Question No 50.Let ‘m’ be any integer such that the quadratic equation 2x2 + mx + 7 = 0 has no real roots and the quadratic equation x2 + (m – 3)x + 2 = 0 has two distinct real roots for x. Then, the number of possible values of ‘m’ is1)  22)  103)  144)  15

The problem involves two quadratic equations and conditions on their roots. Let's solve it step by step.

1. The first equation is 2x^2 + mx + 7 = 0. The condition is that it has no real roots. For a quadratic equation ax^2 + bx + c = 0, the discriminant is b^2 - 4ac. If the discriminant is less than 0, the equation has no real roots. So, for the first equation, we need m^2 - 427 < 0. This simplifies to m^2 < 56.

2. The second equation is x^2 + (m – 3)x + 2 = 0. The condition is that it has two distinct real roots. For a quadratic equation to have two distinct real roots, the discriminant must be greater than 0. So, for the second equation, we need (m-3)^2 - 412 > 0. This simplifies to m^2 - 6m + 9 > 8, or m^2 - 6m + 1 > 0.

3. We need to find the values of m that satisfy both conditions. The first condition gives -sqrt(56) < m < sqrt(56), or approximately -7.48 < m < 7.48. The second condition is a quadratic inequality that can be solved by finding the roots and testing the intervals. The roots are 3 - sqrt(2) and 3 + sqrt(2), or approximately 1.59 and 4.41. Testing the intervals gives m < 1.59 or m > 4.41.

4. Combining the two conditions, we find that the possible values of m are from -7.48 to 1.59 and from 4.41 to 7.48. These are continuous intervals, but the problem asks for the number of integer values of m. Counting the integers in these intervals gives the possible values of m as -7, -6, -5, -4, -3, -2, -1, 0, 1, 4, 5, 6, 7. So, there are 13 possible values of m.

Therefore, none of the given options is correct. The number of possible values of 'm' is 13.