JEE Main 2022 — Mathematics Calculus

Here $f(x)={\begin{matrix}x+3 & ; & x<-3 \\ -(x+3) & ; & -3\leq x<0 \\ {e}^{x} & ; & x\geq 0\end{matrix}}$f(x)=x+3−(x+3)ex​;;;​x<−3−3≤x<0x≥0​

and $g(x)={\begin{matrix}{x}^{2}+{k}_{1}x & ; & x<0 \\ 4x+{k}_{2} & ; & x\geq 0\end{matrix}}$g(x)=x2+k1​x4x+k2​​;;​x<0x≥0​

Now $g(f(x))={\begin{matrix}f{(x)}^{2}+{k}_{1}f(x) & ; & f(x)<0 \\ 4f(x)+{k}_{2} & ; & f(x)\geq 0\end{matrix}}$g(f(x))=f(x)2+k1​f(x)4f(x)+k2​​;;​f(x)<0f(x)≥0​

i.e. $g(f(x))={\begin{matrix}{(x+3)}^{2}+{k}_{1}(x+3) & ; & x<-3 \\ {(x+3)}^{2}-{k}_{1}(x+3) & ; & -3\leq x<0 \\ 4{e}^{x}+{k}_{2} & ; & x\geq 0\end{matrix}}$g(f(x))=(x+3)2+k1​(x+3)(x+3)2−k1​(x+3)4ex+k2​​;;;​x<−3−3≤x<0x≥0​

For continuity at $x=0$x=0,

$$gof(0)=g(f({0}^{-}))=g(f({0}^{+}))$$gof(0)=g(f(0−))=g(f(0+))

i.e. $4+{k}_{2}=9-3{k}_{1}=4+{k}_{2}$4+k2​=9−3k1​=4+k2​

$$\Rightarrow 3{k}_{1}+{k}_{2}=5\cdots (i)$$⇒3k1​+k2​=5⋯(i)

Now differentiating, we get

$${(g(f(x)))}^{'}={\begin{matrix}2(x+3)+{k}_{1} & ; & x<-3 \\ 2(x+3)-{k}_{1} & ; & -3\leq x<0 \\ 4{e}^{x} & ; & x\geq 0\end{matrix}}$$(g(f(x)))′=2(x+3)+k1​2(x+3)−k1​4ex​;;;​x<−3−3≤x<0x≥0​

At $x=0$x=0,

$$6-{k}_{1}=4$$6−k1​=4

$$\Rightarrow {k}_{1}=2\ldots (ii)$$⇒k1​=2…(ii)

$\therefore {k}_{1}=2,{k}_{2}=-1$∴k1​=2,k2​=−1 (from $(i)$(i))

So $gof(x)={\begin{matrix}{(x+3)}^{2}+2(x+3) & ; & x<-3 \\ {(x+3)}^{2}-2(x+3) & ; & -3\leq x<0 \\ 4{e}^{x}-1 & ; & x\geq 0\end{matrix}}$gof(x)=(x+3)2+2(x+3)(x+3)2−2(x+3)4ex−1​;;;​x<−3−3≤x<0x≥0​

Hence, $gof(-4)+gof(4)=4{e}^{4}-2$gof(−4)+gof(4)=4e4−2$=2(2{e}^{4}-1)$=2(2e4−1)