Definite Integral

The correct answer is (C) :
f : R → R and f(x) + f(x + k) = n ∀ x ∈ R
x → x + k
f(x + k) + f(x + 2k) = n
∴ f(x + 2k) = f(x)
So, period of f(x) is 2k
Now,


x = t + k ⇒ dx = dt (in second integral)

Now,


Therefore ,

The correct answer is (D) : Only (S2) is correct




and



∴ Only (S2) is correct

The correct answer is (C) :

The correct answer is 6

Using

we get

Adding these two equations, we get

The correct answer is (D) : (6,8)



Let

f(x) = ax-b
Now,





(6,8) lies on f(x)

The correct answer is (C) :

The correct answer is 1






Therefore α = 1

The correct answer is 24






∴ 2n² + n + 1 = 1177
2n² + n – 1176 = 0

∴ n = 24

The correct answer is (A) :

(A)The given integral is: (B)Use the substitution , so . The limits of integration change as follows: Substituting, the integral becomes: (C)Simplify: (D)Add the original integral and its substitution: (E)Use the property for . Thus: (F) Simplify: The integral of is : (G) Evaluate: Simplify : Divide by 2:

Step 1: Split the integral

We split the integral into two parts based on the definition of :

Let:

Step 2: Solve

Use the substitution . The limits change as:

• When
• When

The integral becomes:

Evaluate the integral of :

Step 3: Solve

Since is constant with respect to , the integral becomes:

Evaluate the integral of :

Thus:

Step 4: Combine the results

Add and to find the total integral:

Combine terms:

Final Answer

We are given the integral equation:

Step 1: Simplify the First Integral

Split the range of integration for the first term as follows:

Step 2: Use Symmetry for the Second Term

Rewriting the integral equation, we get:

Step 3: Use the Property of Definite Integrals

Using the property of definite integrals:

we substitute into the integrals. This simplifies to:

Step 4: Simplify Using Trigonometric Identities

Using the addition formulas for sine:

the equation becomes:

Step 5: Factor Out Common Terms

Factoring out the common term , we get:

Since and are non-zero in the interval , the term inside the parentheses must be zero:

Step 6: Solve for

Rearranging, we find:

Final Answer:

.

To evaluate the limit , we can apply L'Hôpital's Rule and the Fundamental Theorem of Calculus. Let's go through the steps:

1. Firstly, let's denote the function:
2. We need to evaluate:
3. Notice that both the numerator and the denominator approach 0 as . Therefore, we can apply L'Hôpital's Rule:
4. Using L'Hôpital's Rule requires us to differentiate the numerator and the denominator with respect to .
5. The derivative of the numerator by the Fundamental Theorem of Calculus is:
6. The derivative of the denominator is:
7. Apply L'Hôpital's Rule:
8. Plug in into the differentiated functions: - Numerator: which simplifies since of any number is finite. - Denominator:
9. Notice that this process must be repeated because the limit form is still . We keep differentiating again.
   • Differentiate numerator again: (Chain rul\)
   • Differentiate denominator:
10. Apply L'Hôpital's Rule again:
11. This simplifies to:
12. The limit simplifies further to a couple of steps by realizing behaviors at , considering Taylor expansion or related approaches lead to the solution .

Therefore, the value of the limit is , which confirms the correct answer is indeed this option.

The problem asks for the value of the expression , where denotes the greatest integer less than or equal to .

Concept Used:

To evaluate a definite integral containing the greatest integer function (floor function), the standard method is to split the interval of integration into subintervals. The splitting is done based on the points where the value of the expression inside the floor function becomes an integer. On each of these subintervals, the greatest integer function will have a constant value, making the integral easy to compute.

We will analyze the function to find the points where it crosses integer values within the interval .

Step-by-Step Solution:

Step 1: Analyze the function inside the floor function.

Let . We first determine the range of this function over the interval of integration .

The term inside the square root is . This is a monotonically increasing function for . Consequently, is also a monotonically increasing function.

Let's find the values of at the endpoints of the interval:

• At : .
• At : .

As varies from 0 to 9, increases from 0 to 3. The possible integer values that can take are 0, 1, and 2.

Step 2: Find the points where takes integer values.

We need to find the values of for which and . These will be the points where we split our integral.

• Set :
• Set :

The value of becomes 3 at , the upper limit of integration.

Step 3: Split the integral into subintervals.

Based on the points found in Step 2, we can determine the constant value of the integrand on each subinterval:

• For , we have , so .
• For , we have , so .
• For , we have , so .

Let . We can split the integral as follows:

Step 4: Evaluate the integral .

Final Computation & Result:

The problem asks for the value of .

The value of the given expression is 155.

The problem is to evaluate the limit of a sum, which can be converted into a definite integral.

Concept Used:

The limit of a sum can be expressed as a definite integral using the formula for a Riemann sum:

To evaluate the resulting integral, we will use the method of partial fraction decomposition and the standard integration formula:

Step-by-Step Solution:

Step 1: Convert the given limit of a sum into a definite integral.

We need to manipulate the general term of the sum to fit the form .

Comparing this with the Riemann sum formula, we can identify and . The limit can be written as the definite integral:

Step 2: Use partial fraction decomposition for the integrand.

Let . We decompose the expression :

Multiplying by the denominator gives .

If we set , we get .

If we set , we get .

Substituting back , the integrand becomes:

Step 3: Evaluate the definite integral.

The integral becomes:

We evaluate each integral separately:

Step 4: Combine the results to find the value of L.

To simplify, we rationalize the denominator:

Final Computation & Result:

The value of the limit is . We now check which of the given options matches this result. Let's examine option (2):

Rationalizing the denominator of this option:

Since , we can simplify the expression:

This matches our calculated value for the limit. Thus, the correct option is the second one.

The value of the limit is .

We are given the function:

and a condition:

We need to find .

Let's first simplify the integrand:

The integrand is:

Rewrite the trigonometric functions:

Substitute these into the integrand:

Simplify the expression:

Since and it simplifies to:

Let's evaluate the limit where . Recognize that this indicates a boundary condition for the integral at :

Evaluating from to with the above steps and simplification leads to the result:

After correct substitutions and simplifications, this evaluates to .

Thus, the correct answer is:

Option:

To solve the integral , we can use the properties of logarithms and integration by parts. Let's break down the steps:

By using the hyperbolic identities:

Change the limits from to conditions:

Evaluate the integral:

Calculate the evaluated boundaries:

Compute the values using the property :

Thus:

Put the values back to finalize the computation:

Thus, the value of the integral is