JEE Main 2023 (29 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a differentiable function that satisfies the relation $f (x + y) = f (x) + f (y) - 1, \forall x, y \in R$ . If $f^{'} (0) = 2$ , then $∣ f (- 2) ∣$ is equal to

Q: Why does the $f(x)$ term disappear in the derivative definition?

Because the functional equation $f(x + y) = f(x) + f(y) - 1$ is essentially a linear structure. The $f(x)$ terms cancel out because the 'offset' (the $-1$) is independent of $x$, which is exactly what happens when you differentiate a linear function.

Enter Numerical Value:

Visualized Solution

Analyze the Functional Equation

Given: $f (x + y) = f (x) + f (y) - 1$
Given: $f^{'} (0) = 2$
Goal: Find $∣ f (- 2) ∣$

Visual Anchor: Finding $f (0)$

Substitute $x = 0$ and $y = 0$ into the equation:
$f (0 + 0) = f (0) + f (0) - 1$

Compute the Base Point

$f (0) = 2 f (0) - 1$
$f (0) = 1$

Logic Bridge: First Principles

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

Raw Setup: Substitute Relation

Using $f (x + h) = f (x) + f (h) - 1$ :
$f^{'} (x) = lim_{h \to 0} \frac{[ f ( x ) + f ( h ) - 1 ] - f ( x )}{h}$

Atomic Compute: Simplify

Cancel $f (x)$ terms:
$f^{'} (x) = lim_{h \to 0} \frac{f ( h ) - 1}{h}$

Atomic Compute: Relate to $f^{'} (0)$

Since $f (0) = 1$ , substitute it back:
$f^{'} (x) = lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{h}$
This is exactly $f^{'} (0)$ !

Determine the Derivative

Given $f^{'} (0) = 2$
Therefore, $f^{'} (x) = 2$ for all $x \in R$

Logic Bridge: Integrate

Integrate to find $f (x)$ :
$f (x) = \int 2 d x$
$f (x) = 2 x + C$

Solve for Constant $C$

Use the base point $f (0) = 1$ :
$1 = 2 (0) + C ⟹ C = 1$
Exact function: $f (x) = 2 x + 1$

Calculate $f (- 2)$

Substitute $x = - 2$ :
$f (- 2) = 2 (- 2) + 1$
$f (- 2) = - 3$

Calculate Absolute Value

Find $∣ f (- 2) ∣$ :
$∣ f (- 2) ∣ = ∣ - 3∣ = 3$

The Way Forward

Final Answer: $3$
Functional equations like $f (x + y) = f (x) + f (y) + k$ are always linear.

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The Sigma Insight: Techniques of Differentiation

Solution Diagram

Analyzing the Setup

We are tasked with decoding the functional equation defined by the relation:

f (x + y) = f (x) + f (y) - 1

To find our anchor point, we set

x = 0

and

y = 0

. This yields:

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

f (0) = 2 f (0) - 1

f (0) = 1

This is our first concrete piece of information regarding the function.

The Derivative Bridge

Next, we utilize the formal definition of the derivative:

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

Substituting the given relation

f (x + h) = f (x) + f (h) - 1

into the limit, the

f (x)

terms cancel out:

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

Since we established that

f (0) = 1

, we can rewrite the expression as:

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

This is exactly the definition of

f^{'} (0)

. Given that

f^{'} (0) = 2

, we conclude that the derivative is constant:

f^{'} (x) = 2

The Revelation

Geometrically, a constant derivative implies that the function must be a linear equation. Integrating

f^{'} (x) = 2

with respect to

x

gives:

f (x) = 2 x + C

Using our anchor point

f (0) = 1

, we solve for the constant

C

1 = 2 (0) + C \Rightarrow C = 1

Thus, the explicit form of our function is:

f (x) = 2 x + 1

The Final Calculation

Finally, we evaluate the function at

x = - 2

to find the requested value:

f (- 2) = 2 (- 2) + 1 = - 4 + 1 = - 3

Taking the absolute value as required:

∣ f (- 2) ∣ = ∣ - 3∣ = 3

The final result is 3.

Common Traps & FAQs

Wait, why do we substitute

x = 0

and

y = 0

first? Is that always the trick?

In functional equations, finding the value at the origin is the most powerful way to 'anchor' the function. It is like finding the y-intercept of a line before you know the slope. It gives you a concrete point to build upon.

Why does the

f (x)

term disappear in the derivative definition?

Because the functional equation

f (x + y) = f (x) + f (y) - 1

is essentially a linear structure. The

f (x)

terms cancel out because the 'offset' (the

- 1

) is independent of

x

, which is exactly what happens when you differentiate a linear function.

Can I just assume

f (x) = a x + b

and solve for

a

and

b

Yes, you can! This is a valid 'shortcut' for JEE. If you recognize the structure, you can assume linearity, but proving it via first principles is the rigorous path that ensures you do not miss any edge cases.

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a differentiable function that satisfies the relation $f (x + y) = f (x) + f (y) - 1, \forall x, y \in R$ . If $f^{'} (0) = 2$ , then $∣ f (- 2) ∣$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Functional Equation

Visual Anchor: Finding $f (0)$

Compute the Base Point

Logic Bridge: First Principles

Raw Setup: Substitute Relation

Atomic Compute: Simplify

Atomic Compute: Relate to $f^{'} (0)$

Determine the Derivative

Logic Bridge: Integrate

Solve for Constant $C$

Calculate $f (- 2)$

Calculate Absolute Value

The Way Forward

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Derivative Bridge

The Revelation

The Final Calculation

Similar Questions

Let $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$ for all real $x$ and $y$ . If $f^{'} (0)$ exists and equals $- 1$ and $f (0) = 1$ , find $f (2)$ .

For the differentiable function $f : R - {0} \to R$ , let $3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10$ , then $∣ f (3) + f^{'} (1/4) ∣$ is equal to

If $f (x + y) = f (x) f (y) \forall x, y$ and $f (5) = 2, f^{'} (0) = 3$ , then $f^{'} (5)$ is

Let $f : (- 1, 1) \to R$ be a differentiable function with $f (0) = - 1$ and $f^{'} (0) = 1$ . Let $g (x) = [f (2 f (x) + 2)]^{2}$ . Then $g^{'} (0) =$

Let $f : R \to R$ satisfy $f (x + y) = 2^{x} f (y) + 4^{y} f (x), \forall x, y \in R$ . If $f (2) = 3$ , then $14 \cdot \frac{f ^{'} ( 4 )}{f ^{'} ( 2 )}$ is equal to

If $f (x) = ∣ x - 2∣$ and $g (x) = f [f (x)]$ , then $g^{'} (x) = \dots\dots\dots$ for $x > 20$ .

A function $f : R \to R$ satisfies the equation $f (x + y) = f (x) f (y)$ for all $x, y$ in $R$ and $f (x) \neq = 0$ for any $x$ in $R$ . Let the function be differentiable at $x = 0$ and $f^{'} (0) = 2$ . Show that $f^{'} (x) = 2 f (x)$ for all $x$ in $R$ . Hence, determine $f (x)$ .

If the function $f (x) = x^{3} + e^{x /2}$ and $g (x) = f^{- 1} (x)$ , then the value of $g^{'} (1)$ is

If $f (1) = 1, f^{'} (1) = 3$ , then the derivative of $f (f (f (x))) + (f (x))^{2}$ at $x = 1$ is :

Let $g (x)$ be a linear function and $f (x) = {g (x) (\frac{1 + x}{2 + x})^{1/ x}, x \leq 0, x > 0$ , is continuous at $x = 0$ . If $f^{'} (1) = f (- 1)$ , then the value of $g (3)$ is

Enter Numerical Value:

Visualized Solution

Analyze the Functional Equation

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

Compute the Base Point

Logic Bridge: First Principles

Raw Setup: Substitute Relation

Atomic Compute: Simplify

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

Determine the Derivative

Logic Bridge: Integrate

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

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

Calculate Absolute Value

The Way Forward

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Derivative Bridge

The Revelation

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(2x+y​)=2f(x)+f(y)​ for all real x and y. If f′(0) exists and equals −1 and f(0)=1, find f(2).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the differentiable function f:R−{0}→R, let 3f(x)+2f(x1​)=x1​−10, then ∣f(3)+f′(1/4)∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x+y)=f(x)f(y)∀x,y and f(5)=2,f′(0)=3, then f′(5) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(−1,1)→R be a differentiable function with f(0)=−1 and f′(0)=1. Let g(x)=[f(2f(x)+2)]2. Then g′(0)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R satisfy f(x+y)=2xf(y)+4yf(x),∀x,y∈R. If f(2)=3, then 14⋅f′(2)f′(4)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∣x−2∣ and g(x)=f[f(x)], then g′(x)=……… for x>20.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)=x3+ex/2 and g(x)=f−1(x), then the value of g′(1) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(1)=1,f′(1)=3, then the derivative of f(f(f(x)))+(f(x))2 at x=1 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x) be a linear function and f(x)={g(x)(2+x1+x​)1/x​,x≤0,x>0​, is continuous at x=0. If f′(1)=f(−1), then the value of g(3) is

Visual Anchor: Finding $f (0)$

Atomic Compute: Relate to $f^{'} (0)$

Solve for Constant $C$

Calculate $f (- 2)$

Let $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$ for all real $x$ and $y$ . If $f^{'} (0)$ exists and equals $- 1$ and $f (0) = 1$ , find $f (2)$ .

For the differentiable function $f : R - {0} \to R$ , let $3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10$ , then $∣ f (3) + f^{'} (1/4) ∣$ is equal to

If $f (x + y) = f (x) f (y) \forall x, y$ and $f (5) = 2, f^{'} (0) = 3$ , then $f^{'} (5)$ is

Let $f : (- 1, 1) \to R$ be a differentiable function with $f (0) = - 1$ and $f^{'} (0) = 1$ . Let $g (x) = [f (2 f (x) + 2)]^{2}$ . Then $g^{'} (0) =$

Let $f : R \to R$ satisfy $f (x + y) = 2^{x} f (y) + 4^{y} f (x), \forall x, y \in R$ . If $f (2) = 3$ , then $14 \cdot \frac{f ^{'} ( 4 )}{f ^{'} ( 2 )}$ is equal to

If $f (x) = ∣ x - 2∣$ and $g (x) = f [f (x)]$ , then $g^{'} (x) = \dots\dots\dots$ for $x > 20$ .

If the function $f (x) = x^{3} + e^{x /2}$ and $g (x) = f^{- 1} (x)$ , then the value of $g^{'} (1)$ is

If $f (1) = 1, f^{'} (1) = 3$ , then the derivative of $f (f (f (x))) + (f (x))^{2}$ at $x = 1$ is :

Let $g (x)$ be a linear function and $f (x) = {g (x) (\frac{1 + x}{2 + x})^{1/ x}, x \leq 0, x > 0$ , is continuous at $x = 0$ . If $f^{'} (1) = f (- 1)$ , then the value of $g (3)$ is