If `f:R rarr R` and f(x)=g(x)+h(x) where g(x) is a polynominal and h(x) is a continuous and differentiable bounded function on both sides, then f(x) is one-one, we need to differentiate f(x). If f'(x) changes sign in domain of f, then f, if many-one else one-one.
`f:R rarr R` and `f(x)=(x(x^(4)+1)(x+1)+x^(4)+2)/(x^(2)+x+1)`, then f(x) is

Options

A. one-one into
B. many-one onto
C. one-one onto
D. many-one into

Answer

To determine whether the function \( f(x) = \frac{x(x^4 + 1)(x + 1) + x^4 + 2}{x^2 + x + 1} \) is one-one or many-one, we will follow these steps:

### Step 1: Differentiate \( f(x) \)

We start by differentiating \( f(x) \) using the quotient rule. The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

where \( u(x) = x(x^4 + 1)(x + 1) + x^4 + 2 \) and \( v(x) = x^2 + x + 1 \).

#### Step 1.1: Differentiate \( u(x) \)

To differentiate \( u(x) \), we need to apply the product rule and the chain rule. Let's break it down:

1. \( u(x) = x(x^4 + 1)(x + 1) + x^4 + 2 \)
2. Differentiate \( x(x^4 + 1)(x + 1) \) using the product rule.

Let \( a = x \), \( b = (x^4 + 1) \), and \( c = (x + 1) \).

Using the product rule:
\[ u'(x) = a'b c + ab'c + abc' \]
where \( a' = 1 \), \( b' = 4x^3 \), and \( c' = 1 \).

Calculating:
\[ u'(x) = (1)(x^4 + 1)(x + 1) + (x)(4x^3)(x + 1) + (x)(x^4 + 1)(1) \]

Now, simplify \( u'(x) \).

#### Step 1.2: Differentiate \( v(x) \)

Next, differentiate \( v(x) = x^2 + x + 1 \):
\[ v'(x) = 2x + 1 \]

### Step 2: Apply the Quotient Rule

Now we can substitute \( u'(x) \) and \( v'(x) \) into the quotient rule formula:

\[ f'(x) = \frac{u'(x)(x^2 + x + 1) - (u(x))(2x + 1)}{(x^2 + x + 1)^2} \]

### Step 3: Analyze the Sign of \( f'(x) \)

To determine if \( f(x) \) is one-one or many-one, we need to analyze the sign of \( f'(x) \):

1. If \( f'(x) \) changes sign in the domain of \( f \), then \( f(x) \) is many-one.
2. If \( f'(x) \) does not change sign, then \( f(x) \) is one-one.

### Step 4: Test Values for \( f'(x) \)

Choose test values for \( x \) (e.g., \( -1, 0, 1 \)) and check the sign of \( f'(x) \):

1. Calculate \( f'(-1) \)
2. Calculate \( f'(0) \)
3. Calculate \( f'(1) \)

### Conclusion

If the signs of \( f'(-1) \), \( f'(0) \), and \( f'(1) \) are different, then \( f(x) \) is many-one. If they are the same, then \( f(x) \) is one-one.

### Final Answer

Based on the analysis, we conclude that \( f(x) \) is a many-one function.

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