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Joint Entrance Examination
Mathematics
Application of Derivatives
Question 172
Question 172 of 212
Medium
MCQ
If the tangent at the point (x1, y1) on the curve
y
=
x
3
+
3
x
2
+
5
y = {x^3} + 3{x^2} + 5
y
=
x
3
+
3
x
2
+
5
passes through the origin, then (x1, y1) does NOT lie on the curve :
A
\[{x^2} + {{{y^2}} \over {81}} = 2\]
B
\[{{{y^2}} \over 9} - {x^2} = 8\]
C
\[y = 4{x^2} + 5\]
D
\[{x \over 3} - {y^2} = 2\]
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Official solution
Given curve,
y
=
x
3
+
3
x
2
+
5
y = {x^3} + 3{x^2} + 5
y
=
x
3
+
3
x
2
+
5
Slope of tangent, \[{{dy} \over {dx}} = 3{x^2} + 6x\] Slope of line joined by (x1, y1) and (0, 0) is \[ = {{{y_0} - 0} \over {{x_1} - 0}}\] \[\therefore\] \[{{{y_1}} \over {{x_1}}} = 3x_1^2 + 6{x_1}\] \[ \Rightarrow {y_1} = 3x_1^3 + 6x_1^2\] ...... (1) Point (x1, y1) is on the line, \[\therefore\]
y
1
=
x
1
3
+
3
x
1
2
+
5
{y_1} = x_1^3 + 3x_1^2 + 5
y
1
=
x
1
3
+
3
x
1
2
+
5
..... (2) \[\therefore\]
3
x
1
3
+
6
x
1
2
=
x
1
3
+
3
x
1
2
+
6
3x_1^3 + 6x_1^2 = x_1^3 + 3x_1^2 + 6
3
x
1
3
+
6
x
1
2
=
x
1
3
+
3
x
1
2
+
6
\[ \Rightarrow 2x_1^3 + 3x_1^2 - 5 = 0\] x = 1 satisfy the equation Now,
y
1
=
x
1
3
+
3
x
1
2
+
5
{y_1} = x_1^3 + 3x_1^2 + 5
y
1
=
x
1
3
+
3
x
1
2
+
5
=
1
+
3
+
5
= 1 + 3 + 5
=
1
+
3
+
5
=
9
= 9
=
9
\[\therefore\] (x1, y1) = (1, 9) Option D does not satisfy point (1, 9).
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