Limits And Derivatives

To solve the given problem, we first need to analyze the provided determinant condition and the properties of the function . Given the determinant:

This determinant is zero if the following equality holds:

Rearranging, we get:

This implies that we can write:

Separating the variables and integrating both sides, we obtain:

Thus, integrating gives:

Exponentiating both sides leads to:

where . This suggests that is an exponential function. Solving the differential equation , we get:

Using the conditions and , we determine and :

• From .
• From and .

Therefore, the function is:

Now, we calculate :

Since , . This value lies within the interval (6, 9\)

Thus, the value of lies in the interval (6, 9).

The correct answer is (B) : 3
Given :

and


=
Now,

The correct answer is 11



So f(x) = 2x³ – 7x² + ax +b = 0 has x = 1 as repeated root, therefore f(1) = 0 and f ′(1) = 0 gives
a + b + 5 and a = 8
So, a – b = 11

Given:


Now, we need to evaluate the limit:

Rewriting the limit expression:

Simplifying:

Simplifying further:

Therefore, we get:

So, the final answer is .

Let We can express in terms of and :

This gives the equations:

From equation (2), we have . Substitute into equation (1):

Now, rewrite as:

Substitute and :

To solve this problem, we begin by evaluating the given integral:

Notice that the integral involves a square root and exponential function, which suggests a possible substitution to simplify it. Assume the substitution:

.

Differentiating both sides with respect to gives:

This means . Rewrite the integral in terms of :

Back-substitute :

This integral is the standard inverse tangent function:

Now, let's use the limits of the integral. When , , so . When , . Therefore, the definite integral becomes:

Thus,

This indicates that satisfies this equation:

and .

We are told and are roots of a quadratic equation. Using the property of roots:

1. Sum of roots = .
2. Product of roots = .

This matches the quadratic:

Therefore, the correct equation is:

The first integral represents an odd function, so:

Now consider the second integral:

We can rewrite this as:

Using the symmetry properties and integrating by parts, we find:

Thus, the answer is Option (1):

The problem asks for the value of , where and are integers derived from the expression . The complex numbers and are the points in the region where the distance is maximum and minimum, respectively. (Note: There appears to be a typo in the original question image. The expression should be , not , as is a variable. We will proceed with the corrected expression.)

Concept Used:

The solution involves the geometric interpretation of complex numbers in the Argand plane.

1. Set P: represents a closed disk with center and radius .

2. Set Q: . By substituting , this inequality simplifies to define a half-plane bounded by a straight line.

3. Intersection : This is the geometric intersection of the disk and the half-plane, which forms a segment of a circle.

4. Extremum of : This represents the maximum or minimum distance from a point in the region to a fixed point . These extrema will occur at the boundary points of the region .

Step-by-Step Solution:

Step 1: Characterize the sets P and Q geometrically.

For set P: . This inequality describes a closed disk with center (coordinates ) and radius .

For set Q: Let . The inequality is .

This represents a half-plane on and above the line .

Step 2: Analyze the intersection region .

We need to find the points in the disk that also satisfy . Let's check the distance from the disk's center to the line .

Since the distance , the line intersects the circle. The region is a segment of the disk.

Step 3: Identify the points and for maximum and minimum distance from .

We need to find the points in that are farthest from and closest to the point . Let the point corresponding to be .

The point (maximum distance) will be the point on the circle's boundary that is furthest from A, provided it lies in the region Q. This point lies on the line passing through A and . The vector from A to is . The unit vector is .

Let's check if is in Q. For , we check if .

This is true, so is in the region .

The point (minimum distance) will be the point in closest to A. The point on the circle closest to A is , but this point does not satisfy . Therefore, the minimum distance must be to a point on the line segment boundary of the region, which is the chord . The closest point on this line to A is the perpendicular projection of A onto the line. Let this point be M.

The line is . The projection M is . Let's find its coordinates. The projection of on is . . . This point lies on the chord, so .

Final Computation & Result:

Step 4: Calculate and .

And for :

Step 5: Compute the final expression.

Comparing this with , we have and .

The value of is 36.

The problem asks to find the value of the expression , where L and R are the left-hand limit and right-hand limit, respectively, of the function at . Here, denotes the fractional part of .

Concept Used:

1. Fractional Part Function ({x}): The fractional part of is defined as , where is the greatest integer less than or equal to . The behavior of around an integer is key:

• For the right-hand limit as (e.g., x = 0.001), , so .
• For the left-hand limit as (e.g., x = -0.001), , so .

2. Standard Limits in Calculus: We will use the following standard limits:

This second limit can be derived using L'Hôpital's rule or by substituting .

Step-by-Step Solution:

Step 1: Calculate the Right-Hand Limit (R) at .

As , we substitute .

We can factor the denominator and separate the limit into two parts:

For the second part, as , and . So, the second limit is .

For the first part, let . As , and . The limit becomes:

This is a standard limit which evaluates to .

Combining the results:

Step 2: Calculate the Left-Hand Limit (L) at .

As , we substitute . To make the limit easier to evaluate, let , where . Then .

Simplify the terms in the numerator and denominator:

The expression for L becomes:

Separate the limit into parts:

Evaluate each part:

• (standard limit).
• For the second part, substitute : .

Combining the results:

Final Computation & Result:

Step 3: Substitute the values of L and R into the given expression.

We found and .

First, calculate and :

Now, calculate their sum:

Finally, compute the required value:

The value of the expression is 18.