JEE Main 2021 (31 Aug Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then

Q: Does this mean $f''(x)$ is always zero for all $x$?

Not at all! The theorem states that $f''(x) = 0$ for *at least one* point $x$ in the interval $(0, 2)$. It does not imply that the function is linear or that the second derivative is zero everywhere. It simply guarantees the existence of at least one point where the curvature (represented by the second derivative) vanishes.

Q: Why do we need the second derivative? Why not just stop at the first derivative?

The question asks us to evaluate properties of $f''(x)$. By applying Rolle's Theorem to $h(x)$, we find that $h'(x)$ has roots in $(0, 1)$ and $(1, 2)$. Since $h'(x)$ is zero at two distinct points, Rolle's Theorem applied to $h'(x)$ guarantees that its derivative, $h''(x)$, must be zero at some point between those two roots. This is the standard path to proving properties about higher-order derivatives.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the three points

(0, 0)

(1, 1)

, and

(2, 2)

on the Cartesian plane. They are collinear, lying on the line

y = x

. This suggests that the function

f (x)

behaves like

x

at these specific points, which is a strong hint to look at the difference between

f (x)

and

x

Hint 2

Hint 2 (The Arsenal): To prove something about the second derivative

f^{''} (x)

, we need to use Rolle's Theorem. Since Rolle's Theorem requires a function to have equal values at the endpoints of an interval, define an auxiliary function

h (x) = f (x) - x

. What are the values of

h (x)

x = 0, 1, 2

Hint 3

Hint 3 (The Execution): Apply Rolle's Theorem to

h (x)

on the interval

[0, 1]

to find a point

c_{1}

where

h^{'} (c_{1}) = 0

, and on

[1, 2]

to find a point

c_{2}

where

h^{'} (c_{2}) = 0

. Finally, apply Rolle's Theorem to

h^{'} (x)

on the interval

[c_{1}, c_{2}]

to conclude that

h^{''} (x) = 0

for some

x e x t in (c_{1}, c_{2})

Visualized Solution

Visualizing the Given Conditions

Given: $f (x)$ is continuous on $[0, 2]$ and twice differentiable on $(0, 2)$ .
Points: $f (0) = 0$ , $f (1) = 1$ , $f (2) = 2$ .
Observation: All three points lie on the straight line $y = x$ .

Defining the Auxiliary Function $h (x)$

To apply Rolle's Theorem, we need a function that is zero at the boundaries.
Define an auxiliary function: $h (x) = f (x) - x$ .
Since $f (x)$ and $x$ are continuous and differentiable, $h (x)$ is also continuous on $[0, 2]$ and differentiable on $(0, 2)$ .

Evaluating $h (x)$ at Given Points

$h (0) = f (0) - 0 = 0 - 0 = 0$
$h (1) = f (1) - 1 = 1 - 1 = 0$
$h (2) = f (2) - 2 = 2 - 2 = 0$
The function $h (x)$ has roots at $x = 0, 1, 2$ .

Applying Rolle's Theorem on $[0, 1]$

Consider the interval $[0, 1]$ .
Since $h (0) = h (1) = 0$ , Rolle's Theorem applies.
There exists at least one point $c_{1} \in (0, 1)$ such that $h^{'} (c_{1}) = 0$ .
Geometrically, the tangent to $f (x)$ at $c_{1}$ is parallel to $y = x$ .

Applying Rolle's Theorem on $[1, 2]$

Now consider the interval $[1, 2]$ .
Since $h (1) = h (2) = 0$ , Rolle's Theorem applies again.
There exists at least one point $c_{2} \in (1, 2)$ such that $h^{'} (c_{2}) = 0$ .
The tangent to $f (x)$ at $c_{2}$ is also parallel to $y = x$ .

Applying Rolle's Theorem to $h^{'} (x)$

We now know that $h^{'} (c_{1}) = 0$ and $h^{'} (c_{2}) = 0$ .
The function $h^{'} (x)$ is continuous and differentiable.
Apply Rolle's Theorem to $h^{'} (x)$ on the interval $[c_{1}, c_{2}]$ .
There exists some $x \in (c_{1}, c_{2})$ such that $h^{''} (x) = 0$ .

Relating $h^{''} (x)$ back to $f^{''} (x)$

Recall our auxiliary function: $h (x) = f (x) - x$ .
First derivative: $h^{'} (x) = f^{'} (x) - 1$ .
Second derivative: $h^{''} (x) = f^{''} (x) - 0 = f^{''} (x)$ .
Since $h^{''} (x) = 0$ for some $x \in (c_{1}, c_{2})$ , it follows that $f^{''} (x) = 0$ for that same $x$ .

Conclusion and Key Takeaway

Since $c_{1} \in (0, 1)$ and $c_{2} \in (1, 2)$ , the interval $(c_{1}, c_{2})$ is completely inside $(0, 2)$ .
Therefore, $f^{''} (x) = 0$ for some $x \in (0, 2)$ .
Key Takeaway: If a function intersects a line at $n$ points, its $(n - 1)$ -th derivative vanishes at least once.

00:00 / 00:00

The Sigma Insight: Mean Value Theorems

Solution Diagram

Analyzing the Setup

We are given a function

f (x)

that is continuous on

[0, 2]

and twice differentiable on

(0, 2)

. We are provided with three specific coordinates:

(0, 0)

(1, 1)

, and

(2, 2)

If you plot these points, you see they are perfectly collinear, sitting on the line

y = x

. This line serves as our geometric anchor for the problem.

The Magic of the Auxiliary Function

The challenge here is that Rolle's Theorem requires

f (a) = f (b)

. Since

f (0) = 0

and

f (1) = 1

, they are not equal, so we cannot apply the theorem directly to

f (x)

We introduce the Auxiliary Function

h (x) = f (x) - x

. When we evaluate this at our known points, we find:

h (0) = f (0) - 0 = 0

h (1) = f (1) - 1 = 0

h (2) = f (2) - 2 = 0

We have transformed the problem into one where

h (x)

is zero at three distinct points, allowing us to unleash the power of Rolle's Theorem.

The First Layer

Finding the Tangents

Consider the interval

[0, 1]

. Since

h (0) = 0

and

h (1) = 0

, Rolle's Theorem guarantees there exists some

c_{1} \in (0, 1)

such that

h^{'} (c_{1}) = 0

Because

h^{'} (x) = f^{'} (x) - 1

, this implies

f^{'} (c_{1}) = 1

. Geometrically, the tangent to our curve

f (x)

c_{1}

is perfectly parallel to the line

y = x

Now, we repeat this for the interval

[1, 2]

. Since

h (1) = 0

and

h (2) = 0

, there must exist some

c_{2} \in (1, 2)

such that

h^{'} (c_{2}) = 0

The Second Layer

The Curvature

Now, consider the function

g (x) = h^{'} (x)

. We know

g (c_{1}) = 0

and

g (c_{2}) = 0

Since

h (x)

is twice differentiable,

h^{'} (x)

is continuous and differentiable. We apply Rolle's Theorem to

h^{'} (x)

on the interval

[c_{1}, c_{2}]

This guarantees there exists some

x \in (c_{1}, c_{2})

such that the derivative of

h^{'} (x)

, which is

h^{''} (x)

, is zero. Because

h (x) = f (x) - x

, differentiating twice gives:

h^{''} (x) = f^{''} (x)

Therefore, we conclude that $f^{''} (x) = 0$ for some

x \in (0, 2)

The Grand Conclusion

We have proven that

f^{''} (x) = 0

for some

x \in (0, 2)

. This is a powerful result in calculus.

The key takeaway for your JEE preparation is this: if a function intersects a straight line at

n

points, its

(n - 1)

-th derivative will vanish at least once. This is the elegance of calculus—using the structure of derivatives to reveal the hidden geometry of the function.

Common Traps & FAQs

Wait, why do we define

h (x) = f (x) - x

? Can't we just use

f (x)

directly?

Rolle's Theorem requires that the function takes the same value at the endpoints of the interval (i.e.,

f (a) = f (b)

). In our case,

f (0) = 0

f (1) = 1

, and

f (2) = 2

. Since

f (0) \neq = f (1)

and

f (1) \neq = f (2)

, we cannot apply Rolle's Theorem directly to

f (x)

. By subtracting

x

, we create a new function

h (x) = f (x) - x

such that

h (0) = 0

h (1) = 0

, and

h (2) = 0

. Now,

h (x)

satisfies the conditions of Rolle's Theorem on both

[0, 1]

and

[1, 2]

Does this mean

f^{''} (x)

is always zero for all

x

Not at all! The theorem states that

f^{''} (x) = 0

for *at least one* point

x

in the interval

(0, 2)

. It does not imply that the function is linear or that the second derivative is zero everywhere. It simply guarantees the existence of at least one point where the curvature (represented by the second derivative) vanishes.

Why do we need the second derivative? Why not just stop at the first derivative?

The question asks us to evaluate properties of

f^{''} (x)

. By applying Rolle's Theorem to

h (x)

, we find that

h^{'} (x)

has roots in

(0, 1)

and

(1, 2)

. Since

h^{'} (x)

is zero at two distinct points, Rolle's Theorem applied to

h^{'} (x)

guarantees that its derivative,

h^{''} (x)

, must be zero at some point between those two roots. This is the standard path to proving properties about higher-order derivatives.

Animated Solution for Mathematics - Differentiation: Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then

Select Answer:

Visualized Solution

Visualizing the Given Conditions

Defining the Auxiliary Function $h (x)$

Evaluating $h (x)$ at Given Points

Applying Rolle's Theorem on $[0, 1]$

Applying Rolle's Theorem on $[1, 2]$

Applying Rolle's Theorem to $h^{'} (x)$

Relating $h^{''} (x)$ back to $f^{''} (x)$

Conclusion and Key Takeaway

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Magic of the Auxiliary Function

The First Layer

The Second Layer

The Grand Conclusion

Similar Questions

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

Let $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

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

Let $f (x)$ be a non-constant twice differentiable function defined on $(- \infty, \infty)$ such that $f (x) = f (1 - x)$ and $f^{'} (\frac{1}{4}) = 0$ . Then,

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$ . If for all $x \in (a, b)$ , $f^{'} (x) > 0$ and $f^{''} (x) < 0$ , then for any $c \in (a, b)$ , $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than:

For all $x$ in $[0, 1]$ , let the second derivative $f^{''} (x)$ of a function $f (x)$ exist and satisfy $∣ f^{''} (x) ∣ < 1$ . If $f (0) = f (1)$ , then show that $∣ f^{'} (x) ∣ < 1$ for all $x$ in $[0, 1]$ .

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$ , then for some $c \in] 0, 1 [$

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8$ , $f^{'} (2) = 5$ , $f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4$ for all $x \in (1, 6)$ , then :

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be any continuous function on [0,2] and twice differentiable on (0,2). If f(0)=0,f(1)=1 and f(2)=2, then

Select Answer:

Visualized Solution

Visualizing the Given Conditions

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x) at Given Points

Applying Rolle's Theorem on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,1]

Applying Rolle's Theorem on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [1,2]

Applying Rolle's Theorem to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h′(x)

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

Conclusion and Key Takeaway

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Magic of the Auxiliary Function

The First Layer

The Second Layer

The Grand Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For all twice differentiable functions f:R→R, with f(0)=f(1)=f′(0)=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is a twice differentiable function and given that f(1)=1;f(2)=4,f(3)=9, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g:R→R be a non constant twice differentiable such that g′(21​)=g′(23​). If a real valued function f is defined as f(x)=21​[g(x)+g(2−x)], then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:R→R is a twice differentiable function such that f′′(x)>0 for all x∈R, and f(21​)=f(1)=1, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a non-constant twice differentiable function defined on (−∞,∞) such that f(x)=f(1−x) and f′(41​)=0. Then,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is twice differentiable and continuous function in x∈[a,b] also f′(x)>0 and f′′(x)<0 and c∈(a,b) then f(b)−f(c)f(c)−f(a)​ is greater than

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f and g are differentiable functions in [0,1] satisfying f(0)=2=g(1),g(0)=0 and f(1)=6, then for some c∈]0,1[

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a twice differentiable function on (1,6). If f(2)=8, f′(2)=5, f′(x)≥1 and f′′(x)≥4 for all x∈(1,6), then :

Animated Solution for Mathematics - Differentiation: Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then

Defining the Auxiliary Function $h (x)$

Evaluating $h (x)$ at Given Points

Applying Rolle's Theorem on $[0, 1]$

Applying Rolle's Theorem on $[1, 2]$

Applying Rolle's Theorem to $h^{'} (x)$

Relating $h^{''} (x)$ back to $f^{''} (x)$

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

Let $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

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

Let $f (x)$ be a non-constant twice differentiable function defined on $(- \infty, \infty)$ such that $f (x) = f (1 - x)$ and $f^{'} (\frac{1}{4}) = 0$ . Then,

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$ , then for some $c \in] 0, 1 [$

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8$ , $f^{'} (2) = 5$ , $f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4$ for all $x \in (1, 6)$ , then :