JEE Main 2023 — Mathematics Calculus

Given:

$$f(x)=2x+{\mathrm{tan}}^{-1}x$$f(x)=2x+tan−1x

$$\Rightarrow {f}^{'}(x)=2+\frac{1}{1+{x}^{2}}$$⇒f′(x)=2+1+x21​

And,

$$g(x)=\mathrm{ln}(\sqrt{1+{x}^{2}}+x)$$g(x)=ln(1+x2​+x)

$$\Rightarrow {g}^{'}(x)=\frac{1}{\sqrt{1+{x}^{2}}}$$⇒g′(x)=1+x2​1​

Now,

$$0\leq x\leq 3$$0≤x≤3

$$\Rightarrow 0\leq {x}^{2}\leq 9$$⇒0≤x2≤9

$$\Rightarrow 1\leq 1+{x}^{2}\leq 10$$⇒1≤1+x2≤10

So,

$$\frac{1}{10}\leq \frac{1}{1+{x}^{2}}\leq 1$$101​≤1+x21​≤1

$$\Rightarrow 2+\frac{1}{10}\leq 2+\frac{1}{1+{x}^{2}}\leq 3$$⇒2+101​≤2+1+x21​≤3

$$\Rightarrow 2+\frac{1}{10}\leq {f}^{'}(x)\leq 3$$⇒2+101​≤f′(x)≤3

$$\Rightarrow \frac{21}{10}\leq {f}^{'}(x)\leq 3$$⇒1021​≤f′(x)≤3

And,

$$\frac{1}{\sqrt{10}}\leq {g}^{'}(x)\leq 1$$10​1​≤g′(x)≤1

So, $\mathrm{min}{f}^{'}(x)=\frac{21}{10}\neq [1+\mathrm{max}{g}^{'}(x)]$minf′(x)=1021​=[1+maxg′(x)]

Option $(4)$(4) is incorrect

From above,

$${g}^{'}(x)<{f}^{'}(x)\forall x\in [0,3]$$g′(x)<f′(x)∀x∈[0,3]

Option $(1)$(1) is incorrect.

Since, ${f}^{'}(x)$f′(x) and ${g}^{'}(x)$g′(x) are both positive so $f(x)$f(x) and $g(x)$g(x) both are increasing

So,

$\mathrm{max}(f(x))$max(f(x)) at $x=3$x=3 is $6+{\mathrm{tan}}^{-1}3$6+tan−13

$\mathrm{max}(g(x))$max(g(x)) at $x=3$x=3 is $\mathrm{ln}(3+\sqrt{10})$ln(3+10​)

And, $6+{\mathrm{tan}}^{-1}3>\mathrm{ln}(3+\sqrt{10})$6+tan−13>ln(3+10​)

Option $(2)$(2) is correct