Calculus

Step 1: Understanding the Concept:
To find and classify the extrema (maxima, minima) and saddle points of a function of two variables, we use the second partial derivative test. This involves finding critical points where the first partial derivatives are zero, and then using the second partial derivatives to classify them.

Step 2: Key Formula or Approach:
1. Find Critical Points: Solve the system of equations and .
2. Second Derivative Test: Calculate the discriminant (or Hessian determinant) .
- If and , it's a local minimum.
- If and , it's a local maximum.
- If , it's a saddle point.
- If , the test is inconclusive.

Step 3: Detailed Explanation:
The given function is .
Part 1: Find Critical Points
First, find the partial derivatives with respect to and : Set them to zero to find the critical points: The critical points are all possible combinations of these and values: (1, 2), (1, -2), (-1, 2), and (-1, -2).
Part 2: Second Derivative Test
Next, find the second partial derivatives: Calculate the discriminant : Part 3: Classify Critical Points
Now, test each critical point:
At (1, 2):
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Since and , this is a local minimum. Statement A is correct.

At (-1, -2):
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Since and , this is a local maximum. Statement B is correct.

At (1, -2):
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Since , this is a saddle point.

At (-1, 2):
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Since , this is a saddle point.
Part 4: Evaluate Statements C and D

Statement D: "the saddle points are (-1,2) and (1,-2)". This matches our findings. Statement D is correct.

Statement C: "neither a maxima nor a minima at (1,-2) and (-1,2)". A saddle point is by definition a point that is not a local extremum (not a max or min). Since (1,-2) and (-1,2) are saddle points, this statement is also true. Statement C is correct.

Step 4: Final Answer:
All four statements A, B, C, and D are correct descriptions based on our calculations. Therefore, the correct option is (C).

Step 1: Understanding the Concept:
The given limit presents an indeterminate form . The standard method for resolving this form when it involves square roots is to multiply by the conjugate expression.

Step 2: Key Formula or Approach:
We use the algebraic identity . We multiply the expression by .

Step 3: Detailed Explanation:
Now the limit is in the indeterminate form . To evaluate it, we divide the numerator and the denominator by the highest power of , which is . As , the term .
Step 4: Final Answer:
The value of the limit is .

Step 1: Understanding the Concept:
We need to find the surface area of a solid of revolution. The curve is given in parametric form, and it is revolved around the y-axis. The formula for surface area in this case involves integrating multiplied by the arc length element .
Step 2: Key Formula or Approach:
The surface area generated by revolving a parametric curve from to about the y-axis is: where the arc length element .
Step 3: Detailed Explanation:
1. Find Derivatives:


2. Find the Arc Length Element :
3. Set up and Evaluate the Integral:
The integral for the surface area is: We use the standard integral formula: . Here, a=2 and b=1. Now evaluate the definite integral: 4. Calculate the Final Surface Area:
Step 4: Final Answer:
The surface area is sq. unit.

Step 1: Understanding the Concept:
We need to evaluate four limits, most of which involve indeterminate forms. Standard limits and L'Hôpital’s Rule will help resolve them.
Step 2: Detailed Explanation:
(A) :
This is a form. Alternatively, using L'Hôpital’s Rule: Hence, **(A) matches (I)**.

(B) :
This is an form. Let . So, **(B) matches (III)**.

(C) :
This is an form. Apply L'Hôpital’s Rule twice: Thus, **(C) matches (II)**.

(D) :
This is an form. Let the limit be . This is a form. By L'Hôpital’s Rule: Hence, **(D) matches (IV)**.

Step 3: Final Answer:
The correct pairings are: - (A) → (I) - (B) → (III) - (C) → (II) - (D) → (IV) This corresponds to **Option (B)**.

Step 1: Understanding the Concept:
The given inequality is a form of a Lipschitz condition. It places a strong constraint on how fast the function can change. We can use this condition to find the derivative of the function.

Step 2: Key Formula or Approach:
We use the definition of the derivative: The given condition is . Let's apply this to the definition of the derivative.

Step 3: Detailed Explanation:
Let . Then . The given condition becomes . Now let's look at the expression for the derivative's magnitude. For : Using the given inequality: So, we have . Now, take the limit as to find the derivative: Using the squeeze theorem with the inequality we derived: This implies that , which means for all . If the derivative of a function is zero everywhere, the function must be a constant function. This means that options A, C, and D are incorrect.

Step 4: Final Answer:
The function must be a constant function.

Step 1: Understanding the Concept:
This problem involves finding the limit of a product of terms. This type of product is often a "telescoping product," where intermediate terms cancel out, simplifying the expression significantly.

Step 2: Key Formula or Approach:
The key is to rewrite the general term of the product, , using the difference of squares formula: .

Step 3: Detailed Explanation:
Let be the partial product: Using the factorization from Step 2: Let's expand this product to see the telescoping pattern: We can rearrange the terms to group the canceling parts together: In the first parenthesis, terms cancel out, leaving .
In the second parenthesis, terms cancel out, leaving .
So, the partial product simplifies to: Now, we need to find the limit as : Divide the numerator and denominator by :
Step 4: Final Answer:
The value of the limit is .

Step 1: Understanding the Concept:
This problem requires finding the derivative of an integral whose limits are functions of . This is a direct application of the Leibniz integral rule, which is a generalization of the Fundamental Theorem of Calculus.

Step 2: Key Formula or Approach:
The Leibniz integral rule states that for a function , its derivative is given by: In this problem, we have: - is the integrand. - The upper limit is . - The lower limit is .

Step 3: Detailed Explanation:
First, we find the derivatives of the limits of integration: Now, we apply the Leibniz rule formula: Substitute the given functions and their derivatives:
Step 4: Final Answer:
The derivative is .

Step 1: Understanding the Concept:
To determine the nature of a critical point for a function of two variables, we typically use the second partial derivative test. If this test is inconclusive (Discriminant D=0), we must analyze the function's behavior in the neighborhood of the critical point directly.

Step 2: Find Critical Points:
First, find the partial derivatives and set them to zero. From the second equation, either or . Case 1: If , the first equation gives . So is a critical point. Case 2: If , substitute into the first equation: . This again gives . The only critical point is the origin .

Step 3: Second Derivative Test:
Now, find the second partial derivatives: Evaluate at the critical point : Calculate the discriminant : Since , the second derivative test is inconclusive.

Step 4: Analyze the Function Directly:
We must investigate the behavior of near . The value at the origin is . Let's try to rewrite the function by completing the square: The first term, , is a square, so it is always greater than or equal to 0. The second term, , is also always greater than or equal to 0. The sum of two non-negative terms must be non-negative. Therefore, for all . Since , the function has a global (and local) minimum at the origin.

Step 1: Understanding the Concept:
This is a definite integral involving trigonometric functions. A standard technique for integrals of the form where is to use the property .

Step 2: Key Formula or Approach:
Let the integral be . We use the property with : Since and , we have . The second term is just .
Step 3: Detailed Explanation:
From the equation above, we get: To evaluate this integral, let . Then . The limits of integration change as well: When , . When , .
Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
This question tests the concepts of limits and continuity for a function of two variables. For the limit to exist at a point, the function must approach the same value along all possible paths to that point. If the limit does not exist, the function cannot be continuous or differentiable at that point.

Step 2: Key Formula or Approach:
To check for the existence of the limit at , we can test the value of the function as we approach the origin along different paths. A common technique is to use the path or to switch to polar coordinates. Let's use polar coordinates: . As , we have .

Step 3: Detailed Explanation:
Substitute the polar coordinates into the function for : Now, let's evaluate the limit as , which corresponds to : The result of the limit depends on , which represents the angle of approach to the origin. For example: - If we approach along the positive x-axis ( ), the limit is . - If we approach along the positive y-axis ( ), the limit is . Since the limit has different values for different paths of approach, the overall limit does not exist.

Step 4: Analyzing the Consequences:
- Since the limit does not exist at , the function is not continuous at . This rules out options B and C. - A removable discontinuity occurs when the limit exists but is not equal to the function's value. Since the limit does not exist, the discontinuity is non-removable. This rules out option D. - Therefore, statement A is the only correct one.

Final Answer: does not exist.

Concept: Definition of
The number is defined as:
Step 1: Recognize the standard limit
The given expression is a direct standard limit used to define .
Step 2: Intuition behind the limit
This expression arises in:

• Compound interest problems
• Continuous growth models

As increases, the quantity: approaches a fixed number.
Step 3: Numerical idea

Step 4: Final value
Conclusion:

Concept: Radius of convergence using Ratio Test
For a power series: the radius of convergence is found using:
Step 1: Identify coefficients

Step 2: Apply Ratio Test

Step 3: Take the limit

Step 4: Interpretation
Since the limit is 0 for all , the series converges for all real . Conclusion:

Concept: Bolzano–Weierstrass Theorem
This is a fundamental theorem in real analysis which states:

• Every bounded sequence in has at least one convergent subsequence.

Step 1: Understand bounded sequence
A sequence is bounded if:
Step 2: Apply the theorem
If the sequence is bounded, then it cannot “escape to infinity,” so there exists a subsequence that converges.
Step 3: Conclusion

Concept: Heine–Cantor Theorem
A very important result in real analysis states:

• If a function is continuous on a closed and bounded interval ,
• Then it is uniformly continuous on that interval.

Step 1: Understand continuity vs uniform continuity

• Continuity: For every point , we can find a depending on .
• Uniform continuity: A single works for all .

Step 2: Apply the theorem
Since the function is continuous on a closed interval , it satisfies:
Step 3: Why closed interval matters
Closed intervals are:

• Bounded
• Contain all limit points

This ensures no “escape” of function behavior at endpoints. Conclusion:

Concept: Laplace Equation
The Laplace equation is a second-order partial differential equation widely used in physics and engineering.
Step 1: Standard form in two dimensions

Step 2: Interpretation
It describes:

• Steady-state heat distribution
• Electrostatic potential
• Fluid flow

Step 3: Identify correct option
Option (B) matches the standard Laplace equation. Conclusion:

Concept: Cauchy sequence and completeness

• A sequence is Cauchy if its terms get arbitrarily close to each other.
• A sequence converges if it approaches a limit in the space.

Step 1: Key idea
In general metric spaces:
Step 2: Special case
If the space is complete, then:
Step 3: Example
In , a Cauchy sequence may converge to an irrational number, which is not in , so it does not converge in that space. Conclusion:

Concept: Gaussian Integral
The integral: is a standard result known as the Gaussian integral.
Step 1: Define the integral
Let:
Step 2: Square the integral

Step 3: Convert to polar coordinates

Step 4: Evaluate the integral

Step 5: Final result