JEE Advanced 2006

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If $F (x) = (f (x /2))^{2} + (g (x /2))^{2}$ where $f^{''} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

Q: Wait, why is $g'(x) = -f(x)$? I thought $g'(x) = f''(x)$.

You are absolutely correct! $g(x) = f'(x)$ implies $g'(x) = f''(x)$. Since the problem explicitly gives $f''(x) = -f(x)$, it follows by substitution that $g'(x) = -f(x)$. It is just a chain of logical substitutions.

Q: Does the $x/2$ argument change the derivative of the constant? I am worried about the $1/2$ factor.

The chain rule is your best friend here. The $1/2$ factor comes from the derivative of the inner function $x/2$. Because it appears in both terms of the sum, it cancels out perfectly with the $2$ from the power rule. It is designed to be elegant!

Select Answer:

Progressive Hints

Hint 1

Look at the structure of

F (x) = (f (x /2))^{2} + (g (x /2))^{2}

. It is a sum of squares of two functions. In many physics and math problems, a sum of squares often hints at a conserved quantity. Instead of trying to solve for

f (x)

and

g (x)

individually, try to see how

F (x)

changes as

x

varies.

Hint 2

You need to differentiate

F (x)

with respect to

x

. Remember the chain rule! When you differentiate

f (x /2)

, you get

f^{'} (x /2) \cdot (1/2)

. This

1/2

will appear in both terms of your derivative. How can you use the given relations

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

and

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

to simplify the result?

Hint 3

Start by writing

F^{'} (x) = 2 f (x /2) \cdot f^{'} (x /2) \cdot (1/2) + 2 g (x /2) \cdot g^{'} (x /2) \cdot (1/2)

. Now, substitute

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

and

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

into this expression. What do you notice about the two terms?

Visualized Solution

Analyze the Function $F (x)$

Given: $F (x) = (f (\frac{x}{2}))^{2} + (g (\frac{x}{2}))^{2}$
Conditions: $f^{''} (x) = - f (x)$ and $g (x) = f^{'} (x)$
Initial Value: $F (5) = 5$
Goal: Find $F (10)$

Differentiate $F (x)$

Differentiating $F (x)$ with respect to $x$ :
$F^{'} (x) = \frac{d}{d x} [(f (\frac{x}{2}))^{2}] + \frac{d}{d x} [(g (\frac{x}{2}))^{2}]$

Apply the Chain Rule

Applying chain rule to each term:
$F^{'} (x) = 2 f (\frac{x}{2}) \cdot f^{'} (\frac{x}{2}) \cdot \frac{1}{2} + 2 g (\frac{x}{2}) \cdot g^{'} (\frac{x}{2}) \cdot \frac{1}{2}$

Simplify the Derivative

Cancelling the constants $2$ and $\frac{1}{2}$ :
$F^{'} (x) = f (\frac{x}{2}) f^{'} (\frac{x}{2}) + g (\frac{x}{2}) g^{'} (\frac{x}{2})$

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

Using the given condition: $g (x) = f^{'} (x)$
Therefore, $g (\frac{x}{2}) = f^{'} (\frac{x}{2})$
Substitute this into the first term:
$F^{'} (x) = f (\frac{x}{2}) g (\frac{x}{2}) + g (\frac{x}{2}) g^{'} (\frac{x}{2})$

Find $g^{'} (x)$

Differentiate $g (x) = f^{'} (x)$ to get $g^{'} (x) = f^{''} (x)$
Given condition: $f^{''} (x) = - f (x)$
Therefore, $g^{'} (x) = - f (x)$

Substitute $g^{'} (x)$ into $F^{'} (x)$

From previous step, $g^{'} (\frac{x}{2}) = - f (\frac{x}{2})$
Substitute this into the second term:
$F^{'} (x) = f (\frac{x}{2}) g (\frac{x}{2}) + g (\frac{x}{2}) [- f (\frac{x}{2})]$

Evaluate $F^{'} (x)$

Simplifying the expression:
$F^{'} (x) = f (\frac{x}{2}) g (\frac{x}{2}) - f (\frac{x}{2}) g (\frac{x}{2})$
$F^{'} (x) = 0$

Conclusion: $F (x)$ is Constant

Since $F^{'} (x) = 0$ for all $x$ , $F (x)$ must be a constant function.
Let $F (x) = C$

Find the Constant $C$

We are given the initial value: $F (5) = 5$
Since $F (x) = C$ , we have $C = 5$
Therefore, $F (x) = 5$ for all $x$ .

Final Answer: $F (10)$

We need to find the value of $F (10)$ .
Since $F (x) = 5$ for all $x$ , substituting $x = 10$ gives:
$F (10) = 5$
Correct Option: 5

00:00 / 00:00

The Sigma Insight: Higher Order Derivatives

Solution Diagram

Analyzing the Setup

The function is defined as:

F (x) = (f (\frac{x}{2}))^{2} + (g (\frac{x}{2}))^{2}

Our objective is to determine the value of

F (10)

, given the initial condition

F (5) = 5

Differentiating the Function

To understand the behavior of

F (x)

, we differentiate with respect to

x

using the chain rule:

F^{'} (x) = 2 f (\frac{x}{2}) \cdot f^{'} (\frac{x}{2}) \cdot (\frac{1}{2}) + 2 g (\frac{x}{2}) \cdot g^{'} (\frac{x}{2}) \cdot (\frac{1}{2})

Simplifying the expression by canceling the

2

and the

1/2

terms, we obtain:

F^{'} (x) = f (\frac{x}{2}) f^{'} (\frac{x}{2}) + g (\frac{x}{2}) g^{'} (\frac{x}{2})

Applying the Differential Constraints

We are provided with the conditions

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

and

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

First, we substitute

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

into the expression. Next, we determine

g^{'} (x /2)

by differentiating

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

, which yields

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

Given

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

, it follows that

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

, and consequently:

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

The Final Calculation

Substituting these relationships back into our derivative equation, we get:

F^{'} (x) = f (\frac{x}{2}) g (\frac{x}{2}) + g (\frac{x}{2}) (- f (\frac{x}{2}))

The terms cancel out perfectly, resulting in

F^{'} (x) = 0

Because the derivative is zero,

F (x)

must be a constant function. Given that

F (5) = 5

, the constant value is

5

for all

x

Therefore, the final answer is:

$F (10) = 5$

Common Traps & FAQs

Why did we differentiate

F (x)

? Can't we just solve for

f (x)

and

g (x)

You could, but that is the long way! Solving the differential equation

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

leads to

f (x) = A sin (x) + B cos (x)

. While valid, differentiating

F (x)

is a much more elegant and faster way to show that the function is constant, which is a classic JEE strategy to save time.

Wait, why is

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

? I thought

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

You are absolutely correct!

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

implies

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

. Since the problem explicitly gives

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

, it follows by substitution that

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

. It is just a chain of logical substitutions.

Does the

x /2

argument change the derivative of the constant? I am worried about the

1/2

factor.

The chain rule is your best friend here. The

1/2

factor comes from the derivative of the inner function

x /2

. Because it appears in both terms of the sum, it cancels out perfectly with the

2

from the power rule. It is designed to be elegant!

Animated Solution for Mathematics - Differentiation: If $F (x) = (f (x /2))^{2} + (g (x /2))^{2}$ where $f^{''} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

Select Answer:

Visualized Solution

Analyze the Function $F (x)$

Differentiate $F (x)$

Apply the Chain Rule

Simplify the Derivative

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

Find $g^{'} (x)$

Substitute $g^{'} (x)$ into $F^{'} (x)$

Evaluate $F^{'} (x)$

Conclusion: $F (x)$ is Constant

Find the Constant $C$

Final Answer: $F (10)$

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Differentiating the Function

Applying the Differential Constraints

The Final Calculation

Similar Questions

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .

Let $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f^{'} (10)$ is equal to

If $f (x) = x^{2} + g^{'} (1) x + g^{''} (2)$ and $g (x) = f (1) x^{2} + x f^{'} (x) + f^{''} (x)$ , then the value of $f (4) - g (4)$ is equal to ______.

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3), x \in R$ . Then the value of $f^{'} (5)$ is:

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

Let $f : R \to R$ be a twice differentiable function such that $(sin x cos y) (f (2 x + 2 y) - f (2 x - 2 y)) = (cos x sin y) (f (2 x + 2 y) + f (2 x - 2 y))$ for all $x, y \in R$ . If $f^{'} (0) = 1/2$ , then the value of $24 f^{''} (5 π /3)$ is:

If $f (x) = x^{3} - x^{2} f^{'} (1) + x f^{''} (2) - f^{'''} (3), x \in R$ , then

Let $f (x) = \frac{s i n x + c o s x - 2}{s i n x - c o s x}, x \in [0, π] - {\frac{π}{4}}$ , then $f (\frac{7 π}{12}) f^{''} (\frac{7 π}{12})$ is equal to

Let $f$ and $g$ be twice differentiable functions on $R$ such that $f^{''} (x) = g^{''} (x) + 6 x$ , $f^{'} (1) = 4 g^{'} (1) - 3 = 9$ , $f (2) = 3 g (2) = 12$ . Then which of the following is NOT true?

Let $f : S \to S$ where $S = (0, \infty)$ be a twice differentiable function such that $f (x + 1) = x f (x)$ . If $g : S \to R$ be defined as $g (x) = lo g_{e} f (x)$ , then the value of $∣ g^{''} (5) - g^{''} (1) ∣$ is equal to :

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If F(x)=(f(x/2))2+(g(x/2))2 where f′′(x)=−f(x) and g(x)=f′(x) and given that F(5)=5, then F(10) is equal to

Select Answer:

Visualized Solution

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

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

Apply the Chain Rule

Simplify the Derivative

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

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

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

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

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

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

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(10)

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Differentiating the Function

Applying the Differential Constraints

The Final Calculation

Similar Questions

.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 such that f′′(x)=−f(x), and f′(x)=g(x),h(x)=[f(x)]2+[g(x)]2. Find h(10) if h(5)=11.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),x∈R. Then f′(10) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x2+g′(1)x+g′′(2) and g(x)=f(1)x2+xf′(x)+f′′(x), then the value of f(4)−g(4) is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3),x∈R. Then the value of f′(5) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function such that f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),x∈R. Then f(2) equal :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈R, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=sinx−cosxsinx+cosx−2​​,x∈[0,π]−{4π​}, then f(127π​)f′′(127π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f and g be twice differentiable functions on R such that f′′(x)=g′′(x)+6x, f′(1)=4g′(1)−3=9, f(2)=3g(2)=12. Then which of the following is NOT true?

Animated Solution for Mathematics - Differentiation: If $F (x) = (f (x /2))^{2} + (g (x /2))^{2}$ where $f^{''} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

Analyze the Function $F (x)$

Differentiate $F (x)$

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

Find $g^{'} (x)$

Substitute $g^{'} (x)$ into $F^{'} (x)$

Evaluate $F^{'} (x)$

Conclusion: $F (x)$ is Constant

Find the Constant $C$

Final Answer: $F (10)$

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .

Let $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f^{'} (10)$ is equal to

If $f (x) = x^{2} + g^{'} (1) x + g^{''} (2)$ and $g (x) = f (1) x^{2} + x f^{'} (x) + f^{''} (x)$ , then the value of $f (4) - g (4)$ is equal to ______.

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3), x \in R$ . Then the value of $f^{'} (5)$ is:

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

If $f (x) = x^{3} - x^{2} f^{'} (1) + x f^{''} (2) - f^{'''} (3), x \in R$ , then

Let $f (x) = \frac{s i n x + c o s x - 2}{s i n x - c o s x}, x \in [0, π] - {\frac{π}{4}}$ , then $f (\frac{7 π}{12}) f^{''} (\frac{7 π}{12})$ is equal to

Let $f$ and $g$ be twice differentiable functions on $R$ such that $f^{''} (x) = g^{''} (x) + 6 x$ , $f^{'} (1) = 4 g^{'} (1) - 3 = 9$ , $f (2) = 3 g (2) = 12$ . Then which of the following is NOT true?