If `f (x) = cot ^(-1)((3x -x ^(3))/( 1- 3x ^(2)))and g (x) = cos ^(-1) ((1-x ^(2))/(1+x^(2)))` then `lim _(xtoa)(f(x) - f(a))/( g(x) -g (a)), 0 lt 1/2` is :

Options

A. `(3)/(2 (1+a^(2)))`
B. `3/2`
C. `(-3)/(2 (1+a^(2)))`
D. `-3/2`

Answer

To solve the limit problem given by the functions \( f(x) \) and \( g(x) \), we will follow these steps:

### Step 1: Define the functions
We have:
\[ f(x) = \cot^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \]
\[ g(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]

### Step 2: Simplify \( f(x) \)
Using the identity for the tangent of a triple angle, we can rewrite \( f(x) \):
\[ \frac{3x - x^3}{1 - 3x^2} = \tan(3\theta) \quad \text{where } \theta = \tan^{-1}(x) \]
Thus, we have:
\[ f(x) = \cot^{-1}(\tan(3\theta)) = \frac{\pi}{2} - 3\theta = \frac{\pi}{2} - 3\tan^{-1}(x) \]

### Step 3: Simplify \( g(x) \)
For \( g(x) \), we can use the identity for cosine:
\[ \frac{1 - x^2}{1 + x^2} = \cos(2\theta) \quad \text{where } \theta = \tan^{-1}(x) \]
Thus, we have:
\[ g(x) = \cos^{-1}(\cos(2\theta)) = 2\theta = 2\tan^{-1}(x) \]

### Step 4: Find the limit
We need to compute:
\[ \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} \]
Using L'Hôpital's Rule, we differentiate the numerator and denominator:
\[ f'(x) = -3 \cdot \frac{1}{1 + x^2} \]
\[ g'(x) = \frac{2}{1 + x^2} \]

### Step 5: Apply L'Hôpital's Rule
Now we apply L'Hôpital's Rule:
\[ \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} = \frac{-3/(1 + a^2)}{2/(1 + a^2)} = \frac{-3}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = -\frac{3}{2} \]