JEE Main 2021 (01 Sep Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: The function $f (x) = x^{3} - 6 x^{2} + a x + b$ is such that $f (2) = f (4) = 0$ . Consider two statements. \\ (S1) there exists $x_{1}, x_{2} \in (2, 4), x_{1} < x_{2}$ , such that $f^{'} (x_{1}) = - 1$ and $f^{'} (x_{2}) = 0$ . \\ (S2) there exists $x_{3}, x_{4} \in (2, 4), x_{3} < x_{4}$ , such that $f$ is decreasing in $(2, x_{4})$ , increasing in $(x_{4}, 4)$ and $2 f^{'} (x_{3}) = 3 f (x_{4})$ . \\ Then

Q: Wait, why do we assume $f(x)$ is decreasing in $(2, x_4)$ and increasing in $(x_4, 4)$? Can't it be the other way around?

Great observation! We determine this by looking at the second derivative $f''(x) = 6x - 12$. For any $x > 2$, $f''(x)$ is positive, meaning the slope $f'(x)$ is strictly increasing. Since $f'(2) = -4$ (negative) and $f'(4) = 8$ (positive), the slope must cross zero exactly once at $x_4$. Before $x_4$, the slope is negative (decreasing function), and after $x_4$, the slope is positive (increasing function). The math dictates the behavior, not our assumption!

Q: Why are we using the Intermediate Value Theorem (IVT) here? Isn't that for continuous functions?

Exactly! And since $f(x)$ is a polynomial, it is continuous and differentiable everywhere. The derivative $f'(x)$ is also a polynomial, so it is continuous. If a continuous function takes a value $y_1$ at $x_a$ and $y_2$ at $x_b$, it must take every value between $y_1$ and $y_2$ at some point in between. That is the power of IVT—it guarantees existence without requiring us to solve for the exact coordinate.

Q: Is it possible that $x_1$ and $x_2$ in (S1) are the same point?

The statement specifies $x_1 < x_2$. Since $f'(x)$ is strictly increasing on $(2, 4)$, it is a one-to-one function. This means for any two distinct values of the derivative, the corresponding $x$-values must also be distinct. Because $-1 \neq 0$, the $x$-values $x_1$ and $x_2$ must be different. The strict monotonicity of the derivative is the key to this proof.

Q: How do I handle the calculation of $f(x_4)$ without getting lost in the algebra?

The trick is to use the fact that $x_4$ is a root of $f'(x) = 3x^2 - 12x + 8 = 0$. Instead of plugging $x_4 = 2 + \frac{2}{\sqrt{3}}$ directly into the cubic, you can perform polynomial division of $f(x)$ by $f'(x)$ or use the property that $3x_4^2 = 12x_4 - 8$. This reduces the degree of the expression and makes the arithmetic much cleaner.

Q: Why does $2f'(x_3) = \sqrt{3}f(x_4)$ imply $x_3$ exists?

Once you calculate $f(x_4) = -\frac{16}{3\sqrt{3}}$, the right side of the equation becomes $-\frac{16}{3}$. Dividing by 2, we need $f'(x_3) = -\frac{8}{3}$. Since we know $f'(2) = -4$ and $f'(x_4) = 0$, and $-4 < -\frac{8}{3} < 0$, the IVT guarantees that there is some $x_3$ in the interval $(2, x_4)$ where the slope is exactly $-\frac{8}{3}$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Before diving into the calculus, treat the function

f (x) = x^{3} - 6 x^{2} + a x + b

as a geometric object. You have two roots at

x = 2

and

x = 4

. Use these to lock down the values of

a

and

b

immediately. Once you have the full polynomial, you are no longer guessing; you are exploring a fixed landscape.

Hint 2

Hint 2: To evaluate the statements, you need the derivative

f^{'} (x)

. Remember that

f^{'} (x)

tells you the slope of the tangent. Since

f (x)

is a cubic,

f^{'} (x)

is a quadratic. Analyze the range of this quadratic on the interval

(2, 4)

using the Intermediate Value Theorem to see if specific slope values are 'hit' by the function.

Hint 3

Hint 3: For statement (S2), identify the local minimum

x_{4}

by setting

f^{'} (x) = 0

. Once you have

x_{4}

, calculate the function value

f (x_{4})

. The condition

2 f^{'} (x_{3}) = 3 f (x_{4})

then becomes a simple search for a value

f^{'} (x_{3})

within the range of the derivative on the interval

(2, x_{4})

Visualized Solution

$I n t r o d u c t i o n t o t h e F u n c t i o n$

Given function: $f (x) = x^{3} - 6 x^{2} + a x + b$
We are given two roots: $f (2) = 0$ and $f (4) = 0$
Our goal is to analyze the behavior of $f (x)$ and its derivative in the interval $(2, 4)$ .

$S e tt in g u pE q u a t i o n s f or aan d b$

Substitute $x = 2$ : $f (2) = 2^{3} - 6 (2)^{2} + 2 a + b = 0$
$8 - 24 + 2 a + b = 0 \Rightarrow 2 a + b = 16$
Substitute $x = 4$ : $f (4) = 4^{3} - 6 (4)^{2} + 4 a + b = 0$
$64 - 96 + 4 a + b = 0 \Rightarrow 4 a + b = 32$

$S o l v in g f or t h e C o n s t an t s$

Subtract the first equation from the second:
$(4 a + b) - (2 a + b) = 32 - 16$
$2 a = 16 \Rightarrow a = 8$
Substitute $a = 8$ back: $2 (8) + b = 16 \Rightarrow b = 0$
The complete function is: $f (x) = x^{3} - 6 x^{2} + 8 x$

$A na l y z in g t h eD er i v a t i v e f^{'} (x)$

First derivative: $f^{'} (x) = 3 x^{2} - 12 x + 8$
Second derivative: $f^{''} (x) = 6 x - 12$
For $x \in (2, 4)$ , $f^{''} (x) = 6 (x - 2) > 0$
Therefore, $f^{'} (x)$ is strictly increasing in $(2, 4)$ .

$B o u n d a r y V a l u eso f f^{'} (x)$

Evaluate slope at $x = 2$ : $f^{'} (2) = 3 (2)^{2} - 12 (2) + 8 = - 4$
Evaluate slope at $x = 4$ : $f^{'} (4) = 3 (4)^{2} - 12 (4) + 8 = 8$
The range of $f^{'} (x)$ for $x \in [2, 4]$ is $[- 4, 8]$ .

$V er i f y in g S t a t e m e n t (S 1)$

We need $x_{1}, x_{2} \in (2, 4)$ such that $f^{'} (x_{1}) = - 1$ and $f^{'} (x_{2}) = 0$ .
Since $- 4 < - 1 < 8$ and $- 4 < 0 < 8$ , by Intermediate Value Theorem, such points exist.
Because $f^{'} (x)$ is strictly increasing, $f^{'} (x_{1}) < f^{'} (x_{2}) \Rightarrow x_{1} < x_{2}$ .
Conclusion: Statement (S1) is True.

$F in d in g t h e C r i t i c a l P o in t x_{4}$

Statement (S2) mentions $x_{4}$ where monotonicity changes.
Set $f^{'} (x) = 0 \Rightarrow 3 x^{2} - 12 x + 8 = 0$
$x = \frac{12 \pm 144 - 96}{6} = \frac{12 \pm 4 3}{6} = 2 \pm \frac{2}{3}$
In the interval $(2, 4)$ , the critical point is $x_{4} = 2 + \frac{2}{3}$ .

$M o n o t o ni c i t y a r o u n d x_{4}$

For $x \in (2, x_{4})$ , $f^{'} (x) < 0 \Rightarrow f (x)$ is decreasing.
For $x \in (x_{4}, 4)$ , $f^{'} (x) > 0 \Rightarrow f (x)$ is increasing.
This matches the first part of Statement (S2).

$C a l c u l a t in g f (x_{4})$

We need the value of $f (x_{4})$ for the condition in (S2).
$f (x_{4}) = f (2 + \frac{2}{3})$
Using $f (x) = x^{3} - 6 x^{2} + 8 x$ , after careful expansion and simplification:
$f (x_{4}) = - \frac{16}{3 3}$

$S e tt in g u pt h e C o n d i t i o n f or x_{3}$

Statement (S2) requires: $2 f^{'} (x_{3}) = 3 f (x_{4})$
Substitute the value of $f (x_{4})$ :
$2 f^{'} (x_{3}) = 3 (- \frac{16}{3 3})$
$2 f^{'} (x_{3}) = - \frac{16}{3} \Rightarrow f^{'} (x_{3}) = - \frac{8}{3}$

$V er i f y in g t h e E x i s t e n ceo f x_{3}$

We need $x_{3} \in (2, x_{4})$ such that $f^{'} (x_{3}) = - \frac{8}{3}$ .
At the boundaries of this sub-interval:
$f^{'} (2) = - 4$ and $f^{'} (x_{4}) = 0$
Since $- 4 < - \frac{8}{3} < 0$ , by Intermediate Value Theorem, such an $x_{3}$ definitely exists.
Conclusion: Statement (S2) is True.

$F ina l A n s w er$

Statement (S1) is verified to be True.
Statement (S2) is verified to be True.
Both statements are mathematically sound based on the properties of the derivative and the Intermediate Value Theorem.
Final Answer: Both (S1) and (S2) are true.

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

The landscape is defined by the function

f (x) = x^{3} - 6 x^{2} + a x + b

. We are given that the function vanishes at

x = 2

and

x = 4

, which implies

f (2) = 0

and

f (4) = 0

Substituting these values into the function, we obtain the following system of linear equations:

8 - 24 + 2 a + b = 0 \Rightarrow 2 a + b = 16

64 - 96 + 4 a + b = 0 \Rightarrow 4 a + b = 32

Solving this system, we subtract the first equation from the second to find

2 a = 16

, which yields

a = 8

. Substituting

a = 8

back into the first equation, we find

b = 0

. Thus, the function is uniquely determined as:

f (x) = x^{3} - 6 x^{2} + 8 x

The Pulse of the Function

The Derivative

To understand the behavior of the landscape, we examine the first derivative, which represents the slope:

f^{'} (x) = 3 x^{2} - 12 x + 8

The second derivative,

f^{''} (x) = 6 x - 12

, provides insight into the concavity. For any

x > 2

f^{''} (x) > 0

, which implies that the slope

f^{'} (x)

is strictly increasing on the interval

(2, \infty)

Since

f^{'} (2) = - 4

and

f^{'} (4) = 8

, the Intermediate Value Theorem guarantees that the slope takes on every value in the interval

[- 4, 8]

. Consequently, there must exist an

x_{1}

such that

f^{'} (x_{1}) = - 1

and an

x_{2}

such that

f^{'} (x_{2}) = 0

. Because the slope is strictly increasing, it follows that

x_{1} < x_{2}

The Critical Point and the Mystery of $x_{4}$

Statement (S2) requires an analysis of the interval

(2, 4)

. We locate the critical point

x_{4}

by setting the first derivative to zero:

3 x^{2} - 12 x + 8 = 0

Using the quadratic formula, we find the root in the interval

(2, 4)

to be:

x_{4} = 2 + \frac{2}{3}

Evaluating the function at this critical point, we determine the local minimum value:

f (x_{4}) = - \frac{16}{3 3}

The Final Synthesis

We now address the condition

2 f^{'} (x_{3}) = 3 f (x_{4})

. Substituting the value of

f (x_{4})

derived above, the equation becomes:

2 f^{'} (x_{3}) = 3 (- \frac{16}{3 3}) = - \frac{16}{3}

This simplifies to the requirement that

f^{'} (x_{3}) = - \frac{8}{3}

In the interval

(2, x_{4})

, the slope

f^{'} (x)

varies continuously from

f^{'} (2) = - 4

f^{'} (x_{4}) = 0

. Since

- \frac{8}{3}

lies within the interval

(- 4, 0)

, the Intermediate Value Theorem guarantees the existence of a point

x_{3}

satisfying the condition.

Both statements (S1) and (S2) are verified as true.

Common Traps & FAQs

Wait, why do we assume

f (x)

is decreasing in

(2, x_{4})

and increasing in

(x_{4}, 4)

? Can't it be the other way around?

Great observation! We determine this by looking at the second derivative

f^{''} (x) = 6 x - 12

. For any

x > 2

f^{''} (x)

is positive, meaning the slope

f^{'} (x)

is strictly increasing. Since

f^{'} (2) = - 4

(negative) and

f^{'} (4) = 8

(positive), the slope must cross zero exactly once at

x_{4}

. Before

x_{4}

, the slope is negative (decreasing function), and after

x_{4}

, the slope is positive (increasing function). The math dictates the behavior, not our assumption!

Why are we using the Intermediate Value Theorem (IVT) here? Isn't that for continuous functions?

Exactly! And since

f (x)

is a polynomial, it is continuous and differentiable everywhere. The derivative

f^{'} (x)

is also a polynomial, so it is continuous. If a continuous function takes a value

y_{1}

x_{a}

and

y_{2}

x_{b}

, it must take every value between

y_{1}

and

y_{2}

at some point in between. That is the power of IVT—it guarantees existence without requiring us to solve for the exact coordinate.

Is it possible that

x_{1}

and

x_{2}

in (S1) are the same point?

The statement specifies

x_{1} < x_{2}

. Since

f^{'} (x)

is strictly increasing on

(2, 4)

, it is a one-to-one function. This means for any two distinct values of the derivative, the corresponding

x

-values must also be distinct. Because

- 1 \neq = 0

, the

x

-values

x_{1}

and

x_{2}

must be different. The strict monotonicity of the derivative is the key to this proof.

How do I handle the calculation of

f (x_{4})

without getting lost in the algebra?

The trick is to use the fact that

x_{4}

is a root of

f^{'} (x) = 3 x^{2} - 12 x + 8 = 0

. Instead of plugging

x_{4} = 2 + \frac{2}{3}

directly into the cubic, you can perform polynomial division of

f (x)

f^{'} (x)

or use the property that

3 x_{4}^{2} = 12 x_{4} - 8

. This reduces the degree of the expression and makes the arithmetic much cleaner.

Why does

2 f^{'} (x_{3}) = 3 f (x_{4})

imply

x_{3}

exists?

Once you calculate

f (x_{4}) = - \frac{16}{3 3}

, the right side of the equation becomes

- \frac{16}{3}

. Dividing by 2, we need

f^{'} (x_{3}) = - \frac{8}{3}

. Since we know

f^{'} (2) = - 4

and

f^{'} (x_{4}) = 0

, and

- 4 < - \frac{8}{3} < 0

, the IVT guarantees that there is some

x_{3}

in the interval

(2, x_{4})

where the slope is exactly

- \frac{8}{3}

Select Answer:

Visualized Solution

$I n t r o d u c t i o n t o t h e F u n c t i o n$

$S e tt in g u pE q u a t i o n s f or aan d b$

$S o l v in g f or t h e C o n s t an t s$

$A na l y z in g t h eD er i v a t i v e f^{'} (x)$

$B o u n d a r y V a l u eso f f^{'} (x)$

$V er i f y in g S t a t e m e n t (S 1)$

$F in d in g t h e C r i t i c a l P o in t x_{4}$

$M o n o t o ni c i t y a r o u n d x_{4}$

$C a l c u l a t in g f (x_{4})$

$S e tt in g u pt h e C o n d i t i o n f or x_{3}$

$V er i f y in g t h e E x i s t e n ceo f x_{3}$

$F ina l A n s w er$

The Sigma Insight: Monotonicity

Analyzing the Setup

The Pulse of the Function

The Critical Point and the Mystery of $x_{4}$

The Final Synthesis

Similar Questions

Let $f : R \to R$ be defined by $f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$ . Then which of the following statements is (are) TRUE ?

Comprehension Passage

Consider the statements: $P$ : There exists some $x \in R$ such that $f (x) + 2 x = 2 (1 + x^{2})$ , $Q$ : There exists some $x \in R$ such that $2 f (x) + 1 = 2 x (1 + x)$

Which of the following is true?

For the function $f (x) = (cos x) - x + 1, x \in R$ , between the following two statements (S1) $f (x) = 0$ for only one value of $x$ in $[0, π]$ . (S2) $f (x)$ is decreasing in $[0, \frac{π}{2}]$ and increasing in $[\frac{π}{2}, π]$ .

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

If $f : R \to R$ is a differentiable function such that $f^{'} (x) > 2 f (x)$ for all $x \in R$ , and $f (0) = 1$ , then

If the function $f$ given by $f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$ , for some $a \in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$ , then a root of the equation $\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0 (x \neq = 1)$ is :

For the function $f (x) = sin x + 3 x - \frac{2}{π} (x^{2} + x)$ , where $x \in [0, \frac{π}{2}]$ , consider the following two statements : (I) $f$ is increasing in $(0, \frac{π}{2})$ . (II) $f^{'}$ is decreasing in $(0, \frac{π}{2})$ . Between the above two statements,

Consider the following three statements for the function $f : (0, \infty) \to R$ defined by $f (x) = ∣ lo g_{e} x ∣ - ∣ x - 1∣$ : (I) $f$ is differentiable at all $x > 0$ . (II) $f$ is increasing in $(0, 1)$ . (III) $f$ is decreasing in $(1, \infty)$ . Then.

Select Answer:

Visualized Solution

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VerifyingStatement(S1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingtheCriticalPointx4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Monotonicityaroundx4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculatingf(x4​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SettinguptheConditionforx3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VerifyingtheExistenceofx3​

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

The Sigma Insight: Monotonicity

Analyzing the Setup

The Pulse of the Function

The Critical Point and the Mystery of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x4​

The Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined by f(x)=x2+2x+4x2−3x−6​. Then which of the following statements is (are) TRUE ?

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the statements: P: There exists some x∈R such that f(x)+2x=2(1+x2), Q: There exists some x∈R such that 2f(x)+1=2x(1+x)

Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)={−34​x3+2x2+3x3xex​,x>0,x≤0​. Then f is increasing function in the interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:R→R is a differentiable function such that f′(x)>2f(x) for all x∈R, and f(0)=1, then

$I n t r o d u c t i o n t o t h e F u n c t i o n$

$S e tt in g u pE q u a t i o n s f or aan d b$

$S o l v in g f or t h e C o n s t an t s$

$A na l y z in g t h eD er i v a t i v e f^{'} (x)$

$B o u n d a r y V a l u eso f f^{'} (x)$

$V er i f y in g S t a t e m e n t (S 1)$

$F in d in g t h e C r i t i c a l P o in t x_{4}$

$M o n o t o ni c i t y a r o u n d x_{4}$

$C a l c u l a t in g f (x_{4})$

$S e tt in g u pt h e C o n d i t i o n f or x_{3}$

$V er i f y in g t h e E x i s t e n ceo f x_{3}$

$F ina l A n s w er$

The Critical Point and the Mystery of $x_{4}$

Let $f : R \to R$ be defined by $f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$ . Then which of the following statements is (are) TRUE ?

Consider the statements: $P$ : There exists some $x \in R$ such that $f (x) + 2 x = 2 (1 + x^{2})$ , $Q$ : There exists some $x \in R$ such that $2 f (x) + 1 = 2 x (1 + x)$

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

If $f : R \to R$ is a differentiable function such that $f^{'} (x) > 2 f (x)$ for all $x \in R$ , and $f (0) = 1$ , then