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Animated Solution for Mathematics - Differential Equations: $f (x)$ and $g (x)$ are two differentiable functions on $[0, 2]$ such that $f^{''} (x) - g^{''} (x) = 0, f^{'} (1) = 2 g^{'} (1) = 4, f (2) = 3 g (2) = 9$ then $f (x) - g (x)$ at $x = 3/2$ is

Q: How do we know for sure that $h(x)$ is a straight line?

Great intuition! We know $h(x)$ is a straight line because its second derivative is zero. In calculus, the second derivative represents the rate of change of the slope. If the rate of change of the slope is zero, the slope itself must be constant. A function with a constant slope is, by definition, a linear function of the form $h(x) = ax + b$.

Q: Is it possible that $f(x)$ and $g(x)$ are not polynomials?

The problem states that $f(x)$ and $g(x)$ are differentiable. While they might not be simple polynomials, the condition $f''(x) - g''(x) = 0$ is a universal constraint. Regardless of the complexity of $f(x)$ and $g(x)$, their difference $h(x)$ must behave like a linear function within the given interval. The beauty of calculus is that it allows us to ignore the complexity and focus on the underlying structure.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Look closely at the condition

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

. This is not just an equation; it is a profound statement about the nature of the difference between these two functions. If the second derivative of a function is zero, what does that tell you about the function itself? Visualize the shape of the graph.

Hint 2

Hint 2 (The Arsenal): Define a new function

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

. Your goal is to find

h (\frac{3}{2})

. Since

h^{''} (x) = 0

, you know that

h (x)

must be a linear function of the form

h (x) = a x + b

. Your task is to determine the constants

a

and

b

using the given derivative and function values.

Hint 3

Hint 3 (The Execution): Start by calculating

h^{'} (1)

using the given

f^{'} (1) = 4

and

2 g^{'} (1) = 4

. This will give you the slope

a

. Then, calculate

h (2)

using

f (2) = 9

and

3 g (2) = 9

. With

a

and

h (2)

known, you can easily solve for the intercept

b

in the equation

h (2) = a (2) + b

Visualized Solution

Define $h (x) = f (x) - g (x)$

Let $h (x) = f (x) - g (x)$ be the difference function.
The goal is to find the value of $h (\frac{3}{2})$ .

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

Given: $f^{''} (x) - g^{''} (x) = 0$
Since $h (x) = f (x) - g (x)$ , then $h^{''} (x) = f^{''} (x) - g^{''} (x)$
Therefore, $h^{''} (x) = 0$

Integrate $h^{''} (x)$ to find $h^{'} (x)$

Integrating $h^{''} (x) = 0$ with respect to $x$ :
$\int h^{''} (x) d x = \int 0 d x$
$h^{'} (x) = c_{1}$ , where $c_{1}$ is a constant.

Calculate $h^{'} (1)$

Given: $f^{'} (1) = 4$ and $2 g^{'} (1) = 4$
This implies $g^{'} (1) = \frac{4}{2} = 2$
$h^{'} (1) = f^{'} (1) - g^{'} (1) = 4 - 2$

Determine constant $c_{1}$

$h^{'} (1) = 2$
Since $h^{'} (x) = c_{1}$ for all $x$ , then $c_{1} = 2$
Thus, $h^{'} (x) = 2$

Integrate $h^{'} (x)$ to find $h (x)$

Integrating $h^{'} (x) = 2$ with respect to $x$ :
$\int h^{'} (x) d x = \int 2 d x$
$h (x) = 2 x + c_{2}$ , where $c_{2}$ is another constant.

Calculate $h (2)$

Given: $f (2) = 9$ and $3 g (2) = 9$
This implies $g (2) = \frac{9}{3} = 3$
$h (2) = f (2) - g (2) = 9 - 3 = 6$

Determine constant $c_{2}$

Substitute $x = 2$ and $h (2) = 6$ into $h (x) = 2 x + c_{2}$ :
$6 = 2 (2) + c_{2}$
$6 = 4 + c_{2} ⟹ c_{2} = 2$

Final expression for $h (x)$

The complete function is $h (x) = 2 x + 2$
This represents $f (x) - g (x) = 2 x + 2$

Substitute $x = \frac{3}{2}$

Substitute $x = \frac{3}{2}$ into $h (x) = 2 x + 2$ :
$h (\frac{3}{2}) = 2 (\frac{3}{2}) + 2$

Final result

$h (\frac{3}{2}) = 3 + 2 = 5$
The final answer is 5.
Key Takeaway: Integrating $h^{''} (x) = 0$ twice results in a linear function $h (x) = a x + b$ .

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

Imagine you are standing on a vast, complex landscape, trying to understand the relationship between two winding paths,

f (x)

and

g (x)

. The secret to understanding their relationship lies in analyzing the gap between them.

We are given two differentiable functions,

f (x)

and

g (x)

, satisfying the condition:

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

Let us define a new function,

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

. This function

h (x)

represents the vertical distance between our two paths at any point

x

Applying our condition to

h (x)

, we observe:

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

The Geometry of Linearity

Think about what

h^{''} (x) = 0

means geometrically. The second derivative represents the rate of change of the slope.

If the rate of change of the slope is zero, then the slope itself must be constant. A function with a constant slope is, by definition, a straight line.

Thus, our complex difference function

h (x)

is a simple linear equation of the form:

h (x) = a x + b

Unlocking the Constants

We now determine the constants

a

and

b

using the provided clues. First, we analyze the derivatives at

x = 1

f^{'} (1) = 4 and 2 g^{'} (1) = 4 ⟹ g^{'} (1) = 2

Since

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

, at

x = 1

we have:

h^{'} (1) = 4 - 2 = 2

Because

h (x) = a x + b

, its derivative is

h^{'} (x) = a

. Therefore, we find that

a = 2

, and our function simplifies to

h (x) = 2 x + b

Next, we use the function values at

x = 2

f (2) = 9 and 3 g (2) = 9 ⟹ g (2) = 3

Calculating the difference at

x = 2

h (2) = f (2) - g (2) = 9 - 3 = 6

Substituting these values into

h (x) = 2 x + b

6 = 2 (2) + b ⟹ 6 = 4 + b ⟹ b = 2

The Final Revelation

We have successfully decoded the mystery. The difference function is:

h (x) = 2 x + 2

To find the value of this difference at

x = \frac{3}{2}

, we substitute:

h (\frac{3}{2}) = 2 (\frac{3}{2}) + 2

h (\frac{3}{2}) = 3 + 2 = 5

The final answer is 5.

Common Traps & FAQs

Wait, why do we define

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

? Can't we just solve for

f (x)

and

g (x)

separately?

That is a fantastic question! The beauty of this problem lies in the fact that we don't actually need to know the individual forms of

f (x)

and

g (x)

. We only have information about their difference. By defining

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

, we consolidate all our information into one single, manageable function. It is a classic strategy in competitive exams: simplify the problem until the answer reveals itself.

How do we know for sure that

h (x)

is a straight line?

Great intuition! We know

h (x)

is a straight line because its second derivative is zero. In calculus, the second derivative represents the rate of change of the slope. If the rate of change of the slope is zero, the slope itself must be constant. A function with a constant slope is, by definition, a linear function of the form

h (x) = a x + b

Is it possible that

f (x)

and

g (x)

are not polynomials?

The problem states that

f (x)

and

g (x)

are differentiable. While they might not be simple polynomials, the condition

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

is a universal constraint. Regardless of the complexity of

f (x)

and

g (x)

, their difference

h (x)

must behave like a linear function within the given interval. The beauty of calculus is that it allows us to ignore the complexity and focus on the underlying structure.

Animated Solution for Mathematics - Differential Equations: $f (x)$ and $g (x)$ are two differentiable functions on $[0, 2]$ such that $f^{''} (x) - g^{''} (x) = 0, f^{'} (1) = 2 g^{'} (1) = 4, f (2) = 3 g (2) = 9$ then $f (x) - g (x)$ at $x = 3/2$ is

Select Answer:

Visualized Solution

Define $h (x) = f (x) - g (x)$

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

Integrate $h^{''} (x)$ to find $h^{'} (x)$

Calculate $h^{'} (1)$

Determine constant $c_{1}$

Integrate $h^{'} (x)$ to find $h (x)$

Calculate $h (2)$

Determine constant $c_{2}$

Final expression for $h (x)$

Substitute $x = \frac{3}{2}$

Final result

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Geometry of Linearity

Unlocking the Constants

The Final Revelation

Similar Questions

Let $f : R \to R$ and $g : R \to R$ be two non-constant differentiable functions. If $f^{'} (x) = (e^{f (x) - g (x)}) g^{'} (x)$ for all $x \in R$ , and $f (1) = g (2) = 1$ , then which of the following statement(s) is (are) TRUE ?

Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f (0) = 1$ . Let $F (x) = \int_{0}^{x^{2}} f (t) d t$ for $x \in [0, 2]$ . If $F^{'} (x) = f^{'} (x)$ for all $x \in (0, 2)$ , then $f (2)$ equals

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

Let $f$ be a non-negative function in $[0, 1]$ and twice differentiable in $(0, 1)$ . If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ and $f (0) = 0$ , then $lim_{x \to 0} \frac{1}{x ^{2}} \int_{0}^{x} f (t) d t$ :

Let for some function $y = f (x)$ , $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ , $x > 0$ and $f (2) = 3$ Then $f (6)$ is equal to

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

A function $y = f (x)$ has a second order derivative $f^{''} (x) = 6 (x - 1)$ . If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3 x - 5$ , then the function is

If $y = y (x), y \in [0, π /2)$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ , then $5 y^{'} (\frac{π}{2})$ is equal to

Let $f : [0, 1] \to R$ be such that $f (x y) = f (x) . f (y)$ for all $x, y \in [0, 1]$ , and $f (0) \neq = 0$ . If $y = y (x)$ satisfies the differential equation, $\frac{d y}{d x} = f (x)$ with $y (0) = 1$ , then $y (1/4) + y (3/4)$ is equal to

Let $f$ be a differentiable function defined on $[0, π /2]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) 1 - (lo g_{e} f (t))^{2} d t = e, \forall x \in [0, π /2]$ . Then $(6 lo g_{e} f (π /6))^{2}$ is equal to ______.

Select Answer:

Visualized Solution

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

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

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

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

Determine constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c1​

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

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(2)

Determine constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c2​

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

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=23​

Final result

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Geometry of Linearity

Unlocking the Constants

The Final Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∫0x​1−(y′(t))2​dt=∫0x​y(t)dt,0≤x≤3,y≥0,y(0)=0. Then at x=2,y′′+y+1 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let for some function y=f(x), ∫0x​tf(t)dt=x2f(x), x>0 and f(2)=3 Then f(6) is equal to

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x),y∈[0,π/2) is the solution of the differential equation secydxdy​−sin(x+y)−sin(x−y)=0, with y(0)=0, then 5y′(2π​) is equal to

Define $h (x) = f (x) - g (x)$

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

Integrate $h^{''} (x)$ to find $h^{'} (x)$

Calculate $h^{'} (1)$

Determine constant $c_{1}$

Integrate $h^{'} (x)$ to find $h (x)$

Calculate $h (2)$

Determine constant $c_{2}$

Final expression for $h (x)$

Substitute $x = \frac{3}{2}$

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

Let for some function $y = f (x)$ , $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ , $x > 0$ and $f (2) = 3$ Then $f (6)$ is equal to

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

If $y = y (x), y \in [0, π /2)$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ , then $5 y^{'} (\frac{π}{2})$ is equal to