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Let F(x) be a function such that F(x) F(x + 1) = – F(x – 1)F(x–2)F(x–3)F(x–4) for all x ≥ 0.Given the values of If F (83) = 81 and F(77) = 9, then the value of F(81) equals to
- a)27
- b)54
- c)729
- d)Data Insufficient
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let F(x) be a function such that F(x) F(x + 1) = – F(x – 1...
When the value of x = 81 and 82 is substituted in the given expression, we get,
F (81) F (82) = – F (80) F (79) F(78) F(77)
F (82) F (83) = – F (81) F (80) F(79) F(78)
On dividing (i) by (a), we get
F (81) F (82) = – F (80) F (79) F(78) F(77)
F (82) F (83) = – F (81) F (80) F(79) F(78)
On dividing (i) by (a), we get
Option (a) is the correct answer.
Most Upvoted Answer
Let F(x) be a function such that F(x) F(x + 1) = – F(x – 1...
F(x^2 + x) for all real numbers x. Find F(0) and F(1).
Setting x = 0, we have F(0) F(1) = F(0), so either F(0) = 0 or F(1) = 1. If F(0) = 0, then setting x = -1, we have F(1) F(0) = F(0), which implies F(1) = 1. Thus, we have F(0) = 0 and F(1) = 1 as possible solutions.
Now, we will show that these are the only solutions. Suppose F(0) = 0 and F(1) = 1. We will prove by induction that F(n) = n for all non-negative integers n.
Base case: F(0) = 0.
Inductive step: Suppose F(k) = k for some non-negative integer k. Then, setting x = k in the given equation, we have F(k^2 + k) = F(k) F(k + 1) = k(k + 1). Since k^2 + k is a non-negative integer, we have F(k^2 + k) = k^2 + k. Thus, F(k + 1) = (k^2 + k)/k = k + 1.
Therefore, by induction, we have F(n) = n for all non-negative integers n. Now, setting x = -1 in the given equation, we have F(0) F(-1) = F(0), which implies F(-1) = 1. Setting x = 1 in the given equation, we have F(2) = F(1) F(2) = F(1^2 + 1) = F(2), so F(2) = 2. Setting x = 2 in the given equation, we have F(6) = F(2) F(3) = 2F(3), so F(3) = 3/2. Setting x = 3 in the given equation, we have F(12) = F(3) F(4) = 9, so F(4) = 9/4.
Thus, we have F(0) = 0, F(1) = 1, F(2) = 2, F(3) = 3/2, and F(4) = 9/4. We can continue this process to find F(n) for any non-negative integer n. Therefore, the only solutions are F(x) = x and F(x) = 0 for all x.
Setting x = 0, we have F(0) F(1) = F(0), so either F(0) = 0 or F(1) = 1. If F(0) = 0, then setting x = -1, we have F(1) F(0) = F(0), which implies F(1) = 1. Thus, we have F(0) = 0 and F(1) = 1 as possible solutions.
Now, we will show that these are the only solutions. Suppose F(0) = 0 and F(1) = 1. We will prove by induction that F(n) = n for all non-negative integers n.
Base case: F(0) = 0.
Inductive step: Suppose F(k) = k for some non-negative integer k. Then, setting x = k in the given equation, we have F(k^2 + k) = F(k) F(k + 1) = k(k + 1). Since k^2 + k is a non-negative integer, we have F(k^2 + k) = k^2 + k. Thus, F(k + 1) = (k^2 + k)/k = k + 1.
Therefore, by induction, we have F(n) = n for all non-negative integers n. Now, setting x = -1 in the given equation, we have F(0) F(-1) = F(0), which implies F(-1) = 1. Setting x = 1 in the given equation, we have F(2) = F(1) F(2) = F(1^2 + 1) = F(2), so F(2) = 2. Setting x = 2 in the given equation, we have F(6) = F(2) F(3) = 2F(3), so F(3) = 3/2. Setting x = 3 in the given equation, we have F(12) = F(3) F(4) = 9, so F(4) = 9/4.
Thus, we have F(0) = 0, F(1) = 1, F(2) = 2, F(3) = 3/2, and F(4) = 9/4. We can continue this process to find F(n) for any non-negative integer n. Therefore, the only solutions are F(x) = x and F(x) = 0 for all x.
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Let F(x) be a function such that F(x) F(x + 1) = – F(x – 1)F(x–2)F(x–3)F(x–4) for all x ≥ 0.Given the values of If F (83) = 81 and F(77) = 9, then the value of F(81) equals toa)27b)54c)729d)Data InsufficientCorrect answer is option 'A'. Can you explain this answer?
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