Functions

Given: We are analyzing the function to determine where it attains maximum and minimum values.

From the graph or the nature of the function, it is observed that:
- Maximum value occurs at: and
- Minimum value occurs at:

These are the critical points of the function — where the first derivative is zero or undefined, and the second derivative test or graph shape confirms the nature of the extrema.

Correct option(s):
(B): is maximum at two points and ✅
(C): is minimum at ✅

Step 1: Check for Continuity

We need to check the continuity at the point where the function definition changes, which is .

The left-hand limit at is:

The right-hand limit at is:

The function value at is:

Since , the function is discontinuous at .

Step 2: Determine Maximum and Minimum Values

Despite the discontinuity, let's analyze the function in the interval .

For ,

* At , (minimum value)

* At ,

For ,

* As approaches from the right, approaches .

* At , (this one isn't a candidate)

Let then We see there is a min at =-1 and max at 1

Conclusion:

is discontinuous in , but it still has a maximum value (1) and a minimum value (-1) in this interval.

Therefore, the correct option is: is discontinuous in but still has the maximum and minimum value.

Let

Step 1: Check if the function is differentiable at To find , we use the definition of derivative:

Step 2: Find To do that, we evaluate the limit: As :

• is bounded in and oscillates

Hence, So,

Therefore, define

Step 3: Apply derivative definition Substitute : Split the limit:

Step 4: Evaluate the first term Let , then as We know:

So, first term =

Step 5: Evaluate the second term Here,

• is bounded

So, the product tends to

Step 6: Final answer

The given equation represents an ellipse centered at the origin, with its major axis along the y-axis.

The standard form of the ellipse is:

where is the length of the semi-major axis and is the length of the semi-minor axis.

Now, consider the given equation:

To match the standard form, we divide both sides of the equation by 1 (which is already done), and compare:

So, we can see:
,

Now, we are told that is strictly decreasing and for all .
Since the major axis is along the y-axis, it must be that:

Because is strictly decreasing, the above inequality holds when:

Solving this:

This inequality holds when:

Conclusion: For the given equation to represent an ellipse with the major axis along the y-axis, the value of must lie in the intervals:

(A):
(C):

The correct answer is/are option(s):
(A): R^-1
(C): R ∩ R₁

Given, R and R₁ are equivalence relations on set A.

A relation is called an equivalence relation if it is:

• Reflexive: for all
• Symmetric: if , then
• Transitive: if and , then

Since R₁ is an equivalence relation, it satisfies:

• Reflexive: for all
• Symmetric: if , then
• Transitive: if , , then

Similarly, R is also an equivalence relation and hence satisfies the same three properties.

We now prove that is an equivalence relation:

• Reflexive:
  Since and for all ,
  .
  So, is reflexive.
• Symmetric:
  Let .
  Then and .
  Since both are symmetric,
  and .
  Thus, .
  So, is symmetric.
• Transitive:
  Let and .
  Then and .
  Since both R and R₁ are transitive,
  and .
  So, .
  Hence, is transitive.

Conclusion: Since is reflexive, symmetric, and transitive, it is an equivalence relation.

Also, if R is an equivalence relation, its inverse is also an equivalence relation.
So both (A) and (C) are correct.