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27 Feb 2022, 07:51
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Given: Let the function g(a, b) = f(a) + f(b).
Asked: For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?
g(a + b, a + b) = f(a+b) + f(a+b) = 2f(a+b)
g(a, a) + g(b, b) = f(a) + f(a) + f(b) + f(b) = 2f(a) + 2f(b)
For which function f will f(a+b) = f(a) + f(b)
A: x +3: f(a+b) = a+b+3; f(a) + f(b) = a + 3 + b + 3; f(a+b) \(\neq\) f(a) + f(b)
B: x^2 : f(a+b) = (a+b)^2 ; f(a) + f(b) = a^2 + b^2; f(a+b) \(\neq\) f(a) + f(b)
C: |x| : f(a+b) = |a+b|; f(a) + f(b) = |a| + |b|; f(a+b) \(\neq\) f(a) + f(b)
D: 1/x : f(a+b) = 1/(a+b); f(a) + f(b) = 1/a + 1/b; f(a+b) \(\neq\) f(a) + f(b)
E: x/4 : f(a+b) = (a+b)/4 ; f(a) + f(b) = a/4 + b/4; f(a+b) = f(a) + f(b)
IMO E
Asked: For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?
g(a + b, a + b) = f(a+b) + f(a+b) = 2f(a+b)
g(a, a) + g(b, b) = f(a) + f(a) + f(b) + f(b) = 2f(a) + 2f(b)
For which function f will f(a+b) = f(a) + f(b)
A: x +3: f(a+b) = a+b+3; f(a) + f(b) = a + 3 + b + 3; f(a+b) \(\neq\) f(a) + f(b)
B: x^2 : f(a+b) = (a+b)^2 ; f(a) + f(b) = a^2 + b^2; f(a+b) \(\neq\) f(a) + f(b)
C: |x| : f(a+b) = |a+b|; f(a) + f(b) = |a| + |b|; f(a+b) \(\neq\) f(a) + f(b)
D: 1/x : f(a+b) = 1/(a+b); f(a) + f(b) = 1/a + 1/b; f(a+b) \(\neq\) f(a) + f(b)
E: x/4 : f(a+b) = (a+b)/4 ; f(a) + f(b) = a/4 + b/4; f(a+b) = f(a) + f(b)
IMO E
19 Aug 2022, 11:17
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Given that g(a, b) = f(a) + f(b) and we need to find For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)
Let's start by simplifying g(a + b, a + b) = g(a, a) + g(b, b)
To find g(a + b, a + b) we need to compare what is inside the bracket in g(a + b, a + b) and g(a,b)
=> We need to substitute both a and b with a+b in g(a, b) = f(a) + f(b) to get the value of g(a + b, a + b)
=> g(a + b, a + b) = f(a+b) + f(a+b) = 2* f(a+b)
Similarly, g(a, a) = f(a) + f(a) = 2* f(a) and
g(b, b) = f(b) + f(b) = 2* f(b)
=> g(a + b, a + b) = g(a, a) + g(b, b) can be simplified as
2* f(a+b) = 2*f(a) + 2*f(a)
=> f(a+b) = f(a) + f(b)
So, the question becomes. For which function f below will f(a+b) = f(a) + f(b)
Let's solve the problem using two methods
Method 1: Logic (Eliminate Option Choices)
f(a+b) = f(a) + f(b)
Now, this can be true only when
Using above logic we can eliminate the answer choices
(A) x +3
=> Eliminate : Doesn't Satisfy Point 1 above. It has a constant added(+3)
(B) \(x^2\)
=> Eliminate : Doesn't Satisfy Point 3 above. Power of x is \(2\)
(C) |x| (Watch this video to know the Basics of Absolute Value)
=> Now according to point 4 we need to evaluate the option choice using numbers.
Let's say a = 1 and b = 2
=> f(a+b) = f(1+2) = f(3) = |3| = 3
And f(a) + f(b) = f(1) + f(2) = |1| + |2| = 1 + 2 = 3 => TRUE
Let's take another example say a = -1 and b = 2
=> f(a+b) = f(-1+2) = f(1) = |1| = 1
And f(a) + f(b) = f(-1) + f(2) = |-1| + |2| = 1 + 2 = 3 ≠ 1 => FALSE
=> Eliminate
(D) \(\frac{1}{x}\)
=> Eliminate : Doesn't Satisfy Point 2 above. x is in denominator
(E) \(\frac{x}{4}\)
=> POSSIBLE: Satisfies all the conditions above.
So, Answer will be E.
Method 2: Algebra (taking all option choices)
(A) \(x + 3\)
f(x) = \(x + 3\)
To find f(a+b) we need to compare what is inside the bracket in f(a+b) and f(x)
=> We need to substitute x with a+b in f(x) = \(x + 3\) to get the value of f(a+b)
=> f(a+b) = \(a+b + 3\)
=> f(a) + f(b) = \(a + 3\) + \(b + 3\) = \(a+b + 6\) ≠ \(a+b + 3\)
=> f(a+b) ≠ f(a) + f(b) => FALSE
(B) \(x^2\)
f(x) = \(x^2\)
=> f(a+b) = \((a+b)^2\) = \(a^2 + 2ab + b^2\)
f(a) = \(a^2\) and f(b) = \(b^2\)
=> f(a) + f(b) = \(a^2\) + \(b^2\) = \(a^2 + b^2\) ≠ \(a^2 + 2ab + b^2\)
=> f(a+b) ≠ f(a) + f(b) => FALSE
(C) \(|x|\) (Watch this video to know the Basics of Absolute Value)
f(x) = \(|x|\)
=> f(a+b) = |a+b|
f(a) + f(b) = |a| + |b| can be = |a+b| ONLY when both a and b have same sign. But we don't know that.
=> f(a+b) = f(a) + f(b) => CAN BE FALSE
(D) \(\frac{1}{x}\)
f(x) = \(\frac{1}{x}\)
=> f(a+b) = \(\frac{1}{a+b}\)
f(a) + f(b) = \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{a + b }{ ab}\) ≠ \(\frac{1}{a+b}\)
=> f(a+b) ≠ f(a) + f(b) => FALSE
(E) \(\frac{x}{4}\)
f(x) = \(\frac{x}{4}\)
=> f(a+b) = \(\frac{a+b}{4}\)
f(a) + f(b) = \(\frac{a}{4}\) + \(\frac{b}{4}\) = \(\frac{a + b }{ 4}\)
=> f(a+b) = f(a) + f(b) => TRUE
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
Let's start by simplifying g(a + b, a + b) = g(a, a) + g(b, b)
To find g(a + b, a + b) we need to compare what is inside the bracket in g(a + b, a + b) and g(a,b)
=> We need to substitute both a and b with a+b in g(a, b) = f(a) + f(b) to get the value of g(a + b, a + b)
=> g(a + b, a + b) = f(a+b) + f(a+b) = 2* f(a+b)
Similarly, g(a, a) = f(a) + f(a) = 2* f(a) and
g(b, b) = f(b) + f(b) = 2* f(b)
=> g(a + b, a + b) = g(a, a) + g(b, b) can be simplified as
2* f(a+b) = 2*f(a) + 2*f(a)
=> f(a+b) = f(a) + f(b)
So, the question becomes. For which function f below will f(a+b) = f(a) + f(b)
Let's solve the problem using two methods
Method 1: Logic (Eliminate Option Choices)
f(a+b) = f(a) + f(b)
Now, this can be true only when
- 1. We don't have any constant term added or subtracted from any term of x. As if we have one then on Left Hand Side(LHS) that constant term will be added or subtracted only once, but on Right Hand Side(RHS) it will be added or subtracted twice.
2. We don't have x in the denominator (in general) as we wont be able to match the LHS and RHS then.
3. We don't have any power of x ≠ 1 in the numerator. As otherwise (in general) we wont be able to match the LHS and RHS.
4. We have a term of x in the numerator (if it has an absolute value sign then evaluate the option choice by taking numbers) with power of 1 with any positive or negative constant multiplied with it. Ex 2x, -3x, etc
Using above logic we can eliminate the answer choices
(A) x +3
=> Eliminate : Doesn't Satisfy Point 1 above. It has a constant added(+3)
(B) \(x^2\)
=> Eliminate : Doesn't Satisfy Point 3 above. Power of x is \(2\)
(C) |x| (Watch this video to know the Basics of Absolute Value)
=> Now according to point 4 we need to evaluate the option choice using numbers.
Let's say a = 1 and b = 2
=> f(a+b) = f(1+2) = f(3) = |3| = 3
And f(a) + f(b) = f(1) + f(2) = |1| + |2| = 1 + 2 = 3 => TRUE
Let's take another example say a = -1 and b = 2
=> f(a+b) = f(-1+2) = f(1) = |1| = 1
And f(a) + f(b) = f(-1) + f(2) = |-1| + |2| = 1 + 2 = 3 ≠ 1 => FALSE
=> Eliminate
(D) \(\frac{1}{x}\)
=> Eliminate : Doesn't Satisfy Point 2 above. x is in denominator
(E) \(\frac{x}{4}\)
=> POSSIBLE: Satisfies all the conditions above.
So, Answer will be E.
Method 2: Algebra (taking all option choices)
(A) \(x + 3\)
f(x) = \(x + 3\)
To find f(a+b) we need to compare what is inside the bracket in f(a+b) and f(x)
=> We need to substitute x with a+b in f(x) = \(x + 3\) to get the value of f(a+b)
=> f(a+b) = \(a+b + 3\)
=> f(a) + f(b) = \(a + 3\) + \(b + 3\) = \(a+b + 6\) ≠ \(a+b + 3\)
=> f(a+b) ≠ f(a) + f(b) => FALSE
(B) \(x^2\)
f(x) = \(x^2\)
=> f(a+b) = \((a+b)^2\) = \(a^2 + 2ab + b^2\)
f(a) = \(a^2\) and f(b) = \(b^2\)
=> f(a) + f(b) = \(a^2\) + \(b^2\) = \(a^2 + b^2\) ≠ \(a^2 + 2ab + b^2\)
=> f(a+b) ≠ f(a) + f(b) => FALSE
(C) \(|x|\) (Watch this video to know the Basics of Absolute Value)
f(x) = \(|x|\)
=> f(a+b) = |a+b|
f(a) + f(b) = |a| + |b| can be = |a+b| ONLY when both a and b have same sign. But we don't know that.
=> f(a+b) = f(a) + f(b) => CAN BE FALSE
(D) \(\frac{1}{x}\)
f(x) = \(\frac{1}{x}\)
=> f(a+b) = \(\frac{1}{a+b}\)
f(a) + f(b) = \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{a + b }{ ab}\) ≠ \(\frac{1}{a+b}\)
=> f(a+b) ≠ f(a) + f(b) => FALSE
(E) \(\frac{x}{4}\)
f(x) = \(\frac{x}{4}\)
=> f(a+b) = \(\frac{a+b}{4}\)
f(a) + f(b) = \(\frac{a}{4}\) + \(\frac{b}{4}\) = \(\frac{a + b }{ 4}\)
=> f(a+b) = f(a) + f(b) => TRUE
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
21 Oct 2025, 05:04
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