Which of the following function follows the rule: f(a + b) = f(a)*f(b)

Which of the following function follows the rule: f(a+b)=f(a)∗f(b)?

Method 1: As per the rule that needs to be true, we can foresee that if the variable is in exponentials then a mere multiplication can simply lead to addition and the rule can hold true for the equation.

As per this, we see that the options 2 and 5 are the only ones where the variable is completely in the exponentials.

In option 2 the denominator will not allow the rule to hold true as in the RHS it will multiply. Hence option 2 is not correct.

In option 5, we have no other issues and the variable in the exponentials will simply add up on multiplication of the entities and hence the rule will always hold true.

Now, we can plug in some sample values and see if our analysis is correct and mark the correct answer which is option E.

Method 2: Putting values. This method is fool proof but may take you a couple of minutes to solve this question.

Option 1: f(x)=x^2+1

f(1 + 1) = f(1) = 1^2 + 1 = 5
f(1) = 1^2 + 1 = 2

If rule is true here then,
f(1 + 1) = f(1) * f(1)

However,
5 <> 2

Hence,
f(1 + 1) <> f(1) * f(1)

Option 1 is not correct.

Option 2: f(x) = (5^(2x))/3

f(1 + 1) = f(2) = (5^(2*2)) / 3 = (5^4) / 3 = 625/3
f(1) = (5^(2*1)) / 3 = 25/3

If rule is true here then,
f(1 + 1) = f(1) * f(1)

625/3 = 25/3 * 25/3 -> Rule holds here.

f(1 + 0) = f(1) = (5^(2*1)) / 3 = 25/3
f(1) = (5^(2*1)) / 3 = 25/3
f(0) = (5^(2*0)) / 3 = 1/3

However,
25/3 <> 25/3 * 1/3

Now,
f(1 + 1) = f(1) * f(1);
BUT, f(1 + 0) <> f(1) * f(0);

Hence, rule holds true sometimes but not always for this option.

Option 2 is not correct.

Option 3: f(x) = 3x + 2

f(1 + 1) = f(2) = 3*2 + 2 = 8
f(1) = 3*1 + 2 = 5

If rule is true here then,
f(1 + 1) = f(1) * f(1)

However,
8 <> 5 * 5

Hence,
f(1 + 1) <> f(1) * f(1)

Option 3 is not correct.

Option 4: f(x) = sqrt(2*x)

Consider,
f(1 + 1) = f(2) = sqrt(2*2) = 2
f(1) = sqrt(2*1) = sqrt(2)

If rule is true here then,
f(1 + 1) = f(1) * f(1)

And,
2 = sqrt(2) * sqrt(2)

Now consider,
f(2 + 3) = f(5) = sqrt(2*5) = sqrt(10)
f(2) = sqrt(2*2) = 2
f(3) = sqrt(2*3) = sqrt(6)

However,
sqrt(10) <> 2 * sqrt(6)

Hence,
f(1 + 1) <> f(1) * f(1)

Option 4 is not correct.

Option 5: f(x) = 24^x

Consider,
f(1 + 1) = f(2) = 24^2
f(1) = 24^1

If rule is true here then,
f(1 + 1) = f(1) * f(1)

And,
24^2 = 24^1 * 24^1

Consider,
f(2 + 3) = f(5) = 24^5
f(2) = 24^2
f(3) = 24^3

If rule is true here then,
f(1 + 1) = f(1) * f(1)

And,
24^5 = 24^2 * 24^3

Hence Option 5 or Option E is correct.

for \(f(x)=x^2+1\):
f(a+b) = \((a+b)^2 + 1 = a^2 + b^2 + 2ab + 1\)
f(a)∗f(b) = \(a^2 + 1 + b^2 + 1 = a^2 + b^2 + 2\) --> not equal for all values


for \(f(x)=\frac{5^{2x}}{3}\):
f(a+b) = \(\frac{5^{2(a+b)}}{3} = \frac{5^{2a}5^{2b}}{3}\)
f(a)∗f(b) = \(\frac{5^{2a}}{3}*\frac{5^{2b}}{3} = \frac{5^{2a}5^{2b}}{9}\)--> not equal for all values

for \(f(x)=3x+2\):
f(a+b) = \(3(a+b) + 2 = 3a + 3b + 2\)
f(a)∗f(b) = \((3a+2)(3b+2) = 9ab + 6b + 6a + 4\) --> not equal for all values

for f(x)= \(\sqrt{{2x}}\):
f(a+b) = \(\sqrt{{2(a+b)}} = \sqrt{{2a+2b}}\)
f(a)∗f(b) = \(\sqrt{{2a}}* \sqrt{{2b}} = \sqrt{{4ab}} = 2\sqrt{{ab}}\) --> not equal for all values

for \(f(x)= 24^x\):
f(a+b) = \(24^{a+b}\)
f(a)∗f(b) = \(24^{a}*24^{b} = 24^{a+b}\) --> equal

E

Which of the following function follows the rule: f(a+b)=f(a)∗f(b)?

A. f(x)=\(x^2+1\)

B. f(x)=\(5^2^x * 3\)

C. f(x)=3x+2

D. f(x)=\(\sqrt{2x}\)

E. f(x)=\(24^x\)

A - f(a) = \(a^2 +1\) f(b) = \(b^2 +1\) f(a+b) = \((a+b)^2 +1\).
f(a) * f(b) = \(a^2*b^2 +a^2+b^21\)
Clearly f(a+b) is not equal to f(a)∗f(b).

B - f(a)=\(5^2^a * 3\) f(b)=\(5^2^b * 3\) f(a+b)=\(5^2^(a+b) * 3\).
f(a) * f(b) = \(5^2^a+b * 9\)
Clearly f(a+b) is not equal to f(a)∗f(b).

C - f(a)=\(3a+2\) f(b)=\(3b+2\) f(a+b)=\(3(a+b)+2\).
f(a) * f(b) = \(9ab+6a+6b+4\)
Clearly f(a+b) is not equal to f(a)∗f(b).

D - f(a)=\(\sqrt{2a}\)
f(b)=\(\sqrt{2b}\)
f(a+b)=\(\sqrt{2(a+b)}\)
f(a)*f(b)=\(\sqrt{4ab}\)
Clearly f(a+b) is not equal to f(a)∗f(b).

E - f(x)=\(24^x\)

f(a)=\(24^a\)
f(b)=\(24^b\)
f(a+b)=\(24^(a+b)\)
f(a) * f(b)=\(24^(a+b)\).
Clearly f(a+b) is equal to f(a)∗f(b)

Hence E

Which of the following function follows the rule: f(a+b)=f(a)∗f(b)?

We have to look for the option in which addition and multiplication might result in same value.
If we look at the answer choices we can eliminate A, C and D easily. But to make sure we will put value of x and check.

A. f(x)=x^2+1
f(a+b)= (a+b)^2+1
f(a)*f(b) = (a^2 + 1)* (b^2+1)
after solving we do not get similar values.

B. f(x)=5^2x/3
f(a+b)=5^2(a+b)/3
f(a)*f(b)=5^2a/3* 5^2b/3
After solving this we get same numerator but different denominator.

C. f(x)=3x+2
f(a+b)=3(a+b)+2
f(a)*f(b)=(3a+2)*(3b+2)
after solving we do not get similar values.

D. f(x)=sq root of 2x
f(a+b)=sq root of 2(a+b)
f(a)*f(b)=sq root of 2a * sq root of 2b
after solving we do not get similar values.

E. f(x)=24^x
f(a+b)=24^(a+b)
f(a)*f(b)=24^a* 24^b
We get similar values on both sides.
Correct.

Which of the following function follows the rule: f(a+b)=f(a)∗f(b)?

A. \(f(x) = x^2+1\)
\(f(a + b) = (a + b)^2 + 1 = a^2 + b^2 + 2ab + 1\)
\(f(a)*f(b) = (a^2+1)*(b^2+1) = (ab)^2 + a^2 + b^2 + 1\) --> NO

B. \(f(x) = \frac{5^{2x}}{3}\)
\(f(a + b) = \frac{5^{2(a + b)}}{3} = \frac{5^{2a + 2b}}{3}\)
\(f(a)*f(b) = \frac{5^{2a}}{3}*\frac{5^{2b}}{3} = \frac{5^{2a + 2b}}{9}\) --> NO

C. \(f(x) = 3x + 2\)
\(f(a + b) = 3(a + b) + 2 = 3a + 3b + 2\)
\(f(a)*f(b) = (3a + 2)*(3b + 2) = 9ab + 6a + 6b + 4\) --> NO

D. \(f(x) = 2x\)
\(f(a + b) = 2(a + b) = 2a + 2b\)
\(f(a)*f(b) = 2a*2b = 4ab\) --> NO

E. \(f(x) = 24^x\)
\(f(a + b) = 24^{a + b}\)
\(f(a)*f(b) = 24^a*24^b = 24^{a + b}\) --> YES

IMO Option E

Pls Hit Kudos if you like the solution

Which of the following function follows the rule: f(a+b)=f(a)∗f(b)?

Let us try a=1, b=3, a+b= 1+3 = 4

A. f(x)=x^2+1
- f(a+b)=f(4)=17
- f(a)=f(1)=2, f(b)=f(3)=10, then f(a)∗f(b)= f(1)∗f(3) = 20
- the above function does NOT follow the rule: f(a+b)=f(a)∗f(b)

B. f(x)=5^(2x)/3
- f(a+b)=f(4)=(5^8)/3
- f(a)=f(1)=(5^2)/3, f(b)=f(3)=(5^6)/3, then f(a)∗f(b)= f(1)∗f(3) = (5^8)/9
- The above function does NOT follow the rule: f(a+b)=f(a)∗f(b)

C. f(x)=3x+2
- f(a+b)=f(4)=14
- f(a)=f(1)=5, f(b)=f(3)=11, then f(a)∗f(b)= f(1)∗f(3) = 55
- the above function does NOT follow the rule: f(a+b)=f(a)∗f(b)

D. f(x)=√(2x)
- f(a+b)=f(4)=√8
- f(a)=f(1)=√2, f(b)=f(3)=√6, then f(a)∗f(b)= f(1)∗f(3) = √12
- the above function does NOT follow the rule: f(a+b)=f(a)∗f(b)

E. f(x)=24^x
- f(a+b)=f(4)=24^4
- f(a)=f(1)=24^1, f(b)=f(3)=24^3, then f(a)∗f(b)= f(1)∗f(3) = 24^4
- the above function does follow the rule: f(a+b)=f(a)∗f(b)

ANSWER IS (E)

A. f(x)=\(x^2\)+1

f(a+b) = \(a^2\)+\(b^2\)+2ab+1
f(a).f(b) = \(a^2\)\(b^2\)+\(a^2\)+\(b^2\)+1

Not equal

B. f(x)=\(5^{2x}\)/3

f(a+b) = \(5^{2a+2b}\)/3
f(a).f(b) = \(5^{2a+2b}\)/9

Not equal

C. f(x)=3x+2

f(a+b) = 3a+3b+2
f(a).f(b) = (3a+2)(3b+2)

Not equal

D. \(f(x)=\sqrt{2x}\)

\(f(a+b) = \sqrt{2(a+b)}\)
\(f(a).f(b) = 2\sqrt{ab}\)

Not equal

E. f(x)=\(24^x\)

f(a+b) = \(24^{a+b}\)
f(a).f(b) = \(24^{a+b}\)

Equal

Hence, option E.

You can solve this question under 5 seconds without any calculation. Notice the condition is f(a + b) = f(a)*f(b)

We know that we can add the exponents when multiplying two powers with the same base. We can easily do that with option E.

Answer E

can anyone help a=1 and b=2 ?
not able to get the right answer after taking a=1 and b=2