Statement -1 The maximum value of `f(x)=1/(3x^4+8x^3-18x^2+60) "is"1/(53)`
Statement -2 : The function g(x) =`1/(f(x))` attains its minimum value at x=1 and x=-3

Options

A. Statement-1 is True, Statement-2 is True,Statement -2 is a correct explanation for Statement -1
B. Statement -1 True ,Statement -2 is True ,Stament -2 is not a correct explanation for Statement -!
C. Statement -1 is True Statement -2 is False
D. Statement -1 is Flase,Statement -2 is True

Explanation

To solve the given problem, we will analyze the two statements provided and verify their correctness step by step.

### Step 1: Analyze Statement 1
We need to determine the maximum value of the function \( f(x) = \frac{1}{3x^4 + 8x^3 - 18x^2 + 60} \).

1. **Find the critical points of \( f(x) \)**:
- To find the maximum value, we first need to find the minimum value of the denominator \( g(x) = 3x^4 + 8x^3 - 18x^2 + 60 \).
- We will differentiate \( g(x) \) and set the derivative to zero to find critical points.

\[ g'(x) = 12x^3 + 24x^2 - 36x \]

2. **Set the derivative to zero**:
\[ 12x(x^2 + 2x - 3) = 0 \]
This gives us:
\[ x = 0, \quad x^2 + 2x - 3 = 0 \]
Factoring the quadratic:
\[ (x + 3)(x - 1) = 0 \implies x = -3, \quad x = 1 \]

3. **Evaluate \( g(x) \) at critical points**:
- Calculate \( g(-3) \), \( g(0) \), and \( g(1) \) to find the minimum value.
\[ g(-3) = 3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60 = 243 - 216 - 162 + 60 = -75 \]
\[ g(0) = 60 \]
\[ g(1) = 3(1)^4 + 8(1)^3 - 18(1)^2 + 60 = 3 + 8 - 18 + 60 = 53 \]

4. **Determine the minimum value of \( g(x) \)**:
- The minimum value occurs at \( g(1) = 53 \).

5. **Find the maximum value of \( f(x) \)**:
- Since \( f(x) = \frac{1}{g(x)} \), the maximum value of \( f(x) \) occurs at the minimum value of \( g(x) \):
\[ \text{Maximum of } f(x) = \frac{1}{g(1)} = \frac{1}{53} \]

### Conclusion for Statement 1:
- Statement 1 is **true**: The maximum value of \( f(x) \) is indeed \( \frac{1}{53} \).

---

### Step 2: Analyze Statement 2
We need to verify if the function \( g(x) = \frac{1}{f(x)} \) attains its minimum value at \( x = 1 \) and \( x = -3 \).

1. **Recall \( g(x) \)**:
\[ g(x) = \frac{1}{f(x)} = 3x^4 + 8x^3 - 18x^2 + 60 \]

2. **Find the critical points of \( g(x) \)**:
- We already found the derivative \( g'(x) \) and set it to zero:
\[ g'(x) = 12x^3 + 24x^2 - 36x = 12x(x^2 + 2x - 3) = 0 \]
- This gives us \( x = 0, -3, 1 \).

3. **Determine the nature of critical points**:
- We already calculated \( g(-3) = -75 \), \( g(1) = 53 \), and \( g(0) = 60 \).
- Since \( g(-3) \) is the lowest value, \( g(x) \) attains its minimum at \( x = -3 \) and \( x = 1 \) is not a minimum.

### Conclusion for Statement 2:
- Statement 2 is **partially true**: \( g(x) \) attains its minimum at \( x = -3 \), but not at \( x = 1 \).

---

### Final Conclusion:
- Statement 1 is true.
- Statement 2 is partially true (only \( x = -3 \) is a minimum).