If `f(x)={{:(mx^2+x+n,"," x lt 0),(nx +m ,"," 0 le x le 1),(2nx^3+x^2-2x+m,","x gt1):}`
and `lim_(xto0)f(x)` and `lim_(xto1)` f(x) exist then

Options

A. m=2,n=1
B. m=0,n=1
C. m=1,n=2
D. m=1,n=1

Answer

To solve the problem, we need to ensure that the limits of the piecewise function \( f(x) \) exist at \( x = 0 \) and \( x = 1 \). We will analyze both limits step by step.

### Step 1: Finding the limit as \( x \) approaches 0

We need to check the left-hand limit and the right-hand limit at \( x = 0 \):

1. **Left-hand limit**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (mx^2 + x + n) = m(0)^2 + 0 + n = n \]

2. **Right-hand limit**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (nx + m) = n(0) + m = m \]

For the limit to exist at \( x = 0 \), these two limits must be equal:
\[ n = m \quad \text{(Relation 1)} \]

### Step 2: Finding the limit as \( x \) approaches 1

Next, we check the limits at \( x = 1 \):

1. **Left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (nx + m) = n(1) + m = n + m \]

2. **Right-hand limit**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2nx^3 + x^2 - 2x + m) = 2n(1)^3 + (1)^2 - 2(1) + m = 2n + 1 - 2 + m = 2n + m - 1 \]

For the limit to exist at \( x = 1 \), these two limits must also be equal:
\[ n + m = 2n + m - 1 \]

### Step 3: Solving the equation

Now we simplify the equation:
\[ n + m = 2n + m - 1 \]
Subtract \( m \) from both sides:
\[ n = 2n - 1 \]
Rearranging gives:
\[ n - 2n = -1 \implies -n = -1 \implies n = 1 \]

### Step 4: Finding \( m \)

Using Relation 1 (\( n = m \)):
\[ m = n = 1 \]

### Conclusion

Thus, the values of \( n \) and \( m \) are:
\[ n = 1, \quad m = 1 \]