Let f and g be twice differentiable functions on R such that f ^ " (x) = g ^ " (x) + 6x f ^ ' (1) = 4 g ^ ' (1) - 3 = 9 f(2) = 3g(2) = 12 Then which of the following is NOT true ?

Correct answer: B. If - 1 < x < 2 , then |f(x) - g(x)| < 8 Given functions are: f ^ " (x) = g ^ " (x) + 6x (1) f ^ ' (1) = 4 g ^ ' (1) - 3 = 9 (2) f(2) = 3g(2) = 12 (3) By integrating (1) , we get f ^ ' (x) = g ^ ' (x) + 3 x ^ 2 + C At x = 1 f ^ ' (1) = g ^ ' (1) + 3 + C ⇒ 9 = 3 + 3 + C ⇒ C = 3 ∴ f ^ ' (x) = g ^ ' (x) + 3 x ^ 2 + 3 Again by integrating, we get f(x) = g(x) + x ^ 3 + 3x + D At x = 2 , we get f(2) = g(2) + 8 + 3(2) + D ⇒ 12 = 4 + 8 + 6 + D ⇒ D = - 6 So, f(x) = g(x) + x ^ 3 + 3x - 6 Option (A) : At x = - 2 , ⇒ g(- 2) - f(- 2) = 20 So, this option is true. Option (B) If - 1 Let h(x) = f(x) - g(x) = x ^ 3 + 3x - 6 ⇒ h ^ ' (x) = 3 x ^ 2 + 3 ⇒ h ' (x) > 0 for all values of x . So, h(- 1) ⇒ - 10 ⇒ |h(x)| So, this option is NOT true. Option (C) h ^ ' (x) = f ^ ' (x) - g ^ ' (x) = 3 x ^ 2 + 3 If | h ^ ' (x)| ⇒ 3 x ^ 2 + 3 ⇒ x ^ 2 - 3(alwaystrue) ⇒ - 1 So, If x ∈ (- 1, 1) then | f ^ ' (x) - g ^ ' (x)| So, this option is true. Option (D) f(x) - g(x) = 0 ⇒ x ^ 3 + 3x - 6 = 0 h(x) = x ^ 3 + 3x - 6 Here, h(1) = - ve and h((3)/(2)) = + ve So, there exists x _ 0 ∈ (1, (3)/(2)) such that f( x _ 0 ) = g ( x _ 0 ) So, this option is true.

JEE Main 202329 JanuaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2023 · 29 January · Evening Shift · Mathematics · Continuity and Differentiability

Let f and g be twice differentiable functions on R such that f ^ "...

Let $f$ and $g$ be twice differentiable functions on $R$ such that

${f}^{"}\left(x\right) = {g}^{"}\left(x\right) + 6x$

${f}^{'}\left(1\right) = 4{g}^{'}\left(1\right) - 3 = 9$

$f\left(2\right) = 3g\left(2\right) = 12$

Then which of the following is NOT true ?

Options

A $g\left(- 2\right) - f\left(- 2\right) = 20$
BIf $- 1 < x < 2$ , then \left|f\left(x\right) - g\left(x\right)\right| < 8
C $\left|{f}^{'}\left(x\right) - {g}^{'}\left(x\right)\right| < 6 ⇒ - 1 < x < 1$
DThere exists ${x}_{0} \in \left(1, \dfrac{3}{2}\right)$ such that $f\left({x}_{0}\right) = g\left({x}_{0}\right)$

Correct answer

B. If $- 1 < x < 2$ , then \left|f\left(x\right) - g\left(x\right)\right| < 8

Step-by-step solution

Given functions are:

${f}^{"}\left(x\right) = {g}^{"}\left(x\right) + 6x \ldots \left(1\right)$

${f}^{'}\left(1\right) = 4{g}^{'}\left(1\right) - 3 = 9 \ldots \left(2\right)$

$f\left(2\right) = 3g\left(2\right) = 12 \ldots \left(3\right)$

By integrating \left(1\right) $, we get $ {f}^{'}\left(x\right) = {g}^{'}\left(x\right) + 3{x}^{2} + C $At$ x = 1 $$ {f}^{'}\left(1\right) = {g}^{'}\left(1\right) + 3 + C $$ ⇒ 9 = 3 + 3 + C ⇒ C = 3 $$ ∴ {f}^{'}\left(x\right) = {g}^{'}\left(x\right) + 3{x}^{2} + 3 $Again by integrating, we get$ f\left(x\right) = g\left(x\right) + {x}^{3} + 3x + D $At$ x = 2 $, we get$ f\left(2\right) = g\left(2\right) + 8 + 3\left(2\right) + D $$ ⇒ 12 = 4 + 8 + 6 + D ⇒ D = - 6 $So,$ f\left(x\right) = g\left(x\right) + {x}^{3} + 3x - 6 $Option \left(A\right)$ :

At $x = - 2$ ,

$⇒ g\left(- 2\right) - f\left(- 2\right) = 20$

So, this option is true.

Option \left(B\right) $If$ - 1 < x < 2 $Let$ h\left(x\right) = f\left(x\right) - g\left(x\right) = {x}^{3} + 3x - 6 $$ ⇒ {h}^{'}\left(x\right) = 3{x}^{2} + 3 $$ ⇒ h ' \left(x\right) > 0 $for all values of$ x $.So,$ h\left(- 1\right) < h\left(x\right) < h\left(2\right) $$ ⇒ - 10 < h\left(x\right) < 8 $$ ⇒ \left|h\left(x\right)\right| < 10 $So, this option is NOT true.Option \left(C\right)$

${h}^{'}\left(x\right) = {f}^{'}\left(x\right) - {g}^{'}\left(x\right) = 3{x}^{2} + 3$

If \left|{h}^{'}\left(x\right)\right| < 6 ⇒ \left|3{x}^{2} + 3\right| < 6 $$ ⇒ 3{x}^{2} + 3 < 6 $and$ - 6 < 3{x}^{2} + 3 $$ ⇒ {x}^{2} < 1 $and$ {x}^{2} > - 3\left(alwaystrue\right) $$ ⇒ - 1 < x < 1 $So, If$ x \in \left(- 1, 1\right) $then \left|{f}^{'}\left(x\right) - {g}^{'}\left(x\right)\right| < 6$

So, this option is true.

Option \left(D\right) $$ f\left(x\right) - g\left(x\right) = 0 $$ ⇒ {x}^{3} + 3x - 6 = 0 $$ h\left(x\right) = {x}^{3} + 3x - 6 $Here,$ h\left(1\right) = - ve $and$ h\left(\dfrac{3}{2}\right) = + ve $So, there exists$ {x}_{0} \in \left(1, \dfrac{3}{2}\right) $such that$ f\left({x}_{0}\right) = g $\left({x}_{0}\right)$

So, this option is true.

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