JEE Main 2023 — Mathematics Calculus

Given,

Domain and range of function,

$$f:[2,4]\rightarrow \mathbb{R}$$f:[2,4]→R

$$(x{\mathrm{log}}_{e}x){f}^{'}(x)+({\mathrm{log}}_{e}x)f(x)+f(x)\geq 1,x\in [2,4]$$(xloge​x)f′(x)+(loge​x)f(x)+f(x)≥1,x∈[2,4]

$$\Rightarrow (x{\mathrm{log}}_{e}x)\frac{d}{dx}f(x)+({\mathrm{log}}_{e}x)f(x)\frac{d}{dx}(x)+xf(x)\frac{d}{dx}({\mathrm{log}}_{e}x)\geq 1$$⇒(xloge​x)dxd​f(x)+(loge​x)f(x)dxd​(x)+xf(x)dxd​(loge​x)≥1

$$\Rightarrow \frac{d}{dx}[x\mathrm{ln}xf(x)]\geq 1$$⇒dxd​[xlnxf(x)]≥1

$$\Rightarrow \frac{d}{dx}[x\mathrm{ln}xf(x)]\geq \frac{d}{dx}(x)$$⇒dxd​[xlnxf(x)]≥dxd​(x)

$$\Rightarrow \frac{d}{dx}[x\mathrm{ln}xf(x)-x]\geq 0$$⇒dxd​[xlnxf(x)−x]≥0

Hence, $h(x)=x(\mathrm{ln}x)f(x)-x$h(x)=x(lnx)f(x)−x is a increasing function,

$$\therefore h(x)\geq h(2),x\in [2,4]$$∴h(x)≥h(2),x∈[2,4]

$$\Rightarrow x\mathrm{ln}x\times f(x)-x\geq 2\mathrm{ln}2\times f(2)-2$$⇒xlnx×f(x)−x≥2ln2×f(2)−2

$$\Rightarrow x\mathrm{ln}xf(x)-x\geq \mathrm{ln}2-2$$⇒xlnxf(x)−x≥ln2−2

Similarly, $h(x)\leq h(4)$h(x)≤h(4)

$$\Rightarrow x\mathrm{ln}xf(x)-x\leq \mathrm{ln}4-4$$⇒xlnxf(x)−x≤ln4−4

So, $\frac{\mathrm{ln}2-2}{x\mathrm{ln}x}+\frac{1}{\mathrm{ln}x}\leq f(x)\leq \frac{\mathrm{ln}4-4}{x\mathrm{ln}x}+\frac{1}{\mathrm{ln}x}$xlnxln2−2​+lnx1​≤f(x)≤xlnxln4−4​+lnx1​

Now for $x\in [2,4]$x∈[2,4]

$$\frac{\mathrm{ln}4-4}{x\mathrm{ln}x}+\frac{1}{\mathrm{ln}x}\leq \frac{\mathrm{ln}4-4}{2\mathrm{ln}2}+\frac{1}{\mathrm{ln}2}=1-\frac{1}{\mathrm{ln}2}<1$$xlnxln4−4​+lnx1​≤2ln2ln4−4​+ln21​=1−ln21​<1

$$\Rightarrow f(x)\leq 1\text{for}x\in [2,4]$$⇒f(x)≤1forx∈[2,4]

Now for $x\in [2,4]$x∈[2,4]

$$\frac{\mathrm{ln}2-2}{x\mathrm{ln}x}+\frac{1}{\mathrm{ln}x}\geq \frac{\mathrm{ln}2-2}{4\mathrm{ln}4}+\frac{1}{\mathrm{ln}4}=\frac{\mathrm{ln}2-2}{4\mathrm{ln}4}+\frac{1}{\mathrm{ln}4}=\frac{1}{8}+\frac{1}{4\mathrm{ln}2}>\frac{1}{8}$$xlnxln2−2​+lnx1​≥4ln4ln2−2​+ln41​=4ln4ln2−2​+ln41​=81​+4ln21​>81​

Hence, $f(x)\geq \frac{1}{8}$f(x)≥81​

Hence both $A$A & $B$B are correct.

Note this question was bonus in Jee Main 2023 April session, as LMVT on $f(x)\cdot x\mathrm{ln}x$f(x)⋅xlnx can't be satisfied.

Hence, no such $f(x)$f(x) exist.