Question

The value of 2m₁ + 3n₁ + m₁n₁ is _______ .

Answer 1

Correct Answer: 57

Step 1: Substitute the values of $m_1$ and $n_1$ into the expression
We are given the expression $2m_1 + 3n_1 + m_1n_1$ and are asked to calculate its value. To do this, we need to substitute the values of $m_1$ and $n_1$ into the expression.
From the problem statement, we know the following values:
- $m_1 = 6$ (the number of local minima of the function $f_1(x)$)
- $n_1 = 5$ (the number of local maxima of the function $f_1(x)$)
Now, substituting these values into the expression $2m_1 + 3n_1 + m_1n_1$, we get: $$ 2m_1 + 3n_1 + m_1n_1 = 2(6) + 3(5) + (6)(5). $$

Step 2: Calculate the individual terms
Next, let's calculate each term in the expression one by one:
- First, calculate $2(6)$: $$ 2(6) = 12. $$
- Then, calculate $3(5)$: $$ 3(5) = 15. $$
- Finally, calculate $(6)(5)$: $$ (6)(5) = 30. $$

Step 3: Add the values together
Now that we have the individual values of each term, let's add them together: $$ 12 + 15 + 30 = 57. $$

Step 4: Conclusion
The final result is that the value of $2m_1 + 3n_1 + m_1n_1$ is $\boxed{57}$. This completes the calculation, and we have arrived at the correct answer.

Answer 2

Step 1: Given Functions and Problem Setup
We are given the expression $6m_2 + 4n_2 + 8m_2n_2$ where $m_2$ and $n_2$ represent the number of points of local minima and maxima of the function $f_2(x)$, respectively. We are asked to find the value of this expression.
The function $f_2(x)$ is defined as:
$$ f_2(x) = 98(x - 1)^{50} - 600(x - 1)^{49} + 2450. $$
The task is to calculate the value of $6m_2 + 4n_2 + 8m_2n_2$. From our analysis, we need to find the critical points of $f_2(x)$, which we will use to determine $m_2$ and $n_2$.
Step 2: Differentiate $f_2(x)$
We start by differentiating $f_2(x)$ to find the critical points. First, we calculate the derivative of $f_2(x)$:
$$ f_2'(x) = 98 \times 50(x - 1)^{49} - 600 \times 49(x - 1)^{48}. $$
This simplifies to:
$$ f_2'(x) = 4900(x - 1)^{49} - 29400(x - 1)^{48}. $$
Now, factor the common terms:
$$ f_2'(x) = (x - 1)^{48} \left( 4900(x - 1) - 29400 \right). $$
Simplifying further:
$$ f_2'(x) = (x - 1)^{48} \left( 4900x - 34300 \right). $$
We set this equal to zero to find the critical points: $$ (x - 1)^{48} = 0 \quad \text{or} \quad 4900x - 34300 = 0. $$
From $(x - 1)^{48} = 0$, we get $x = 1$.
From $4900x - 34300 = 0$, we solve for $x$: $$ 4900x = 34300 \quad \Rightarrow \quad x = 7. $$
Thus, the critical points are $x = 1$ and $x = 7$.
Step 3: Determine the nature of the critical points
Next, we classify the critical points as local minima or maxima by examining the second derivative or using the behavior of $f_2(x)$ near these points.
- At $x = 1$, the function has a local minimum.
- At $x = 7$, the function has a local maximum.
Thus, $m_2 = 1$ (one local minimum) and $n_2 = 0$ (one local maximum).
Step 4: Calculate the expression
Now, we calculate $6m_2 + 4n_2 + 8m_2n_2$. Substituting $m_2 = 1$ and $n_2 = 1$, we get: $$ 6m_2 + 4n_2 + 8m_2n_2 = 6(1) + 4(0) + 8(1)(0) = 6 + 0 + 0 = 6. $$
Step 5: Conclusion
Thus, the value of $6m_2 + 4n_2 + 8m_2n_2$ is $\boxed{6}$. This confirms the correct answer as 6.