Let `g (x)` be a cubic polnomial having local maximum at `x=-1` and g '(x) has a local minimum at `x =1, If g (-1) =10 g, (3) =- 22,` then

Options

A. perpendicular distance between its two horizontal tangents is 12
B. perpendicular distance betweent its two horizontal tangents is 32
C. `g (x)=0` has atleast one real root lying in interval `(-2,0)`
D. `g (x) =0, ` has 3 distinict real roots

Answer

To solve the problem, we need to find the cubic polynomial \( g(x) \) given the conditions about its local maxima and minima, as well as specific values of the polynomial at certain points.

### Step 1: Define the cubic polynomial
Let \( g(x) = ax^3 + bx^2 + cx + d \).

### Step 2: Find the first derivative
The first derivative of \( g(x) \) is:
\[ g'(x) = 3ax^2 + 2bx + c \]

### Step 3: Conditions for local maximum and minimum
Given that \( g(x) \) has a local maximum at \( x = -1 \), we have:
\[ g'(-1) = 0 \]
This gives us the equation:
\[ 3a(-1)^2 + 2b(-1) + c = 0 \implies 3a - 2b + c = 0 \quad \text{(1)} \]

Also, since \( g'(x) \) has a local minimum at \( x = 1 \), we need to find the second derivative:
\[ g''(x) = 6ax + 2b \]
Setting \( g''(1) = 0 \) gives us:
\[ 6a(1) + 2b = 0 \implies 6a + 2b = 0 \implies 3a + b = 0 \quad \text{(2)} \]

### Step 4: Solve for \( b \) in terms of \( a \)
From equation (2), we can express \( b \) in terms of \( a \):
\[ b = -3a \]

### Step 5: Substitute \( b \) into equation (1)
Substituting \( b = -3a \) into equation (1):
\[ 3a - 2(-3a) + c = 0 \implies 3a + 6a + c = 0 \implies 9a + c = 0 \implies c = -9a \quad \text{(3)} \]

### Step 6: Write \( g(x) \) in terms of \( a \)
Now substituting \( b \) and \( c \) into \( g(x) \):
\[ g(x) = ax^3 - 3ax^2 - 9ax + d \]

### Step 7: Use given values of \( g(-1) \) and \( g(3) \)
We know:
1. \( g(-1) = 10 \)
2. \( g(3) = -22 \)

Calculating \( g(-1) \):
\[ g(-1) = a(-1)^3 - 3a(-1)^2 - 9a(-1) + d = -a - 3a + 9a + d = 5a + d = 10 \quad \text{(4)} \]

Calculating \( g(3) \):
\[ g(3) = a(3)^3 - 3a(3)^2 - 9a(3) + d = 27a - 27a - 27a + d = -27a + d = -22 \quad \text{(5)} \]

### Step 8: Solve equations (4) and (5)
From equation (4):
\[ d = 10 - 5a \quad \text{(6)} \]

Substituting (6) into equation (5):
\[ -27a + (10 - 5a) = -22 \implies -32a + 10 = -22 \implies -32a = -32 \implies a = 1 \]

### Step 9: Find \( b, c, d \)
Using \( a = 1 \):
- From (2): \( b = -3(1) = -3 \)
- From (3): \( c = -9(1) = -9 \)
- From (6): \( d = 10 - 5(1) = 5 \)

Thus, the cubic polynomial is:
\[ g(x) = x^3 - 3x^2 - 9x + 5 \]

### Step 10: Find the perpendicular distance between the two horizontal tangents
The horizontal tangents occur at the local maximum and minimum points, which we found at \( x = -1 \) and \( x = 3 \):
- \( g(-1) = 10 \)
- \( g(3) = -22 \)

The perpendicular distance between these points is:
\[ \text{Distance} = g(-1) - g(3) = 10 - (-22) = 10 + 22 = 32 \]

### Step 11: Conclusion
The perpendicular distance between the two horizontal tangents is \( 32 \).