IGDTUW 2026 — Mathematics PYQ

Step 1: Use the First Principle of Derivative

By the definition of the derivative (First Principle), the derivative of a function $f(x)$ at any point $x$ is given by:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Step 2: Apply the Functional Rule

We are given the functional equation:

$$f(x+y)=f(x)\cdot f(y)$$

Replacing $y$ with $h$ in the limit formula:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x)\cdot f(h)-f(x)}{h}$$

Factor out $f(x)$ from the numerator:

$$f^{\prime }(x)=f(x)\cdot \underset{h\to 0}{\lim }\frac{f(h)-1}{h}$$

Step 3: Find the Value of $f(0)$

To simplify the expression further, let's find $f(0)$ by substituting $x=0$ and $y=0$ into the original functional equation:

$$f(0+0)=f(0)\cdot f(0)$$

$$f(0)=[f(0){]}^2$$

$$f(0)[f(0)-1]=0$$

This gives either $f(0)=0$ or $f(0)=1$.

Note: If $f(0)=0$, then $f(x)=f(x+0)=f(x)\cdot f(0)=0$ for all $x$, which contradicts the given condition that $f(3)=3$. Therefore, $f(0)=1$.

Step 4: Express $f^{\prime }(0)$

Using the limit definition for $f^{\prime }(0)$:

$$f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{f(0+h)-f(0)}{h}$$

$$f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{f(h)-1}{h}$$

We are given that $f^{\prime }(0)=11$. Therefore:

$$\underset{h\to 0}{\lim }\frac{f(h)-1}{h}=11$$

Step 5: Substitute Back to Find $f^{\prime }(3)$

From Step 2, we established that:

$$f^{\prime }(x)=f(x)\cdot \underset{h\to 0}{\lim }\frac{f(h)-1}{h}$$

Substituting the limit value from Step 4:

$$f^{\prime }(x)=f(x)\cdot 11$$

Now, evaluate this at $x=3$:

$$f^{\prime }(3)=f(3)\cdot 11$$

Given that $f(3)=3$:

$$f^{\prime }(3)=3\cdot 11=33$$

Correct Answer:

(a) $33$