The differentiable function `y= f(x)` has a property that the chord joining any two points `A (x _(1), f (x_(1)) and B (x_(2), f (x _(2)))` always intersects y-axis at `(0,2 x _(1)x _(2)).` Given that `f (1) =-1.` then:
In which of the following intervals, the Rolle's theorem is applicable to the function `F (x) =f (x) + x` ? (a) `[-1,0]` (b) `[0,1]` (c) `[-1,1]` (d) `[0,2]`

Options

A. `0-1,0]`
B. `[0,1]`
C. `[-1,1]`
D. `[0,2]`

Answer

To solve the problem, we need to analyze the function \( f(x) \) based on the given conditions and then determine the intervals where Rolle's theorem is applicable to the function \( F(x) = f(x) + x \).

### Step 1: Find the function \( f(x) \)

We are given that the chord joining any two points \( A(x_1, f(x_1)) \) and \( B(x_2, f(x_2)) \) intersects the y-axis at the point \( (0, 2x_1x_2) \).

The equation of the chord joining points \( A \) and \( B \) can be expressed as:
\[ y - f(x_1) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}(x - x_1) \]
Substituting \( x = 0 \) (the y-axis intersection), we have:
\[ 2x_1x_2 - f(x_1) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}(-x_1) \]

Rearranging gives:
\[ 2x_1x_2 - f(x_1) = -\frac{x_1(f(x_2) - f(x_1))}{x_2 - x_1} \]

### Step 2: Substitute \( x_2 = 1 \)

Let’s substitute \( x_2 = 1 \):
\[ 2x_1 - f(x_1) = -\frac{x_1(f(1) - f(x_1))}{1 - x_1} \]
Given that \( f(1) = -1 \):
\[ 2x_1 - f(x_1) = -\frac{x_1(-1 - f(x_1))}{1 - x_1} \]

### Step 3: Solve for \( f(x_1) \)

Now, simplifying:
\[ 2x_1 - f(x_1) = \frac{x_1(1 + f(x_1))}{1 - x_1} \]
Multiplying through by \( (1 - x_1) \):
\[ (2x_1 - f(x_1))(1 - x_1) = x_1(1 + f(x_1)) \]
Expanding gives:
\[ 2x_1 - 2x_1^2 - f(x_1) + x_1f(x_1) = x_1 + x_1f(x_1) \]
Rearranging results in:
\[ 2x_1 - 2x_1^2 - f(x_1) = x_1 \]
Thus:
\[ f(x_1) = x_1 - 2x_1^2 \]

### Step 4: Define \( F(x) \)

Now, we have:
\[ f(x) = x - 2x^2 \]
So,
\[ F(x) = f(x) + x = (x - 2x^2) + x = 2x - 2x^2 \]

### Step 5: Check the intervals for Rolle's theorem

Rolle's theorem states that if \( F(a) = F(b) \) for \( a \) and \( b \) in the interval, and \( F(x) \) is continuous and differentiable in that interval, then there exists at least one \( c \) in \( (a, b) \) such that \( F'(c) = 0 \).

1. **Option (a):** Interval \([-1, 0]\)
- \( F(-1) = 2(-1) - 2(-1)^2 = -2 - 2 = -4 \)
- \( F(0) = 2(0) - 2(0)^2 = 0 \)
- \( F(-1) \neq F(0) \) → Not applicable.

2. **Option (b):** Interval \([0, 1]\)
- \( F(0) = 0 \)
- \( F(1) = 2(1) - 2(1)^2 = 2 - 2 = 0 \)
- \( F(0) = F(1) \) → Applicable.

3. **Option (c):** Interval \([-1, 1]\)
- \( F(-1) = -4 \)
- \( F(1) = 0 \)
- \( F(-1) \neq F(1) \) → Not applicable.

4. **Option (d):** Interval \([0, 2]\)
- \( F(2) = 2(2) - 2(2)^2 = 4 - 8 = -4 \)
- \( F(0) = 0 \)
- \( F(0) \neq F(2) \) → Not applicable.

### Conclusion

The only interval where Rolle's theorem is applicable is \([0, 1]\).