JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding Difference of Two Functions Given Equal Second Derivatives

Q: Why do we define h(x) = f(x) - g(x)?

Because f''(x) = g''(x) implies (f - g)''(x) = 0, drastically simplifying the system to a linear polynomial in one variable.

Let f(x) and g(x) be twice differentiable functions satisfying f''(x) = g''(x) for all x ∈ R, f'(1) = 2g'(1) = 4 and g(2) = 3f(2) = 9. Then f(25) - g(25) is equal to:

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A20B40C-20D-40

Step-by-step solutionView

Correct answer

Defining $h (x) = f (x) - g (x)$ , the condition $h^{''} (x) = 0$ implies $h (x)$ is linear, leading directly to $f (25) - g (25) = 40$ .

Option analysis

Why each option works or fails

A · 20

The student calculates the slope as $m = f^{'} (1) - g^{'} (1) = 2$ , but erroneously computes the constant term by evaluating $2 (25 - 2)$ instead of accounting for the initial offset $f (2) - g (2) = - 6$ , or sets $h (x) = 2 x - 30$ . Consistently write $h (x) = h (2) + h^{'} (1) (x - 2)$ and substitute the correct values: $- 6 + 2 (23) = 40$ .

B · 40

Correct option. Defining $h (x) = f (x) - g (x)$ gives $h^{''} (x) = 0$ , so $h (x) = m x + c$ . With $h^{'} (1) = 4 - 2 = 2$ and $h (2) = 3 - 9 = - 6$ , we get $h (x) = 2 x - 10$ , so $h (25) = 50 - 10 = 40$ .

C · -20

The student inverts the sign of the constant term or computes $h (25)$ as $- 6 - 2 (23)$ or similar arithmetic error. Ensure the sign of the slope $h^{'} (x) = f^{'} (x) - g^{'} (x) = 4 - 2 = + 2$ is maintained when extrapolating from $x = 2$ to $x = 25$ .

D · -40

The student calculates $g (25) - f (25)$ instead of $f (25) - g (25)$ , reversing the order of subtraction. Check the requested quantity in the question stem: $f (25) - g (25)$ , not $g (25) - f (25)$ .

Reviewed route

Solution

StepWorking

01Given

$f^{''} (x) = g^{''} (x)$ , $f^{'} (1) = 4$ , $g^{'} (1) = 2$ , $g (2) = 9$ , $f (2) = 3$ .

02Goal

Find the value of $f (25) - g (25)$ .

03Approach

Define $h (x) = f (x) - g (x)$ . Then $h^{''} (x) = 0$ , which means $h^{'} (x)$ is a constant and $h (x)$ is a linear function $A x + B$ . Determine $A$ and $B$ from the initial values.

04Execute

Since $h^{''} (x) = 0$ , $h^{'} (x) = C_{1}$ . At $x = 1$ , $h^{'} (1) = f^{'} (1) - g^{'} (1) = 4 - 2 = 2$ , so $h^{'} (x) = 2$ . Integrating gives $h (x) = 2 x + C_{2}$ . At $x = 2$ , $h (2) = f (2) - g (2) = 3 - 9 = - 6$ . Substitute $x = 2$ : $- 6 = 2 (2) + C_{2} ⟹ C_{2} = - 10$ . Thus, $h (x) = f (x) - g (x) = 2 x - 10$ . For $x = 25$ : $h (25) = 2 (25) - 10 = 40$ .

✓Verify

Check at $x = 2$ : $h (2) = 2 (2) - 10 = - 6 = f (2) - g (2)$ , and $h^{'} (1) = 2 = f^{'} (1) - g^{'} (1)$ , which matches the given data.

Hints that build this answer step by step

Let $h (x) = f (x) - g (x)$ . What does the condition $f^{''} (x) = g^{''} (x)$ for all $x \in R$ imply about $h (x)$ ?
$h^{''} (x) = 0$ , which means $h (x)$ is a linear polynomial of the form $A x + B$ .
What are the values of $h^{'} (1)$ and $h (2)$ ?
$h^{'} (1) = 2$ and $h (2) = - 6$
Using $h (x) = h (2) + h^{'} (1) (x - 2)$ , what is the value of $h (25) = f (25) - g (25)$ ?
$40$

✓ Source and academic review

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

Quick checks

Students also ask

Why do we define

h (x) = f (x) - g (x)

Because $f^{''} (x) = g^{''} (x)$ implies $(f - g)^{''} (x) = 0$ , drastically simplifying the system to a linear polynomial in one variable.

Answer

Defining $h (x) = f (x) - g (x)$ , the condition $h^{''} (x) = 0$ implies $h (x)$ is linear, leading directly to $f (25) - g (25) = 40$ .

Why each option works or fails

A: 20 - The student calculates the slope as $m = f^{'} (1) - g^{'} (1) = 2$ , but erroneously computes the constant term by evaluating $2 (25 - 2)$ instead of accounting for the initial offset $f (2) - g (2) = - 6$ , or sets $h (x) = 2 x - 30$ . Consistently write $h (x) = h (2) + h^{'} (1) (x - 2)$ and substitute the correct values: $- 6 + 2 (23) = 40$ .
B · correct: 40 - Correct option. Defining $h (x) = f (x) - g (x)$ gives $h^{''} (x) = 0$ , so $h (x) = m x + c$ . With $h^{'} (1) = 4 - 2 = 2$ and $h (2) = 3 - 9 = - 6$ , we get $h (x) = 2 x - 10$ , so $h (25) = 50 - 10 = 40$ .
C: -20 - The student inverts the sign of the constant term or computes $h (25)$ as $- 6 - 2 (23)$ or similar arithmetic error. Ensure the sign of the slope $h (x) = h (2) + h^{'} (1) (x - 2)$ is maintained when extrapolating from $x = 2$ to $x = 25$ .
D: -40 - The student calculates $g (25) - f (25)$ instead of $f (25) - g (25)$ , reversing the order of subtraction. Check the requested quantity in the question stem: $f (25) - g (25)$ , not $g (25) - f (25)$ .

Step-by-step solution

Given: $f^{''} (x) = g^{''} (x)$ , $f^{'} (1) = 4$ , $g^{'} (1) = 2$ , $g (2) = 9$ , $f (2) = 3$ .
Goal: Find the value of $f (25) - g (25)$ .
Approach: Define $h (x) = f (x) - g (x)$ . Then $h^{''} (x) = 0$ , which means $x = 25$ is a constant and $h (x)$ is a linear function $A x + B$ . Determine $A$ and $B$ from the initial values.
Execute: Since $h^{''} (x) = 0$ , $h^{'} (x) = C_{1}$ . At $x = 1$ , $h^{'} (1) = f^{'} (1) - g^{'} (1) = 4 - 2 = 2$ , so $h^{'} (x) = 2$ . Integrating gives $h (x) = 2 x + C_{2}$ . At $x = 2$ , $h (2) = f (2) - g (2) = 3 - 9 = - 6$ . Substitute $x = 2$ : $- 6 = 2 (2) + C_{2} ⟹ C_{2} = - 10$ . Thus, $h (x) = f (x) - g (x) = 2 x - 10$ . For $x = 25$ : $h (25) = 2 (25) - 10 = 40$ .
Verify: Check at $x = 2$ : $h (2) = 2 (2) - 10 = - 6 = f (2) - g (2)$ , and $h^{'} (1) = 2 = f^{'} (1) - g^{'} (1)$ , which matches the given data.

Finding Difference of Two Functions Given Equal Second Derivatives

What feels right?

Defining $h (x) = f (x) - g (x)$ , the condition $h^{''} (x) = 0$ implies $h (x)$ is linear, leading directly to $f (25) - g (25) = 40$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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