Let `f(x) = x^5 + 2x^3 + 3x + 1`, `x in R`, and `g(x)` be a function such that `g(f(x)) = x` for all `x in R`. Then `frac{g(7)}{g’(7)}` is equal to

Options

A. 14
B. 1
C. 7
D. 42

Answer

To solve the problem, we need to find the value of \(\frac{g(7)}{g'(7)}\) given that \(g(f(x)) = x\) for all \(x \in \mathbb{R}\) and \(f(x) = x^5 + 2x^3 + 3x + 1\).

### Step 1: Evaluate \(f(1)\)

First, we calculate \(f(1)\):
\[ f(1) = 1^5 + 2 \cdot 1^3 + 3 \cdot 1 + 1 = 1 + 2 + 3 + 1 = 7 \]
Thus, we have \(f(1) = 7\).

### Step 2: Use the property of \(g\)

Since \(g(f(x)) = x\), we can substitute \(x = 1\):
\[ g(f(1)) = g(7) = 1 \]
This tells us that \(g(7) = 1\).

### Step 3: Differentiate both sides of \(g(f(x)) = x\)

Next, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}[g(f(x))] = \frac{d}{dx}[x] \]
Using the chain rule on the left side, we get:
\[ g'(f(x)) \cdot f'(x) = 1 \]

### Step 4: Evaluate \(f'(x)\)

Now, we need to find \(f'(x)\):
\[ f'(x) = 5x^4 + 6x^2 + 3 \]
Substituting \(x = 1\):
\[ f'(1) = 5 \cdot 1^4 + 6 \cdot 1^2 + 3 = 5 + 6 + 3 = 14 \]

### Step 5: Find \(g'(7)\)

From the differentiated equation \(g'(f(x)) \cdot f'(x) = 1\), substituting \(x = 1\) gives:
\[ g'(f(1)) \cdot f'(1) = 1 \implies g'(7) \cdot 14 = 1 \]
Thus, we can solve for \(g'(7)\):
\[ g'(7) = \frac{1}{14} \]

### Step 6: Calculate \(\frac{g(7)}{g'(7)}\)

Now, we can find \(\frac{g(7)}{g'(7)}\):
\[ \frac{g(7)}{g'(7)} = \frac{1}{\frac{1}{14}} = 14 \]

### Final Answer

Thus, the value of \(\frac{g(7)}{g'(7)}\) is:
\[ \boxed{14} \]