If `f(x)=2x^(3)-3x^(2)+1` then number of distinct real solution (s) of the equation `f(f(x))=0` is/(are) `k` then `(7k)/(10^(2))` is equal to_______

Explanation

To solve the problem step by step, we need to find the number of distinct real solutions of the equation \( f(f(x)) = 0 \), where \( f(x) = 2x^3 - 3x^2 + 1 \).

### Step 1: Find the roots of \( f(x) = 0 \)

We start by solving the equation \( f(x) = 0 \):

\[ f(x) = 2x^3 - 3x^2 + 1 = 0 \]

### Step 2: Find the derivative \( f'(x) \)

To determine the nature of the roots, we first find the derivative:

\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 1) = 6x^2 - 6x = 6x(x - 1) \]

### Step 3: Find critical points

Setting the derivative equal to zero gives us the critical points:

\[ 6x(x - 1) = 0 \implies x = 0 \text{ or } x = 1 \]

### Step 4: Analyze the behavior of \( f(x) \)

Now we analyze the function \( f(x) \) at the critical points and the limits:

- As \( x \to -\infty \), \( f(x) \to -\infty \)
- At \( x = 0 \):
\[ f(0) = 2(0)^3 - 3(0)^2 + 1 = 1 \]
- At \( x = 1 \):
\[ f(1) = 2(1)^3 - 3(1)^2 + 1 = 2 - 3 + 1 = 0 \]
- As \( x \to +\infty \), \( f(x) \to +\infty \)

### Step 5: Determine the number of roots

Since \( f(x) \) is a cubic polynomial, it can have at most 3 real roots. We know:

- \( f(0) = 1 > 0 \)
- \( f(1) = 0 \) (one root)
- The function changes from negative to positive as \( x \) increases from \( -\infty \) to \( 0 \) and again from \( 1 \) to \( +\infty \).

Using the Intermediate Value Theorem, we can conclude that there is one root between \( -\infty \) and \( 0 \) and one root between \( 0 \) and \( 1 \). Therefore, \( f(x) = 0 \) has 3 distinct real roots.

### Step 6: Find \( k \)

Let \( k \) be the number of distinct real solutions of \( f(x) = 0 \):

\[ k = 3 \]

### Step 7: Find \( f(f(x)) = 0 \)

Since \( f(x) = 0 \) has 3 distinct roots, we need to find \( f(f(x)) = 0 \). Each root of \( f(x) = 0 \) can be plugged back into \( f(x) \), which means we need to find the number of distinct solutions to \( f(x) = r \) for each root \( r \).

Since \( f(x) \) has 3 distinct roots, and since \( f(x) \) is a cubic polynomial, it can take each value (including the roots) at most 3 times. Therefore, the total number of distinct solutions to \( f(f(x)) = 0 \) is:

\[ 3 \text{ (roots of } f(x) = 0\text{)} \times 3 \text{ (each root can be achieved 3 times)} = 9 \]

### Step 8: Calculate \( \frac{7k}{10^2} \)

Now we calculate:

\[ \frac{7k}{10^2} = \frac{7 \times 3}{100} = \frac{21}{100} = 0.21 \]

### Final Answer

Thus, the value of \( \frac{7k}{10^2} \) is:

\[ \boxed{0.21} \]