Consider the function `f (x) and g (x),` both defined from `R to R`
`f (x) = (x ^(3))/(2 )+1 -x int _(0)^(x) g (t) dt and g (x) =x - int _(0) ^(1) f (t) dt, ` then
The number of points of intersection of `f (x) and g (x)` is/are:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of points of intersection of the functions \( f(x) \) and \( g(x) \), we start with the definitions of the functions:

1. **Given Functions:**
- \( f(x) = \frac{x^3}{2} + 1 - x \int_0^x g(t) dt \)
- \( g(x) = x - \int_0^1 f(t) dt \)

2. **Analyzing \( g(x) \):**
- The term \( \int_0^1 f(t) dt \) is a constant since it does not depend on \( x \).
- Let \( a = \int_0^1 f(t) dt \). Therefore, we can rewrite \( g(x) \) as:
\[ g(x) = x - a \]

3. **Substituting \( g(t) \) into \( f(x) \):**
- We now substitute \( g(t) = t - a \) into the integral in \( f(x) \):
\[ f(x) = \frac{x^3}{2} + 1 - x \int_0^x (t - a) dt \]
- Evaluating the integral:
\[ \int_0^x (t - a) dt = \left[\frac{t^2}{2} - at\right]_0^x = \frac{x^2}{2} - ax \]
- Thus, we have:
\[ f(x) = \frac{x^3}{2} + 1 - x\left(\frac{x^2}{2} - ax\right) \]
- Simplifying this:
\[ f(x) = \frac{x^3}{2} + 1 - \left(\frac{x^3}{2} - ax^2\right) = 1 + ax^2 \]

4. **Setting \( f(x) \) equal to \( g(x) \):**
- Now, we need to find the points of intersection by solving:
\[ f(x) = g(x) \]
- This gives us:
\[ 1 + ax^2 = x - a \]
- Rearranging this equation:
\[ ax^2 - x + (a + 1) = 0 \]

5. **Finding the Discriminant:**
- The discriminant \( D \) of the quadratic equation \( ax^2 - x + (a + 1) = 0 \) is given by:
\[ D = b^2 - 4ac = (-1)^2 - 4a(a + 1) = 1 - 4a^2 - 4a \]

6. **Finding the value of \( a \):**
- We need to calculate \( a = \int_0^1 f(t) dt \):
\[ a = \int_0^1 \left(1 + at^2\right) dt = \left[t + \frac{at^3}{3}\right]_0^1 = 1 + \frac{a}{3} \]
- Rearranging gives:
\[ a - \frac{a}{3} = 1 \implies \frac{2a}{3} = 1 \implies a = \frac{3}{2} \]

7. **Substituting \( a \) back into the Discriminant:**
- Now substituting \( a = \frac{3}{2} \) into the discriminant:
\[ D = 1 - 4\left(\frac{3}{2}\right)^2 - 4\left(\frac{3}{2}\right) = 1 - 4 \cdot \frac{9}{4} - 6 = 1 - 9 - 6 = -14 \]

8. **Conclusion:**
- Since the discriminant \( D < 0 \), the quadratic equation has no real roots, which means there are no points of intersection between \( f(x) \) and \( g(x) \).

Thus, the number of points of intersection of \( f(x) \) and \( g(x) \) is **0**.