JEE Main 2024 — 31 January, Shift 2 — Question 25

$a x^{2}\left(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots ..\right)-b\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\ldots \ldots.\right)$

Sol.- $\lim _{x \rightarrow 0} \frac{+c x\left(1-x+\frac{x^{2}}{x!}-\frac{x^{3}}{3!}+\ldots \ldots .\right)}{x^{3} \cdot \frac{\sin x}{x}}$

$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^{2}+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^{3}+\ldots \ldots .}{x^{3}}=1$

$c-b=0, \quad \frac{b}{2}-c+a=0$

$a-\frac{b}{3}+\frac{c}{2}=1$

$\quad a=\frac{3}{4} \quad b=c=\frac{3}{2}$

$a^{2}+b^{2}+c^{2}=\frac{9}{16}+\frac{9}{4}+\frac{9}{4}$

$16\left(a^{2}+b^{2}+c^{2}\right)=81$.