Q no 1 )Range of the function f(x) = {x}+1+{x} where { } fractional part functionQ no 2) if 2 f(x²)+3 f(1+x²)= x²-1 ( x is not equal to 0) then f(x²) isQ no 3) the function f:[2,infinity) y defined by f(x) = x²+4x+5 is both one one and onto if

Let's dive into each of your questions one by one, starting with the first one regarding the function f(x) = {x} + 1 + {x}, where { } denotes the fractional part function.

Finding the Range of f(x) = {x} + 1 + {x}

The fractional part function, denoted as {x}, is defined as {x} = x - ⌊x⌋, where ⌊x⌋ is the greatest integer less than or equal to x. This means that {x} always yields a value in the interval [0, 1). Now, let's break down the function f(x).

Breaking Down the Function

We can rewrite f(x) as follows:

• f(x) = {x} + 1 + {x} = 2{x} + 1

Since {x} ranges from 0 to just below 1, we can substitute this into our function:

Calculating the Range

When {x} = 0, f(x) becomes:

• f(x) = 2(0) + 1 = 1

When {x} approaches 1 (but never actually reaches it), f(x) approaches:

• f(x) = 2(1) + 1 = 3

Thus, the function f(x) takes values in the interval [1, 3) as {x} varies from 0 to just below 1. Therefore, the range of f(x) is:

Solving the Equation 2f(x²) + 3f(1 + x²) = x² - 1

Now, let's tackle your second question regarding the equation 2f(x²) + 3f(1 + x²) = x² - 1, where x is not equal to 0. We need to find f(x²).

Analyzing the Equation

To find f(x²), we can assume a form for f(x) and see if it fits the equation. A common approach is to assume that f(x) is a linear function, say f(x) = ax + b. However, without loss of generality, let's analyze the structure of the equation.

Substituting Values

Let's substitute x² into the function:

• Let y = x², then the equation becomes 2f(y) + 3f(1 + y) = y - 1.

Now, we can express f(y) and f(1 + y) in terms of y. If we assume f(y) = ky + c, we can substitute this into our equation:

Setting Up the System

Substituting gives:

• 2(ky + c) + 3(k(1 + y) + c) = y - 1

Expanding this leads to:

• 2ky + 2c + 3ky + 3k + 3c = y - 1

Combining like terms results in:

• (5k)y + (5c + 3k) = y - 1

For this to hold for all y, we equate coefficients:

• 5k = 1, which gives k = 1/5
• 5c + 3(1/5) = -1, solving gives c = -2/5

Thus, we find:

Conditions for One-to-One and Onto Functions

Lastly, let's discuss the function f: [2, ∞) → ℝ defined by f(x) = x² + 4x + 5. We want to determine the conditions under which this function is both one-to-one and onto.

Understanding One-to-One

A function is one-to-one if it never assigns the same value to two different domain elements. For a quadratic function like f(x), we can check its derivative:

• f'(x) = 2x + 4

Since f'(x) is always positive for x ≥ 2, f(x) is strictly increasing in this interval, confirming that it is one-to-one.

Exploring Onto

A function is onto if every possible output in the codomain is covered. The minimum value of f(x) occurs at x = -2, but since our domain starts at x = 2, we calculate:

• f(2) = 2² + 4(2) + 5 = 4 + 8 + 5 = 17

As x approaches infinity, f(x) also approaches infinity. Therefore, the range of f(x) is [17, ∞), which means it is onto if we restrict the codomain to [17, ∞).