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A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)
- a)128
- b)256
- c)512
- d)1024
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A certain function always obeys the rule: If f (x.y) = f(x). f(y) wher...
Since f (128) = 4, we can see that the product of f (256). f (0.5) = f (256 × 0.5) = f (128) = 4.
Similarly, the products f (1). f (128) = f (2). f (64)
= f (4). f (32) = f (8). f (16) = 4.
Thus, M = 4 × 4 × 4 × 4 × 4 = 1024.
Option (d) is the correct answer.
Similarly, the products f (1). f (128) = f (2). f (64)
= f (4). f (32) = f (8). f (16) = 4.
Thus, M = 4 × 4 × 4 × 4 × 4 = 1024.
Option (d) is the correct answer.
Most Upvoted Answer
A certain function always obeys the rule: If f (x.y) = f(x). f(y) wher...
Given information:
- f(x.y) = f(x).f(y) for positive real numbers x and y
- f(128) = 4
To find:
M = f(0.5) . f(1) . f(2) . f(4) . f(8) . f(16) . f(32) . f(64) . f(128) . f(256)
Solution:
We can use the given rule to simplify the expression for M. We start with f(1) and use it to find f(2), then use f(2) to find f(4), and so on, until we reach f(128). Then we can substitute the given value of f(128) and simplify the expression for M.
1. f(1) = f(0.5 . 2) = f(0.5) . f(2)
2. f(2) = f(1 . 2) = f(1) . f(2) = (f(0.5) . f(2)) . f(2) = f(0.5) . f(2)^2
3. f(4) = f(2 . 2) = f(2) . f(2) = f(0.5) . f(2)^2 . f(0.5) . f(2)^2 = f(0.5)^2 . f(2)^4
4. f(8) = f(4 . 2) = f(4) . f(2) = f(0.5)^2 . f(2)^4 . f(0.5) . f(2)^2 = f(0.5)^3 . f(2)^6
5. f(16) = f(8 . 2) = f(8) . f(2) = f(0.5)^3 . f(2)^6 . f(0.5) . f(2)^2 = f(0.5)^4 . f(2)^8
6. f(32) = f(16 . 2) = f(16) . f(2) = f(0.5)^4 . f(2)^8 . f(0.5) . f(2)^2 = f(0.5)^5 . f(2)^10
7. f(64) = f(32 . 2) = f(32) . f(2) = f(0.5)^5 . f(2)^10 . f(0.5) . f(2)^2 = f(0.5)^6 . f(2)^12
8. f(128) = f(64 . 2) = f(64) . f(2) = f(0.5)^6 . f(2)^12 . f(0.5) . f(2)^2 = f(0.5)^7 . f(2)^14 = 4
Substituting f(128) = 4 into the expression for M, we get:
M = f(0.5) . f(1) . f(2) . f(4) . f(8) . f(16) . f(32)
- f(x.y) = f(x).f(y) for positive real numbers x and y
- f(128) = 4
To find:
M = f(0.5) . f(1) . f(2) . f(4) . f(8) . f(16) . f(32) . f(64) . f(128) . f(256)
Solution:
We can use the given rule to simplify the expression for M. We start with f(1) and use it to find f(2), then use f(2) to find f(4), and so on, until we reach f(128). Then we can substitute the given value of f(128) and simplify the expression for M.
1. f(1) = f(0.5 . 2) = f(0.5) . f(2)
2. f(2) = f(1 . 2) = f(1) . f(2) = (f(0.5) . f(2)) . f(2) = f(0.5) . f(2)^2
3. f(4) = f(2 . 2) = f(2) . f(2) = f(0.5) . f(2)^2 . f(0.5) . f(2)^2 = f(0.5)^2 . f(2)^4
4. f(8) = f(4 . 2) = f(4) . f(2) = f(0.5)^2 . f(2)^4 . f(0.5) . f(2)^2 = f(0.5)^3 . f(2)^6
5. f(16) = f(8 . 2) = f(8) . f(2) = f(0.5)^3 . f(2)^6 . f(0.5) . f(2)^2 = f(0.5)^4 . f(2)^8
6. f(32) = f(16 . 2) = f(16) . f(2) = f(0.5)^4 . f(2)^8 . f(0.5) . f(2)^2 = f(0.5)^5 . f(2)^10
7. f(64) = f(32 . 2) = f(32) . f(2) = f(0.5)^5 . f(2)^10 . f(0.5) . f(2)^2 = f(0.5)^6 . f(2)^12
8. f(128) = f(64 . 2) = f(64) . f(2) = f(0.5)^6 . f(2)^12 . f(0.5) . f(2)^2 = f(0.5)^7 . f(2)^14 = 4
Substituting f(128) = 4 into the expression for M, we get:
M = f(0.5) . f(1) . f(2) . f(4) . f(8) . f(16) . f(32)
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A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)a)128b)256c)512d)1024Correct answer is option 'D'. Can you explain this answer?
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A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)a)128b)256c)512d)1024Correct answer is option 'D'. Can you explain this answer? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)a)128b)256c)512d)1024Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)a)128b)256c)512d)1024Correct answer is option 'D'. Can you explain this answer?.
A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)a)128b)256c)512d)1024Correct answer is option 'D'. Can you explain this answer? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)a)128b)256c)512d)1024Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive realnumbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of thevariable M = f (0.5). f (1). f (2). f (4). f (8). f (16). f (32). f (64). f (128). f (256)a)128b)256c)512d)1024Correct answer is option 'D'. Can you explain this answer?.
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