Question

If the roots of the equation $=\underset{x\to \infty }{\lim }\frac{2+1/x}{1+2/x}=e^{-2}$ are real and less than 3,then

• A. $[-1,\infty )-\{0\}$
• B. $\text{x}=0$
• C. $\therefore$
• D. $R{f}^{\prime }(0)=\underset{h\to 0}{\lim }\frac{f(0+h)-f(0)}{h}$

Answer 1

Answer: (A)

Given equation is $\upsilon /\text{1}0$ . If roots are real, then $f$ $1.11f$ $1.22f$ $f$ $1.27f$ $\text{1}.0\text{1}\times \text{1}{0}^{\text{5}}\text{ N}/{\text{m}}^{\text{2}}$ $\text{9}.\text{13}\times \text{1}{0}^{\text{4}}\text{ N}/{\text{m}}^{\text{2}}$ Also roots are less than 3, hence $\text{9}.\text{13}\times \text{1}{0}^{\text{3}}\text{N}/{\text{m}}^{\text{2}}$ $\text{18}.\text{26 N}/{\text{m}}^{\text{2}}$ $\text{2}.\text{25}\times \text{1}{0}^{\text{3}}\text{min}$ $\text{3}.\text{97}\times \text{1}{0}^{\text{3}}\text{min}$ $9.13\times {10}^{3}N/m^2$ $\text{5}.\text{25}\times \text{1}{0}^{\text{3}}\text{min}$ $\left[\text{FL}{\text{T}}^{-\text{2}}\right]$ Either $\left[\text{F}{\text{L}}^{\text{2}}{T}^{-\text{2}}\right]$ or $\left[\text{F}{\text{L}}^{-\text{1}}{\text{T}}^{\text{2}}\right]$ Hence, only $\left[{\text{F}}^2\text{L}{\text{T}}^{\text{-2}}\right]$ satisfy.