JEE Advanced 2020

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $b$ be a nonzero real number. Suppose $f : R \to R$ is a differentiable function such that $f (0) = 1$ . If the derivative $f^{'}$ of $f$ satisfies the equation $f^{'} (x) = \frac{f ( x )}{b ^{2} + x ^{2}}$ for all $x \in R$ , then which of the following statements is/are TRUE?

Q: Why is Option (B) false? If $b < 0$, shouldn't the function decrease?

It is a common trap! Look at the derivative $f'(x) = \frac{f(x)}{b^2 + x^2}$. Since $f(x) > 0$ and $b^2 + x^2 > 0$ (because $b \neq 0$), the derivative is always positive regardless of the sign of $b$. A positive derivative always implies an increasing function. The sign of $b$ affects the 'steepness' or the shape of the curve, but it does not change the fact that the function is climbing.

Q: How do I prove $f(x)f(-x) = 1$ without getting lost in the algebra?

Think about the structure of the function. We found $f(x) = e^{\frac{1}{b} \tan^{-1}(\frac{x}{b})}$. When you calculate $f(-x)$, you get $e^{\frac{1}{b} \tan^{-1}(\frac{-x}{b})}$. Since $\tan^{-1}$ is an odd function, this becomes $e^{-\frac{1}{b} \tan^{-1}(\frac{x}{b})}$. When you multiply $f(x)$ and $f(-x)$, the exponents add up to zero, and $e^0 = 1$. It is a beautiful cancellation!

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

The Spark: Look at the structure of the differential equation

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

. This is a classic separable differential equation. Do not let the

b^{2} + x^{2}

term intimidate you; it is a standard integral form you have encountered in your calculus journey. Your first goal is to isolate the variables

f

and

x

on opposite sides of the equation.

Hint 2

The Arsenal: Recall the standard integral formula for

\int \frac{1}{a ^{2} + x ^{2}} d x

, which is

\frac{1}{a} tan^{- 1} (\frac{x}{a}) + C

. Once you integrate both sides, use the initial condition

f (0) = 1

to determine the value of the constant

C

. Remember that

ln (1) = 0

, which will simplify your life significantly.

Hint 3

The Execution: After finding the explicit function

f (x) = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})}

, analyze the derivative

f^{'} (x)

to determine monotonicity. For the symmetry options, evaluate

f (- x)

by substituting

- x

into your derived function and utilize the property that

tan^{- 1} (- z) = - tan^{- 1} (z)

to see how the function behaves under reflection.

Visualized Solution

Given Differential Equation

Given differential equation:
$f^{'} (x) = \frac{f ( x )}{b ^{2} + x ^{2}}$

Separating Variables

Rearrange to separate variables $f (x)$ and $x$ :
$\frac{f ^{'} ( x )}{f ( x )} = \frac{1}{b ^{2} + x ^{2}}$

Integration Setup

Integrate both sides with respect to $x$ :
$\int \frac{f ^{'} ( x )}{f ( x )} d x = \int \frac{1}{b ^{2} + x ^{2}} d x$

Evaluating Integrals

Evaluate the integrals on both sides:
$ln ∣ f (x) ∣ = \frac{1}{b} tan^{- 1} (\frac{x}{b}) + C$

Applying Initial Condition $f (0) = 1$

Use the given initial condition to find $C$ :
Given: $f (0) = 1$
$ln ∣1∣ = \frac{1}{b} tan^{- 1} (0) + C$

Finding Constant $C$

Evaluate the terms:
$0 = \frac{1}{b} (0) + C$
$C = 0$

Isolating $∣ f (x) ∣$

Substitute $C = 0$ back and exponentiate:
$ln ∣ f (x) ∣ = \frac{1}{b} tan^{- 1} (\frac{x}{b})$
$∣ f (x) ∣ = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})}$

Explicit Function $f (x)$

Since $e^{y} > 0$ for all real $y$ , $f (x)$ must be strictly positive.
$f (x) = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})}$

Checking Monotonicity

To check if $f (x)$ is increasing or decreasing, analyze the sign of $f^{'} (x)$ .
$f^{'} (x) = \frac{f ( x )}{b ^{2} + x ^{2}}$

Sign of $f^{'} (x)$

Numerator: $f (x) = e^{\dots} > 0$
Denominator: $b^{2} + x^{2} > 0$ (since $b \neq = 0$ )
$⟹ f^{'} (x) > 0$

Conclusion for Options A & B

Since $f^{'} (x) > 0$ for all $x \in R$ , $f (x)$ is strictly increasing for all $b \neq = 0$ .
Option (A) is True.
Option (B) is False.

Evaluating $f (- x)$

To check options C and D, substitute $- x$ in place of $x$ :
$f (- x) = e^{\frac{1}{b} t a n^{- 1} (\frac{- x}{b})}$
Using $tan^{- 1} (- θ) = - tan^{- 1} (θ)$ :
$f (- x) = e^{- \frac{1}{b} t a n^{- 1} (\frac{x}{b})}$

Checking Option C

Evaluate the product $f (x) \cdot f (- x)$ :
$f (x) f (- x) = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})} \cdot e^{- \frac{1}{b} t a n^{- 1} (\frac{x}{b})}$
$f (x) f (- x) = e^{0} = 1$
Option (C) is True.

Checking Option D

Evaluate the difference $f (x) - f (- x)$ :
$f (x) - f (- x) = f (x) - \frac{1}{f ( x )} \neq = 0$
Option (D) is False.
Final Answer: Statements (A) and (C) are True.

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

We are given the differential equation

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

with the initial condition

f (0) = 1

. This is a classic setup where the rate of change of a function is proportional to the function itself, scaled by a geometric factor.

To solve this, we separate the variables by grouping all terms involving

f

on one side and all terms involving

x

on the other. By dividing both sides by

f (x)

and multiplying by

d x

, we obtain:

\frac{f ^{'} ( x )}{f ( x )} = \frac{1}{b ^{2} + x ^{2}}

The Integration Journey

Now, we integrate both sides of the equation:

\int \frac{f ^{'} ( x )}{f ( x )} d x = \int \frac{1}{b ^{2} + x ^{2}} d x

The left side yields

ln ∣ f (x) ∣

, while the right side is the standard integral of the inverse tangent function. This gives us:

ln ∣ f (x) ∣ = \frac{1}{b} tan^{- 1} (\frac{x}{b}) + C

We use the initial condition

f (0) = 1

to determine the constant

C

. Substituting

x = 0

and

f (0) = 1

into the equation:

ln (1) = \frac{1}{b} tan^{- 1} (0) + C

Since

ln (1) = 0

and

tan^{- 1} (0) = 0

, we find that

C = 0

. Our function simplifies to:

ln ∣ f (x) ∣ = \frac{1}{b} tan^{- 1} (\frac{x}{b})

Exponentiating both sides and removing the absolute value bars (as the exponential function is always positive), we arrive at the final expression:

f (x) = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})}

Analyzing the Behavior

To determine monotonicity, we examine

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

. Since

f (x) > 0

and

b^{2} + x^{2} > 0

for all

x

f^{'} (x)

is strictly positive. This confirms that the function is always increasing.

Finally, we check for symmetry by evaluating

f (- x)

f (- x) = e^{\frac{1}{b} t a n^{- 1} (\frac{- x}{b})} = e^{- \frac{1}{b} t a n^{- 1} (\frac{x}{b})}

Multiplying

f (x)

and

f (- x)

yields:

f (x) \cdot f (- x) = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})} \cdot e^{- \frac{1}{b} t a n^{- 1} (\frac{x}{b})} = e^{0} = 1

This confirms the property

f (x) f (- x) = 1

. You have successfully navigated the logic of this function; keep this analytical rigor in your toolkit to conquer any problem.

Common Traps & FAQs

Wait, are we allowed to divide by

f (x)

? What if

f (x) = 0

That is a brilliant question! In the separation of variables method, we technically assume

f (x) \neq = 0

. However, notice that our final solution is

f (x) = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})}

. Since the exponential function

e^{u}

is strictly positive for all real inputs

u

f (x)

can never be zero. Thus, our division was perfectly valid and safe.

Why is Option (B) false? If

b < 0

, shouldn't the function decrease?

It is a common trap! Look at the derivative

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

. Since

f (x) > 0

and

b^{2} + x^{2} > 0

(because

b \neq = 0

), the derivative is always positive regardless of the sign of

b

. A positive derivative always implies an increasing function. The sign of

b

affects the 'steepness' or the shape of the curve, but it does not change the fact that the function is climbing.

How do I prove

f (x) f (- x) = 1

without getting lost in the algebra?

Think about the structure of the function. We found

f (x) = e^{\frac{1}{b} t a n^{- 1} (\frac{x}{b})}

. When you calculate

f (- x)

, you get

e^{\frac{1}{b} t a n^{- 1} (\frac{- x}{b})}

. Since

tan^{- 1}

is an odd function, this becomes

e^{- \frac{1}{b} t a n^{- 1} (\frac{x}{b})}

. When you multiply

f (x)

and

f (- x)

, the exponents add up to zero, and

e^{0} = 1

. It is a beautiful cancellation!

Select Answer:

Visualized Solution

Given Differential Equation

Separating Variables

Integration Setup

Evaluating Integrals

Applying Initial Condition $f (0) = 1$

Finding Constant $C$

Isolating $∣ f (x) ∣$

Explicit Function $f (x)$

Checking Monotonicity

Sign of $f^{'} (x)$

Conclusion for Options A & B

Evaluating $f (- x)$

Checking Option C

Checking Option D

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Integration Journey

Analyzing the Behavior

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 $f$ 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 a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

Let $f (x) = lim_{r \to x} {\frac{2 r ^{2} [( f ( r ) ) ^{2} - f ( x ) f ( r )]}{r ^{2} - x ^{2}} - r^{3} e^{\frac{f ( r )}{r}}}$ be differentiable in $(- \infty, 0) \cup (0, \infty)$ and $f (1) = 1$ . Then the value of $e a$ , such that $f (a) = 0$ , 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 :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ 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 ______.

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$

Select Answer:

Visualized Solution

Given Differential Equation

Separating Variables

Integration Setup

Evaluating Integrals

Applying Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)=1

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

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

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

Checking Monotonicity

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

Conclusion for Options A & B

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

Checking Option C

Checking Option D

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Integration Journey

Analyzing the Behavior

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a curve y=f(x) pass through the point (2,(loge​2)2) and have slope xloge​x2y​ for all positive real value of x. Then the value of f(e) 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; } Let y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0,y(1)=e. Then limx→0+​f(x) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is true for y(x) that satisfies the differential equation dxdy​=xy−1+x−y;y(0)=0

Applying Initial Condition $f (0) = 1$

Finding Constant $C$

Isolating $∣ f (x) ∣$

Explicit Function $f (x)$

Sign of $f^{'} (x)$

Evaluating $f (- x)$

Let a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ 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 :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ is equal to

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$