Question

If the interval of differencing being unity and $\Delta$ be the forward difference operator, then for constants $a$ and $b$, the value of $\Delta (ab^x)$ is given by

• A. $a b^x (b - 1)$
• B. $a^x b (b - 1)$
• C. $a b^x (a - 1)$
• D. $a^x b (a - 1)$

Hint:

The forward difference of any exponential function $k^x$ is always proportional to the original function: $\Delta (k^x) = k^x(k-1)$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The forward difference operator $\Delta$ calculates the difference between successive values of a function over a defined step size $h$.
Key Formula or Approach:
For an interval of differencing of unity ($h = 1$), the forward difference of a function $f(x)$ is:
\[ \Delta f(x) = f(x+1) - f(x) \]

Step 2: Detailed Explanation:
Let the function be:
\[ f(x) = a b^x \]
Apply the forward difference definition:
\[ \Delta(a b^x) = a b^{x+1} - a b^x \]
Using exponential laws, we can write $b^{x+1} = b^x \cdot b$:
\[ \Delta(a b^x) = a \cdot b^x \cdot b - a \cdot b^x \]
Factor out the common term $a b^x$ from both terms:
\[ \Delta(a b^x) = a b^x (b - 1) \]
Therefore, the value is $a b^x (b - 1)$.

Step 3: Final Answer
The correct option is (A).