A) \[[-1,\infty )-\{0\}\]
B) \[\text{x}=0\]
C) \[\therefore \]
D) \[Rf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}\]
Correct Answer: A
Solution :
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.
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