If `f(x)={{:(,x,x le 1, and f'x(x)),(,x^(2)+bx+c, , x gt 1):}` and exists finetely for all `x in R`, then

Options

A. `b=-1, c in R`
B. `c=1, b in R`
C. `b=1, c=-1`
D. `b=-1, c=1`

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at the point where the definition of the function changes, which is at \( x = 1 \).

### Step 1: Define the Function
The function is defined as:
\[ f(x) = \begin{cases} x & \text{if } x \leq 1 \\ x^2 + bx + c & \text{if } x > 1 \end{cases} \]

### Step 2: Check Continuity at \( x = 1 \)
For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit (LHL) as \( x \) approaches 1 from the left must equal the right-hand limit (RHL) as \( x \) approaches 1 from the right, and both must equal \( f(1) \).

Calculating the left-hand limit:
\[ \lim_{x \to 1^-} f(x) = f(1) = 1 \]

Calculating the right-hand limit:
\[ \lim_{x \to 1^+} f(x) = 1^2 + b(1) + c = 1 + b + c \]

Setting the two limits equal for continuity:
\[ 1 + b + c = 1 \]
This simplifies to:
\[ b + c = 0 \quad \text{(Equation 1)} \]

### Step 3: Check Differentiability at \( x = 1 \)
For \( f(x) \) to be differentiable at \( x = 1 \), the left-hand derivative must equal the right-hand derivative.

Calculating the left-hand derivative:
\[ f'(x) = 1 \quad \text{for } x < 1 \]
Thus,
\[ \lim_{x \to 1^-} f'(x) = 1 \]

Calculating the right-hand derivative:
\[ f'(x) = 2x + b \quad \text{for } x > 1 \]
Thus,
\[ \lim_{x \to 1^+} f'(x) = 2(1) + b = 2 + b \]

Setting the two derivatives equal for differentiability:
\[ 2 + b = 1 \]
This simplifies to:
\[ b = -1 \quad \text{(Equation 2)} \]

### Step 4: Solve for \( c \)
Now, substitute \( b = -1 \) into Equation 1:
\[ -1 + c = 0 \]
This gives us:
\[ c = 1 \]

### Conclusion
The values of \( b \) and \( c \) that make \( f(x) \) continuous and differentiable for all \( x \in \mathbb{R} \) are:
\[ b = -1, \quad c = 1 \]

### Final Answer
Thus, the final values are:
\[ b = -1, \quad c = 1 \]

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