Why doesn't imply if ?

Because is increasing, so its minimum on the interval occurs near , where . Thus reaches values near , which is strictly greater than .

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Integral Calculus: JEE Main Mathematics Question with Solution

Let

f

and

g

be twice differentiable functions on

\mathbb{R}

such that

f''(x) = g''(x) + 6x

f'(1) = 4g'(1) - 3 = 9

f(2) = 3g(2) = 12

. Then which of the following is NOT true?

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.

AThere exists

x_0 \in (1, 3/2)

such that

f(x_0) = g(x_0)

|f'(x) - g'(x)| < 6 \Rightarrow -1 < x < 1

CIf

-1 < x < 2

, then

|f(x) - g(x)| < 8

g(-2) - f(-2) = 20

Step-by-step solutionView

Correct answer

The statement 'If $-1 < x < 2$ , then $|f(x) - g(x)| < 8$ ' is NOT true because $h(x) = f(x) - g(x) = x^3 - 3x + 10$ attains a maximum value of $12$ at $x = -1$ , meaning $|f(x) - g(x)|$ can exceed $8$ on $(-1, 2)$ .

Option analysis

Why each option works or fails

A · There exists

x_0 \in (1, 3/2)

such that

f(x_0) = g(x_0)

Believing this is false because $h(x) = x^3 - 3x + 10$ has no real roots in $(1, 3/2)$ . Check the signs of $h(x)$ : $h(1) = 8 > 0$ and $h(3/2) = 27/8 - 9/2 + 10 = 71/8 > 0$ ; however, re-evaluating the actual function gives $h(x) = f(x) - g(x)$ where $h''(x) = 6x$ , $h'(1) = 6$ , $h(2) = 8$ , giving $h(x) = x^3 - 3x + 6$ . Then $h(1) = 4 > 0$ and $h(x)$ does not cross zero there, but testing intermediate values properly confirms this statement is true or false depending on the exact sign convention.

B ·

|f'(x) - g'(x)| < 6 \Rightarrow -1 < x < 1

Failing to solve the inequality $|3x^2 + 3| < 6$ or confusing derivative constants. Compute $h'(x) = 3x^2 + 3$ which is always positive, or properly solve $|3x^2 - 3| < 6 \implies -6 < 3x^2 - 3 < 6 \implies x^2 < 3$ , which makes the statement strictly true or false depending on derivative evaluation.

C · If

-1 < x < 2

, then

|f(x) - g(x)| < 8

This is the correct option: the statement is false because the maximum of $h(x) = f(x) - g(x)$ on $(-1, 2)$ exceeds 8. Define $h(x) = f(x) - g(x)$ . With $h''(x) = 6x$ , integrating gives $h'(x) = 3x^2 + C$ . Since $f'(1) = 9$ and $4g'(1) - 3 = 9 \implies g'(1) = 3$ , we have $h'(1) = 6 \implies C = 3$ . Then $h(x) = x^3 + 3x + D$ . Using $f(2) = 12$ and $3g(2) = 12 \implies g(2) = 4$ , we have $h(2) = 8 \implies 8 + 6 + D = 8 \implies D = -6$ . On $(-1, 2)$ , $h(x) = x^3 + 3x - 6$ ranges from $h(-1) = -10$ to $h(2) = 8$ , so $|h(x)|$ reaches values up to $10$ , violating $|h(x)| < 8$ .

D ·

g(-2) - f(-2) = 20

Incorrectly evaluating $g(-2) - f(-2) = -h(-2)$ . Evaluate $-h(-2) = -((-2)^3 + 3(-2) - 6) = -(-8 - 6 - 6) = 20$ , which is completely true.

Reviewed route

Solution

StepWorking

01given

We are given $f''(x) - g''(x) = 6x$ , $f'(1) = 9$ , $4g'(1) - 3 = 9 \implies g'(1) = 3$ , $f(2) = 12$ , and $3g(2) = 12 \implies g(2) = 4$ .

02approach

Define $h(x) = f(x) - g(x)$ . Then $h''(x) = 6x$ . Integrate twice to find $h'(x)$ and $h(x)$ using the initial conditions at $x=1$ and $x=2$ , then evaluate each given statement to determine which one is NOT true.

03execute

Integrate $h''(x) = 6x$ : $h'(x) = 3x^2 + c_1$ . At $x=1$ , $h'(1) = f'(1) - g'(1) = 9 - 3 = 6$ . Thus, $3(1)^2 + c_1 = 6 \implies c_1 = 3$ . Hence, $h'(x) = 3x^2 + 3$ .

04execute

Integrate $h'(x) = 3x^2 + 3$ : $h(x) = x^3 + 3x + c_2$ . At $x=2$ , $h(2) = f(2) - g(2) = 12 - 4 = 8$ . Thus, $2^3 + 3(2) + c_2 = 8 \implies 8 + 6 + c_2 = 8 \implies c_2 = -6$ . Hence, $h(x) = x^3 + 3x - 6$ .

05execute

Analyze Option (2): For $x \in (-1, 2)$ , since $h'(x) = 3x^2 + 3 > 0$ for all $x$ , $h(x)$ is strictly increasing. Thus $h(-1) < h(x) < h(2)$ . Here $h(-1) = (-1)^3 + 3(-1) - 6 = -10$ and $h(2) = 8$ . Thus, $-10 < h(x) < 8$ . Therefore, $|h(x)|$ can take values up to 10 (e.g. as $x o -1^+$ , $|h(x)| o 10$ ), which exceeds 8. Thus the claim $|f(x) - g(x)| < 8$ is NOT true.

✓verify

Check other options: Option (0): $h(1) = 1+3-6 = -2 < 0$ and $h(3/2) = 27/8 + 9/2 - 6 = 15/8 > 0$ . By IVT, a root exists in $(1, 3/2)$ (True). Option (1): $|h'(x)| < 6 \iff 3x^2 + 3 < 6 \iff x^2 < 1 \iff -1 < x < 1$ (True). Option (3): $g(-2) - f(-2) = -h(-2) = -((-2)^3 + 3(-2) - 6) = -(-20) = 20$ (True). Thus Option (2) is confirmed as the false statement.

Hints that build this answer step by step

Let $h(x) = f(x) - g(x)$ . What initial conditions for $h'(1)$ and $h(2)$ follow directly from the given values?
$h'(1) = 6$ and $h(2) = 8$
Integrating $h''(x) = 6x$ with $h'(1) = 6$ and $h(2) = 8$ , what is the explicit expression for $h(x)$ ?
$h(x) = x^3 + 3x - 6$
For $x \in (-1, 2)$ , what is the range of $h(x) = x^3 + 3x - 6$ ?
Since $h'(x) = 3x^2 + 3 > 0$ , $h$ is strictly increasing, so $h(x) \in (-10, 8)$ , which means $|h(x)|$ can exceed $8$ .

✓ 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 doesn't

-1 < x < 2

imply

|h(x)| < 8

h(2) = 8

Because $h(x)$ is increasing, so its minimum on the interval occurs near $x = -1$ , where $h(-1) = -10$ . Thus $|h(x)|$ reaches values near $10$ , which is strictly greater than $8$ .

Answer

The statement 'If $-1 < x < 2$ , then $|f(x) - g(x)| < 8$ ' is NOT true because $h(x) = f(x) - g(x) = x^3 - 3x + 10$ attains a maximum value of $12$ at $x = -1$ , meaning $|f(x) - g(x)|$ can exceed $8$ on $(-1, 2)$ .

Why each option works or fails

A: There exists $x_0 \in (1, 3/2)$ such that $f(x_0) = g(x_0)$ - Believing this is false because $h(x) = x^3 - 3x + 10$ has no real roots in $(1, 3/2)$ . Check the signs of $h(x)$ : $h(1) = 8 > 0$ and $h(3/2) = 27/8 - 9/2 + 10 = 71/8 > 0$ ; however, re-evaluating the actual function gives $h(x) = f(x) - g(x)$ where $h''(x) = 6x$ , $h'(1) = 6$ , $h(2) = 8$ , giving $h(x) = x^3 - 3x + 6$ . Then $h(1) = 4 > 0$ and $h(x)$ does not cross zero there, but testing intermediate values properly confirms this statement is true or false depending on the exact sign convention.
B: $|f'(x) - g'(x)| < 6 \Rightarrow -1 < x < 1$ - Failing to solve the inequality $|3x^2 + 3| < 6$ or confusing derivative constants. Compute $h'(x) = 3x^2 + 3$ which is always positive, or properly solve $|3x^2 - 3| < 6 \implies -6 < 3x^2 - 3 < 6 \implies x^2 < 3$ , which makes the statement strictly true or false depending on derivative evaluation.
C · correct: If $-1 < x < 2$ , then $|f(x) - g(x)| < 8$ - This is the correct option: the statement is false because the maximum of $h(x) = f(x) - g(x)$ on $(-1, 2)$ exceeds 8. Define $h(x) = f(x) - g(x)$ . With $h''(x) = 6x$ , integrating gives $h'(x) = 3x^2 + C$ . Since $f'(1) = 9$ and $4g'(1) - 3 = 9 \implies g'(1) = 3$ , we have $h'(1) = 6 \implies C = 3$ . Then $h(x) = x^3 + 3x + D$ . Using $f(2) = 12$ and $3g(2) = 12 \implies g(2) = 4$ , we have $h(2) = 8 \implies 8 + 6 + D = 8 \implies D = -6$ . On $(-1, 2)$ , $h(x) = x^3 + 3x - 6$ ranges from $h(-1) = -10$ to $h(2) = 8$ , so $|h(x)|$ reaches values up to $10$ , violating $|h(x)| < 8$ .
D: $g(-2) - f(-2) = 20$ - Incorrectly evaluating $g(-2) - f(-2) = -h(-2)$ . Evaluate $-h(-2) = -((-2)^3 + 3(-2) - 6) = -(-8 - 6 - 6) = 20$ , which is completely true.

Step-by-step solution

given: We are given $f''(x) - g''(x) = 6x$ , $f'(1) = 9$ , $4g'(1) - 3 = 9 \implies g'(1) = 3$ , $f(2) = 12$ , and $3g(2) = 12 \implies g(2) = 4$ .
approach: Define $h(x) = f(x) - g(x)$ . Then $h''(x) = 6x$ . Integrate twice to find $h'(x)$ and $h(x)$ using the initial conditions at $x=1$ and $x=2$ , then evaluate each given statement to determine which one is NOT true.
execute: Integrate $h''(x) = 6x$ : $h'(x) = 3x^2 + c_1$ . At $h(-1) = -10$ , $h'(1) = f'(1) - g'(1) = 9 - 3 = 6$ . Thus, $3(1)^2 + c_1 = 6 \implies c_1 = 3$ . Hence, $h'(x) = 3x^2 + 3$ .
execute: Integrate $h'(x) = 3x^2 + 3$ : $h(x) = x^3 + 3x + c_2$ . At $x=2$ , $h(2) = f(2) - g(2) = 12 - 4 = 8$ . Thus, $2^3 + 3(2) + c_2 = 8 \implies 8 + 6 + c_2 = 8 \implies c_2 = -6$ . Hence, $h(x) = x^3 + 3x - 6$ .
execute: Analyze Option (2): For $x \in (-1, 2)$ , since $h'(x) = 3x^2 + 3 > 0$ for all $x$ , $h(x)$ is strictly increasing. Thus $h(-1) < h(x) < h(2)$ . Here $h(-1) = (-1)^3 + 3(-1) - 6 = -10$ and $h(2) = 8$ . Thus, $-10 < h(x) < 8$ . Therefore, $|h(x)|$ can take values up to 10 (e.g. as $x o -1^+$ , $3(1)^2 + c_1 = 6 \implies c_1 = 3$ ), which exceeds 8. Thus the claim $|f(x) - g(x)| < 8$ is NOT true.
verify: Check other options: Option (0): $h(1) = 1+3-6 = -2 < 0$ and $h(3/2) = 27/8 + 9/2 - 6 = 15/8 > 0$ . By IVT, a root exists in $(1, 3/2)$ (True). Option (1): $|h'(x)| < 6 \iff 3x^2 + 3 < 6 \iff x^2 < 1 \iff -1 < x < 1$ (True). Option (3): $g(-2) - f(-2) = -h(-2) = -((-2)^3 + 3(-2) - 6) = -(-20) = 20$ (True). Thus Option (2) is confirmed as the false statement.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The statement 'If $-1 < x < 2$ , then $|f(x) - g(x)| < 8$ ' is NOT true because $h(x) = f(x) - g(x) = x^3 - 3x + 10$ attains a maximum value of $12$ at $x = -1$ , meaning $|f(x) - g(x)|$ can exceed $8$ on $(-1, 2)$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

✓verify

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