Let's analyze the given curve and line to determine the correct statements.
The curve is given by:
Γ:y=be−x∕a The straight line is given by:
L:+=1 First, let's find where the curve
Γ crosses the y -axis. This occurs when
x=0 :
y=be0=bSo, the curve crosses the
y-axis at the point
(0,b).
Next, let's check whether the line
L touches the curve
Gamma at this point. Substituting
x=0 and
y=b into the equation of the line
L :
+=1This is simplified to:
0+1=1Thus, the point
(0,b) lies on the line
L.
To determine if the line is tangent to the curve at this point, we need to check if they have the same slope at that point. First, find the derivative of the curve
Γ with respect to
x :
At the point
(0,b), the derivative (slope of the curve) is:
|x=0=−e0=− Now, find the slope of the line
L. The line equation
+=1 can be rearranged to the slope-intercept form:
=−+1or:
y=−x+b The slope of the line
L is:
−Since both the curve
Γ and the line
L have the same slope
− at the point
(0,b), the line is tangent to the curve at that point.
Therefore, the correct answer is:
Option A: L touches the curve
Γ at the point where the curve crosses the axis of
y.
Finally, for completeness, let's evaluate the other options:
Option B: This is incorrect because L does indeed touch the curve at the point where the curve crosses the
y-axis.
Option C: The curve
Γ never touches the x -axis because as
x⟶∞,y⟶0, but it never equals zero.
Option D: This is correct because as analyzed, the curve
Γ asymptotically approaches the x-axis but does not touch it.