JEE MainMathematics

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Determining Function Differences from Differential and Boundary Conditions

Q: Why doesn't -1 < x < 2 imply |h(x)| < 8 if 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.

Let f and g be twice differentiable functions on 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?

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AThere exists x_0 ∈ (1, 3/2) such that f(x_0) = g(x_0)B|f'(x) - g'(x)| < 6 ⇒ -1 < x < 1CIf -1 < x < 2, then |f(x) - g(x)| < 8Dg(-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 $h (x) = f (x) - g (x) = x^{3} - 3 x + 10$ 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 ∈ (1, 3/2) such that f(x_0) = g(x_0)

Believing this is false because $h (x) = x^{3} - 3 x + 10$ 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 $h^{''} (x) = 6 x$ , $h^{'} (1) = 6$ , $h (2) = 8$ , giving $h (x) = x^{3} - 3 x + 6$ . 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 ⇒ -1 < x < 1

Failing to solve the inequality $∣ 3 x^{2} + 3 ∣ < 6$ or confusing derivative constants. Compute $h^{'} (x) = 3 x^{2} + 3$ which is always positive, or properly solve $∣ 3 x^{2} - 3 ∣ < 6 ⟹ - 6 < 3 x^{2} - 3 < 6 ⟹ x^{2} < 3$ , 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 $h^{''} (x) = 6 x$ , integrating gives $h^{'} (x) = 3 x^{2} + C$ . Since $f^{'} (1) = 9$ and $4 g^{'} (1) - 3 = 9 ⟹ g^{'} (1) = 3$ , we have $h^{'} (1) = 6 ⟹ C = 3$ . Then $h (x) = x^{3} + 3 x + D$ . Using $f (2) = 12$ and $3 g (2) = 12 ⟹ g (2) = 4$ , we have $h (2) = 8 ⟹ 8 + 6 + D = 8 ⟹ D = - 6$ . On $(- 1, 2)$ , $h (x) = x^{3} + 3 x - 6$ 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 $f^{''} (x) - g^{''} (x) = 6 x$ , $f^{'} (1) = 9$ , $4 g^{'} (1) - 3 = 9 ⟹ g^{'} (1) = 3$ , $f (2) = 12$ , and $3 g (2) = 12 ⟹ g (2) = 4$ .

02Approach

Define $h (x) = f (x) - g (x)$ . Then $h^{''} (x) = 6 x$ . 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 $h^{''} (x) = 6 x$ : $h^{'} (x) = 3 x^{2} + c_{1}$ . At $x = 1$ , $h^{'} (1) = f^{'} (1) - g^{'} (1) = 9 - 3 = 6$ . Thus, $3 (1)^{2} + c_{1} = 6 ⟹ c_{1} = 3$ . Hence, $h^{'} (x) = 3 x^{2} + 3$ .

04Execute

Integrate $h^{'} (x) = 3 x^{2} + 3$ : $h (x) = x^{3} + 3 x + c_{2}$ . At $x = 2$ , $h (2) = f (2) - g (2) = 12 - 4 = 8$ . Thus, $2^{3} + 3 (2) + c_{2} = 8 ⟹ 8 + 6 + c_{2} = 8 ⟹ c_{2} = - 6$ . Hence, $h (x) = x^{3} + 3 x - 6$ .

05Execute

Analyze Option C: For $x \in (- 1, 2)$ , since $h^{'} (x) = 3 x^{2} + 3 > 0$ 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 A: $h (1) = 1 + 3 - 6 = - 2 < 0$ and $h (3 / 2) = 27 / 8 + 9 / 2 - 6 = 15 / 8 > 0$ . By IVT, a root exists in $(1, 3 / 2)$ (True). Option B: $∣ h^{'} (x) ∣ < 6 ⟺ 3 x^{2} + 3 < 6 ⟺ x^{2} < 1 ⟺ - 1 < x < 1$ (True). Option D: $g (- 2) - f (- 2) = - h (- 2) = - ((- 2)^{3} + 3 (- 2) - 6) = - (- 20) = 20$ (True). Thus Option C 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 $h^{''} (x) = 6 x$ with $h^{'} (1) = 6$ and $h (2) = 8$ , what is the explicit expression for $h (x)$ ?
$h (x) = x^{3} + 3 x - 6$
For $x \in (- 1, 2)$ , what is the range of $h (x) = x^{3} + 3 x - 6$ ?
Since $h^{'} (x) = 3 x^{2} + 3 > 0$ , $h$ 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $h (x) = f (x) - g (x) = x^{3} - 3 x + 10$ 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 ∈ (1, 3/2) such that f(x_0) = g(x_0) - Believing this is false because $h (x) = x^{3} - 3 x + 10$ 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 $h^{''} (x) = 6 x$ , $h^{'} (1) = 6$ , $h (2) = 8$ , giving $h (x) = x^{3} - 3 x + 6$ . 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 ⇒ -1 < x < 1 - Failing to solve the inequality $∣ 3 x^{2} + 3 ∣ < 6$ or confusing derivative constants. Compute $h^{'} (x) = 3 x^{2} + 3$ which is always positive, or properly solve $∣ 3 x^{2} - 3 ∣ < 6 ⟹ - 6 < 3 x^{2} - 3 < 6 ⟹ x^{2} < 3$ , 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 $h^{''} (x) = 6 x$ , integrating gives $h^{'} (x) = 3 x^{2} + C$ . Since $f^{'} (1) = 9$ and $4 g^{'} (1) - 3 = 9 ⟹ g^{'} (1) = 3$ , we have $h^{'} (1) = 6 ⟹ C = 3$ . Then $h (x) = x^{3} + 3 x + D$ . Using $f (2) = 12$ and $3 g (2) = 12 ⟹ g (2) = 4$ , we have $h (2) = 8 ⟹ 8 + 6 + D = 8 ⟹ D = - 6$ . On $(- 1, 2)$ , $h (x) = x^{3} + 3 x - 6$ 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 $f^{''} (x) - g^{''} (x) = 6 x$ , $f^{'} (1) = 9$ , $4 g^{'} (1) - 3 = 9 ⟹ g^{'} (1) = 3$ , $f (2) = 12$ , and $3 g (2) = 12 ⟹ g (2) = 4$ .
Approach: Define $h (x) = f (x) - g (x)$ . Then $h^{''} (x) = 6 x$ . Integrate twice to find $h^{'} (x)$ and $h (x)$ using the initial conditions at $x = 1$ and $f^{''} (x) - g^{''} (x) = 6 x$ , then evaluate each given statement to determine which one is NOT true.
Execute: Integrate $h^{''} (x) = 6 x$ : $h^{'} (x) = 3 x^{2} + c_{1}$ . At $x = 1$ , $h^{'} (1) = f^{'} (1) - g^{'} (1) = 9 - 3 = 6$ . Thus, $3 (1)^{2} + c_{1} = 6 ⟹ c_{1} = 3$ . Hence, $h^{'} (x) = 3 x^{2} + 3$ .
Execute: Integrate $h^{'} (x) = 3 x^{2} + 3$ : $h (x) = x^{3} + 3 x + c_{2}$ . At $x = 2$ , $h (2) = f (2) - g (2) = 12 - 4 = 8$ . Thus, $2^{3} + 3 (2) + c_{2} = 8 ⟹ 8 + 6 + c_{2} = 8 ⟹ c_{2} = - 6$ . Hence, $h (x) = x^{3} + 3 x - 6$ .
Execute: Analyze Option C: For $x \in (- 1, 2)$ , since $h^{'} (x) = 3 x^{2} + 3 > 0$ 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 A: $h (1) = 1 + 3 - 6 = - 2 < 0$ and $h (3 / 2) = 27 / 8 + 9 / 2 - 6 = 15 / 8 > 0$ . By IVT, a root exists in $(1, 3 / 2)$ (True). Option B: $∣ h^{'} (x) ∣ < 6 ⟺ 3 x^{2} + 3 < 6 ⟺ x^{2} < 1 ⟺ - 1 < x < 1$ (True). Option D: $g (- 2) - f (- 2) = - h (- 2) = - ((- 2)^{3} + 3 (- 2) - 6) = - (- 20) = 20$ (True). Thus Option C is confirmed as the false statement.

Determining Function Differences from Differential and Boundary Conditions

What feels right?

The statement 'If $- 1 < x < 2$ , then $∣ f (x) - g (x) ∣ < 8$ ' is NOT true because $h (x) = f (x) - g (x) = x^{3} - 3 x + 10$ 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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Students also ask

Determining Function Differences from Differential and Boundary Conditions

The statement 'If −1<x<2−1<x<2, then ∣f(x)−g(x)∣<8∣f(x)−g(x)∣<8' is NOT true because h(x)=f(x)−g(x)=x3−3x+10h(x)=f(x)−g(x)=x3−3x+10 attains a maximum value of 1212 at x=−1x=−1, meaning ∣f(x)−g(x)∣∣f(x)−g(x)∣ can exceed 88 on (−1,2)(−1,2).

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

The same chapter, from another paper — Integral Calculus, 2025

The value of the integral…

Find out the major product for the following reaction.

Students also ask

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