Trigonometric Functions

Given equation:

We know that:

at in the interval .

Setting equal to these values:

Solving for :

Thus, the correct answer is:

Given expression:

Using the identity:

We rewrite:

Since , we simplify to:

Using the identity:

Substituting :

Since , we get:

Given:

Using the identity:

, we assume and for some .

Using the Pythagorean identity:

Substituting the values:

(taking the positive value as )

Thus,

Therefore, the correct answer is:

Given equation:

We use the identity:

Rewriting the equation:

Taking common denominator:

Rearranging:

Squaring both sides:

Expanding:

Using the identity :

Expanding:

Rearranging:

Solving the quadratic equation using the quadratic formula:

Solving for :

(not valid as must be in range )

Thus, .

Finding using :

(as is in the first quadrant, is positive)

Finding :

Thus, the correct answer is:

Given equation:

Using the identity:

, we rewrite the equation as:

For , the general solution is:

or for some integer .

Applying this to our equation:

Solving for :

Since lies in the given range , the valid solution is:

We are given the equation: and the condition that .
Step 1: Use the trigonometric identity We can use the following trigonometric identity to simplify the expression for : Since , we know that . Therefore: Thus, the expression becomes:
Step 2: Maximize the expression To maximize the value of , we need to maximize . The maximum value of is 1, which occurs when . Thus, when , the expression becomes: Therefore, the maximum value of is 5.

The correct option is (C) :

We are given the function defined by .

Step 1: Check if is even.

A function is even if for all . For , we have:

Therefore, is an even function.

Step 2: Check if is one-one (injective).

A function is one-one if for every distinct pair and , implies . For , this is not true because is periodic and takes the same value for different values (for example, ). Hence, is not one-one.

Step 3: Check if is onto (surjective).

A function is onto if for every element in the codomain, there is a corresponding element in the domain. The range of is limited to the interval , so it does not cover all real numbers. Therefore, is not onto.

Thus, the function is even but not onto, which matches option (E).

The correct option is (E) : f is even but not onto

We are given the expression . The range of the inverse cosine function, , is . However, is outside this range. To find the value, we first calculate the cosine of . We know that: Now, we want to find the angle in the range whose cosine is . This angle is . Thus, the value of is .

The correct option is (D) :

We are given: We need to find . We will use the half-angle identity for sine: Since , we are in the third quadrant, where sine is negative. Thus, we will take the negative sign in the half-angle formula: Step 1: Substitute Simplifying the expression inside the square root: Step 2: Simplify

The correct option is (A) :

The period of the cotangent function is . For a function of the form , the period is given by: In this case, the argument of the cotangent is , where . Thus, the period of is:

The correct option is (B) :

The function given is . We know that the sine function has a range of .
Therefore, the expression will have a range of because multiplying by 2 scales the range.
Now, adding 1 to this range shifts the entire range upward by 1 unit. Thus, the range of is:

The correct option is (E) :

Given:

• The expression for any real number

Step 1: Express the trigonometric combination in amplitude-phase form: where is an angle satisfying and , and the amplitude .

Step 2: Determine the range of the expression: Adding 5:

Step 3: Identify the least value: The minimum value occurs when :

The least value of the expression is (C) 0.

Given:

• Equation:

Step 1: Minimum LHS value:

Step 2: RHS range:

Step 3: Find critical point:

Solution:

Given:

• Equation:

Step 1: Infinite geometric series:

Step 2: Simplify equation:

Step 3: Solve for :

Solution:

Given:

Step 1: Find from :

Step 2: Determine using :

Step 3: Calculate :

Step 4: Compute :

The correct answer is (E) 0.

Given:

• The function for any real

Step 1: Express in terms of :

Step 2: Let (note ):

Step 3: Analyze the quadratic function :

• The vertex occurs at
• Minimum value:
• Maximum value: (at the endpoints)

Step 4: Determine the range of :

The correct option is (E) .

Step 1: Understanding the range of the cosine function The standard cosine function, , has a range of: Step 2: Scaling by the coefficient
Given that the function is , multiplying the cosine function by 7 scales the range by 7: Step 3: Determining the correct range From the above calculation, the range of is , which matches option (D).