QJEEM24Ma45cf: JEE Main 2024 Mathematics - Class 12 JEEM24Ma45cf Solution: If lim x → 0 a x 2 e x − b log e ⁡ ( 1 + x ) + c x e − x x 2…

$\mathop {\lim }\limits_{x \to 0} {{a{x^2}\left( {1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ....} \right) - b\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - .....} \right) + cx\left( {1 - x + {{{x^2}} \over {x!}} - {{{x^3}} \over {3!}} + .....} \right)} \over {{x^3}\,.\,{{\sin x} \over x}}}$

$$=\underset{x\to \mathrm{\infty }}{lim}\frac{(c-b)x+(\frac{b}{2}-c+a)x^2+(a-\frac{b}{3}+\frac{c}{2})x^3+\dots \dots }{x^3}=1$$

$\begin{aligned}& \mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0 \\& \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} \\& \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} \\& 16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81\end{aligned}$