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Consider a function f satisfying f (x + y) = f (x) f (y) where x, y are positive integers, and f (1) = 2. If f (a + 1) + f (a + 2) + ... + f (a + n) = 16(2n – 1) then a is equal to
(2019)
- a)3
- b)4
- c)6
- d)5
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider a function f satisfying f (x + y) = f (x) f (y) where x, y ar...
Given,
f(a + 1) + f(a + 2) + .. + f(a + n) = 16(2n – 1)
⇒ f(a) f(1) + f(a) f(2) + ... + f(a) f(n) = 16(2n – 1)
⇒ f(a) (f(1) + f(2) + ... + f(n)) = 16 (2n – 1)
When n = 1, then f(a)f(1) = 16(21 – 1) = 16
⇒ f(a) × 2 = 16 ⇒ f(a) = 8
∴ f(a) (f(1) + f(2) + ... + f(n)) = 16(2n – 1)
⇒ f(1) + f(2) + ... + f(n) = 2(2n – 1)
When n = 2, then f(1) + f(2) = 2(22 – 1) = 6
⇒ f(2) = 6 – f(1) = 6 – 2 = 4
When n = 3, then f(1) + f(2) + f(3) = 2(23 – 1) = 14
⇒ f(3) = 14 – f(1) – f(2) = 14 – 2 – 4 = 8 = f(a)
∴ a = 3
f(a + 1) + f(a + 2) + .. + f(a + n) = 16(2n – 1)
⇒ f(a) f(1) + f(a) f(2) + ... + f(a) f(n) = 16(2n – 1)
⇒ f(a) (f(1) + f(2) + ... + f(n)) = 16 (2n – 1)
When n = 1, then f(a)f(1) = 16(21 – 1) = 16
⇒ f(a) × 2 = 16 ⇒ f(a) = 8
∴ f(a) (f(1) + f(2) + ... + f(n)) = 16(2n – 1)
⇒ f(1) + f(2) + ... + f(n) = 2(2n – 1)
When n = 2, then f(1) + f(2) = 2(22 – 1) = 6
⇒ f(2) = 6 – f(1) = 6 – 2 = 4
When n = 3, then f(1) + f(2) + f(3) = 2(23 – 1) = 14
⇒ f(3) = 14 – f(1) – f(2) = 14 – 2 – 4 = 8 = f(a)
∴ a = 3
Most Upvoted Answer
Consider a function f satisfying f (x + y) = f (x) f (y) where x, y ar...
We can start by finding the value of f(2). From the given equation, we have:
f(2) = f(1) f(2) = 2 f(2)
Subtracting f(2) from both sides, we get:
0 = f(2) - 2 f(2)
Combining like terms, we have:
0 = -f(2)
Therefore, f(2) = 0.
Next, let's find the value of f(3). Again from the given equation, we have:
f(3) = f(1) f(3) = 2 f(3)
Subtracting f(3) from both sides, we get:
0 = f(3) - 2 f(3)
Combining like terms, we have:
0 = -f(3)
Therefore, f(3) = 0.
Continuing this pattern, we can see that for any positive integer n, f(n) = 0.
Now, let's consider the expression f(a1) f(a2) ... f(an). From our previous findings, we know that each term in this product is 0. Therefore, the product itself is also 0.
However, the given equation states that this product is equal to 16(2n). Setting this equal to 0, we have:
16(2n) = 0
Dividing both sides by 16, we get:
2n = 0
This equation has no solution for n, as 2n can never equal 0 for any positive integer n. Therefore, there is no value of n that satisfies the given equation.
In conclusion, there is no function f that satisfies the given conditions.
f(2) = f(1) f(2) = 2 f(2)
Subtracting f(2) from both sides, we get:
0 = f(2) - 2 f(2)
Combining like terms, we have:
0 = -f(2)
Therefore, f(2) = 0.
Next, let's find the value of f(3). Again from the given equation, we have:
f(3) = f(1) f(3) = 2 f(3)
Subtracting f(3) from both sides, we get:
0 = f(3) - 2 f(3)
Combining like terms, we have:
0 = -f(3)
Therefore, f(3) = 0.
Continuing this pattern, we can see that for any positive integer n, f(n) = 0.
Now, let's consider the expression f(a1) f(a2) ... f(an). From our previous findings, we know that each term in this product is 0. Therefore, the product itself is also 0.
However, the given equation states that this product is equal to 16(2n). Setting this equal to 0, we have:
16(2n) = 0
Dividing both sides by 16, we get:
2n = 0
This equation has no solution for n, as 2n can never equal 0 for any positive integer n. Therefore, there is no value of n that satisfies the given equation.
In conclusion, there is no function f that satisfies the given conditions.
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Consider a function f satisfying f (x + y) = f (x) f (y) where x, y are positive integers, and f (1) = 2. If f (a + 1) + f (a + 2) + ... + f (a + n) = 16(2n – 1) then a is equal to(2019)a)3b)4c)6d)5Correct answer is option 'A'. Can you explain this answer?
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Consider a function f satisfying f (x + y) = f (x) f (y) where x, y are positive integers, and f (1) = 2. If f (a + 1) + f (a + 2) + ... + f (a + n) = 16(2n – 1) then a is equal to(2019)a)3b)4c)6d)5Correct answer is option 'A'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Consider a function f satisfying f (x + y) = f (x) f (y) where x, y are positive integers, and f (1) = 2. If f (a + 1) + f (a + 2) + ... + f (a + n) = 16(2n – 1) then a is equal to(2019)a)3b)4c)6d)5Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a function f satisfying f (x + y) = f (x) f (y) where x, y are positive integers, and f (1) = 2. If f (a + 1) + f (a + 2) + ... + f (a + n) = 16(2n – 1) then a is equal to(2019)a)3b)4c)6d)5Correct answer is option 'A'. Can you explain this answer?.
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