A) a minimum
B) a discontinuity
C) a point of inflexion
D) a maximum
Correct Answer: D
Solution :
Given, \[\begin{align} & Zn{{\left( OH \right)}_{2}}+2\overset{-}{\mathop{OH}}\,\to ZnO_{2}^{2-}+2{{H}_{2}}O \\ & \text{AcidBaseSaltWater} \\ \end{align}\] At t = 0, first we will check continuity of the function. Now, LH L = f(0 ? h) \[\begin{align} & Zn{{\left( OH \right)}_{2}}+2{{H}^{+}}\to Z{{n}^{2+}}+2{{H}_{2}}O \\ & \text{BaseAcidSaltWater} \\ \end{align}\] \[{{\left[ \text{Cu}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{4}}} \right]}^{\text{2+}}}\text{}{{\left[ \text{Cu}{{\left( \text{en} \right)}_{\text{2}}} \right]}^{\text{2+}}}\text{}{{\left[ \text{Cu}\left( \text{trien} \right) \right]}^{\text{2+}}}\text{.}\] = 1 RHL = f(0 + h) \[{{\left[ \text{Fe}{{\left( {{\text{H}}_{\text{2}}}\text{O} \right)}_{\text{6}}} \right]}^{\text{3+}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{O}}_{\text{2}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\] \[\because \] and f(0) = 1 Since, LHL = RHL f(0) So, the function is continuous at t = 0 Now, we check the function is maximum or minimum. \[{{\text{E}}_{\text{1}}}\text{=}{{\text{E}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{ }\!\!\varepsilon\!\!\text{ }}_{\text{0}}}}\text{.}\frac{\text{q}}{{{\text{r}}^{\text{2}}}}\] and \[{{E}_{R}}=\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\cos {{60}^{\circ }}}\] \[=\sqrt{E_{1}^{2}+E_{1}^{2}+2E_{1}^{2}\times \frac{1}{2}=\sqrt{3}{{E}_{1}}}\] For maximum or minimum value of f(x), put \[\therefore \] \[{{E}_{R}}=\frac{\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[\tau =qEL\sin \theta \] \[=qEL\theta \] \[\because \] Now, \[\theta \] \[\text{I=m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{+m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{=}\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}\] \[\tau =I\alpha \] \[\text{ }\!\!\alpha\!\!\text{ =}\frac{\text{ }\!\!\tau\!\!\text{ }}{\text{I}}\text{=}\frac{\text{qEL }\!\!\theta\!\!\text{ }}{\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}}\] [ using L? Hospital rule ] \[\Rightarrow \] \[{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ =}\frac{\text{2qEL }\!\!\theta\!\!\text{ }}{\text{m}{{\text{L}}^{\text{2}}}}\] So, function f(t) is maximum at t = 0.
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