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# extract all questions and give solutions

Sure! Here are all the extracted questions from the provided image, each followed by a friendly, step-by-step solution with explanations and relevant emojis:

## 17️⃣ What is the value of \$ f(0) \$?

### **Given:**

- \$ f \$ is a real-valued function on \$ \mathbb{R} \$.
- \$ f(x + y) = f(x)f(y) \$ for all \$ x, y \in \mathbb{R} \$.
- \$ f(1) = 7 \$, \$ f'(0) = 2 \$.


### **Solution:**

Let’s use the property: \$ f(0 + 0) = f(0) f(0) \implies f(0)^2 = f(0) \$

So either \$ f(0) = 0 \$ or \$ f(0) = 1 \$.

But, let's check if \$ f(0) = 0 \$ is possible:

If \$ f(0) = 0 \$, then for any \$ x \$,

$$
f(x) = f(x + 0) = f(x) f(0) = f(x) \times 0 = 0
$$

This would mean \$ f(x) = 0 \$ for all \$ x \$, but we are given \$ f(1) = 7 \neq 0 \$.

Thus, **\$ f(0) = 1 \$**. ✅✨

## 18️⃣ What is the value of \$ f'(1) \$?

### **Solution:**

Let's hypothesize the general form for \$ f \$ using the given functional equation.

Whenever \$ f(x+y) = f(x)f(y) \$, the continuous solutions are of the form \$ f(x) = a^x \$.

We know \$ f(1) = 7 \Rightarrow a^1 = 7 \implies a = 7 \$.

So, \$ f(x) = 7^x \$.

Now differentiate:

$$
f'(x) = 7^x \ln 7
$$

So,

$$
f'(1) = 7^1 \ln 7 = 7 \ln 7
$$

The answer: **\$ f'(1) = 7\ln 7 \$**. 🧮

## 19️⃣ Consider a function \$ f \$ defined as,

$$
f(x) = 
\begin{cases}
3mx + n, & x < 1 \\
11, & x = 1 \\
5mx + 2n, & x > 1
\end{cases}
$$

If \$ f \$ is continuous at \$ x = 1 \$, then the value of \$ m + n \$ is?

### **Solution:**

For continuity at \$ x = 1 \$:

$$
\lim_{x \to 1^-} f(x) = 3m(1) + n = 3m + n
$$

$$
f(1) = 11
$$

$$
\lim_{x \to 1^+} f(x) = 5m(1) + 2n = 5m + 2n
$$

Set all equal to 11:

$$
3m + n = 11 \\
5m + 2n = 11
$$

Let’s solve the system:
Subtract the first from the second:

$$
(5m + 2n) - (3m + n) = 0 \implies 2m + n = 0 \implies n = -2m
$$

Substitute \$ n = -2m \$ into the first equation:

$$
3m + (-2m) = 11 \implies m = 11
$$

So \$ n = -2 \times 11 = -22 \$.

Sum: \$ m + n = 11 - 22 = \boxed{-11} \$ 🔢

## 20️⃣ Maximizing Profit for an LED Manufacturer

**Given:**

- Price per LED: \$ p(x) = 1000 - x \$ (when \$ x \leq 800 \$).
- Total cost to produce \$ x \$ LEDs:

$$
g(x) =
\begin{cases}
    30000 + 30x & x \leq 400 \\
    100x + 11000 & 400 < x \leq 800
\end{cases}
$$
- Maximum production possible due to a constraint: \$ x = 400 \$.

Find the number of LEDs the company should produce and sell to maximize profit.

### **Solution:**

Profit is maximized where the difference between revenue and cost is the greatest.

- Revenue: \$ R(x) = x \times (1000 - x) \$
- Cost: \$ C(x) = 30000 + 30x \$ (since \$ x \leq 400 \$)
- Profit: \$ P(x) = x(1000 - x) - (30000 + 30x) = 1000x - x^2 - 30000 - 30x = 970x - x^2 - 30000 \$

Take the derivative and find the critical point within \$ 0 < x \leq 400 \$:

$$
\frac{dP}{dx} = 970 - 2x
$$

Set to 0:

$$
970 - 2x = 0 \implies x = 485
$$

But \$ x \leq 400 \$ (due to production issue), so the maximum profit occurs at \$ x = 400 \$.

**Answer:** \$ \boxed{400} \$ LEDs should be produced and sold to maximize profit 💡💰

Let me know if you’d like even more step-by-step details or a graphical explanation for any part! 😁✨

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[^1]: Screenshot-2025-07-29-160848.jpg