JEE Advanced 2005

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: $f (x)$ 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 $g (c) g^{''} (c) < 0$ .

Q: Wait, why do we assume $g(0)$ is positive? What if it's negative?

That is a brilliant observation! If $g(0)$ is negative, the logic still holds perfectly. If $g(0)$ is negative, the function $g(x)$ would have a local minimum instead of a maximum. At a local minimum, the first derivative $g'(c)$ is still $0$, but the second derivative $g''(c)$ would be positive. Since $g(c)$ is negative and $g''(c)$ is positive, their product $g(c)g''(c)$ is still negative! The proof is robust regardless of the sign.

Q: Why did we choose the interval $[-3, 3]$ for the Mean Value Theorem?

The choice of interval is dictated by the bounds. We know $|f(x)| \le 1$. When we apply the Mean Value Theorem, we get a denominator of $3$ (from $0 - (-3)$ or $3 - 0$). This makes the bound $\frac{2}{3}$, which is clearly smaller than our $g(0)$ value of $\approx 2.82$. If we chose a smaller interval, the bound might not be small enough to guarantee a peak, and the proof would fall apart. We chose the interval that makes the math work!

Q: Why can't we just use energy conservation or something simpler?

Energy conservation is a physical principle, but here we are dealing with pure analysis. The problem is about the existence of a point $c$ where a specific inequality holds. This is a classic 'Existence Theorem' problem. We aren't looking for a specific value; we are looking for a property of the function's shape. The Mean Value Theorem and the Second Derivative Test are the precise tools designed to extract information about the shape of a function from its bounds.

Progressive Hints

Hint 1

Hint 1: The Spark. Look closely at the constraint

∣ f (x) ∣ \leq 1

. If you plug

x = 0

into the given equation

f^{2} (0) + g^{2} (0) = 9

, what is the minimum possible value for

g^{2} (0)

? This tells you that

g (0)

is quite large, specifically

∣ g (0) ∣ \geq 8

. This is your starting point.

Hint 2

Hint 2: The Arsenal. Now, consider the behavior of

g (x)

away from the origin. Since

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

, apply the Mean Value Theorem to

f (x)

on the intervals

[- 3, 0]

and

[0, 3]

. This will give you values for

g (x)

at some points

x_{1}

and

x_{2}

that are significantly smaller than

g (0)

Hint 3

Hint 3: The Execution. You have a function

g (x)

that is large at

x = 0

and small at

x_{1}

and

x_{2}

. This forces

g (x)

to have a local maximum in the interval

(- 3, 3)

. At this local maximum

c

, what do you know about

g^{'} (c)

and

g^{''} (c)

? Combine these to prove the inequality

g (c) g^{''} (c) < 0

Visualized Solution

Given Conditions

We are given a differentiable function $f (x)$ and a double differentiable function $g (x)$ .
The key relationship is: $f^{'} (x) = g (x)$ .
We are also given a strict bound: $∣ f (x) ∣ \leq 1$ for all $x$ .

Analyzing the Condition at $x = 0$

We are given a specific value equation at $x = 0$ :
$f^{2} (0) + g^{2} (0) = 9$
We need to find how large $g (0)$ can be.

Bounding $g (0)$

Since $∣ f (x) ∣ \leq 1$ , we know that $f^{2} (0) \leq 1$ .
Substituting this into our equation:
$g^{2} (0) = 9 - f^{2} (0) \geq 9 - 1 = 8$
Therefore, $∣ g (0) ∣ \geq 8 = 22 \approx 2.82$

Applying Mean Value Theorem (MVT)

To understand $g (x)$ away from $x = 0$ , we apply the Mean Value Theorem on $f (x)$ .
Let's apply MVT on the interval $[- 3, 0]$ .
There exists some $x_{1} \in (- 3, 0)$ such that:
$f^{'} (x_{1}) = \frac{f ( 0 ) - f ( - 3 )}{0 - ( - 3 )}$

Bounding $g (x_{1})$

Since $f^{'} (x) = g (x)$ , we have $g (x_{1}) = \frac{f ( 0 ) - f ( - 3 )}{3}$ .
Using the triangle inequality and $∣ f (x) ∣ \leq 1$ :
$∣ g (x_{1}) ∣ \leq \frac{∣ f ( 0 ) ∣ + ∣ f ( - 3 ) ∣}{3} \leq \frac{1 + 1}{3} = \frac{2}{3}$

Applying MVT on $[0, 3]$

Similarly, applying MVT on the interval $[0, 3]$ :
There exists $x_{2} \in (0, 3)$ such that $g (x_{2}) = \frac{f ( 3 ) - f ( 0 )}{3}$ .
By the same logic: $∣ g (x_{2}) ∣ \leq \frac{2}{3}$ .

Visualizing the Peak

We have $∣ g (0) ∣ \geq 2.82$ .
And $∣ g (x_{1}) ∣ \leq 0.67$ and $∣ g (x_{2}) ∣ \leq 0.67$ .
Assuming $g (0) > 0$ , the function must rise from $x_{1}$ , reach a high value at $x = 0$ , and fall towards $x_{2}$ .

Existence of a Local Maximum

Because $g (x)$ is continuous and differentiable, Extreme Value Theorem guarantees a local maximum.
There must exist some point $c \in (x_{1}, x_{2}) \subset (- 3, 3)$ where $g (x)$ attains this local maximum.

Properties at the Local Maximum $c$

At this local maximum $c$ :
1. $g (c) \geq g (0) > 0$ (The peak is at least as high as $g (0)$ )
2. $g^{'} (c) = 0$ (By Fermat's Theorem, the tangent is horizontal)

Second Derivative Test

By the Second Derivative Test for a local maximum:
If $g^{'} (c) = 0$ and the curve is concave downwards at the peak, then $g^{''} (c) < 0$ .

Final Conclusion

We have established two facts at $x = c$ :
$g (c) > 0$ (Positive value)
$g^{''} (c) < 0$ (Negative second derivative)
Multiplying them together: $g (c) \cdot g^{''} (c) < 0$ .
Hence proved!

00:00 / 00:00

The Sigma Insight: Mean Value Theorems

Solution Diagram

Analyzing the Constraints

We are given a differentiable function

f (x)

and a twice-differentiable function

g (x)

such that

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

. We are constrained by the condition

∣ f (x) ∣ \leq 1

for all

x

At the origin, we are given the relationship:

f^{2} (0) + g^{2} (0) = 9

Since

∣ f (0) ∣ \leq 1

, it follows that

f^{2} (0) \leq 1

. Substituting this into our equation, we find:

g^{2} (0) = 9 - f^{2} (0) \geq 9 - 1 = 8

Thus,

∣ g (0) ∣ \geq 8 \approx 2.82

Applying the Mean Value Theorem

To understand the behavior of

g (x)

, we apply the Mean Value Theorem (MVT) on the interval

[- 3, 0]

. There exists some

x_{1} \in (- 3, 0)

such that:

g (x_{1}) = f^{'} (x_{1}) = \frac{f ( 0 ) - f ( - 3 )}{0 - ( - 3 )} = \frac{f ( 0 ) - f ( - 3 )}{3}

Applying the triangle inequality to the numerator, we observe:

∣ g (x_{1}) ∣ = \frac{∣ f ( 0 ) - f ( - 3 ) ∣}{3} \leq \frac{∣ f ( 0 ) ∣ + ∣ f ( - 3 ) ∣}{3} \leq \frac{1 + 1}{3} = \frac{2}{3}

Similarly, applying the MVT on the interval

[0, 3]

, there exists some

x_{2} \in (0, 3)

such that:

∣ g (x_{2}) ∣ \leq \frac{2}{3}

The Geometric Conclusion

We have established that

∣ g (0) ∣ \geq 8

and

∣ g (x_{1}) ∣, ∣ g (x_{2}) ∣ \leq \frac{2}{3}

. Since

g (x)

is continuous, it must transition from a large value at the origin to smaller values at

x_{1}

and

x_{2}

This implies the existence of a local maximum

c

for the function

g (x)

within the interval

(- 3, 3)

. At this local maximum

c

, the following conditions must hold: 1. The first derivative vanishes:

g^{'} (c) = 0

. 2. The function is concave downwards:

g^{''} (c) < 0

Since

g (c)

is a local maximum and

g (0)

is a value within the interval, we have

g (c) \geq g (0) > 0

. Therefore, the product of the function value and its second derivative at this point is:

g (c) \cdot g^{''} (c) < 0

Conclusion: There exists at least one point

c \in (- 3, 3)

such that $g (c) g^{''} (c) < 0$ .

Common Traps & FAQs

Wait, why do we assume

g (0)

is positive? What if it's negative?

That is a brilliant observation! If

g (0)

is negative, the logic still holds perfectly. If

g (0)

is negative, the function

g (x)

would have a local minimum instead of a maximum. At a local minimum, the first derivative

g^{'} (c)

is still

0

, but the second derivative

g^{''} (c)

would be positive. Since

g (c)

is negative and

g^{''} (c)

is positive, their product

g (c) g^{''} (c)

is still negative! The proof is robust regardless of the sign.

Why did we choose the interval

[- 3, 3]

for the Mean Value Theorem?

The choice of interval is dictated by the bounds. We know

∣ f (x) ∣ \leq 1

. When we apply the Mean Value Theorem, we get a denominator of

3

(from

0 - (- 3)

3 - 0

). This makes the bound

\frac{2}{3}

, which is clearly smaller than our

g (0)

value of

\approx 2.82

. If we chose a smaller interval, the bound might not be small enough to guarantee a peak, and the proof would fall apart. We chose the interval that makes the math work!

Why can't we just use energy conservation or something simpler?

Energy conservation is a physical principle, but here we are dealing with pure analysis. The problem is about the existence of a point

c

where a specific inequality holds. This is a classic 'Existence Theorem' problem. We aren't looking for a specific value; we are looking for a property of the function's shape. The Mean Value Theorem and the Second Derivative Test are the precise tools designed to extract information about the shape of a function from its bounds.

Visualized Solution

Given Conditions

Analyzing the Condition at $x = 0$

Bounding $g (0)$

Applying Mean Value Theorem (MVT)

Bounding $g (x_{1})$

Applying MVT on $[0, 3]$

Visualizing the Peak

Existence of a Local Maximum

Properties at the Local Maximum $c$

Second Derivative Test

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Constraints

Applying the Mean Value Theorem

The Geometric Conclusion

Similar Questions

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

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

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

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:

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 $g (x)$ .

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:

Let $f : [0, 1] \to R$ be a twice differentiable function in $(0, 1)$ such that $f (0) = 3$ and $f (1) = 5$ . If the line $y = 2 x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$ , then the least number of points $x \in (0, 1)$ , at which $f^{''} (x) = 0$ , is ________

Visualized Solution

Given Conditions

Analyzing the Condition at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

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

Applying Mean Value Theorem (MVT)

Bounding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x1​)

Applying MVT on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,3]

Visualizing the Peak

Existence of a Local Maximum

Properties at the Local Maximum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Second Derivative Test

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Constraints

Applying the Mean Value Theorem

The Geometric Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) and g(x) are differentiable function for 0≤x≤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)=2g′(c).

.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 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; } 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

Analyzing the Condition at $x = 0$

Bounding $g (0)$

Bounding $g (x_{1})$

Applying MVT on $[0, 3]$

Properties at the Local Maximum $c$

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

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

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