Let `f(x)=(x^3+2)^(30)` If `f^n (x)` is a polynomial of degree 20 where `f^n(x)` denotes the `n^(th)` derivativeof `f(x)` w.r.t `x` then then value of `n` is

Options

A. 60
B. 40
C. 70
D. 50

Answer

To solve the problem, we need to determine the value of \( n \) such that the \( n^{th} \) derivative of the function \( f(x) = (x^3 + 2)^{30} \) is a polynomial of degree 20.

### Step-by-Step Solution:

1. **Identify the Degree of the Original Function:**
The function \( f(x) = (x^3 + 2)^{30} \) is a polynomial. The highest degree term in \( f(x) \) is obtained by considering the term \( (x^3)^{30} \), which gives us:
\[ \text{Degree of } f(x) = 3 \times 30 = 90. \]

2. **Calculate the Degree of the First Derivative:**
When we differentiate \( f(x) \) once, we apply the chain rule:
\[ f'(x) = 30(x^3 + 2)^{29} \cdot 3x^2. \]
The degree of \( f'(x) \) is:
\[ \text{Degree of } f'(x) = 29 \times 3 + 2 = 87 + 2 = 89. \]

3. **Calculate the Degree of the Second Derivative:**
For the second derivative, we differentiate \( f'(x) \):
\[ f''(x) = \text{(derivative of first term)} + \text{(derivative of second term)}. \]
The highest degree term will still be dominated by the first term, giving us:
\[ \text{Degree of } f''(x) = 28 \times 3 + 2 + 1 = 86 + 2 + 1 = 88. \]

4. **Calculate the Degree of the Third Derivative:**
Continuing this process, we find the degree of the third derivative:
\[ f'''(x) = \text{(derivative of the second derivative)}. \]
The highest degree term will yield:
\[ \text{Degree of } f'''(x) = 27 \times 3 + 2 + 2 = 87. \]

5. **General Pattern for the \( n^{th} \) Derivative:**
We observe that the degree of the \( n^{th} \) derivative can be expressed as:
\[ \text{Degree of } f^{(n)}(x) = 90 - n. \]

6. **Set Up the Equation:**
We want the degree of the \( n^{th} \) derivative to equal 20:
\[ 90 - n = 20. \]

7. **Solve for \( n \):**
Rearranging the equation gives:
\[ n = 90 - 20 = 70. \]

### Conclusion:
The value of \( n \) is \( 70 \).