`f(x) = _/int_{0}^x(t-1)(t-2)^(2)(t-4)^(5) dt (xgt0)` then numb er of points of extremum of `f(x)` is

Options

A. `4`
B. `3`
C. `2`
D. `1`

Answer

To solve the problem, we need to find the number of points of extremum of the function defined by the integral:

\[ f(x) = \int_{0}^{x} (t - 1)(t - 2)^2(t - 4)^5 \, dt \]

### Step 1: Find the derivative of \( f(x) \)

Using the Fundamental Theorem of Calculus (Newton's Leibniz formula), we can differentiate \( f(x) \):

\[ f'(x) = (x - 1)(x - 2)^2(x - 4)^5 \]

**Hint:** Remember that the derivative of an integral with respect to its upper limit is simply the integrand evaluated at that limit.

### Step 2: Set the derivative equal to zero

To find the points of extremum, we need to set \( f'(x) = 0 \):

\[ (x - 1)(x - 2)^2(x - 4)^5 = 0 \]

**Hint:** A product is zero if at least one of the factors is zero.

### Step 3: Solve for \( x \)

The equation \( (x - 1)(x - 2)^2(x - 4)^5 = 0 \) gives us the following roots:

1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( (x - 2)^2 = 0 \) → \( x = 2 \) (with multiplicity 2)
3. \( (x - 4)^5 = 0 \) → \( x = 4 \) (with multiplicity 5)

Thus, the critical points are \( x = 1, 2, 4 \).

**Hint:** Check the multiplicity of the roots as they affect the behavior of the function around those points.

### Step 4: Analyze the sign of \( f'(x) \)

To determine the nature of these critical points (whether they are maxima or minima), we analyze the sign of \( f'(x) \) around the critical points:

- For \( x < 1 \): \( f'(x) < 0 \) (negative)
- For \( 1 < x < 2 \): \( f'(x) > 0 \) (positive)
- For \( 2 < x < 4 \): \( f'(x) < 0 \) (negative)
- For \( x > 4 \): \( f'(x) > 0 \) (positive)

**Hint:** A change from negative to positive indicates a local minimum, while a change from positive to negative indicates a local maximum.

### Step 5: Identify the points of extremum

From the sign analysis:
- At \( x = 1 \): \( f'(x) \) changes from negative to positive → local minimum.
- At \( x = 2 \): \( f'(x) \) does not change sign (it remains zero) → not an extremum.
- At \( x = 4 \): \( f'(x) \) changes from negative to positive → local minimum.

Thus, the points of extremum are at \( x = 1 \) and \( x = 4 \).

### Conclusion

The total number of points of extremum of \( f(x) \) is **2** (at \( x = 1 \) and \( x = 4 \)).

**Final Answer:** 2

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