Suppose `g(x)=2x+1` and `h(x)=4x^(2)+4x+5` and `h(x)=(fog)(x)`. The area enclosed by the graph of the function `y=f(x)` and the pair of tangents drawn to it from the origin is:

Options

A. (a) `8/3`
B. (b) `16/3`
C. (c) `32/3`
D. (d) None of these

Answer

To solve the problem step-by-step, we will follow the instructions given in the video transcript and derive the area enclosed by the graph of the function \( y = f(x) \) and the pair of tangents drawn from the origin.

### Step 1: Define the functions
We have two functions:
- \( g(x) = 2x + 1 \)
- \( h(x) = 4x^2 + 4x + 5 \)

We also know that \( h(x) = f(g(x)) \).

### Step 2: Substitute \( g(x) \) into \( h(x) \)
Substituting \( g(x) \) into \( h(x) \):
\[ h(x) = f(2x + 1) \]
We also have:
\[ h(x) = 4x^2 + 4x + 5 \]

### Step 3: Express \( h(x) \) in a different form
We can rewrite \( h(x) \):
\[ h(x) = 4x^2 + 4x + 1 + 4 = (2x + 1)^2 + 4 \]

### Step 4: Set the two expressions for \( h(x) \) equal
Now we can equate the two expressions for \( h(x) \):
\[ f(2x + 1) = (2x + 1)^2 + 4 \]

### Step 5: Let \( t = 2x + 1 \)
Now, let \( t = 2x + 1 \). Then we can express \( x \) in terms of \( t \):
\[ x = \frac{t - 1}{2} \]
Substituting this into the function \( f(t) \):
\[ f(t) = t^2 + 4 \]

### Step 6: Identify the function \( f(x) \)
Thus, we have:
\[ f(x) = x^2 + 4 \]

### Step 7: Find the points of tangency
To find the area enclosed by the graph and the tangents from the origin, we need to find the points where the tangents touch the curve. The slope of the tangent line at a point \( (t, f(t)) \) is given by:
\[ \text{slope} = \frac{f(t) - 0}{t - 0} = \frac{t^2 + 4}{t} \]
The derivative of \( f(x) \) is:
\[ f'(x) = 2x \]
At the point \( t \), the slope of the tangent is:
\[ 2t \]
Setting these equal gives:
\[ \frac{t^2 + 4}{t} = 2t \]
Multiplying through by \( t \) (assuming \( t \neq 0 \)):
\[ t^2 + 4 = 2t^2 \]
This simplifies to:
\[ t^2 = 4 \quad \Rightarrow \quad t = \pm 2 \]

### Step 8: Find the coordinates of points A and B
For \( t = 2 \):
\[ A(2, f(2)) = (2, 2^2 + 4) = (2, 8) \]
For \( t = -2 \):
\[ B(-2, f(-2)) = (-2, (-2)^2 + 4) = (-2, 8) \]

### Step 9: Calculate the area under the curve
We will calculate the area under the curve from \( x = 0 \) to \( x = 2 \):
\[ \text{Area} = \int_0^2 (x^2 + 4) \, dx \]
Calculating the integral:
\[ \int (x^2 + 4) \, dx = \frac{x^3}{3} + 4x \]
Evaluating from 0 to 2:
\[ \left[ \frac{2^3}{3} + 4(2) \right] - \left[ \frac{0^3}{3} + 4(0) \right] = \left[ \frac{8}{3} + 8 \right] = \frac{8}{3} + \frac{24}{3} = \frac{32}{3} \]

### Step 10: Calculate the area of the triangle
The area of triangle \( OAB \) (where \( O \) is the origin) is:
\[ \text{Area}_{\triangle OAB} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 8 = 8 \]

### Step 11: Find the required area
The area enclosed by the curve and the tangents is:
\[ \text{Required Area} = \text{Area under curve} - \text{Area of triangle} = \frac{32}{3} - 8 = \frac{32}{3} - \frac{24}{3} = \frac{8}{3} \]
Since the graph is symmetric, we multiply this area by 2:
\[ \text{Total Area} = 2 \times \frac{8}{3} = \frac{16}{3} \]

### Final Answer
The area enclosed by the graph of the function \( y = f(x) \) and the pair of tangents drawn from the origin is:
\[ \frac{16}{3} \text{ square units} \]