Let f and g be twice differentiable functions on R such that $\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g}^{\prime \prime}(\mathrm{x})+6 \mathrm{x}$ $\mathrm{f}^{\prime}(1)=4 \mathrm{~g}^{\prime}(1)-3=9$ $\mathrm{f}(2)=3 \mathrm{~g}(2)=12$ Then which of the following is NOT true ?

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

(C) $\left|\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})\right|<6 \Rightarrow-1<\mathrm{x}<1 \mid$

(D) There exists $\mathrm{x}_{0} \in\left(1, \frac{3}{2}\right)$ such that $\mathrm{f}\left(\mathrm{x}_{0}\right)=\mathrm{g}\left(\mathrm{x}_{0}\right)$

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Application of Derivatives

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

f^{''} (x) = g^{''} (x) + 6 x

f^{'} (1) = 4 g^{'} (1) - 3 = 9

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

Then which of the following is NOT true ?

(A) $g (- 2) - f (- 2) = 20$
(B)If $- 1 < x < 2$ , then $∣ f (x) - g (x) ∣ < 8$ ✓ Correct
(C) $∣ f^{'} (x) - g^{'} (x) ∣ < 6 \Rightarrow - 1 < x < 1 ∣$
(D)There exists $x_{0} \in (1, \frac{3}{2})$ such that $f (x_{0}) = g (x_{0})$

Answer: (B)

Solution

Integrating

f^{''} (x) = g^{''} (x) + 6 x

gives

f^{'} (x) - g^{'} (x) = 3 x^{2} + C

; since

f^{'} (1) = 9

and

g^{'} (1) = 3

6 = 3 + C

C = 3

. Integrating again,

f (x) - g (x) = x^{3} + 3 x + D

; from

f (2) = 12

and

g (2) = 4

8 = 8 + 6 + D

D = - 6

. Thus

h (x) = f (x) - g (x) = x^{3} + 3 x - 6

. Here

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

holds;

∣ h^{'} (x) ∣ = 3 x^{2} + 3 < 6 ⟺ x^{2} < 1

holds; and

h (1) = - 2 < 0

with

h (3/2) > 0

gives a root in

(1, \frac{3}{2})

by the intermediate value theorem. However

h

increases on

- 1 < x < 2

from

h (- 1) = - 10

h (2) = 8

, so

∣ f (x) - g (x) ∣

climbs close to

10

there; the statement that fails is that

∣ f (x) - g (x) ∣ < 8

for all

- 1 < x < 2

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Application of Derivatives

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