Functions

The given function is .

We need to determine whether the function is one-one (injective), onto (surjective), even, or odd.

1. Checking if the function is one-one (injective):

A function is one-one if implies .

Assume , i.e., .

Simplifying: which gives .

Since , the function is one-one.

2. Checking if the function is onto (surjective):

A function is onto if for every element , there exists such that .

We solve for : , so , and .

Since this is valid for any , the function is onto.

3. Checking if the function is even:

A function is even if for all .

We check: , which is not equal to .

Thus, the function is not even.

4. Checking if the function is odd:

A function is odd if for all .

We check: and .

Since , the function is not odd.

The function is one-one and onto.

For the function , the domain requires the expression inside the square root, , to be non-negative. Thus, we need to solve the inequality: Factoring the quadratic expression: The solution to this inequality is

The correct option is (E) :

Step 1: Understand the problem and given function.

We are tasked with finding the domain of the real-valued function:

The domain of is the set of all for which the function is well-defined. This requires both terms in the function to be defined:

• : The expression inside the square root must be non-negative, i.e., .
• : The denominator must not be zero, and the expression inside the square root must be positive, i.e., .

Step 2: Analyze .

The inequality can be rewritten as:

Taking the square root on both sides:

This implies:

Step 3: Analyze .

The denominator must be positive, so we solve:

Factorize the quadratic:

The inequality becomes:

Using the sign chart method, the critical points are and . Testing intervals around these points:

• For , (positive).
• For , (negative).
• For , (positive).

Thus:

Step 4: Combine the conditions.

To find the domain of , we combine the two conditions:

• from .
• from .

The intersection of these two sets is:

Final Answer:

The domain of is:

.

We are given the function . To find its range:

Step 1: Analyze the function

The function is a quadratic function opening upwards with its vertex at .

Step 2: Determine the minimum value

The minimum value of occurs at :

Step 3: Determine the range

Since for all , . Thus, the range of is all real numbers such that .

The range of is , which corresponds to option (C).

Step 1: Analyze the function where is the greatest integer function and .

Step 2: Evaluate option (A) - Continuity at :
Compute
Left limit:
Right limit:
Since , is not continuous at . Thus, (A) is false.

Step 3: Evaluate option (B) - Continuity at :
Compute
Left limit:
Right limit:
Since , is not continuous at . Thus, (B) is true.

Step 4: Evaluate option (C) - Continuity at :
Compute
Left limit:
Right limit:
Since , is not continuous at . Thus, (C) is false.

Step 5: Evaluate option (D) - Differentiability at :
First check continuity at :
Compute
Left limit:
Right limit:
Since , is not continuous (and thus not differentiable) at . Thus, (E) is false.

Conclusion: The only correct statement is .

Given the functions:

We want to find . Let's break it down:

1. : Since , .
2. : .
3. when : .
4. :
5. Since , .
6. : .
7. : .

Therefore, .

The function is , and its domain excludes .

The inverse function is , and its domain excludes .

Compute : Simplify: Thus, for all .

The set contains points where , which are and . Hence: The cardinality of is 2.

We are given: and we need to find . First, compute : Now, we substitute into the expression for : Next, we compute : Since is a real value, the exact value of is .
However, simplifying further we observe that at , we have:
Thus, the value of is .