JEE Main 2023 — Mathematics Calculus

Given,

$$f(x)={\int }_{0}^{2}{e}^{|x-t|}dt$$f(x)=∫02​e∣x−t∣dt

Now For $x\leq 0$x≤0

$$f(x)={\int }_{0}^{2}{e}^{t-x}dt={e}^{-x}({e}^{2}-1)$$f(x)=∫02​et−xdt=e−x(e2−1)

And for $0<x<2$0<x<2

$$f(x)={\int }_{0}^{x}{e}^{x-t}dt+{\int }_{x}^{2}{e}^{t-x}dt={e}^{x}+{e}^{2-x}-2$$f(x)=∫0x​ex−tdt+∫x2​et−xdt=ex+e2−x−2

For $x\geq 2$x≥2

$$f(x)={\int }_{0}^{2}{e}^{x-t}dt={e}^{x-2}({e}^{2}-1)$$f(x)=∫02​ex−tdt=ex−2(e2−1)

Now for $x\leq 0,f(x)$x≤0,f(x) is decreasing as ${e}^{-x}$e−x is decreasing function and $x\geq 2,f(x)$x≥2,f(x) is increasing as ${e}^{x-2}$ex−2 is increasing function,

So, minimum value of $f(x)$f(x) lies in $x\in (0,2)$x∈(0,2)

Applying A.M $\geq$≥ G.M in ${e}^{x}+{e}^{2-x}$ex+e2−xwe get,

$$\frac{{e}^{x}+{e}^{2-x}}{2}\geq \sqrt{{e}^{x}\times {e}^{2-x}}$$2ex+e2−x​≥ex×e2−x​

$$\Rightarrow {e}^{x}+{e}^{2-x}\geq 2e$$⇒ex+e2−x≥2e

Hence, the minimum value of $f(x)$f(x) is $2e-2=2(e-1)$2e−2=2(e−1)