JEE Main 2022 — Mathematics Calculus

Given,

$$\frac{dy}{dx}+\frac{2y}{x}={e}^{x}$$dxdy​+x2y​=ex

$$I.F.={e}^{\int \frac{2}{x}dx}={e}^{2\mathrm{ln}x}={x}^{2}$$I.F.=e∫x2​dx=e2lnx=x2

$\therefore$∴ General solution: $\displaystyle y({x}^{2})=\int {e}^{x}({x}^{2})dx$y(x2)=∫ex(x2)dx

We know that

$$\int {e}^{x}f(x)dx=={e}^{x}[f(x)-{f}^{'}(x)+{f}^{''}(x)-{f}^{'''}(x)+\cdots +{(-1)}^{n}{f}^{n}(x))]+C$$∫exf(x)dx==ex[f(x)−f′(x)+f′′(x)−f′′′(x)+⋯+(−1)nfn(x))]+C

So, $y({x}^{2})={e}^{x}[{x}^{2}-2x+2]+C...(i)$y(x2)=ex[x2−2x+2]+C...(i)

Given $y(1)=0\Rightarrow (0)(1)={e}^{1}[1-2(1)+2]+C$y(1)=0⇒(0)(1)=e1[1−2(1)+2]+C

$$\Rightarrow C=-e$$⇒C=−e

$\therefore y=\frac{{e}^{x}}{{x}^{2}}[{x}^{2}-2x+2]-e$∴y=x2ex​[x2−2x+2]−e (from eq $(i)$(i))

Hence, $z(x)={x}^{2}(\frac{{e}^{x}}{{x}^{2}})[{x}^{2}-2x+2]-e-{e}^{x}$z(x)=x2(x2ex​)[x2−2x+2]−e−ex

$$={e}^{x}[{x}^{2}-2x+2]-e-{e}^{x}$$=ex[x2−2x+2]−e−ex

$$z(x)={e}^{x}[{x}^{2}-2x+1]-e$$z(x)=ex[x2−2x+1]−e

$${z}^{'}(x)={e}^{x}(2x-2)+({x}^{2}-2x+1){e}^{x}$$z′(x)=ex(2x−2)+(x2−2x+1)ex

$${z}^{'}(x)={e}^{x}[{x}^{2}-1]$$z′(x)=ex[x2−1]

To find local maxima, put ${z}^{'}(x)=0$z′(x)=0

$\Rightarrow {e}^{x}({x}^{2}-1)=0$⇒ex(x2−1)=0 $\Rightarrow (x-1)(x+1)=0$⇒(x−1)(x+1)=0 $(\because {e}^{x}>0,\forall x\in R)$(∵ex>0,∀x∈R)

$$\Rightarrow x=1,-1$$⇒x=1,−1

It is clear from the sign scheme method, $z(x)$z(x) has local maximum value at $x=-1$x=−1 and has local minimum value at $x=1$x=1.

$\therefore$∴ The local maximum value is

$$z(-1)={e}^{-1}[{(-1)}^{2}-2(-1)+1]-e$$z(−1)=e−1[(−1)2−2(−1)+1]−e

$=\frac{1}{e}[1+2+1]-e$=e1​[1+2+1]−e $=\frac{4}{e}-e$=e4​−e