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Limits, Continuity and Differentiability: Mathematics | JEE Main

Let

f: \mathbb{R} \rightarrow \mathbb{R}

be a differentiable function that satisfies the relation

f(x+y)=f(x)+f(y)-1, \forall x, y \in \mathbb{R}

. If

f^{\prime}(0)=2

, then

|f(-2)|

is equal to

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Is partial differentiation valid for functional equations in single-variable calculus?

Yes, because the relation holds for all independent real variables $x$ and $y$ . Differentiating with respect to $x$ while holding $y$ fixed is equivalent to evaluating $\lim_{h \to 0} \frac{f(x+h+y)-f(x+y)}{h}$ .

Answer

The value of $|f(-2)|$ is 3.

Why each option works or fails

Step-by-step solution

given: $f(x+y) = f(x) + f(y) - 1$ for all $x, y \in \mathbb{R}$ , $f$ is differentiable, and $f'(0) = 2$ .
goal: Find the value of $|f(-2)|$ .
approach: Find $f(0)$ by setting $x = y = 0$ . Then differentiate the relation partially with respect to $x$ treating $y$ as a constant to find $f'(y)$ , integrate to get $f(x)$ , and evaluate $|f(-2)|$ .
execute: Substitute $x = 0, y = 0$ into the equation: $f(0) = f(0) + f(0) - 1 \implies f(0) = 1$
execute: Differentiating both sides with respect to $x$ treating $y$ as a constant: $f'(x+y) = f'(x)$ Setting $x = 0$ : $f'(y) = f'(0) = 2$
execute: Integrating $f'(y) = 2$ gives $f(y) = 2y + c$ . Since $f(0) = 1$ , we have $c = 1$ . Thus, $f(x) = 2x + 1$ .
execute: Evaluating at $x = -2$ : $f(-2) = 2(-2) + 1 = -3$ Taking absolute value: $|f(-2)| = |-3| = 3$
verify: Check: $f(x+y) = 2(x+y) + 1 = 2x + 2y + 1$ . Also $f(x) + f(y) - 1 = (2x+1) + (2y+1) - 1 = 2x + 2y + 1$ . The functional relation holds identically.

Shortcut:

When to use it: Rigorous textbook method using the limit definition of derivative.

given: $f(x+y) = f(x) + f(y) - 1$ with $f'(0) = 2$ .

goal: Determine $|f(-2)|$ .

approach: Define $g(x) = f(x) - 1$ . Then $g(x+y) = g(x) + g(y)$ , which is Cauchy's functional equation.

execute: Let $g(x) = f(x) - 1$ . Then: $g(x+y) = f(x+y) - 1 = f(x) + f(y) - 2 = (f(x)-1) + (f(y)-1) = g(x) + g(y)$ Since $f$ is differentiable, $g$ is differentiable.

execute: For Cauchy's equation $g(x+y) = g(x) + g(y)$ with differentiable $g$ , $g(x) = kx$ where $k = g'(0) = f'(0) = 2$ . Hence, $g(x) = 2x$ , which implies $f(x) = 2x + 1$ .

execute: $f(-2) = 2(-2) + 1 = -3$ , hence $|f(-2)| = 3$ .

verify: $y$ , which matches $f(0) = f(0)+f(0)-1$ .

Limits, Continuity and Differentiability: Mathematics | JEE Main

What feels right?

The value of $|f(-2)|$ is 3.

Why each option works or fails

Solution

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02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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