JEE Advanced 2005

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If $p (x)$ be a polynomial of degree 3 satisfying $p (- 1) = 10, p (1) = - 6$ and $p (x)$ has maxima at $x = - 1$ and $p^{'} (x)$ has minima at $x = 1$ . Find the distance between the local maxima and local minima of the curve.

Q: Wait, why is $p''(1) = 0$? I thought the minimum of $p(x)$ was at $x=1$.

That is a common trap! The problem states that $p'(x)$ has a minimum at $x=1$, not $p(x)$. In calculus, the minimum of a function $f(x)$ occurs where $f'(x) = 0$. Here, our function is $p'(x)$, so its minimum occurs where its derivative, which is $p''(x)$, is equal to zero. Thus, $p''(1) = 0$ is correct.

Q: Why do we need two constants of integration, $C$ and $D$?

A cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ has four unknown coefficients. When we integrate $p''(x)$ twice, we introduce two constants of integration. These, combined with the scaling factor $k$ and the original structure, allow us to satisfy the four conditions given in the problem: $p(-1)=10$, $p(1)=-6$, the maximum at $x=-1$, and the minimum of the derivative at $x=1$.

Progressive Hints

Hint 1

The Spark: Focus on the relationship between the derivative's extrema and the second derivative. If

p^{'} (x)

has a minimum at

x = 1

, what does that imply about the value of

p^{''} (1)

? Remember that for any differentiable function, the extremum of its derivative occurs where the second derivative is zero.

Hint 2

The Arsenal: Since

p (x)

is a cubic polynomial of the form

p (x) = a x^{3} + b x^{2} + c x + d

, its second derivative

p^{''} (x)

must be a linear function. Given that

p^{''} (1) = 0

, you can express the second derivative as

p^{''} (x) = k (x - 1)

for some constant

k

Hint 3

The Execution: Integrate

p^{''} (x)

to find

p^{'} (x) = \frac{k}{2} (x - 1)^{2} + C

. Use the condition that

p (x)

has a local maximum at

x = - 1

(meaning

p^{'} (- 1) = 0

) to find the relationship between

C

and

k

. Then, integrate

p^{'} (x)

to find

p (x)

and use the given values

p (- 1) = 10

and

p (1) = - 6

to solve for the constants.

Visualized Solution

Defining the Cubic Polynomial $p (x)$

Let $p (x)$ be a cubic polynomial: $p (x) = a x^{3} + b x^{2} + c x + d$ .
Given conditions:
1. $p (- 1) = 10$
2. $p (1) = - 6$
3. Local maxima at $x = - 1$
4. $p^{'} (x)$ has a minima at $x = 1$

Analyzing the Second Derivative $p^{''} (x)$

Since $p^{'} (x)$ has a minima at $x = 1$ , its derivative must vanish at this point.
Therefore, $p^{''} (1) = 0$ .
This point represents the inflection point of the cubic curve.

Formulating $p^{''} (x)$

Since $p (x)$ is a cubic polynomial (degree 3), its second derivative $p^{''} (x)$ must be a linear polynomial (degree 1).
Using the root $x = 1$ , we can write:
$p^{''} (x) = k (x - 1)$ for some non-zero constant $k$ .

Integrating to Find $p^{'} (x)$

Integrate $p^{''} (x)$ with respect to $x$ :
$p^{'} (x) = \int k (x - 1) d x = \frac{k}{2} (x - 1)^{2} + C$
Here, $C$ is the constant of integration.

Applying the Maxima Condition

Since $p (x)$ has a local maxima at $x = - 1$ , the slope at this point must be zero:
$p^{'} (- 1) = 0$
Substitute $x = - 1$ into the derivative equation:
$\frac{k}{2} (- 1 - 1)^{2} + C = 0 ⟹ 2 k + C = 0 ⟹ C = - 2 k$

Finding the Minima Location

Substitute $C = - 2 k$ back into $p^{'} (x)$ :
$p^{'} (x) = \frac{k}{2} (x - 1)^{2} - 2 k = \frac{k}{2} (x^{2} - 2 x - 3)$
Factoring the expression:
$p^{'} (x) = \frac{k}{2} (x - 3) (x + 1)$
The critical points are $x = - 1$ (maxima) and $x = 3$ (minima).

Integrating to Find $p (x)$

Integrate $p^{'} (x)$ to obtain the polynomial $p (x)$ :
$p (x) = \int \frac{k}{2} (x^{2} - 2 x - 3) d x$
$p (x) = \frac{k}{2} (\frac{x ^{3}}{3} - x^{2} - 3 x) + D$
Here, $D$ is another integration constant.

Using Given Values to Find Constants

Use $p (- 1) = 10$ :
$\frac{k}{2} (- \frac{1}{3} - 1 + 3) + D = 10 ⟹ \frac{5 k}{6} + D = 10$ (Eq. 1)
Use $p (1) = - 6$ :
$\frac{k}{2} (\frac{1}{3} - 1 - 3) + D = - 6 ⟹ - \frac{11 k}{6} + D = - 6$ (Eq. 2)

Solving for $k$ and $D$

Subtract Eq. 2 from Eq. 1:
$(\frac{5 k}{6} + D) - (- \frac{11 k}{6} + D) = 10 - (- 6)$
$\frac{16 k}{6} = 16 ⟹ k = 6$
Substitute $k = 6$ into Eq. 1:
$\frac{5 ( 6 )}{6} + D = 10 ⟹ 5 + D = 10 ⟹ D = 5$
The polynomial is: $p (x) = x^{3} - 3 x^{2} - 9 x + 5$

Coordinates of Maxima and Minima

Local Maxima point $A$ : $(- 1, p (- 1)) = (- 1, 10)$
Local Minima point $B$ occurs at $x = 3$ :
$p (3) = (3)^{3} - 3 (3)^{2} - 9 (3) + 5 = 27 - 27 - 27 + 5 = - 22$
So, the minima point is $B (3, - 22)$ .

Calculating the Distance

Use the distance formula between $A (- 1, 10)$ and $B (3, - 22)$ :
$d = (3 - (- 1))^{2} + (- 22 - 10)^{2}$
$d = 4^{2} + (- 32)^{2} = 16 + 1024$
$d = 1040 = 16 \times 65 = 465$
Key Takeaway: The distance between the local extrema is $465$ units.

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

The cubic polynomial

p (x)

defines a terrain whose peaks and valleys are governed by the principles of calculus. To determine the distance between these points, we must first reconstruct the polynomial using the provided constraints.

The Inflection Point

Our journey begins with the clue that the first derivative

p^{'} (x)

has a minimum at

x = 1

. In calculus, the minimum of a function occurs where its derivative is zero; thus, the second derivative must satisfy:

p^{''} (1) = 0

Since

p (x)

is a cubic polynomial,

p^{''} (x)

is a linear function. Given the root at

x = 1

, we can express the second derivative as:

p^{''} (x) = k (x - 1)

where

k

is a non-zero constant. This point represents the inflection point where the curve's concavity shifts.

The Slope Landscape

We climb back up the ladder of derivatives by integrating

p^{''} (x)

with respect to

x

p^{'} (x) = \int k (x - 1) d x = \frac{k}{2} (x - 1)^{2} + C

We know the curve has a local maximum at

x = - 1

, implying

p^{'} (- 1) = 0

. Substituting this into our slope equation:

\frac{k}{2} (- 1 - 1)^{2} + C = 0 \Rightarrow 2 k + C = 0 \Rightarrow C = - 2 k

Substituting

C

back into the expression for

p^{'} (x)

, we obtain:

p^{'} (x) = \frac{k}{2} (x - 1)^{2} - 2 k = \frac{k}{2} (x^{2} - 2 x - 3)

Factoring this quadratic reveals the critical points:

p^{'} (x) = \frac{k}{2} (x - 3) (x + 1)

The critical points are located at

x = - 1

(the maximum) and

x = 3

(the minimum).

Reconstructing the Polynomial

Integrating the slope function

p^{'} (x)

allows us to find the original polynomial

p (x)

p (x) = \int \frac{k}{2} (x^{2} - 2 x - 3) d x = \frac{k}{2} (\frac{x ^{3}}{3} - x^{2} - 3 x) + D

Using the conditions

p (- 1) = 10

and

p (1) = - 6

, we solve the resulting system of linear equations to find

k = 6

and

D = 5

. The resulting polynomial is:

p (x) = x^{3} - 3 x^{2} - 9 x + 5

Final Calculation

The local maximum point

A

is at

(- 1, 10)

. To find the local minimum point

B

, we evaluate the polynomial at

x = 3

p (3) = (3)^{3} - 3 (3)^{2} - 9 (3) + 5 = 27 - 27 - 27 + 5 = - 22

Thus, point

B

is at

(3, - 22)

. We now apply the distance formula to find the separation between

A

and

B

d = (3 - (- 1))^{2} + (- 22 - 10)^{2}

d = 4^{2} + (- 32)^{2} = 16 + 1024 = 1040

Simplifying the radical, we arrive at the final distance:

$d = 465$

Common Traps & FAQs

Wait, why is

p^{''} (1) = 0

? I thought the minimum of

p (x)

was at

x = 1

That is a common trap! The problem states that

p^{'} (x)

has a minimum at

x = 1

, not

p (x)

. In calculus, the minimum of a function

f (x)

occurs where

f^{'} (x) = 0

. Here, our function is

p^{'} (x)

, so its minimum occurs where its derivative, which is

p^{''} (x)

, is equal to zero. Thus,

p^{''} (1) = 0

is correct.

Why do we need two constants of integration,

C

and

D

A cubic polynomial

p (x) = a x^{3} + b x^{2} + c x + d

has four unknown coefficients. When we integrate

p^{''} (x)

twice, we introduce two constants of integration. These, combined with the scaling factor

k

and the original structure, allow us to satisfy the four conditions given in the problem:

p (- 1) = 10

p (1) = - 6

, the maximum at

x = - 1

, and the minimum of the derivative at

x = 1

Can we just use the formula for the distance between extrema of a cubic?

While there are shortcut formulas for the distance between local extrema of a cubic, they are often prone to memorization errors and don't help you understand the underlying calculus. Deriving the polynomial step-by-step as we did ensures you understand the geometry of the curve and is much safer for JEE Advanced.

Animated Solution for Mathematics - Differentiation: If $p (x)$ be a polynomial of degree 3 satisfying $p (- 1) = 10, p (1) = - 6$ and $p (x)$ has maxima at $x = - 1$ and $p^{'} (x)$ has minima at $x = 1$ . Find the distance between the local maxima and local minima of the curve.

Visualized Solution

Defining the Cubic Polynomial $p (x)$

Analyzing the Second Derivative $p^{''} (x)$

Formulating $p^{''} (x)$

Integrating to Find $p^{'} (x)$

Applying the Maxima Condition

Finding the Minima Location

Integrating to Find $p (x)$

Using Given Values to Find Constants

Solving for $k$ and $D$

Coordinates of Maxima and Minima

Calculating the Distance

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Inflection Point

The Slope Landscape

Reconstructing the Polynomial

Final Calculation

Similar Questions

Let $f (x)$ be a polynomial of degree 3 such that $f (- 1) = 10, f (1) = - 6, f (x)$ has a critical point at $x = - 1$ and $f^{'} (x)$ has a critical point at $x = 1$ . Then the local minima at $x = ______$

Let $p (x)$ be a real polynomial of least degree which has a local maximum at $x = 1$ and a local minimum at $x = 3$ . If $p (1) = 6$ and $p (3) = 2$ , then $p^{'} (0)$ is

Let $f (x)$ be a polynomial of degree $3$ such that $f (- 1) = 10$ , $f (1) = - 6$ , $f (x)$ has a critical point at $x = - 1$ and $f^{'} (x)$ has a critical point at $x = 1$ . Then $f (x)$ has a local minima at $x =$

Let $f (x)$ be a cubic polynomial with $f (1) = - 10, f (- 1) = 6$ , and has a local minima at $x = 1$ , and $f^{'} (x)$ has a local minima at $x = - 1$ . Then $f (3)$ is equal to .

$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

If $p (x)$ be a polynomial of degree three that has a local maximum value $8$ at $x = 1$ and a local minimum value $4$ at $x = 2$ ; then $p (0)$ is equal to :

Let $f (x) = ∣ (x - 1) (x^{2} - 2 x - 3) ∣ + x - 3, x \in R$ . If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ______

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where $a > 0$ , attains its local maximum and local minimum values at $p$ and $q$ , respectively, such that $p^{2} = q$ , then $f (3)$ is equal to:

Let $f (x)$ be a polynomial of degree $6$ in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x = - 1$ and $x = 1$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$ , then $5 \cdot f (2)$ is equal to

Visualized Solution

Defining the Cubic Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p(x)

Analyzing the Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p′′(x)

Formulating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p′′(x)

Integrating to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p′(x)

Applying the Maxima Condition

Finding the Minima Location

Integrating to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p(x)

Using Given Values to Find Constants

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

Coordinates of Maxima and Minima

Calculating the Distance

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Inflection Point

The Slope Landscape

Reconstructing the Polynomial

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a cubic polynomial with f(1)=−10,f(−1)=6, and has a local minima at x=1, and f′(x) has a local minima at x=−1. Then f(3) is equal to .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) is cubic polynomial with f(2)=18 and f(1)=−1. Also f(x) has local maxima at x=−1 and f′(x) has local minima at x=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If p(x) be a polynomial of degree three that has a local maximum value 8 at x=1 and a local minimum value 4 at x=2; then p(0) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)=2x3−9ax2+12a2x+1, where a>0, attains its local maximum and local minimum values at p and q, respectively, such that p2=q, then f(3) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x=−1 and x=1. If limx→0​x3f(x)​=1, then 5⋅f(2) is equal to

Defining the Cubic Polynomial $p (x)$

Analyzing the Second Derivative $p^{''} (x)$

Formulating $p^{''} (x)$

Integrating to Find $p^{'} (x)$

Integrating to Find $p (x)$

Solving for $k$ and $D$

Let $p (x)$ be a real polynomial of least degree which has a local maximum at $x = 1$ and a local minimum at $x = 3$ . If $p (1) = 6$ and $p (3) = 2$ , then $p^{'} (0)$ is

Let $f (x)$ be a cubic polynomial with $f (1) = - 10, f (- 1) = 6$ , and has a local minima at $x = 1$ , and $f^{'} (x)$ has a local minima at $x = - 1$ . Then $f (3)$ is equal to .

$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

If $p (x)$ be a polynomial of degree three that has a local maximum value $8$ at $x = 1$ and a local minimum value $4$ at $x = 2$ ; then $p (0)$ is equal to :

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where $a > 0$ , attains its local maximum and local minimum values at $p$ and $q$ , respectively, such that $p^{2} = q$ , then $f (3)$ is equal to:

Let $f (x)$ be a polynomial of degree $6$ in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x = - 1$ and $x = 1$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$ , then $5 \cdot f (2)$ is equal to