If `f(x)=a^x,` which of the following equalities do not hold ? (i) `f(x+2)-2f(x+1)+f(x)=(a-1)^2f(x)` (ii) `f(-x)f(x)-1=0` (iii) `f(x+y)=f(x)f(y)` (iv) `f(x+3)-2f(x+2)+f(x+1)=(a-2)^2f(x+1)`

Options

A. `f(x+2)-2f(x+1)+f(x)=(a-1)^(2)f(x)`
B. `f(-x)f(x)+1=0`
C. `f(x+y)=f(x)+f(y)`
D. `f(x+3)-2f(x+2)+f(x+1)=(a-2)^(2)f(x+1)`

Answer

To determine which of the given equalities do not hold for the function \( f(x) = a^x \), we will evaluate each equality step by step.

### Step 1: Evaluate the first equality
**Equality (i):**
\[ f(x+2) - 2f(x+1) + f(x) = (a-1)^2 f(x) \]

**Left-hand side:**
\[ f(x+2) = a^{x+2} = a^2 a^x \]
\[ f(x+1) = a^{x+1} = a a^x \]
\[ f(x) = a^x \]
Now substituting these into the left-hand side:
\[ a^2 a^x - 2(a a^x) + a^x = a^x (a^2 - 2a + 1) \]
Factoring the quadratic:
\[ a^x ((a-1)^2) = (a-1)^2 f(x) \]
**Conclusion:** The first equality holds true.

### Step 2: Evaluate the second equality
**Equality (ii):**
\[ f(-x)f(x) - 1 = 0 \]

**Left-hand side:**
\[ f(-x) = a^{-x} = \frac{1}{a^x} \]
Now substituting into the left-hand side:
\[ f(-x)f(x) = \left(\frac{1}{a^x}\right)(a^x) = 1 \]
Thus:
\[ 1 - 1 = 0 \]
**Conclusion:** The second equality holds true.

### Step 3: Evaluate the third equality
**Equality (iii):**
\[ f(x+y) = f(x)f(y) \]

**Left-hand side:**
\[ f(x+y) = a^{x+y} \]
**Right-hand side:**
\[ f(x)f(y) = a^x a^y = a^{x+y} \]
**Conclusion:** The third equality holds true.

### Step 4: Evaluate the fourth equality
**Equality (iv):**
\[ f(x+3) - 2f(x+2) + f(x+1) = (a-2)^2 f(x+1) \]

**Left-hand side:**
\[ f(x+3) = a^{x+3} = a^3 a^x \]
\[ f(x+2) = a^{x+2} = a^2 a^x \]
\[ f(x+1) = a^{x+1} = a a^x \]
Now substituting these into the left-hand side:
\[ a^3 a^x - 2(a^2 a^x) + (a a^x) = a^x (a^3 - 2a^2 + a) \]
Factoring out \( a^x \):
\[ = a^x (a^3 - 2a^2 + a) = a^x (a^2(a - 2) + a) = a^x (a^2 - 2a + 2) \]

**Right-hand side:**
\[ (a-2)^2 f(x+1) = (a-2)^2 (a a^x) = a^x (a-2)^2 a \]

Now, we need to check if:
\[ a^x (a^3 - 2a^2 + a) = a^x (a-2)^2 a \]
This simplifies to:
\[ a^3 - 2a^2 + a = (a-2)^2 a \]
Expanding the right-hand side:
\[ = a(a^2 - 4a + 4) = a^3 - 4a^2 + 4a \]
Setting the two sides equal:
\[ a^3 - 2a^2 + a = a^3 - 4a^2 + 4a \]
Simplifying gives:
\[ 2a^2 - 3a = 0 \]
Factoring out \( a \):
\[ a(2a - 3) = 0 \]
This holds true only for \( a = 0 \) or \( a = \frac{3}{2} \), thus it does not hold for all \( a \).

### Conclusion
The equality that does not hold for all values of \( a \) is:
**(iv)** \( f(x+3) - 2f(x+2) + f(x+1) = (a-2)^2 f(x+1) \)