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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why do we define ?

Because implies , 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 \in R

f'(1) = 2g'(1) = 4

and

g(2) = 3f(2) = 9

. Then

f(25) - g(25)

is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

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 $h(x) = 2x - 30$ . 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 $h(x) = mx + c$ . With $h'(1) = 4 - 2 = 2$ and $h(2) = 3 - 9 = -6$ , we get $h(x) = 2x - 10$ , 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 $Ax + B$ . 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 $h(x) = 2x + C_2$ . At $x=2$ , $h(2) = f(2) - g(2) = 3 - 9 = -6$ . Substitute $x=2$ : $-6 = 2(2) + C_2 \implies C_2 = -10$ . Thus, $h(x) = f(x) - g(x) = 2x - 10$ . 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 $x \in \mathbb{R}$ imply about $h(x)$ ?
$h''(x) = 0$ , which means $h(x)$ is a linear polynomial of the form $Ax + B$ .
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $h(x) = 2x - 30$ . 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 $h(x) = mx + c$ . With $h'(1) = 4 - 2 = 2$ and $h(2) = 3 - 9 = -6$ , we get $h(x) = 2x - 10$ , 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 $g(25) - f(25)$ .

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 $h'(x)$ is a constant and $h(x)$ is a linear function $Ax + B$ . 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 $h(x) = 2x + C_2$ . At $x=2$ , $h(2) = f(2) - g(2) = 3 - 9 = -6$ . Substitute $x=2$ : $-6 = 2(2) + C_2 \implies C_2 = -10$ . Thus, $h(x) = f(x) - g(x) = 2x - 10$ . 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.

Integral Calculus: JEE Main Mathematics Question with Solution

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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