Let `f(x)` be a 4 degree polynomial satisfying `f(r)=(1)/(r)` for `r=1,2,3,4,5` then
Q. Value of `f(6)` is

Options

A. 6
B. `(1)/(6)`
C. `(1)/(3)`
D. `f(3)`

Answer

To find the value of \( f(6) \) for the polynomial \( f(x) \) of degree 4 that satisfies \( f(r) = \frac{1}{r} \) for \( r = 1, 2, 3, 4, 5 \), we can follow these steps:

### Step 1: Define the polynomial
Since \( f(x) \) is a polynomial of degree 4, we can express it as:
\[ f(x) = ax^4 + bx^3 + cx^2 + dx + e \]
However, we also know that \( f(r) = \frac{1}{r} \) for \( r = 1, 2, 3, 4, 5 \).

### Step 2: Construct a new function
Let’s define a new function:
\[ g(x) = x f(x) - 1 \]
This function \( g(x) \) will also be a polynomial of degree 5 because \( x f(x) \) is of degree 5.

### Step 3: Identify the roots of \( g(x) \)
From the condition \( f(r) = \frac{1}{r} \), we can see that:
\[ g(r) = r f(r) - 1 = r \cdot \frac{1}{r} - 1 = 0 \]
Thus, \( g(x) \) has roots at \( x = 1, 2, 3, 4, 5 \). Therefore, we can express \( g(x) \) as:
\[ g(x) = k(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) \]
for some constant \( k \).

### Step 4: Find the constant \( k \)
To find \( k \), we can evaluate \( g(0) \):
\[ g(0) = 0 \cdot f(0) - 1 = -1 \]
Now substituting \( x = 0 \) into the expression for \( g(x) \):
\[ g(0) = k(-1)(-2)(-3)(-4)(-5) = k \cdot (-1) \cdot (-120) = 120k \]
Setting these equal gives:
\[ 120k = -1 \implies k = -\frac{1}{120} \]

### Step 5: Write the expression for \( g(x) \)
Now we have:
\[ g(x) = -\frac{1}{120}(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) \]

### Step 6: Find \( f(6) \)
Now we need to find \( f(6) \):
\[ g(6) = 6f(6) - 1 \]
Calculating \( g(6) \):
\[ g(6) = -\frac{1}{120}(6 - 1)(6 - 2)(6 - 3)(6 - 4)(6 - 5) = -\frac{1}{120}(5)(4)(3)(2)(1) = -\frac{120}{120} = -1 \]
Thus, we have:
\[ 6f(6) - 1 = -1 \implies 6f(6) = 0 \implies f(6) = 0 \]

### Conclusion
The value of \( f(6) \) is:
\[ \boxed{0} \]