JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Limits

Expand each term in the numerator as a Maclaurin series about $x=0$:

$a{x}^2{e}^x=a{x}^2+a{x}^3+O(x^4)$

$b\ln (1+x)=bx-\frac{b}{2}{x}^2+\frac{b}{3}{x}^3+O(x^4)$

$cx{e}^{-x}=cx-c{x}^2+\frac{c}{2}{x}^3+O(x^4)$

Numerator $=(c-b)x+\left(a+\frac{b}{2}-c\right)x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right)x^3+O(x^4)$.

The denominator $x^2\sin x=x^3+O(x^5)$ starts at $x^3$, so for the limit to be finite, the coefficients of $x^1$ and $x^2$ in the numerator must vanish:

$c-b=0\Rightarrow b=c$, and $a+\frac{b}{2}-c=0\Rightarrow a=\frac{b}{2}$ (using $c=b$), i.e. $b=c=2a$.

The surviving $x^3$ coefficient is

$a-\frac{b}{3}+\frac{c}{2}=a-\frac{2a}{3}+a=\frac{4a}{3}$.

So the limit equals $\frac{4a/3\cdot x^3}{x^3}=\frac{4a}{3}$ (since $\sin x/x\to 1$). Setting this equal to 1:

$\frac{4a}{3}=1\Rightarrow a=\frac{3}{4}$, and $b=c=2a=\frac{3}{2}$.

$16(a^2+b^2+c^2)=16\left[{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2\right]=16\left[\frac{9}{16}+\frac{9}{4}+\frac{9}{4}\right]=9+36+36=81$.