JEE Main 2021 (18 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ satisfy the equation $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in R$ and $f (x) \neq = 0$ for any $x \in R$ . If the function $f$ is differentiable at $x = 0$ and $f^{'} (0) = 3$ , then $lim_{h \to 0} \frac{1}{h} (f (h) - 1)$ is equal to ___

Q: Why is $f(0) = 1$?

Great question! If $f(x + y) = f(x) \cdot f(y)$, then setting $x = 0$ and $y = 0$ gives $f(0) = f(0) \cdot f(0)$, or $f(0) = [f(0)]^2$. This implies $f(0)(f(0) - 1) = 0$. Since the problem explicitly states $f(x) \neq 0$ for any $x$, we must have $f(0) = 1$. It is a crucial anchor point for the function.

Q: Is the limit $\lim_{h \to 0} \frac{f(h) - 1}{h}$ always equal to $f'(0)$?

Yes, absolutely! By definition, $f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h}$. Since $f(0) = 1$ and $f(0 + h) = f(h)$, the expression simplifies perfectly to the derivative definition. It is a beautiful connection between the functional equation and calculus.

Enter Numerical Value:

Visualized Solution

The Functional Equation

Given functional equation: $f (x + y) = f (x) \cdot f (y)$
Condition: $f (x) \neq = 0$ for all $x \in R$

Setting up for $f (0)$

To find the anchor point, substitute $x = 0$ and $y = 0$ .

Algebraic Substitution

$f (0 + 0) = f (0) \cdot f (0)$
$f (0) = [f (0)]^{2}$

Evaluating $f (0)$

Since $f (x) \neq = 0$ , we can divide by $f (0)$ :
$f (0) = 1$

The Derivative Information

Given: $f^{'} (0) = 3$

First Principles Definition

Definition of derivative at $x = 0$ :
$f^{'} (0) = lim_{h \to 0} \frac{f ( 0 + h ) - f ( 0 )}{h}$

Simplifying the Numerator

Simplify the term inside the function:
$f (0 + h) = f (h)$

Substituting $f (0)$

Substitute $f (0) = 1$ into the limit:
$f^{'} (0) = lim_{h \to 0} \frac{f ( h ) - 1}{h}$

Connecting to the Question

Notice the structure:
$lim_{h \to 0} \frac{1}{h} (f (h) - 1) = f^{'} (0)$

The Final Result

Since $f^{'} (0) = 3$ , we conclude:
$lim_{h \to 0} \frac{1}{h} (f (h) - 1) = 3$

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The Sigma Insight: Differentiability of a Function

Solution Diagram

Analyzing the Functional DNA

You are given a function

f : R \to R

that obeys the rule

f (x + y) = f (x) \cdot f (y)

. In the context of JEE Advanced, this functional equation acts as the structural blueprint for the function.

Our objective is to evaluate the limit:

h \to 0 lim \frac{1}{h} (f (h) - 1)

given the condition that

f^{'} (0) = 3

Finding the Anchor Point

To understand the behavior of the function, we must first determine its value at the origin. We substitute

x = 0

and

y = 0

into the functional equation:

f (0 + 0) = f (0) \cdot f (0)

This simplifies to the algebraic relation

f (0) = [f (0)]^{2}

, or

f (0) - [f (0)]^{2} = 0

. Factoring this expression yields

f (0) (1 - f (0)) = 0

Since the problem states that $f(x) eq 0$ for any

x \in R

, we must reject the possibility that

f (0) = 0

. Thus, we conclude that $f (0) = 1$ .

The Bridge to Calculus

We are given the derivative at the origin,

f^{'} (0) = 3

. By the first principles of calculus, the derivative at

x = 0

is defined as:

f^{'} (0) = h \to 0 lim \frac{f ( 0 + h ) - f ( 0 )}{h}

Using the functional property

f (0 + h) = f (h)

and our anchor point

f (0) = 1

, we substitute these values into the limit definition:

f^{'} (0) = h \to 0 lim \frac{f ( h ) - 1}{h}

The Final Revelation

By comparing our derived expression to the limit requested in the problem, we see they are identical. Since we are given

f^{'} (0) = 3

, it follows directly that:

h \to 0 lim \frac{1}{h} (f (h) - 1) = 3

The final answer is $3$ . This result demonstrates how the fundamental definition of a derivative serves as the bridge between abstract functional equations and concrete numerical values.

Common Traps & FAQs

Why is

f (0) = 1

Great question! If

f (x + y) = f (x) \cdot f (y)

, then setting

x = 0

and

y = 0

gives

f (0) = f (0) \cdot f (0)

, or

f (0) = [f (0)]^{2}

. This implies

f (0) (f (0) - 1) = 0

. Since the problem explicitly states

f (x) \neq = 0

for any

x

, we must have

f (0) = 1

. It is a crucial anchor point for the function.

Can we just assume

f (x) = e^{3 x}

and solve it?

That is a brilliant intuition! You have recognized that exponential functions satisfy this property. However, in a formal proof, we must derive the result from the definition to ensure we are not missing any edge cases. The first principles approach is the rigorous path that works for all differentiable functions, and it is the standard way to solve such problems in JEE Advanced.

Is the limit

lim_{h \to 0} \frac{f ( h ) - 1}{h}

always equal to

f^{'} (0)

Yes, absolutely! By definition,

f^{'} (0) = lim_{h \to 0} \frac{f ( 0 + h ) - f ( 0 )}{h}

. Since

f (0) = 1

and

f (0 + h) = f (h)

, the expression simplifies perfectly to the derivative definition. It is a beautiful connection between the functional equation and calculus.

Enter Numerical Value:

Visualized Solution

The Functional Equation

Setting up for $f (0)$

Algebraic Substitution

Evaluating $f (0)$

The Derivative Information

First Principles Definition

Simplifying the Numerator

Substituting $f (0)$

Connecting to the Question

The Final Result

The Sigma Insight: Differentiability of a Function

Analyzing the Functional DNA

Finding the Anchor Point

The Bridge to Calculus

The Final Revelation

Similar Questions

Let $f : R \to R$ and $g : R \to R$ be functions satisfying $f (x + y) = f (x) + f (y) + f (x) f (y)$ and $f (x) = xg (x)$ for all $x, y \in R$ . If $lim_{x \to 0} g (x) = 1$ , then which of the following statements is/are TRUE?

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

Let $f$ be a differentiable function from $R$ to $R$ such that $∣ f (x) - f (y) ∣ \leq 2∣ x - y ∣^{3/2}$ , for all $x, y \in R$ . If $f (0) = 1$ then $\int_{0}^{1} f^{2} (x) d x$ is equal to

If $f (x - y) = f (x) \cdot g (y) - f (y) \cdot g (x)$ and $g (x - y) = g (x) \cdot g (y) + f (x) \cdot f (y)$ for all $x, y \in R$ . If right hand derivative at $x = 0$ exists for $f (x)$ . Find derivative of $g (x)$ at $x = 0$ .

Let $f (x)$ be a function satisfying the condition $f (- x) = f (x)$ for all real $x$ . If $f^{'} (0)$ exists, find its value.

If $f$ is a real valued differentiable function satisfying $∣ f (x) - f (y) ∣ \leq (x - y)^{2}, x, y \in R$ and $f (0) = 0$ , then $f (1)$ equals

Let $α \in R$ . Prove that a function $f : R \to R$ is differentiable at $α$ if and only if there is a function $g : R \to R$ which is continuous at $α$ and satisfies $f (x) - f (α) = g (x) (x - α)$ for all $x \in R$ .

Let $f : R \to R$ and $g : R \to R$ be respectively given by $f (x) = ∣ x ∣ + 1$ and $g (x) = x^{2} + 1$ . Define $h : R \to R$ by $h (x) = {max {f (x), g (x)} min {f (x), g (x)} if x \leq 0, if x > 0.$ The number of points at which $h (x)$ is not differentiable is .........

Suppose $f (x) = \frac{( 2 ^{x} + 2 ^{- x} ) t a n x t a n ^{- 1} ( x ^{2} - x + 1 )}{( 7 x ^{2} + 3 x + 1 ) ^{3}}$ . Then the value of $f^{'} (0)$ is equal to

Enter Numerical Value:

Visualized Solution

The Functional Equation

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

Algebraic Substitution

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

The Derivative Information

First Principles Definition

Simplifying the Numerator

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

Connecting to the Question

The Final Result

The Sigma Insight: Differentiability of a Function

Analyzing the Functional DNA

Finding the Anchor Point

The Bridge to Calculus

The Final Revelation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x−y)=f(x)⋅g(y)−f(y)⋅g(x) and g(x−y)=g(x)⋅g(y)+f(x)⋅f(y) for all x,y∈R. If right hand derivative at x=0 exists for f(x). Find derivative of g(x) at x=0.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a function satisfying the condition f(−x)=f(x) for all real x. If f′(0) exists, find its value.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f is a real valued differentiable function satisfying ∣f(x)−f(y)∣≤(x−y)2,x,y∈R and f(0)=0, then f(1) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose f(x)=(7x2+3x+1)3(2x+2−x)tanxtan−1(x2−x+1)​​. Then the value of f′(0) is equal to

Setting up for $f (0)$

Evaluating $f (0)$

Substituting $f (0)$

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

Let $f$ be a differentiable function from $R$ to $R$ such that $∣ f (x) - f (y) ∣ \leq 2∣ x - y ∣^{3/2}$ , for all $x, y \in R$ . If $f (0) = 1$ then $\int_{0}^{1} f^{2} (x) d x$ is equal to

If $f (x - y) = f (x) \cdot g (y) - f (y) \cdot g (x)$ and $g (x - y) = g (x) \cdot g (y) + f (x) \cdot f (y)$ for all $x, y \in R$ . If right hand derivative at $x = 0$ exists for $f (x)$ . Find derivative of $g (x)$ at $x = 0$ .

Let $f (x)$ be a function satisfying the condition $f (- x) = f (x)$ for all real $x$ . If $f^{'} (0)$ exists, find its value.

If $f$ is a real valued differentiable function satisfying $∣ f (x) - f (y) ∣ \leq (x - y)^{2}, x, y \in R$ and $f (0) = 0$ , then $f (1)$ equals

Suppose $f (x) = \frac{( 2 ^{x} + 2 ^{- x} ) t a n x t a n ^{- 1} ( x ^{2} - x + 1 )}{( 7 x ^{2} + 3 x + 1 ) ^{3}}$ . Then the value of $f^{'} (0)$ is equal to