JEE Main 2021 — Mathematics Calculus

We know that, $\frac{d}{dx}({x}^{n})=n{x}^{n-1}$dxd​(xn)=nxn−1 and $\frac{d}{dx}({e}^{x})={e}^{x}$dxd​(ex)=ex

Hence, $\frac{d}{dx}(-\frac{4}{3}{x}^{3}+2{x}^{2}+3x)=-\frac{4}{3}\times 3{x}^{2}+2\times 2x+3=-4{x}^{2}+4x+3$dxd​(−34​x3+2x2+3x)=−34​×3x2+2×2x+3=−4x2+4x+3

And, using product rule, we get$\frac{d}{dx}(3x{e}^{x})=3x\frac{d}{dx}({e}^{x})+3{e}^{x}\frac{d}{dx}(x)=3x{e}^{x}+3{e}^{x}=3{e}^{x}(x+1)$dxd​(3xex)=3xdxd​(ex)+3exdxd​(x)=3xex+3ex=3ex(x+1)

\Rightarrow {f}^{'}(x)={\begin{matrix}-4{x}^{2}+4x+3, & x>0 \\ 3{e}^{x}(1+x), & x\leq 0\end{matrix}

For $x>0,{f}^{'}(x)=-4{x}^{2}+4x+3=-(2x+1)(2x-3)$x>0,f′(x)=−4x2+4x+3=−(2x+1)(2x−3)

The sign scheme of ${f}^{'}(x)$f′(x) is

We know that, if ${f}^{'}(x)>0$f′(x)>0 then for those value of $x,f(x)$x,f(x) is increasing.

$\Rightarrow f(x)$⇒f(x) is increasing in $(-\frac{1}{2},\frac{3}{2}).$(−21​,23​).

For $x\leq 0,{f}^{'}(x)=3{e}^{x}(1+x)$x≤0,f′(x)=3ex(1+x)

As ${e}^{x}>0,\forall x\in R$ex>0,∀x∈R

$$\Rightarrow {f}^{'}(x)>0\forall x\in (-1,0).$$⇒f′(x)>0∀x∈(−1,0).

So, in complete domain, $f(x)$f(x) is increasing in $(-1,0)\cup (-\frac{1}{2},\frac{3}{2})=(-1,\frac{3}{2}).$(−1,0)∪(−21​,23​)=(−1,23​).