JEE Main 2022 — Mathematics Calculus

Given,

$$f(x)={\int }_{0}^{x}{e}^{x-t}{f}^{'}(t)dt-({x}^{2}-x+1){e}^{x}$$f(x)=∫0x​ex−tf′(t)dt−(x2−x+1)ex

$$\Rightarrow f(x)={e}^{x}{\int }_{0}^{x}{e}^{-t}{f}^{'}(t)dt-({x}^{2}-x+1){e}^{x}$$⇒f(x)=ex∫0x​e−tf′(t)dt−(x2−x+1)ex

$$\Rightarrow {e}^{-x}f(x)={\int }_{0}^{x}{e}^{-t}{f}^{'}(t)dt-({x}^{2}-x+1)$$⇒e−xf(x)=∫0x​e−tf′(t)dt−(x2−x+1)

Differentiate on both side w.r.t $x$x we get,

$${e}^{-x}{f}^{'}(x)+(-f(x){e}^{-x})={e}^{-x}{f}^{'}(x)-2x+1$$e−xf′(x)+(−f(x)e−x)=e−xf′(x)−2x+1

$$\Rightarrow f(x)={e}^{x}(2x-1)$$⇒f(x)=ex(2x−1)

$$\Rightarrow {f}^{'}(x)={e}^{x}(2)+{e}^{x}(2x-1)$$⇒f′(x)=ex(2)+ex(2x−1)

$$\Rightarrow {f}^{'}(x)={e}^{x}(2x+1)......(1)$$⇒f′(x)=ex(2x+1)......(1)

Now finding critical point we get,

$$2x+1=0\Rightarrow x=-\frac{1}{2}$$2x+1=0⇒x=−21​

Now differentiating equation $(1)$(1) to check maxima and minima we get,

$${f}^{"}(x)={e}^{x}(2)+(2x+1){e}^{x}$$f"(x)=ex(2)+(2x+1)ex

$$\Rightarrow {f}^{"}(x)={e}^{x}(2x+3)$$⇒f"(x)=ex(2x+3)

For $x=-\frac{1}{2},{f}^{"}(\frac{-1}{2})>0$x=−21​,f"(2−1​)>0, so it will give point of minima,

Now minimum value will be given by, $f(\frac{-1}{2})={e}^{-\frac{1}{2}}(-1-1)=-\frac{2}{\sqrt{e}}$f(2−1​)=e−21​(−1−1)=−e​2​