JEE Advanced 1982

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If $f (x)$ and $g (x)$ are differentiable function for $0 \leq x \leq 1$ such that $f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2$ , then show that there exist $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

Q: Why did we choose $h(x) = f(x) - 2g(x)$? Could we have used $2g(x) - f(x)$?

Absolutely! Both functions would work perfectly. If you chose $h(x) = 2g(x) - f(x)$, the derivative would be $2g'(x) - f'(x) = 0$, which is equivalent to $f'(x) = 2g'(x)$. The constant factor doesn't change the existence of the root, so you are free to choose the form that feels most intuitive to you.

Q: Do we really need Rolle's Theorem? Can't we just use the Mean Value Theorem?

That is a brilliant observation! Rolle's Theorem is actually a special case of the Mean Value Theorem (MVT). While you could certainly use MVT, Rolle's Theorem is the more direct tool here because it specifically addresses the case where the function values at the endpoints are equal, which is exactly what we have.

Q: What if the problem didn't state that $f$ and $g$ are differentiable? Could we still solve it?

Great question! No, we could not. The requirement of differentiability is crucial for Rolle's Theorem. Without it, the function could have sharp corners or breaks, meaning the derivative might not exist at all, and the theorem would not apply. Always check your conditions before applying a theorem!

Visualized Solution

Visualizing the Functions $f (x)$ and $g (x)$

We are given two differentiable functions $f (x)$ and $g (x)$ on the interval $[0, 1]$ .
Let's plot their boundary values:
For $f (x)$ : $f (0) = 2$ and $f (1) = 6$ .
For $g (x)$ : $g (0) = 0$ and $g (1) = 2$ .

Analyzing the Target Equation $f^{'} (c) = 2 g^{'} (c)$

We need to prove that there exists some $c \in (0, 1)$ such that $f^{'} (c) = 2 g^{'} (c)$ .
Let's rewrite this equation by moving all terms to one side:
$f^{'} (c) - 2 g^{'} (c) = 0$
This form strongly suggests the derivative of a combined function!

Constructing the Auxiliary Function $h (x)$

To find a point where $f^{'} (c) - 2 g^{'} (c) = 0$ , we define a new function:
$h (x) = f (x) - 2 g (x)$
Since $f (x)$ and $g (x)$ are differentiable on $[0, 1]$ , $h (x)$ must also be differentiable on $[0, 1]$ .

Evaluating $h (x)$ at the Left Boundary

Let's find the value of $h (x)$ at the starting point $x = 0$ :
$h (0) = f (0) - 2 g (0)$
Substitute the given values: $f (0) = 2$ and $g (0) = 0$ :
$h (0) = 2 - 2 (0) = 2$

Evaluating $h (x)$ at the Right Boundary

Let's find the value of $h (x)$ at the ending point $x = 1$ :
$h (1) = f (1) - 2 g (1)$
Substitute the given values: $f (1) = 6$ and $g (1) = 2$ :
$h (1) = 6 - 2 (2) = 6 - 4 = 2$

Recalling Rolle's Theorem

Recall Rolle's Theorem: If a function $h (x)$ is:
1. Continuous on $[a, b]$
2. Differentiable on $(a, b)$
3. $h (a) = h (b)$
Then, there exists at least one $c \in (a, b)$ such that $h^{'} (c) = 0$ .

Applying Rolle's Theorem to $h (x)$

For $h (x)$ on $[0, 1]$ :
$h (x)$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ .
We found $h (0) = h (1) = 2$ .
Therefore, there exists $c \in (0, 1)$ such that $h^{'} (c) = 0$ .

Proving $f^{'} (c) = 2 g^{'} (c)$

We have $h^{'} (x) = f^{'} (x) - 2 g^{'} (x)$ .
At $x = c$ , we know $h^{'} (c) = 0$ .
Substitute $x = c$ :
$f^{'} (c) - 2 g^{'} (c) = 0 ⟹ f^{'} (c) = 2 g^{'} (c)$
This completes the proof!

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The Sigma Insight: Mean Value Theorems

Solution Diagram

Analyzing the Setup

Imagine you are standing on a graph, looking at two distinct paths,

f (x)

and

g (x)

, both winding their way from

x = 0

x = 1

You are given the boundary conditions:

f (0) = 2, g (0) = 0, f (1) = 6,

and

g (1) = 2

Your mission is to prove that at some point

c

between

0

and

1

, the slope of

f

is exactly twice the slope of

g

. This is a beautiful dance of calculus.

The Spark of Insight

The equation

f^{'} (c) = 2 g^{'} (c)

is our North Star. If we rearrange it, we get

f^{'} (c) - 2 g^{'} (c) = 0

This is the key! It looks exactly like the derivative of a combined function.

If we define a new function

h (x) = f (x) - 2 g (x)

, then its derivative is simply:

h^{'} (x) = f^{'} (x) - 2 g^{'} (x)

Suddenly, the problem transforms from a complex relationship between two functions into a simple search for a root of the derivative of one auxiliary function.

Constructing the Bridge

Let us build this auxiliary function,

h (x) = f (x) - 2 g (x)

. Since

f (x)

and

g (x)

are differentiable,

h (x)

inherits this smoothness.

Now, let us test the boundaries: At

x = 0

, we have:

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

x = 1

, we have:

h (1) = f (1) - 2 g (1) = 6 - 2 (2) = 6 - 4 = 2

Look at that!

h (0) = h (1) = 2

. This is the moment where the magic happens.

The Climax

Rolle's Theorem

We have a function

h (x)

that is continuous on

[0, 1]

, differentiable on

(0, 1)

, and satisfies

h (0) = h (1)

. These are the exact conditions required for Rolle's Theorem.

Rolle's Theorem tells us that there must exist some

c \in (0, 1)

such that

h^{'} (c) = 0

Since

h^{'} (c) = f^{'} (c) - 2 g^{'} (c)

, this implies:

f^{'} (c) - 2 g^{'} (c) = 0 \Rightarrow f^{'} (c) = 2 g^{'} (c)

You have just proven that such a point

c

must exist. It is not just a coincidence; it is the inevitable consequence of the geometry of smooth functions.

Common Traps & FAQs

Why did we choose

h (x) = f (x) - 2 g (x)

? Could we have used

2 g (x) - f (x)

Absolutely! Both functions would work perfectly. If you chose

h (x) = 2 g (x) - f (x)

, the derivative would be

2 g^{'} (x) - f^{'} (x) = 0

, which is equivalent to

f^{'} (x) = 2 g^{'} (x)

. The constant factor doesn't change the existence of the root, so you are free to choose the form that feels most intuitive to you.

Do we really need Rolle's Theorem? Can't we just use the Mean Value Theorem?

That is a brilliant observation! Rolle's Theorem is actually a special case of the Mean Value Theorem (MVT). While you could certainly use MVT, Rolle's Theorem is the more direct tool here because it specifically addresses the case where the function values at the endpoints are equal, which is exactly what we have.

What if the problem didn't state that

f

and

g

are differentiable? Could we still solve it?

Great question! No, we could not. The requirement of differentiability is crucial for Rolle's Theorem. Without it, the function could have sharp corners or breaks, meaning the derivative might not exist at all, and the theorem would not apply. Always check your conditions before applying a theorem!

Animated Solution for Mathematics - Differentiation: If $f (x)$ and $g (x)$ are differentiable function for $0 \leq x \leq 1$ such that $f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2$ , then show that there exist $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

Visualized Solution

Visualizing the Functions $f (x)$ and $g (x)$

Analyzing the Target Equation $f^{'} (c) = 2 g^{'} (c)$

Constructing the Auxiliary Function $h (x)$

Evaluating $h (x)$ at the Left Boundary

Evaluating $h (x)$ at the Right Boundary

Recalling Rolle's Theorem

Applying Rolle's Theorem to $h (x)$

Proving $f^{'} (c) = 2 g^{'} (c)$

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Spark of Insight

Constructing the Bridge

The Climax

Similar Questions

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 $c \in] 0, 1 [$

$f (x)$ is a differentiable function and $g (x)$ is a double differentiable function such that $∣ f (x) ∣ \leq 1$ and $f^{'} (x) = g (x)$ . If $f^{2} (0) + g^{2} (0) = 9$ . Prove that there exists some $c \in (- 3, 3)$ such that $g (c) g^{''} (c) < 0$ .

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 $[0, 1]$ .

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

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 :

For a twice differentiable function $f (x), g (x)$ is defined as $g (x) = (f^{'} (x))^{2} + f^{''} (x) f (x)$ on $[a, e]$ . If for $a < b < c < d < e, f (a) = 0, f (b) = 2, f (c) = - 1, f (d) = 2, f (e) = 0$ then find the minimum number of zeros of $g (x)$ .

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

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:

Let $S$ be the set of all functions $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ , depending on $f$ , such that:

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

Visualized Solution

Visualizing the Functions .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Analyzing the Target Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(c)=2g′(c)

Constructing 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 the Left Boundary

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

Recalling Rolle's Theorem

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)

Proving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(c)=2g′(c)

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Spark of Insight

Constructing the Bridge

The Climax

Similar Questions

.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 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; } 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 :

.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; } For all twice differentiable functions f:R→R, with f(0)=f(1)=f′(0)=0

Visualizing the Functions $f (x)$ and $g (x)$

Analyzing the Target Equation $f^{'} (c) = 2 g^{'} (c)$

Constructing the Auxiliary Function $h (x)$

Evaluating $h (x)$ at the Left Boundary

Evaluating $h (x)$ at the Right Boundary

Applying Rolle's Theorem to $h (x)$

Proving $f^{'} (c) = 2 g^{'} (c)$

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 $g : R \to R$ 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

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 :

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

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