Inverse Trigonometric Functions

Given that:

x >0

For x>1
=

⇒ f(x) =

⇒ f(x) =

The correct answer is (B) :


.
Now



Let

The correct answer is (C) :

The correct asnwer is (B) : 11–π


So f(x) is decreasing function and range of f(x) is
[f(1), f(-1)], which is [π + 5, 5π + 9]
Now 4a – b = 4(π + 5) – (5π + 9)
= 11 – π

To solve the given problem, we first address the equation involving x:
. Let the roots be and . By Vieta's formulas, we know:

Next, we proceed with solving the trigonometric and inverse trigonometric functions:

1. We have .
   Denoting , it's clear , so the calculations stay within bounds.
   Given that the principal values for inverse functions are considered, simplify: as it's multiplied by equivalent (in this configuration).
2. It's specified as a solution:
   
   • Using \)

, implying , and since
Therefore, .

Final Calculation:

• , thus yielding \) given . Simplifying provides , a valid integer and falls within the specified 12,12 range.

confirming the condition.

The correct answer is (C) : 3

= 2
So, the correct is (B): 2

Understanding the Problem

We need to find the domain of the function , determine α and β, and then calculate

Solution

1. Domain of :

The logarithm is defined when

Factoring the quadratic:

The solution is

2. Domain of :

The arcsine is defined when

Thus,

3. Domain of :

The arccosine is defined when

Thus,

4. Combining the Domains:

We need the intersection of the three domains.

From and , we get

From and , we get

5. Finding α and β:

and

6. Calculating α + β:

7. Calculating 36|α + β|:

Final Answer

The correct answer is 43.2.

Step 1: Write the given function.
Step 2: Analyze the expression inside the inverse sine function. This means that the expression inside is valid as the sine function only takes values between 0 and 1.
Step 3: Apply the inverse sine function.
Step 4: Multiply by 4.
Step 5: Conclude the range. Thus, the range of is .

To solve the equation , we will use the identity for the sum of two inverse tangents:

Given: and , thus the condition must hold true.

Using the identity, substitute the values of and :

Given that:

This implies that:

Solving the equation :

Multiply both sides by :

Rearrange it as a quadratic equation:

To find the roots, use the quadratic formula , where , , and :

Since we are looking for positive real values, consider:

Checking whether :

Calculate for :

After simplification, it can be verified that . Thus, the condition holds.

Therefore, there is exactly 1 positive real value satisfying the given equation, which makes the answer 1.

Given function:

For to be defined, the following conditions must hold:

and

⇒

⇒

Now, solving :

Critical points:

Sign analysis gives:

Similarly, for :

Critical points:

Sign analysis gives:

Taking the intersection of the two intervals, the domain is:

Thus, the range is:

Now, let and

Then,

Hence, the final answer is:

Domain of the Function:
For to be defined, the argument must satisfy:

Solving this inequality involves two cases:

Case 1:

Case 2:

Therefore, the solution for the function domain is:

Domain of the Function:
For to be defined, the argument must be positive:

Factorize both the numerator and denominator:
Determine the intervals where this inequality holds by testing values between the critical points .

The valid intervals are:

Intersection of the Domains:
The domain of the combined function is the intersection of the two domains:
This simplifies to:

Calculate :
Here, and .

Then:

To solve the equation , we start by noting a key identity: . Let's assign variables for simplicity:

• Let and .

According to the identity, .

The given equation is:

Substitute from the identity:

Expanding this, we get:

Rearranging gives:

Simplify to find:

Thus:

Now, since , it must be true that . However, does not fit this range. Therefore, there are no possible values of satisfying the original equation within the range for inverse sine.

Given this, the number of real solutions is:

0

Step 1: Simplify the expression. Let , where . Then, . Substitute these into the given expression:
Step 2: Find the minimum value. To find the minimum value, take the derivative with respect to and set it to zero: Substitute to find the minimum value:
Step 3: Find the maximum value. The maximum value occurs at the endpoints of the interval for , which are and . At : At : The maximum value is .
Step 4: Find the sum of the maximum and minimum values.

Given: The function is

Domain of the function is: , where we need to find the value of .

Step 1: Domain of

The domain of the function is . In the given function, we have:

For this expression to lie in the domain of , we must have:

We will solve both inequalities: First Inequality:

Second Inequality:

Thus, the values of must satisfy:

Step 2: Domain of

The domain of the function is . For the given function, we have:

To find the domain, we require:

Solving these two inequalities: First Inequality:

Solving the quadratic inequality: Since the discriminant is negative, this inequality holds for all real values of . Second Inequality: Solving this inequality:

Step 3: Intersection of Domains

To find the domain of the entire function, we need the intersection of the two domains: The intersection is: So, we have and .

Step 4: Calculate

Now, we compute :

Concept: For inverse tangent functions, the identity is valid when , with the principal value lying in .
Step 1: Apply the identity. Let Then, Given:
Step 2: Take tangent on both sides.
Step 3: Solve for . Solving:
Step 4: Check interval restriction. Given interval: On numerical evaluation:
One root lies within the given interval
The other root lies outside the given interval Hence, only one value of satisfies both the equation and the interval condition.