Let f ( x ) and g ( x ) be twice differentiable functions satisfying f prime prime ( x )= g prime prime ( x ) for all x in R , f prime (1)=2 g prime (1)=4 and g (2)=3 f (2)=9 . Then f (25)- g (25) is equal to :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 5 April, Evening Shift — Question 43

Let $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ be twice differentiable functions satisfying $\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g}^{\prime \prime}(\mathrm{x})$ for all $\mathrm{x} \in \mathrm{R}$ , $\mathrm{f}^{\prime}(1)=2 \mathrm{~g}^{\prime}(1)=4$ and $\mathrm{g}(2)=3 \mathrm{f}(2)=9$ . Then $\mathrm{f}(25)-\mathrm{g}(25)$ is equal to :

Option A:
$20$
Option B:
$40$
Correct
Option C:
$-20$
Option D:
$-40$

Answer: B

Step-by-step solution

$\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g}^{\prime \prime}(\mathrm{x})$ } Integrate $\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{C}_{1}$ Put $\mathrm{x}=1$ $\mathrm{f}^{\prime}(1)=\mathrm{g}^{\prime}(1)+\mathrm{C}_{1}$ $4=2+\mathrm{C}_{1}$ $\mathrm{C}_{1}=2$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+2$ $\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+2 \mathrm{x}+\mathrm{C}_{2}$ Put $\mathrm{x}=2$ $\mathrm{f}(2)=\mathrm{g}(2)+4+\mathrm{C}_{2}$ $3=9+4+\mathrm{C}_{2}$ $\mathrm{C}_{2}=-10$ $\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+2 \mathrm{x}-10$ Put $x=25$ $\mathrm{f}(\mathrm{x})-\mathrm{g}(25)=40$

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Differentiability

Let f(x)=lim y arrow 0 frac(1-cos (x y)) tan (x y)y^3 . Then the number of solutions of the equation f( x)=sin x , x in R is :

Answer: B

Step-by-step solution

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