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We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).
Correct answer is '50'. Can you explain this answer?
Most Upvoted Answer
We are told that f(x) is a polynomial function such that f(a)f(b) = f(...
f(a)f(b) = f(a) + f(b) + f(ab) - 2
Put a = b = 1.
[f(1)] 2 =3f(1)−2 ⇒ f(1) = 1 (or) 2.
Let's assume f(1) = 1
Now, put b = 1.
f(a) = 2f(a) - 1
⇒ f(a) = 1 ⇒ For all values of a, f(a) = 1.
This is false because f(4) = 17.
⇒ f(1) = 2 is the correct value.
Now put b = 1/a
f(a)f(1/a) = f(a) + f(1/a) + 2 - 2
⇒ f(a)f(1/a) = f(a) + f(1/a)
So taking RHS terms to LHS and adding 1 to both sides we get
f(a)f(1/a) - f(a) - f(1/a) +1 = 1
(f(a) - 1) (f(1/a)-1) = 1
Let g(x) = f(x)-1
So g(x)*g(1/x) = 1
So g(x) is of the form + xn
So f(x) is of the form xn + 1.
f(a) = + an + 1 satisfies the above condition.
- 4n + 1 = 17 ⇒ 4n = 16 which is not possible.
4a +1 = 17 ⇒ n = 2
⇒ f(a) = a2 + 1
⇒ f(7) = 72 + 1 = 50
Put a = b = 1.
[f(1)] 2 =3f(1)−2 ⇒ f(1) = 1 (or) 2.
Let's assume f(1) = 1
Now, put b = 1.
f(a) = 2f(a) - 1
⇒ f(a) = 1 ⇒ For all values of a, f(a) = 1.
This is false because f(4) = 17.
⇒ f(1) = 2 is the correct value.
Now put b = 1/a
f(a)f(1/a) = f(a) + f(1/a) + 2 - 2
⇒ f(a)f(1/a) = f(a) + f(1/a)
So taking RHS terms to LHS and adding 1 to both sides we get
f(a)f(1/a) - f(a) - f(1/a) +1 = 1
(f(a) - 1) (f(1/a)-1) = 1
Let g(x) = f(x)-1
So g(x)*g(1/x) = 1
So g(x) is of the form + xn
So f(x) is of the form xn + 1.
f(a) = + an + 1 satisfies the above condition.
- 4n + 1 = 17 ⇒ 4n = 16 which is not possible.
4a +1 = 17 ⇒ n = 2
⇒ f(a) = a2 + 1
⇒ f(7) = 72 + 1 = 50
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Community Answer
We are told that f(x) is a polynomial function such that f(a)f(b) = f(...
Solution:
Given that f(x) is a polynomial function, we can write it in the form:
f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀
where aₙ, aₙ₋₁, ..., a₁, a₀ are the coefficients of the polynomial and n is the degree of the polynomial.
Step 1: Finding the degree of the polynomial
Since we are given that f(a)f(b) = f(a)f(b)f(ab) - 2, let's substitute x = a and x = b to get two equations:
f(a)f(b) = f(a)f(b)f(ab) - 2 ...(1)
f(a) = 17 ...(2)
Substituting f(a) = 17 in equation (1), we get:
17f(b) = 17f(b)f(ab) - 2
Simplifying the equation, we have:
17f(b)f(ab) - 17f(b) - 2 = 0
Now, consider this equation as a quadratic equation in f(b). Let f(ab) = k, where k is a constant. Then we have:
17kf(b) - 17f(b) - 2 = 0
This is a quadratic equation in f(b) with coefficients 17k, -17, and -2. For this equation to have real roots, the discriminant (b² - 4ac) must be greater than or equal to 0. So we have:
(-17)² - 4(17k)(-2) ≥ 0
289 + 136k ≥ 0
289 ≥ -136k
-289/136 ≤ k
Since k = f(ab), it means that f(ab) ≤ -289/136.
Step 2: Finding the value of f(7)
Now, using the given equation f(a)f(b) = f(a)f(b)f(ab) - 2, let's substitute a = 4 and b = 7 to get:
f(4)f(7) = f(4)f(7)f(28) - 2
Substituting f(4) = 17 from the given information, we have:
17f(7) = 17f(7)f(28) - 2
Dividing both sides by 17, we get:
f(7) = f(7)f(28) - 2
Since f(ab) ≤ -289/136, we know that f(28) ≤ -289/136.
Substituting this inequality in the equation above, we have:
f(7) = f(7)(-289/136) - 2
Multiplying through by 136 to clear the fraction, we get:
136f(7) = -289f(7) - 272
Rearranging the equation, we have:
425f(7) = -272
Dividing both sides by 425, we get:
f(7) = -272/425
Calculating the value, we find:
f(7) ≈ -0.639
Therefore, the value of f(7) is approximately -0.639 and not 50 as mentioned in the
Given that f(x) is a polynomial function, we can write it in the form:
f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀
where aₙ, aₙ₋₁, ..., a₁, a₀ are the coefficients of the polynomial and n is the degree of the polynomial.
Step 1: Finding the degree of the polynomial
Since we are given that f(a)f(b) = f(a)f(b)f(ab) - 2, let's substitute x = a and x = b to get two equations:
f(a)f(b) = f(a)f(b)f(ab) - 2 ...(1)
f(a) = 17 ...(2)
Substituting f(a) = 17 in equation (1), we get:
17f(b) = 17f(b)f(ab) - 2
Simplifying the equation, we have:
17f(b)f(ab) - 17f(b) - 2 = 0
Now, consider this equation as a quadratic equation in f(b). Let f(ab) = k, where k is a constant. Then we have:
17kf(b) - 17f(b) - 2 = 0
This is a quadratic equation in f(b) with coefficients 17k, -17, and -2. For this equation to have real roots, the discriminant (b² - 4ac) must be greater than or equal to 0. So we have:
(-17)² - 4(17k)(-2) ≥ 0
289 + 136k ≥ 0
289 ≥ -136k
-289/136 ≤ k
Since k = f(ab), it means that f(ab) ≤ -289/136.
Step 2: Finding the value of f(7)
Now, using the given equation f(a)f(b) = f(a)f(b)f(ab) - 2, let's substitute a = 4 and b = 7 to get:
f(4)f(7) = f(4)f(7)f(28) - 2
Substituting f(4) = 17 from the given information, we have:
17f(7) = 17f(7)f(28) - 2
Dividing both sides by 17, we get:
f(7) = f(7)f(28) - 2
Since f(ab) ≤ -289/136, we know that f(28) ≤ -289/136.
Substituting this inequality in the equation above, we have:
f(7) = f(7)(-289/136) - 2
Multiplying through by 136 to clear the fraction, we get:
136f(7) = -289f(7) - 272
Rearranging the equation, we have:
425f(7) = -272
Dividing both sides by 425, we get:
f(7) = -272/425
Calculating the value, we find:
f(7) ≈ -0.639
Therefore, the value of f(7) is approximately -0.639 and not 50 as mentioned in the
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We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).Correct answer is '50'. Can you explain this answer?
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We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).Correct answer is '50'. Can you explain this answer? for CAT 2025 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).Correct answer is '50'. Can you explain this answer? covers all topics & solutions for CAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).Correct answer is '50'. Can you explain this answer?.
We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).Correct answer is '50'. Can you explain this answer? for CAT 2025 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).Correct answer is '50'. Can you explain this answer? covers all topics & solutions for CAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for We are told that f(x) is a polynomial function such that f(a)f(b) = f(a) + f(b) + f(ab) - 2 and f(4) = 17, find the value of f(7).Correct answer is '50'. Can you explain this answer?.
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