Given: Equation of curve is
y=2x3−15x2+36x−21 and the tangents to the curve
y=2x3−15x2+36x−21 are parallel to the X - axis.
Let point of the contact be
(x1,y1) As we know that, if a tangent of any curve is parallel to
X− axis then
=0 i.e slope of the tangent is zero.
i.e
[](x1,y1)=0 Now by differentiating equation of curve
y=2x3−15x2+36x−21 with respect to
x, we get
⇒=6x2−30x+36 =6.(x2−5x+6) ⇒[](x1,y1)=6.(x12−5x1+6) ∵[](x1,y1)=0 ⇒6.(x12−5x1+6)=0 ⇒x1=2 or 3
When
x1=2 then by substituting
x1=2 in the equation
y=2x3−15x2+36x−21 we get
y1=7 So, the point (2, 7) is a point of contact.
Similarly, when
x1=3 then by substituting
x1=3 in the equation
y=2x3−15x2+36x−21 we get
y1=6 So, the point (3, 6) is a point of contact.
As we know that, Equation of tangent at any point say
(x1,y1) is given by:
y−y1=[](x1,y1).(x−x1) So, the equation of tangent to the given curve at the point (2, 7) is:
y−7=0⋅(x−2)⇒y=7. Similarly, the equation of tangent to the given curve at the point (3, 6) is
:y−6=0⋅(x−3)⇒y=6. Hence, the correct option is 3.