form a basis for V.

Since we can have at most one term for each degree of x, we can choose any combination of the coefficients a₀, a₁, a₂, a₃, a₄, and a₅ to form a polynomial function in V. Therefore, the dimension of V is 6.

Hence, the dimension of V is 6.
Explore Courses for Mathematics exam
Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer?
Question Description
Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. 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 V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. 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 V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer?.
Solutions for Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.
Here you can find the meaning of Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1,f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis isa)(1, - 1, 1)b)(-1 . - 1, 0)c)(0 , - 1. 2)d)( 1, 0, 1)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Mathematics tests.
Explore Courses for Mathematics exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev