for all `x in R` if `mx^2-9mx+5m+1gt0` then m lies in the interval

Options

A. `( - 61/4,0)`
B. `(4/61, 61/4)`
C. `( 0, 4/61)`
D. `((-4)/61 , 0)`

Answer

To solve the problem, we need to determine the values of \( m \) for which the quadratic expression \( mx^2 - 9mx + (5m + 1) > 0 \) holds for all \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Identify the Quadratic Coefficients**:
The given quadratic expression is:
\[ f(x) = mx^2 - 9mx + (5m + 1) \]
Here, the coefficients are:
- \( a = m \)
- \( b = -9m \)
- \( c = 5m + 1 \)

2. **Condition for the Quadratic to be Positive**:
For the quadratic to be positive for all \( x \), two conditions must be satisfied:
- The leading coefficient \( a \) must be positive: \( m > 0 \)
- The discriminant \( D \) must be less than zero: \( D < 0 \)

3. **Calculate the Discriminant**:
The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-9m)^2 - 4(m)(5m + 1) \]
Simplifying this:
\[ D = 81m^2 - 4m(5m + 1) = 81m^2 - 20m^2 - 4m = 61m^2 - 4m \]

4. **Set the Discriminant Less Than Zero**:
We need:
\[ 61m^2 - 4m < 0 \]
Factoring out \( m \):
\[ m(61m - 4) < 0 \]

5. **Find Critical Points**:
The critical points occur when:
\[ m = 0 \quad \text{or} \quad 61m - 4 = 0 \Rightarrow m = \frac{4}{61} \]

6. **Test Intervals**:
We will test the intervals determined by the critical points \( m = 0 \) and \( m = \frac{4}{61} \):
- Interval 1: \( (-\infty, 0) \)
- Interval 2: \( (0, \frac{4}{61}) \)
- Interval 3: \( (\frac{4}{61}, \infty) \)

We can choose test points from each interval:
- For \( m = -1 \) (Interval 1): \( -1(61(-1) - 4) = -1(-61 - 4) = 65 > 0 \) (not valid)
- For \( m = 1 \) (Interval 2): \( 1(61(1) - 4) = 1(61 - 4) = 57 > 0 \) (valid)
- For \( m = 1 \) (Interval 3): \( 1(61(1) - 4) = 1(61 - 4) = 57 > 0 \) (not valid)

7. **Conclusion**:
The valid interval where \( 61m^2 - 4m < 0 \) and \( m > 0 \) is:
\[ m \in \left(0, \frac{4}{61}\right) \]

### Final Answer:
The values of \( m \) lie in the interval \( \left(0, \frac{4}{61}\right) \).