JEE MainMathematics

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Finding the value of a functional equation using differentiability

Let f: ℝ → ℝ be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)-1, ∀ x, y ∈ ℝ. If f'(0)=2, then |f(-2)| is equal to

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

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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 $x, y \in R$ , $f$ 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: $f (0) = f (0) + f (0) - 1 ⟹ f (0) = 1$
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 $f (y) = 2 y + c$ . Since $f (0) = 1$ , we have $c = 1$ . Thus, $f (x) = 2 x + 1$ .
Execute: Evaluating at $x = - 2$ : $f (- 2) = 2 (- 2) + 1 = - 3$ Taking absolute value: $∣ f (- 2) ∣ = ∣ - 3 ∣ = 3$
Verify: Check: $f (x + y) = 2 (x + y) + 1 = 2 x + 2 y + 1$ . Also $f (x) + f (y) - 1 = (2 x + 1) + (2 y + 1) - 1 = 2 x + 2 y + 1$ . 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 $g$ , $g (x) = k x$ where $k = g^{'} (0) = f^{'} (0) = 2$ . Hence, $g (x) = 2 x$ , which implies $f (x) = 2 x + 1$ .

Execute: $f (- 2) = 2 (- 2) + 1 = - 3$ , hence $∣ f (- 2) ∣ = 3$ .

Verify: $g (0) = 0 ⟹ f (0) = 1$ , which matches $f (0) = f (0) + f (0) - 1$ .

Finding the value of a functional equation using differentiability

What feels right?

The value of $∣ f (- 2) ∣$ is 3.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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