If `A` is domain of `f(x)=lntan^(-1)((x^3-6x^2+11 x-6)(x)(e^x-5)` and `B` is the range of `g(x)=sin^2x/4+cosx/4dot` Then find `AnnB`

Options

A. (0,2)
B. (0,1)
C. (1,2)
D. None of these

Answer

To solve the problem, we need to find the intersection of the domain of the function \( f(x) \) and the range of the function \( g(x) \).

### Step 1: Find the Domain of \( f(x) \)

The function is given as:

\[ f(x) = \ln\left(\tan^{-1}\left((x^3 - 6x^2 + 11x - 6)(x)(e^x - 5)\right)\right) \]

For \( f(x) \) to be defined, the argument of the logarithm must be positive:

\[ \tan^{-1}\left((x^3 - 6x^2 + 11x - 6)(x)(e^x - 5)\right) > 0 \]

This implies:

\[ (x^3 - 6x^2 + 11x - 6)(x)(e^x - 5) > 0 \]

### Step 2: Factor the Polynomial

First, we need to factor the polynomial \( x^3 - 6x^2 + 11x - 6 \).

By testing possible rational roots, we find that \( x = 1 \), \( x = 2 \), and \( x = 3 \) are roots. Thus, we can factor it as:

\[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) \]

### Step 3: Analyze the Sign of the Expression

Now, we analyze the sign of the expression:

\[ (x - 1)(x - 2)(x - 3)(x)(e^x - 5) > 0 \]

The critical points are \( x = 0, 1, 2, 3 \) and the point where \( e^x - 5 = 0 \) which gives \( x = \ln(5) \).

### Step 4: Test Intervals

We will test the sign of the expression in the intervals determined by these critical points:

1. \( (-\infty, 0) \)
2. \( (0, 1) \)
3. \( (1, 2) \)
4. \( (2, 3) \)
5. \( (3, \ln(5)) \)
6. \( (\ln(5), \infty) \)

By testing points in each interval, we find:

- For \( x < 0 \): Negative
- For \( 0 < x < 1 \): Positive
- For \( 1 < x < 2 \): Negative
- For \( 2 < x < 3 \): Positive
- For \( 3 < x < \ln(5) \): Negative
- For \( x > \ln(5) \): Positive

### Step 5: Determine the Domain

From the sign analysis, the intervals where the expression is positive are:

\[ (0, 1) \cup (2, 3) \cup (\ln(5), \infty) \]

Thus, the domain \( A \) of \( f(x) \) is:

\[ A = (0, 1) \cup (2, 3) \cup (\ln(5), \infty) \]

### Step 6: Find the Range of \( g(x) \)

The function is given as:

\[ g(x) = \frac{\sin^2 x}{4} + \frac{\cos x}{4} \]

To find the range, we rewrite it as:

\[ g(x) = \frac{1}{4}(\sin^2 x + \cos x) \]

The maximum value of \( \sin^2 x \) is 1 and the maximum value of \( \cos x \) is also 1. Thus, the maximum value of \( g(x) \) is:

\[ g(x) \leq \frac{1}{4}(1 + 1) = \frac{1}{2} \]

The minimum value occurs when \( \sin^2 x = 0 \) and \( \cos x = -1 \):

\[ g(x) \geq \frac{1}{4}(0 - 1) = -\frac{1}{4} \]

Thus, the range \( B \) of \( g(x) \) is:

\[ B = \left[-\frac{1}{4}, \frac{1}{2}\right] \]

### Step 7: Find the Intersection \( A \cap B \)

Now we find the intersection of \( A \) and \( B \):

- \( A = (0, 1) \cup (2, 3) \cup (\ln(5), \infty) \)
- \( B = \left[-\frac{1}{4}, \frac{1}{2}\right] \)

The intersection is:

\[ A \cap B = (0, \frac{1}{2}] \]

### Final Answer

Thus, the intersection \( A \cap B \) is:

\[ A \cap B = (0, \frac{1}{2}] \]