JEE Main 2023 (Online) 1st February Morning ShiftQuestion 1
Let $f(x)=2x+\tan^{-1}x$ and $g(x)=\log_e\sqrt{1+x^2}+x$, $x\in(0,3)$. Then
Differentiation and its Application
Solution
Concept. $f'-g'>0$ on $(0,3)$ and $f(0)=g(0)=0$, so $f>g$. Both functions increase, so compare their values near $3$. Formulas used. (1) $f'(x)=2+\dfrac{1}{1+x^2}$, $g'(x)=\dfrac{x}{1+x^2}+1$Step 1 - (A) and (C). By (1), $$f'-g'=1+\frac{1-x}{1+x^2}=\frac{x^2-x+2}{1+x^2}>0.$$ So $f'>g'$ everywhere and (A) is false. Since $f(0)=g(0)=0$, $f>g$ on $(0,3)$, so (C) is false. Step 2 - (D). $f'$ decreases on $(0,3)$ towards $2+\frac{1}{10}$, and $\frac{x}{1+x^2}$ is largest at $x=1$, so $\max g'=\frac32$. $1+\frac32=\frac52\ne\frac{21}{10}$. (D) is false. Step 3 - (B). Both increase, so compare at $x\to3$: $6+\tan^{-1}3\approx7.25$ against $3+\ln\sqrt{10}\approx4.15$. (B) is true. Answer: (B)