JEE Main 2020 - 8 Jan (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: 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 $x =$

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Do not confuse the critical points of the function

f (x)

with the critical points of its derivative

f^{'} (x)

. A critical point of

f (x)

x = - 1

means the slope is zero there, i.e.,

f^{'} (- 1) = 0

. However, a critical point of

f^{'} (x)

x = 1

means the slope of the slope is zero, which is the definition of the second derivative being zero, i.e.,

f^{''} (1) = 0

. Visualize the curve: one is a stationary point, the other is an inflection point.

Hint 2

Hint 2: The Arsenal. Since

f (x)

is a cubic polynomial,

f^{'} (x)

must be a quadratic polynomial. You know one root of

f^{'} (x)

x = - 1

. Therefore, you can express

f^{'} (x)

in the factored form

f^{'} (x) = A (x + 1) (x - r_{2})

, where

r_{2}

is the unknown second root. Use the condition

f^{''} (1) = 0

to find

r_{2}

by differentiating this expression.

Hint 3

Hint 3: The Execution. Once you have

f^{'} (x) = A (x^{2} - 2 x - 3)

, integrate it to find

f (x) = A (\frac{x ^{3}}{3} - x^{2} - 3 x) + B

. Now, use the two given coordinate points

f (- 1) = 10

and

f (1) = - 6

to set up a system of two linear equations in terms of

A

and

B

. Solve for these constants to fully define your polynomial.

Visualized Solution

Given Data & Coordinates

Let $f (x)$ be a cubic polynomial.
Given points: $f (- 1) = 10$ and $f (1) = - 6$ .

First Critical Point

Critical point of $f (x)$ at $x = - 1$ .
This implies the tangent is horizontal: $f^{'} (- 1) = 0$ .

Inflection Point Condition

Critical point of $f^{'} (x)$ at $x = 1$ .
This implies the second derivative is zero: $f^{''} (1) = 0$ .

Formulating $f^{'} (x)$

Since $f^{'} (- 1) = 0$ , $(x + 1)$ is a factor.
Let $f^{'} (x) = A (x + 1) (x - r_{2})$ .',' $r_{2}$ is the second critical point.'],
voiceover_hindi":
तो आगे बढ़ते हैं... अब हम एफ डैश एक्स की इक्वेशन बनाते हैं। हमें पता है कि एक्स इक्वल्स माइनस वन इसका एक रूट है, क्योंकि एफ डैश ऑफ माइनस वन ज़ीरो है। तो हम एफ डैश एक्स को ए टाइम्स एक्स प्लस वन, टाइम्स एक्स माइनस आर टू लिख सकते हैं। यहाँ आर टू हमारा दूसरा क्रिटिकल पॉइंट है जो हमें अभी ढूँढना है। बात समझ आ रही है?
voiceover_english":
So, let's move forward... Now we will construct the equation for f prime of x. We know that x equals minus one is one of its roots, because f prime of minus one is zero. So we can write f prime of x as A times x plus one, times x minus r two. Here, r two is our second critical point which we need to find. Are you getting the point?
active_animations":
[{
id":
max_tangent",
effect":
highlight"}]},{
index":
4
title":
Finding the Second Derivative",
content":
[
Differentiate $f^{'} (x)$ using the product rule.
$f^{''} (x) = A [1 \cdot (x - r_{2}) + (x + 1) \cdot 1]$ .
$f^{''} (x) = A (2 x + 1 - r_{2})$ .

Locating the Second Root

Substitute $x = 1$ into $f^{''} (x) = 0$ .
$A (2 (1) + 1 - r_{2}) = 0$ .
$3 - r_{2} = 0 ⟹ r_{2} = 3$ .

The Derivative Function

Substitute $r_{2} = 3$ back into $f^{'} (x)$ .
$f^{'} (x) = A (x + 1) (x - 3) = A (x^{2} - 2 x - 3)$ .
Critical points are at $x = - 1$ and $x = 3$ .

Integrating for $f (x)$

Integrate $f^{'} (x)$ to find the original function.
$f (x) = A \int (x^{2} - 2 x - 3) d x + B$ .
$f (x) = A (\frac{x ^{3}}{3} - x^{2} - 3 x) + B$ .

Applying Boundary Conditions

Substitute $x = - 1$ and $f (- 1) = 10$ .
$A (- \frac{1}{3} - 1 + 3) + B = 10$ .
$\frac{5 A}{3} + B = 10$ .

Applying Second Condition

Substitute $x = 1$ and $f (1) = - 6$ .
$A (\frac{1}{3} - 1 - 3) + B = - 6$ .
$- \frac{11 A}{3} + B = - 6$ .

Solving for Constants

Subtract the two equations: $\frac{16 A}{3} = 16 ⟹ A = 3$ .
Substitute $A = 3$ to get $B = 5$ .
The function is $f (x) = x^{3} - 3 x^{2} - 9 x + 5$ .

First Derivative Test

$f^{'} (x) = 3 (x + 1) (x - 3)$ .
For $x \in (- 1, 3)$ , $f^{'} (x) < 0$ (Function is decreasing).
For $x > 3$ , $f^{'} (x) > 0$ (Function is increasing).

Conclusion: Local Minima

The sign of $f^{'} (x)$ changes from $-$ to $+$ at $x = 3$ .
Therefore, $f (x)$ has a local minima at $x = 3$ .

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a math problem; we are going to map the terrain of a cubic polynomial. Imagine you are standing on a landscape defined by the function

f (x)

This landscape has hills (maxima) and valleys (minima). Our mission is to find the exact location of one of these valleys by reconstructing the function from the ground up.

Decoding the Clues

We are given a cubic polynomial

f (x)

with two anchor points:

f (- 1) = 10

and

f (1) = - 6

The problem states

f (x)

has a critical point at

x = - 1

. In the language of calculus, this means the tangent is horizontal, so

f^{'} (- 1) = 0

We are also told

f^{'} (x)

has a critical point at

x = 1

. This implies that the derivative of the derivative—the second derivative—must be zero at this point:

f^{''} (1) = 0

. This is the point of inflection, the 'hinge' of our cubic curve.

Constructing the Derivative

Since

f (x)

is a cubic, its derivative

f^{'} (x)

must be a quadratic. We know one root of this quadratic is

x = - 1

because

f^{'} (- 1) = 0

. Let the other root be

r_{2}

We can write the derivative in its factored form:

f^{'} (x) = A (x + 1) (x - r_{2})

To find

r_{2}

, we differentiate

f^{'} (x)

to obtain

f^{''} (x)

f^{''} (x) = A \cdot \frac{d}{d x} [(x + 1) (x - r_{2})] = A [(x - r_{2}) + (x + 1)] = A (2 x + 1 - r_{2})

Given

f^{''} (1) = 0

, we substitute

x = 1

A (2 (1) + 1 - r_{2}) = 0 \Rightarrow 3 - r_{2} = 0 \Rightarrow r_{2} = 3

Thus, our derivative is

f^{'} (x) = A (x + 1) (x - 3) = A (x^{2} - 2 x - 3)

The Detective Work

We find the original function

f (x)

by integrating the derivative:

f (x) = \int A (x^{2} - 2 x - 3) d x = A (\frac{x ^{3}}{3} - x^{2} - 3 x) + B

We use our anchor points to solve for the constants

A

and

B

. For

f (- 1) = 10

A (\frac{( - 1 ) ^{3}}{3} - (- 1)^{2} - 3 (- 1)) + B = 10 \Rightarrow \frac{5 A}{3} + B = 10

For

f (1) = - 6

A (\frac{1 ^{3}}{3} - 1^{2} - 3 (1)) + B = - 6 \Rightarrow - \frac{11 A}{3} + B = - 6

Subtracting these equations eliminates

B

to yield

\frac{16 A}{3} = 16

, which means

A = 3

. Substituting

A = 3

back into the first equation gives

B = 5

The resulting function is:

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

The Final Verdict

We identify the local minima by examining the derivative

f^{'} (x) = 3 (x + 1) (x - 3)

. The critical points are

x = - 1

and

x = 3

Testing the slope around

x = 3

: - If

x < 3

(and

x > - 1

f^{'} (x)

is negative, meaning the function is decreasing. - If

x > 3

f^{'} (x)

is positive, meaning the function is increasing.

When a function stops decreasing and starts increasing, it has reached the bottom of a valley. Thus, the local minima occurs at $x = 3$ .

Common Traps & FAQs

Wait, why is

f^{''} (1) = 0

just because

f^{'} (x)

has a critical point at

x = 1

That is a fantastic question! Think about the definition of a critical point. A function

g (x)

has a critical point where its derivative

g^{'} (x)

is zero. Here, our function is

f^{'} (x)

. So, if

f^{'} (x)

has a critical point at

x = 1

, then the derivative of

f^{'} (x)

—which is

f^{''} (x)

—must be zero at

x = 1

. It is just applying the definition of a critical point to a different function!

Why can't we just assume

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

and solve for four variables?

You absolutely could! That is a valid, brute-force method. However, in JEE Advanced, time is your most precious resource. By using the properties of the derivative and the factored form, we reduced the problem to finding only two constants (

A

and

B

) instead of four. Always look for the 'geometric shortcut'—it is usually faster and less prone to calculation errors.

How do I know if

x = 3

is a local minima or a local maxima?

Great intuition! We use the First Derivative Test. We found

f^{'} (x) = 3 (x + 1) (x - 3)

. If you pick a value slightly less than

3

(like

x = 2

f^{'} (x)

is negative. If you pick a value slightly greater than

3

(like

x = 4

f^{'} (x)

is positive. Since the slope changes from negative to positive, the function is 'dipping' down and then coming back up—that is the definition of a local minima.

Animated Solution for Mathematics - Differentiation: 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 $x =$

Enter Numerical Value:

Visualized Solution

Given Data & Coordinates

First Critical Point

Inflection Point Condition

Formulating $f^{'} (x)$

Locating the Second Root

The Derivative Function

Integrating for $f (x)$

Applying Boundary Conditions

Applying Second Condition

Solving for Constants

First Derivative Test

Conclusion: Local Minima

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Decoding the Clues

Constructing the Derivative

The Detective Work

The Final Verdict

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 $f (x)$ 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 .

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

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.

The function $f (x) = \int_{- 1}^{x} t (e^{t} - 1) (t - 1) (t - 2)^{3} (t - 3)^{5} d t$ has a local minimum at $x =$

If $x = 1$ is a critical point of the function $f (x) = (3 x^{2} + a x - 2- a) e^{x}$ , then :

The function $f (x) = 2 x + 3 (x)^{\frac{2}{3}}, x \in R$ , has

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Enter Numerical Value:

Visualized Solution

Given Data & Coordinates

First Critical Point

Inflection Point Condition

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

Locating the Second Root

The Derivative Function

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

Applying Boundary Conditions

Applying Second Condition

Solving for Constants

First Derivative Test

Conclusion: Local Minima

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Decoding the Clues

Constructing the Derivative

The Detective Work

The Final Verdict

Similar Questions

.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; } 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=∫−1x​t(et−1)(t−1)(t−2)3(t−3)5dt has a local minimum at x=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=1 is a critical point of the function f(x)=(3x2+ax−2–a)ex, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=2x+3(x)32​,x∈R, has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=2x​+x2​ has a local minimum at

.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 5 such that x=±1 are its critical points. If limx→0​(2+x3f(x)​)=4, then which one of the following is not true?

.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 5 such that x=±1 are its critical points. If limx→0​(2+x3f(x)​)=4, then which one of the following is not true?

Formulating $f^{'} (x)$

Integrating for $f (x)$

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 .

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

The function $f (x) = \int_{- 1}^{x} t (e^{t} - 1) (t - 1) (t - 2)^{3} (t - 3)^{5} d t$ has a local minimum at $x =$

If $x = 1$ is a critical point of the function $f (x) = (3 x^{2} + a x - 2- a) e^{x}$ , then :

The function $f (x) = 2 x + 3 (x)^{\frac{2}{3}}, x \in R$ , has

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?