JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 26)
Explanation
$$\mathop {\lim }\limits_{x \to 0} {{a{x^2}\left( {1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ....} \right) - b\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - .....} \right) + cx\left( {1 - x + {{{x^2}} \over {x!}} - {{{x^3}} \over {3!}} + .....} \right)} \over {{x^3}\,.\,{{\sin x} \over x}}}$$
$$=\lim _\limits{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots}{x^3}=1$$
$$\begin{aligned} & \mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0 \\ & \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} \\ & \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} \\ & 16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81 \end{aligned}$$
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