If `f(x)=(x^2)/(2-2cosx);g(x)=(x^2)/(6x-6sinx)` where `0 < x < 1,` then

Options

A. f is increasing function
B. g is increasing function
C. f is decreasing function
D. g is decreasing function

Explanation

To determine whether the functions \( f(x) \) and \( g(x) \) are increasing or decreasing in the interval \( 0 < x < 1 \), we need to find their derivatives and analyze their signs.

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

Given:
\[ f(x) = \frac{x^2}{2 - 2 \cos x} \]

Using the quotient rule for differentiation:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
where \( u = x^2 \) and \( v = 2 - 2 \cos x \).

Calculating \( u' \) and \( v' \):
- \( u' = 2x \)
- \( v' = 2 \sin x \)

Now applying the quotient rule:
\[ f'(x) = \frac{(2x)(2 - 2 \cos x) - (x^2)(2 \sin x)}{(2 - 2 \cos x)^2} \]

Simplifying the numerator:
\[ = \frac{4x - 4x \cos x - 2x^2 \sin x}{(2 - 2 \cos x)^2} \]

### Step 2: Analyze the sign of \( f'(x) \)

We need to check if \( f'(x) > 0 \) in the interval \( 0 < x < 1 \).

1. The denominator \( (2 - 2 \cos x)^2 \) is always positive for \( 0 < x < 1 \).
2. Now, let's analyze the numerator:
\[ 4x - 4x \cos x - 2x^2 \sin x \]

For \( 0 < x < 1 \):
- \( 4x \) is positive.
- \( -4x \cos x \) is negative but less than \( 4x \).
- \( -2x^2 \sin x \) is also negative but less than \( 4x \).

Since both negative terms are less than \( 4x \), we conclude that:
\[ f'(x) > 0 \quad \text{for} \quad 0 < x < 1 \]
Thus, \( f(x) \) is an increasing function on \( (0, 1) \).

### Step 3: Find the derivative of \( g(x) \)

Given:
\[ g(x) = \frac{x^2}{6x - 6 \sin x} \]

Using the quotient rule again:
\[ g'(x) = \frac{(2x)(6x - 6 \sin x) - (x^2)(6 - 6 \cos x)}{(6x - 6 \sin x)^2} \]

Simplifying the numerator:
\[ = \frac{12x^2 - 12x \sin x - 6x^2 + 6x^2 \cos x}{(6x - 6 \sin x)^2} \]
\[ = \frac{6x^2 - 12x \sin x + 6x^2 \cos x}{(6x - 6 \sin x)^2} \]
\[ = \frac{6x^2(1 + \cos x - 2 \sin x)}{(6x - 6 \sin x)^2} \]

### Step 4: Analyze the sign of \( g'(x) \)

We need to check if \( g'(x) < 0 \) in the interval \( 0 < x < 1 \).

1. The denominator \( (6x - 6 \sin x)^2 \) is always positive for \( 0 < x < 1 \).
2. Now, let's analyze the numerator:
\[ 6x^2(1 + \cos x - 2 \sin x) \]

For \( 0 < x < 1 \):
- \( 6x^2 \) is positive.
- We need to check \( 1 + \cos x - 2 \sin x \):
- At \( x = 0 \), \( 1 + 1 - 0 = 2 \) (positive).
- At \( x = 1 \), \( 1 + \cos(1) - 2\sin(1) \) needs to be checked numerically.

Since \( \sin x \) grows faster than \( \cos x \) in this interval, \( 1 + \cos x - 2 \sin x \) will eventually become negative. Thus, \( g'(x) < 0 \) for \( 0 < x < 1 \).

### Conclusion

- \( f(x) \) is an **increasing function** on \( (0, 1) \).
- \( g(x) \) is a **decreasing function** on \( (0, 1) \).