If `8f(x)+6f(1/x)=x+5` and `y=x^2(f(x),` then `(dy)/(dx)` at `x=-1` is equal to 0 (b) `-1/(14)` (c) `-1/4` (d) None of these

Options

A. 0
B. `(1)/(14)`
C. `-(1)/(14)`
D. `(1)/(7)`

Answer

To solve the problem step by step, we need to find \( \frac{dy}{dx} \) at \( x = -1 \) given the equations:

1. \( 8f(x) + 6f\left(\frac{1}{x}\right) = x + 5 \)
2. \( y = x^2 f(x) \)

### Step 1: Find \( f(-1) \)

We start by substituting \( x = -1 \) into the first equation:

\[ 8f(-1) + 6f(-1) = -1 + 5 \]

This simplifies to:

\[ 14f(-1) = 4 \]

Now, solving for \( f(-1) \):

\[ f(-1) = \frac{4}{14} = \frac{2}{7} \]

### Step 2: Differentiate the equation

Next, we differentiate the equation \( 8f(x) + 6f\left(\frac{1}{x}\right) = x + 5 \) with respect to \( x \):

Using the product and chain rule, we differentiate:

\[ 8f'(x) + 6\left(-\frac{1}{x^2}f'\left(\frac{1}{x}\right)\right) = 1 \]

This simplifies to:

\[ 8f'(x) - \frac{6}{x^2}f'\left(\frac{1}{x}\right) = 1 \]

### Step 3: Substitute \( x = -1 \)

Now we substitute \( x = -1 \):

\[ 8f'(-1) - 6f'(-1) = 1 \]

This simplifies to:

\[ 2f'(-1) = 1 \]

Thus,

\[ f'(-1) = \frac{1}{2} \]

### Step 4: Differentiate \( y = x^2 f(x) \)

Now we differentiate \( y = x^2 f(x) \) using the product rule:

\[ \frac{dy}{dx} = x^2 f'(x) + 2x f(x) \]

### Step 5: Substitute \( x = -1 \) into \( \frac{dy}{dx} \)

Now we substitute \( x = -1 \):

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

Substituting the values we found earlier:

\[ \frac{dy}{dx} = 1 \cdot \frac{1}{2} + 2(-1) \cdot \frac{2}{7} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{1}{2} - \frac{4}{7} \]

### Step 6: Find a common denominator and simplify

The common denominator between 2 and 7 is 14:

\[ \frac{dy}{dx} = \frac{7}{14} - \frac{8}{14} = \frac{-1}{14} \]

### Final Answer

Thus, \( \frac{dy}{dx} \) at \( x = -1 \) is:

\[ \frac{dy}{dx} = -\frac{1}{14} \]