JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: The curve $y = a x^{3} + b x^{2} + c x + 5$ , touches the $x$ -axis at $P (- 2, 0)$ and cuts the $y$ axis at a point $Q$ , where its gradient is 3. Find $a, b, c$ .

Q: Wait, why do we need two conditions for the point where it 'touches' the x-axis? Isn't $y=0$ enough?

Great question! If the curve simply crossed the x-axis, $y=0$ would be sufficient. But 'touching' implies tangency. A tangent line at a point on the x-axis is horizontal, meaning its slope is zero. Thus, $y'=0$ is the second, crucial condition.

Q: Is there a faster way to solve this using the roots of the cubic?

Yes! Since it touches at $x=-2$, $(x+2)^2$ is a factor. You could write the cubic as $y = (x+2)^2(ax + k)$. This is a powerful alternative, but the system of equations method is often more robust for JEE problems.

Q: Why did we get two equations for $a$ and $b$?

Because we have two unknowns ($a$ and $b$) remaining after finding $c$. To solve for two variables, we need two independent linear equations. We derived these from the two conditions at point $P$: the position ($y=0$) and the slope ($y'=0$).

Visualized Solution

Visualizing the Cubic Curve

Curve Equation: $y = a x^{3} + b x^{2} + c x + 5$
Point $P (- 2, 0)$ is where the curve touches the $x$ -axis.
Point $Q$ is the $y$ -intercept where the curve cuts the $y$ -axis.

Coordinates of Point $Q$

The curve cuts the $y$ -axis at point $Q$ .
On the $y$ -axis, the $x$ -coordinate is always $0$ .
Substitute $x = 0$ into $y = a x^{3} + b x^{2} + c x + 5$ :
$y = a (0)^{3} + b (0)^{2} + c (0) + 5 = 5$
Therefore, point $Q$ is $(0, 5)$ .

Finding $c$ using Gradient at $Q$

The gradient of the curve at $Q$ is given as $3$ .
Gradient means the derivative, $\frac{d y}{d x}$ .
Let's differentiate the curve: $y^{'} = 3 a x^{2} + 2 b x + c$

Calculating the Value of $c$

At $Q (0, 5)$ , the $x$ -coordinate is $0$ .
Substitute $x = 0$ into the derivative:
$y^{'} (0) = 3 a (0)^{2} + 2 b (0) + c = 3$
This simplifies directly to $c = 3$ .

Using Point $P (- 2, 0)$

The curve passes through $P (- 2, 0)$ .
Substitute $x = - 2$ and $y = 0$ into the curve equation.
$0 = a (- 2)^{3} + b (- 2)^{2} + c (- 2) + 5$

First Equation in $a$ and $b$

We know $c = 3$ . Let's substitute it in.
$0 = - 8 a + 4 b - 2 (3) + 5$
$0 = - 8 a + 4 b - 6 + 5$
$- 8 a + 4 b = 1 — (Eq. 1)$

The Meaning of "Touches the $x$ -axis"

The problem states the curve touches the $x$ -axis at $P$ .
"Touching" means the $x$ -axis is a tangent to the curve at $P$ .
Therefore, the tangent is horizontal, and its slope is $0$ .
$y^{'} (- 2) = 0$

Second Equation in $a$ and $b$

Substitute $x = - 2$ and $c = 3$ into the derivative $y^{'} = 3 a x^{2} + 2 b x + c$ .
$3 a (- 2)^{2} + 2 b (- 2) + 3 = 0$
$12 a - 4 b + 3 = 0$
$12 a - 4 b = - 3 — (Eq. 2)$

Solving for $a$

We have a system of linear equations:
1) $- 8 a + 4 b = 1$
2) $12 a - 4 b = - 3$
Add Equation 1 and Equation 2 to eliminate $b$ :
$(- 8 a + 12 a) + (4 b - 4 b) = 1 - 3$
$4 a = - 2$

Final Values of $a$ and $b$

From $4 a = - 2$ , we get $a = - \frac{1}{2}$ .
Substitute $a = - \frac{1}{2}$ into Equation 1:
$- 8 (- \frac{1}{2}) + 4 b = 1$
$4 + 4 b = 1 ⟹ 4 b = - 3 ⟹ b = - \frac{3}{4}$

Conclusion

The coefficients are:
$a = - \frac{1}{2}$
$b = - \frac{3}{4}$
$c = 3$
Key Takeaway: "Touching" the $x$ -axis provides two conditions: the point lies on the curve ( $y = 0$ ), and the derivative is zero ( $y^{'} = 0$ ).

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The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Landscape of the Cubic Curve

The cubic curve is defined by the equation:

y = a x^{3} + b x^{2} + c x + 5

We are tasked with determining the unknown parameters

a

b

, and

c

using the geometric properties provided.

The Gate of the Y-Intercept

At the y-intercept, the

x

-coordinate is

0

. Substituting

x = 0

into the equation, the terms involving

x

vanish, leaving

y = 5

. Thus, the point

Q

(0, 5)

The gradient of the curve is given by the derivative:

y^{'} = 3 a x^{2} + 2 b x + c

We are given that the gradient at the y-intercept (

x = 0

) is

3

. Substituting these values into the derivative expression:

3 = 3 a (0)^{2} + 2 b (0) + c \Rightarrow c = 3

The Geometry of the Touch

We are given that the curve touches the x-axis at

P (- 2, 0)

. In calculus, "touching" the x-axis implies that the x-axis is tangent to the curve at that point.

Because the x-axis is a horizontal line, its slope is

0

. This provides two critical conditions at

x = - 2

1. The curve passes through the point:

y (- 2) = 0

. 2. The slope at the point is zero:

y^{'} (- 2) = 0

The Algebraic Resolution

Using the condition

y (- 2) = 0

with

c = 3

0 = a (- 2)^{3} + b (- 2)^{2} + 3 (- 2) + 5

- 8 a + 4 b - 6 + 5 = 0 \Rightarrow - 8 a + 4 b = 1

Next, we apply the slope condition

y^{'} (- 2) = 0

3 a (- 2)^{2} + 2 b (- 2) + 3 = 0

12 a - 4 b + 3 = 0 \Rightarrow 12 a - 4 b = - 3

We now solve the system of linear equations:

- 8 a + 4 b = 1

12 a - 4 b = - 3

Adding these two equations eliminates

b

4 a = - 2 \Rightarrow a = - \frac{1}{2}

Substituting

a = - \frac{1}{2}

into the first equation:

- 8 (- \frac{1}{2}) + 4 b = 1 \Rightarrow 4 + 4 b = 1 \Rightarrow 4 b = - 3 \Rightarrow b = - \frac{3}{4}

Final Result

By translating the geometric constraints into algebraic conditions, we have determined the coefficients of the cubic curve:

$a = - \frac{1}{2}$ , $b = - \frac{3}{4}$ , and $c = 3$ .

Common Traps & FAQs

Wait, why do we need two conditions for the point where it 'touches' the x-axis? Isn't

y = 0

enough?

Great question! If the curve simply crossed the x-axis,

y = 0

would be sufficient. But 'touching' implies tangency. A tangent line at a point on the x-axis is horizontal, meaning its slope is zero. Thus,

y^{'} = 0

is the second, crucial condition.

Is there a faster way to solve this using the roots of the cubic?

Yes! Since it touches at

x = - 2

(x + 2)^{2}

is a factor. You could write the cubic as

y = (x + 2)^{2} (a x + k)

. This is a powerful alternative, but the system of equations method is often more robust for JEE problems.

Why did we get two equations for

a

and

b

Because we have two unknowns (

a

and

b

) remaining after finding

c

. To solve for two variables, we need two independent linear equations. We derived these from the two conditions at point

P

: the position (

y = 0

) and the slope (

y^{'} = 0

Animated Solution for Mathematics - Differentiation: The curve $y = a x^{3} + b x^{2} + c x + 5$ , touches the $x$ -axis at $P (- 2, 0)$ and cuts the $y$ axis at a point $Q$ , where its gradient is 3. Find $a, b, c$ .

Visualized Solution

Visualizing the Cubic Curve

Coordinates of Point $Q$

Finding $c$ using Gradient at $Q$

Calculating the Value of $c$

Using Point $P (- 2, 0)$

First Equation in $a$ and $b$

The Meaning of "Touches the $x$ -axis"

Second Equation in $a$ and $b$

Solving for $a$

Final Values of $a$ and $b$

Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Landscape of the Cubic Curve

The Gate of the Y-Intercept

The Geometry of the Touch

The Algebraic Resolution

Final Result

Similar Questions

the tangent at the point $(x_{1}, y_{1})$ on the curve $y = x^{3} + 3 x^{2} + 5$ passes through the origin, then $(x_{1}, y_{1})$ does NOT lie on the curve :

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Let $C$ be the curve $y^{3} - 3 x y + 2 = 0$ . If $H$ is the set of points on the curve $C$ where the tangent is horizontal and $V$ is the set of the point on the curve $C$ where the tangent is vertical then $H = .........$ and $V = .........$

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Let the normal at a point $P$ on the curve $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the $y$ -axis at $(0, \frac{3}{2})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $∣ m ∣$ is equal to

The equation of the tangent to the curve $y = x + \frac{4}{x ^{2}}$ , that is parallel to the $x$ -axis, is

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If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x = 12 (t + sin t cos t)$ , $y = 12 (1 + sin t)^{2}, 0 < t < \frac{π}{2}$ , with the positive x-axis is $\frac{π}{3}$ , then $y_{0}$ is equal to

Visualized Solution

Visualizing the Cubic Curve

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c using Gradient at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Using Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(−2,0)

First Equation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Meaning of "Touches the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis"

Second Equation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Final Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Landscape of the Cubic Curve

The Gate of the Y-Intercept

The Geometry of the Touch

The Algebraic Resolution

Final Result

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } the tangent at the point (x1​,y1​) on the curve y=x3+3x2+5 passes through the origin, then (x1​,y1​) does NOT lie on the curve :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The normal to the curve y(x−2)(x−3)=x+6 at the point where the curve intersects the y-axis passes through the point:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines x+y=a and x−y=b touch the curve y=x2−3x+2 at the points where the curve intersects the x-axis, then ba​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the curve y=x3 at the point P(t,t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the curve y=x2−3x​,x∈R,x=±3​, at a point (α,β)=(0,0) on it is parallel to the line 2x+6y−11=0, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the normal at a point P on the curve y2−3x2+y+10=0 intersect the y-axis at (0,23​). If m is the slope of the tangent at P to the curve, then ∣m∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the tangent to the curve y=x+x24​, that is parallel to the x-axis, is

Coordinates of Point $Q$

Finding $c$ using Gradient at $Q$

Calculating the Value of $c$

Using Point $P (- 2, 0)$

First Equation in $a$ and $b$

The Meaning of "Touches the $x$ -axis"

Second Equation in $a$ and $b$

Solving for $a$

Final Values of $a$ and $b$

the tangent at the point $(x_{1}, y_{1})$ on the curve $y = x^{3} + 3 x^{2} + 5$ passes through the origin, then $(x_{1}, y_{1})$ does NOT lie on the curve :

The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point where the curve intersects the y-axis passes through the point:

If the lines $x + y = a$ and $x - y = b$ touch the curve $y = x^{2} - 3 x + 2$ at the points where the curve intersects the $x$ -axis, then $\frac{a}{b}$ is equal to

If the tangent to the curve $y = x^{3}$ at the point $P (t, t^{3})$ meets the curve again at $Q$ , then the ordinate of the point which divides $P Q$ internally in the ratio $1 : 2$ is :

If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, x \neq = \pm 3$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :

Let the normal at a point $P$ on the curve $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the $y$ -axis at $(0, \frac{3}{2})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $∣ m ∣$ is equal to

The equation of the tangent to the curve $y = x + \frac{4}{x ^{2}}$ , that is parallel to the $x$ -axis, is