Verified answer

[tex]{\huge{\underline{\underline{\boxed{\blue{\mathscr{ Answer }}}}}}}[/tex]

To find the value of m for which the line y = mx + 1 is a tangent to the curve [tex]{\sf \[ \ y^2 = 4x\] }[/tex], we should set the discriminant of the quadratic equation to zero because a tangent touches the curve at a single point.

The equation for the curve is [tex]{\sf \[ \ y^2 = 4x\] }[/tex], and we have the equation for the line [tex]{\sf \[ \ y = mx + 1\] }[/tex]. Substituting the line equation into the curve equation:

[tex]{\sf \[ \ (mx + 1)^2 = 4x\] }[/tex]

Now, expand and simplify:

[tex]{\sf \[ \ m^2x^2 + 2mx + 1 = 4x\] }[/tex]

This equation represents the intersection points of the line and the curve. To find the point of tangency, we require this equation to have a single solution, which means its discriminant [tex]{\sf \[ \ (b^2 - 4ac) \] }[/tex] should be zero.

For this equation, a = m², b = 2m - 4, and c = 1. So, the discriminant is:

[tex]{\sf \[\Delta = (2m - 4)^2 - 4(m^2)(1)\] }[/tex]

Now, set [tex]{\sf \[\Delta\] }[/tex] to zero:

[tex]{\sf \[ \ (2m - 4)^2 - 4(m^2)(1) = 0\] }[/tex]

Solve for m:

[tex]{\sf \[ \ (2m - 4)^2 = 4m^2\] }[/tex]

[tex]{\sf \[ \ 4m^2 - 16m + 16 = 4m^2\] }[/tex]

[tex]{\sf \[ \ -16m + 16 = 0\] }[/tex]

Now, add 16 to both sides:

[tex]{\sf \[ \ -16m = -16\] }[/tex]

Finally, divide by -16:

[tex]\pink{\underline{\boxed{\sf \[ \ m = 1\] }}}[/tex]

b) 1 is correct