Let `f (x) = [{:(1-x,,, 0 le x le 1),(0,,, 1 lt x le 2 and g (x) = int_(0)^(x) f (t)dt.),( (2-x)^(2),,, 2 lt x le 3):}`
Let the tangent to the curve `y = g( x)` at point P whose abscissa is `5/2` cuts x-axis in point Q.
Let the prependicular from point Q on x-axis meets the curve `y =g (x)` in point R .Find equation of tangent at to y=g(x) at P .Also the value of
`g (1) =`

Options

A. (a) 0
B. (b)`1/2`
C. (c) 1
D. (d) 2

Answer

To solve the problem step by step, we will first define the functions \( f(x) \) and \( g(x) \), then find the equation of the tangent line to \( y = g(x) \) at the point \( P \) where \( x = \frac{5}{2} \), and finally compute the value of \( g(1) \).

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

The function \( f(x) \) is defined piecewise as follows:
- \( f(x) = 1 - x \) for \( 0 \leq x \leq 1 \)
- \( f(x) = 0 \) for \( 1 < x \leq 2 \)
- \( f(x) = (2 - x)^2 \) for \( 2 < x \leq 3 \)

### Step 2: Define the function \( g(x) \)

The function \( g(x) \) is defined as the integral of \( f(t) \) from 0 to \( x \):
\[ g(x) = \int_0^x f(t) \, dt \]

We will compute \( g(x) \) for each interval.

#### For \( 0 \leq x \leq 1 \):
\[ g(x) = \int_0^x (1 - t) \, dt = \left[t - \frac{t^2}{2}\right]_0^x = x - \frac{x^2}{2} \]

#### For \( 1 < x \leq 2 \):
\[ g(x) = \int_0^1 (1 - t) \, dt + \int_1^x 0 \, dt \]
\[ = \left[t - \frac{t^2}{2}\right]_0^1 + 0 = 1 - \frac{1}{2} = \frac{1}{2} \]

#### For \( 2 < x \leq 3 \):
\[ g(x) = \int_0^1 (1 - t) \, dt + \int_1^2 0 \, dt + \int_2^x (2 - t)^2 \, dt \]
\[ = \frac{1}{2} + 0 + \int_2^x (2 - t)^2 \, dt \]
Calculating the last integral:
\[ \int (2 - t)^2 \, dt = \frac{(2 - t)^3}{3} \]
Evaluating from 2 to \( x \):
\[ = \frac{(2 - x)^3}{3} \quad \text{(since at } t=2, (2-2)^3 = 0\text{)} \]
Thus,
\[ g(x) = \frac{1}{2} - \frac{(2 - x)^3}{3} \]

### Step 3: Find \( g\left(\frac{5}{2}\right) \)

Since \( \frac{5}{2} \) lies in the interval \( 2 < x \leq 3 \):
\[ g\left(\frac{5}{2}\right) = \frac{1}{2} - \frac{(2 - \frac{5}{2})^3}{3} = \frac{1}{2} - \frac{(-\frac{1}{2})^3}{3} \]
Calculating:
\[ = \frac{1}{2} - \frac{-\frac{1}{8}}{3} = \frac{1}{2} + \frac{1}{24} = \frac{12}{24} + \frac{1}{24} = \frac{13}{24} \]

### Step 4: Find the slope of the tangent line at \( P \)

To find the slope of the tangent line at \( P \) where \( x = \frac{5}{2} \), we need to differentiate \( g(x) \):
\[ g'(x) = \frac{d}{dx}\left(\frac{1}{2} - \frac{(2 - x)^3}{3}\right) = \frac{(2 - x)^2}{3} \]
Evaluating at \( x = \frac{5}{2} \):
\[ g'\left(\frac{5}{2}\right) = \frac{(2 - \frac{5}{2})^2}{3} = \frac{(-\frac{1}{2})^2}{3} = \frac{\frac{1}{4}}{3} = \frac{1}{12} \]

### Step 5: Write the equation of the tangent line

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Where \( (x_1, y_1) = \left(\frac{5}{2}, \frac{13}{24}\right) \) and \( m = \frac{1}{12} \):
\[ y - \frac{13}{24} = \frac{1}{12}\left(x - \frac{5}{2}\right) \]
Multiplying through by 24 to eliminate the fraction:
\[ 24y - 13 = 2(x - \frac{5}{2}) \]
\[ 24y - 13 = 2x - 5 \]
Rearranging gives:
\[ 2x - 24y + 8 = 0 \quad \text{or} \quad 2x - 24y = -8 \]

### Step 6: Find \( g(1) \)

For \( x = 1 \):
\[ g(1) = 1 - \frac{1^2}{2} = 1 - \frac{1}{2} = \frac{1}{2} \]

### Final Answers
- The equation of the tangent line at point \( P \) is \( 2x - 24y + 8 = 0 \).
- The value of \( g(1) \) is \( \frac{1}{2} \).