Let `S_(1)` is set of minima and `S_(2)` is set of maxima for the curve `y=9x^(4)+12x^(3)-36x^(2)-25` then (A) `S_(1)={-2,-1},S_(2)={0}` (B) `S_(1){-2,1},S_(2)={0}` (C) `S_(1)={-2,1}:S_(2)={-1}` (D) `S_(1)={-2,2},S_(2)={0}`

Options

A. `S_(1)=(-2),S_(2)=(0.1)`
B. `S_(1)=(-2,0),S_(2)=(1)`
C. `S_(1)=(-2,1),S_(2)=(0)`
D. `S_(1)=(-1),S_(2)=(0,2)`

Answer

To solve the problem, we need to find the local minima and maxima of the function \( y = 9x^4 + 12x^3 - 36x^2 - 25 \).

### Step-by-Step Solution:

1. **Define the function:**
Let \( f(x) = 9x^4 + 12x^3 - 36x^2 - 25 \).

2. **Find the first derivative:**
To find the critical points, we first need to compute the first derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(9x^4) + \frac{d}{dx}(12x^3) - \frac{d}{dx}(36x^2) - \frac{d}{dx}(25) \]
\[ f'(x) = 36x^3 + 36x^2 - 72x \]

3. **Set the first derivative to zero:**
We set \( f'(x) = 0 \):
\[ 36x^3 + 36x^2 - 72x = 0 \]
Factor out \( 36x \):
\[ 36x(x^2 + x - 2) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x^2 + x - 2 = 0 \]

4. **Solve the quadratic equation:**
We can factor the quadratic:
\[ x^2 + x - 2 = (x + 2)(x - 1) = 0 \]
Thus, the critical points are:
\[ x = 0, \quad x = -2, \quad x = 1 \]

5. **Determine the nature of critical points:**
We will use the first derivative test. We will check the sign of \( f'(x) \) in the intervals determined by the critical points \( -2, 0, \) and \( 1 \).

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ f'(-3) = 36(-3)^3 + 36(-3)^2 - 72(-3) = -1080 + 324 + 216 = -540 \quad (\text{negative}) \]

- For \( -2 < x < 0 \) (e.g., \( x = -1 \)):
\[ f'(-1) = 36(-1)^3 + 36(-1)^2 - 72(-1) = -36 + 36 + 72 = 72 \quad (\text{positive}) \]

- For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)):
\[ f'(0.5) = 36(0.5)^3 + 36(0.5)^2 - 72(0.5) = 4.5 + 9 - 36 = -22.5 \quad (\text{negative}) \]

- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ f'(2) = 36(2)^3 + 36(2)^2 - 72(2) = 288 + 144 - 144 = 288 \quad (\text{positive}) \]

6. **Identify local minima and maxima:**
- At \( x = -2 \): \( f'(x) \) changes from negative to positive → Local minimum.
- At \( x = 0 \): \( f'(x) \) changes from positive to negative → Local maximum.
- At \( x = 1 \): \( f'(x) \) changes from negative to positive → Local minimum.

7. **Conclusion:**
The set of minima \( S_1 = \{-2, 1\} \) and the set of maxima \( S_2 = \{0\} \).

### Final Answer:
Thus, the correct option is (B) \( S_1 = \{-2, 1\}, S_2 = \{0\} \).