Let `f (x) =x ^(2) + bx + c AA `x`in R, (b,c, in R) ` attains its least value at `x =-1` and the graph of `f (x)` cuts y-axis at `y =2.`
The least value of `f (x) AA x in ` R is :

Options

A. (a) `-1`
B. (b) 0
C. (c) 1
D. (d) `3//2`

Answer

To solve the problem step by step, we will follow the conditions given in the question regarding the function \( f(x) = x^2 + bx + c \).

### Step 1: Identify the conditions
The function \( f(x) \) attains its least value at \( x = -1 \) and cuts the y-axis at \( y = 2 \). This means:
1. The vertex of the parabola represented by \( f(x) \) is at \( x = -1 \).
2. The value of \( f(0) = 2 \).

### Step 2: Use the vertex formula
For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex (where the minimum or maximum occurs) is given by the formula:
\[ x = -\frac{b}{2a} \]
Since \( a = 1 \) in our case, we have:
\[ -1 = -\frac{b}{2 \cdot 1} \implies b = 2 \]

### Step 3: Find the value of \( c \)
Since the function cuts the y-axis at \( y = 2 \), we can find \( c \) by evaluating \( f(0) \):
\[ f(0) = 0^2 + b \cdot 0 + c = c \]
Given that \( f(0) = 2 \), we have:
\[ c = 2 \]

### Step 4: Write the function
Now we can substitute the values of \( b \) and \( c \) into the function:
\[ f(x) = x^2 + 2x + 2 \]

### Step 5: Find the least value of \( f(x) \)
To find the least value of the function, we can substitute \( x = -1 \) into \( f(x) \):
\[ f(-1) = (-1)^2 + 2(-1) + 2 \]
Calculating this gives:
\[ f(-1) = 1 - 2 + 2 = 1 \]

### Conclusion
Thus, the least value of the function \( f(x) \) is:
\[ \boxed{1} \]