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)-2(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 option step by step.

### Step 1: Evaluate Option (i)
**Expression:**
\[ f(x+2) - 2f(x+1) + f(x) \]

**Substituting \( f(x) = a^x \):**
\[ f(x+2) = a^{x+2} \]
\[ f(x+1) = a^{x+1} \]
\[ f(x) = a^x \]

**Now substituting into the expression:**
\[ a^{x+2} - 2a^{x+1} + a^x \]
Factoring out \( a^x \):
\[ = a^x (a^2 - 2a + 1) = a^x (a-1)^2 \]

**Conclusion for (i):**
\[ f(x+2) - 2f(x+1) + f(x) = (a-1)^2 f(x) \]
This equality holds true.

### Step 2: Evaluate Option (ii)
**Expression:**
\[ f(-x)f(x) - 1 = 0 \]

**Substituting \( f(x) = a^x \):**
\[ f(-x) = a^{-x} = \frac{1}{a^x} \]
Now substituting into the expression:
\[ f(-x)f(x) = \frac{1}{a^x} \cdot a^x = 1 \]
Thus,
\[ f(-x)f(x) - 1 = 1 - 1 = 0 \]

**Conclusion for (ii):**
This equality holds true.

### Step 3: Evaluate Option (iii)
**Expression:**
\[ f(x+y) = f(x)f(y) \]

**Substituting \( f(x) = a^x \):**
\[ f(x+y) = a^{x+y} \]
And,
\[ f(x)f(y) = a^x \cdot a^y = a^{x+y} \]

**Conclusion for (iii):**
\[ f(x+y) = f(x)f(y) \]
This equality holds true.

### Step 4: Evaluate Option (iv)
**Expression:**
\[ f(x+3) - 2f(x+2) + f(x+1) \]

**Substituting \( f(x) = a^x \):**
\[ f(x+3) = a^{x+3}, \quad f(x+2) = a^{x+2}, \quad f(x+1) = a^{x+1} \]

**Now substituting into the expression:**
\[ a^{x+3} - 2a^{x+2} + a^{x+1} \]
Factoring out \( a^{x+1} \):
\[ = a^{x+1} (a^2 - 2a + 1) = a^{x+1} (a-1)^2 \]

**Now we compare with the right-hand side:**
The right-hand side is:
\[ (a-2)^2 f(x+1) = (a-2)^2 a^{x+1} \]

**Conclusion for (iv):**
The left-hand side is \( a^{x+1}(a-1)^2 \) and the right-hand side is \( a^{x+1}(a-2)^2 \). These are not equal unless \( a-1 = a-2 \), which is not generally true.

Thus, this equality does not hold.

### Final Conclusion:
The equality that does not hold is **Option (iv)**.

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