Explanation:

The function f(x) = x – [x] represents the fractional part of x.

We need to find f(-1/2). First, we find the value of [-1/2].

The greatest integer less than or equal to -0.5 is -1. So, [-1/2] = -1.

Now, substitute this value into the function:

f(-1/2) = -1/2 – [-1/2] = -1/2 – (-1) = -1/2 + 1 = 1/2.

Explanation:

For the function to be defined, the expression inside the square root must be strictly positive (since it’s in the denominator).

So, |x| – x > 0, which means |x| > x.

We analyze this condition in two cases:

1. If x ≥ 0, then |x| = x. The inequality becomes x > x, which is false for all x.
2. If x < 0, then |x| = -x. The inequality becomes -x > x. Adding x to both sides gives 0 > 2x, or x < 0.

The condition |x| > x is only true when x is a negative number. Therefore, the domain is x < 0, or (-∞, 0).

Explanation:

The period of the standard sine function, sin(x), is 2π. For a function of the form f(x) = sin(bx), the period is given by the formula T = 2π / |b|.

In this case, the function is sin(x/3), so b = 1/3.

Therefore, the period T = 2π / (1/3) = 2π * 3 = 6π.

Explanation:

We are given f(x) = cos(log x). Let’s evaluate each term in the expression.

• f(x)f(y) = cos(log x)cos(log y)
• f(x/y) = cos(log(x/y)) = cos(log x – log y)
• f(xy) = cos(log(xy)) = cos(log x + log y)

Now substitute these into the main expression:

Expression = cos(log x)cos(log y) – 1/2[cos(log x – log y) + cos(log x + log y)]

We use the trigonometric identity: cos(A – B) + cos(A + B) = 2cosAcosB.

Let A = log x and B = log y. Then the expression in the bracket becomes:

[cos(log x – log y) + cos(log x + log y)] = 2cos(log x)cos(log y).

Substituting this back:

Expression = cos(log x)cos(log y) – 1/2[2cos(log x)cos(log y)]

Expression = cos(log x)cos(log y) – cos(log x)cos(log y) = 0.

Explanation:

We are given the function f(x) = 4^x / (4^x + 2).

First, let’s find the expression for f(1 – x):

f(1 – x) = 4^(1-x) / (4^(1-x) + 2) = (4/4^x) / ((4/4^x) + 2)

To simplify, multiply the numerator and denominator by 4^x:

f(1 – x) = (4) / (4 + 2 * 4^x) = 4 / (2(2 + 4^x)) = 2 / (4^x + 2)

Now, let’s add f(x) and f(1 – x):

f(x) + f(1 – x) = [4^x / (4^x + 2)] + [2 / (4^x + 2)]

Since they have a common denominator, we can add the numerators:

f(x) + f(1 – x) = (4^x + 2) / (4^x + 2) = 1.

Explanation:

To find f(f(x)), we substitute the expression for f(x) into the function itself.

f(f(x)) = f( (a – xⁿ)^1/n )

Now, replace ‘x’ in the original function definition with ‘(a – xⁿ)^1/n‘:

f(f(x)) = [a – { (a – xⁿ)^1/n }ⁿ ]^1/n

The powers 1/n and n cancel each other out:

f(f(x)) = [a – (a – xⁿ)]^1/n

Simplify the expression inside the brackets:

f(f(x)) = [a – a + xⁿ]^1/n = [xⁿ]^1/n

Again, the powers n and 1/n cancel out, leaving:

f(f(x)) = x.

Explanation:

We can simplify the expression f(x) = sin⁶x + cos⁶x.

1. Rewrite using a³ + b³ identity, where a = sin²x and b = cos²x: f(x) = (sin²x)³ + (cos²x)³
2. Using a³ + b³ = (a + b)³ – 3ab(a + b): f(x) = (sin²x + cos²x)³ – 3(sin²x)(cos²x)(sin²x + cos²x)
3. Since sin²x + cos²x = 1: f(x) = (1)³ – 3sin²x cos²x(1) = 1 – 3sin²x cos²x
4. Use the double angle identity sin(2x) = 2sinxcosx, so sin²x cos²x = (1/4)sin²(2x): f(x) = 1 – 3 * (1/4)sin²(2x) = 1 – (3/4)sin²(2x)
5. To find the minimum value of f(x), we must subtract the maximum possible value of (3/4)sin²(2x). The maximum value of sin²(2x) is 1. Minimum value of f(x) = 1 – (3/4) * 1 = 1/4.

Explanation:

1. Periodicity: The function has a period of 2. This means f(x + 2) = f(x) for all x. We can use this to relate f(4) to f(0): f(4) = f(2 + 2) = f(2) = f(0 + 2) = f(0). So, f(4) = f(0).
2. Odd Function Property: The function is odd. This means f(-x) = -f(x) for all x. Let’s apply this property for x = 0: f(-0) = -f(0) f(0) = -f(0) 2f(0) = 0 f(0) = 0.
3. Conclusion: Since f(4) = f(0) and f(0) = 0, it follows that f(4) = 0.

Explanation:

1. Use the identity cos²θ = (1 + cos(2θ))/2.
   f(x) = (1 + cos(2x))/2 + (1 + cos(2(π/3 + x)))/2 + (1 + cos(2(π/3 – x)))/2
2. Combine terms:
   f(x) = 3/2 + 1/2 [cos(2x) + cos(2π/3 + 2x) + cos(2π/3 – 2x)]
3. Use the identity cos(A + B) + cos(A – B) = 2cosAcosB. Let A = 2π/3 and B = 2x.
   cos(2π/3 + 2x) + cos(2π/3 – 2x) = 2cos(2π/3)cos(2x)
4. We know cos(2π/3) = -1/2.
   So, 2(-1/2)cos(2x) = -cos(2x).
5. Substitute this back into the expression for f(x):
   f(x) = 3/2 + 1/2 [cos(2x) – cos(2x)] = 3/2 + 1/2 [0] = 3/2.

The function is constant with a value of 3/2. Since this is not an option, there might be a typo in the question or options provided.

Explanation:

Let’s simplify each term separately.

1. First term: sin^-1(cos(sin^-1x))
   Let sin^-1x = θ, so x = sinθ.
   We need sin^-1(cosθ). We know cosθ = sin(π/2 – θ).
   So, sin^-1(cosθ) = sin^-1(sin(π/2 – θ)) = π/2 – θ = π/2 – sin^-1x.
2. Second term: cos^-1(sin(cos^-1x))
   Let cos^-1x = φ, so x = cosφ.
   We need cos^-1(sinφ). We know sinφ = cos(π/2 – φ).
   So, cos^-1(sinφ) = cos^-1(cos(π/2 – φ)) = π/2 – φ = π/2 – cos^-1x.
3. Adding the terms:
   Expression = (π/2 – sin^-1x) + (π/2 – cos^-1x)
   = π – (sin^-1x + cos^-1x)
4. Using the identity sin^-1x + cos^-1x = π/2:
   Expression = π – (π/2) = π/2.

Explanation:

To find the inverse function f^-1(x), we follow these steps:

1. Let y = f(x). So, y = (3x + 2) / (5x – 3).
2. Swap x and y to start finding the inverse: x = (3y + 2) / (5y – 3).
3. Solve for y:
   x(5y – 3) = 3y + 2
   5xy – 3x = 3y + 2
   5xy – 3y = 3x + 2
   y(5x – 3) = 3x + 2
   y = (3x + 2) / (5x – 3).
4. So, f^-1(x) = (3x + 2) / (5x – 3).

Comparing f^-1(x) with the original function f(x), we see that they are identical. Thus, f^-1(x) = f(x).

Explanation:

We must satisfy the conditions for both the numerator and the denominator.

1. Numerator: sin^-1(x – 3)
   The argument of arcsin must be in [-1, 1].
   -1 ≤ x – 3 ≤ 1
   2 ≤ x ≤ 4. So, the interval is [2, 4].
2. Denominator: √(9 – x²)
   The expression inside the square root must be strictly positive.
   9 – x² > 0
   x² < 9
   -3 < x < 3. So, the interval is (-3, 3).
3. Intersection: We need to find the intersection of [2, 4] and (-3, 3).
   The common values are those that are greater than or equal to 2 AND less than 3.
   Therefore, the domain is [2, 3).

Explanation:

This is a system of equations problem in disguise.

1. Let x = 7:
   3f(7) + 2f((7 + 59)/(7 – 1)) = 10(7) + 30
   3f(7) + 2f(66/6) = 70 + 30
   3f(7) + 2f(11) = 100 —(Eq. 1)
2. Let x = 11: Notice that if we substitute x=11, the term (x+59)/(x-1) becomes (11+59)/(11-1) = 70/10 = 7. This will give us a second equation with f(7) and f(11).
   3f(11) + 2f((11 + 59)/(11 – 1)) = 10(11) + 30
   3f(11) + 2f(7) = 110 + 30
   2f(7) + 3f(11) = 140 —(Eq. 2)
3. Solve the system:
   Multiply Eq. 1 by 3: 9f(7) + 6f(11) = 300
   Multiply Eq. 2 by 2: 4f(7) + 6f(11) = 280
   Subtract the new second equation from the new first one:
   (9f(7) – 4f(7)) = 300 – 280
   5f(7) = 20
   f(7) = 4.