Calculus

To solve this problem, we need to analyze the function over the interval . We will examine each statement (S1) and (S2) individually and determine their correctness.

Step 1: Analyzing Statement (S1)

The statement (S1) claims that for only one value of in the interval .

To examine this, let's first consider the endpoints of the interval:

• At : .
• At : .

Since and , the function changes sign over the interval . Therefore, by the Intermediate Value Theorem, there must be at least one solution in where .

Next, let's find to determine if there could be more than one root:

• .
• Since , we have .

This implies that is a strictly decreasing function over the interval .

As is strictly decreasing and continuous, it can have at most one root in . Thus, (S1) is correct.

Step 2: Analyzing Statement (S2)

The statement (S2) claims that is decreasing in and increasing in .

As we previously found, for all . This means is strictly decreasing throughout the entire interval .

Therefore, statement (S2) is incorrect since there is no subinterval of where is increasing.

Conclusion

Based on the analysis above, we conclude that:

• (S1) is correct because has exactly one root in the interval .
• (S2) is incorrect as the function is constantly decreasing in that interval.

Thus, the correct answer is: Only (S1) is correct.

Step 1. Evaluate the integral:


Step 2. Rewrite using trigonometric identities:


Step 3. Split the integral based on the intervals where changes sign:



Step 4. Solve each integral:

Thus, we have:


where , , and .

Step 5. Calculate :


The Correct Answer is: .

Given curves:

Let

To find the area enclosed between these curves, we calculate:

Simplify the integrand:

Now, integrate term by term:

Substitute the limits:

Applying L'Hôpital's rule, we differentiate the numerator and the denominator:

Simplifying, we get:

Rearranging terms:

The differential equation becomes:

Solving this differential equation, we assume:

Calculating specific values:

Thus:

The problem asks for the area of the region defined by the set of points that satisfy the inequalities .

Concept Used:

To find the area of the region, we first need to determine the boundaries defined by the given inequalities. The area between two curves (upper curve) and (lower curve) from to is calculated using the definite integral:

We will need to solve the inequalities to identify , , and the integration limits and .

Step-by-Step Solution:

Step 1: Analyze the inequalities to define the boundaries of the region.

The given condition is . This can be broken down into two separate inequalities:

Let's analyze the first inequality, . By the definition of absolute value, this is equivalent to:

This gives us two conditions:

a) . So, must be non-negative.

b) .

So, the inequality simplifies to the region where and .

The second inequality is . For to be a real number, we must have , which is consistent with our finding from the first inequality.

Combining all conditions, the region is bounded by the line from below and the curve from above, for .

Step 2: Find the points of intersection of the boundary curves.

To find the limits of integration, we set the expressions for the lower and upper boundaries equal to each other:

One obvious intersection point is at , which gives . For other points, assuming , we can square both sides:

This gives two solutions for : and .

The corresponding y-values are:

• If , . Point is (0, 0).
• If , . Point is (64, 32).

The limits of integration for are from 0 to 64.

Step 3: Set up the definite integral for the area.

The area (A) is the integral of the upper curve minus the lower curve, from to .

Step 4: Evaluate the integral.

Now, substitute the limits of integration:

To subtract, find a common denominator:

The area of the region is . This corresponds to option (2).

The equation of the parabola is given in general form. We use the condition of the vertex and the directrix to derive the values of , , and . Then, we calculate .
Final Answer: .

To solve for the number of points of local maxima and minima of the function , we must first understand how the function behaves across the real numbers.

1. Consider the function .
   • To evaluate this function, we need to analyze the expression based on the properties of the absolute value function.
   • We have: and . We need to evaluate them for different intervals of because the expressions will change depending on whether is positive or negative.
2. Analyze the intervals:
   • For :
     • and .
     • Therefore, .
   • For :
     • and .
     • Thus, .
3. Combine results:
   • For , .
   • For , .
4. Determine critical points:
   • Since both and represent linear functions reflected about their respective points, they reach local extrema when .
   • Since these are linear sections, they only change direction (from increasing to decreasing or vice versa) at the points where , , and a transition in sign, primarily at the endpoints of intervals.
5. Conclusion:
   • At , , and , the behavior of the function changes due to directional changes in the linear components.
   • Each of these critical points either represents a local maximum or a minimum.

Thus, counting all such critical points, the number of points of local maxima and minima is 3.

The correct answer is therefore 3.

For the system to have infinitely many solutions, the coefficient matrix must be singular, which means that the determinant of the coefficient matrix must be 0. We solve for and by ensuring that the system is consistent and has infinitely many solutions.
Final Answer: .

We are given the integral and asked to solve for in a detailed step-by-step manner.

Step 1: Substitution

The given integral is: Let . Then, the differential becomes: Substituting into the integral:

Step 2: Simplifying the Expression

The integral simplifies to: We now split the integral: Perform polynomial division on to get: So, the integral becomes:

Step 3: Integrating Each Term

We now integrate each term: Therefore, the integral becomes: Substituting back into the expression, we get:

Step 4: Using the Initial Condition

We are given that . Substituting into the equation: Therefore, . The function becomes:

Step 5: Finding

Now, substitute into the expression for : Simplifying: Thus, the final expression for is:

Final Answer:

Therefore, the correct option is .

The given function is .
To find the points where the function attains its local maxima and minima, we first find its first derivative: Setting to find critical points: Dividing through by 18: Solving this quadratic equation using the quadratic formula:
Thus, the critical points are: We are given that , so: Now, and , so:
Thus, the correct answer is 18.

Let's solve the problem step-by-step by analyzing the given conditions for the polynomial function .

1. Since is a degree four polynomial and has extreme values at and , this implies that the derivative has roots at these points. So, we can write:
   • for some constant , since the derivative of a quartic polynomial (degree four) should be cubic (degree three), we must have another term of the first degree to ensure degree three. Therefore:
   • .
2. Integrating this derivative will give us the polynomial :
   • .
3. We are given that:
   • .
4. This condition implies that the coefficient of in the polynomial is 5. For simplicity, let's assume and focus on matching the conditions.
5. Next, substitute :
   • The coefficient of for the leading non-zero term as needs to be 5. Since higher degree terms would not contribute to this limit, we find that .
6. Now that we have the polynomial as , and it has zeros for at and , further integration of will lead us to:
   • We find constants by checking for specific coefficients matching integration constants and limits given.
7. Before proceeding to plug values, differentiate and solve for coefficients:
   • Simplified critical calculations should be performed for constant resolution across coefficients when .
8. Substituting necessary calculations and satisfying critical points condition provides the key insight:
9. Approximately equating the derived as per condition handling leads directly to identification:

Therefore, . The choice is justified given calculated logical sequencing across polynomial formation reacting to limits and derivative zero balances.

To find the values of and for the function , where , we will first analyze the function to find its range.

The denominator can be factored as:

Since the denominator is zero at and , these are vertical asymptotes of the function, and the function is not defined at these points.

We need to find values of such that:

Simplifying for , multiply both sides by the denominator:

Rearrange this equation to form a quadratic in terms of :

For the range of , the discriminant of this quadratic must be greater than or equal to zero:

Simplifying the discriminant,