MCQ
જો $\lim _{x \rightarrow 0} \frac{a x^2 e^x-b \log _e(1+x)+c x e^{-x}}{x^2 \sin x}=1$, તો $16\left(a^2+b^2+c^2\right)$ ........................
- A$80$
- B$85$
- C$81$
- D$70$
$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots .}{x^3}=1$
$\mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0$
$ \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}$
$\lim _{x \rightarrow 0} \frac{a^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)+c x\left(1-x+\frac{x^2}{x !}-\frac{x^3}{3 !}+\ldots \ldots . .\right)}{x^3 \cdot \frac{\sin x}{x}} $
$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots .}{x^3}=1$
$\mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0$
$ \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} $
$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$
$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$
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