A) f(6) < 8
B) \[f\left( 6 \right)\ge 8\]
C) \[f\left( 6 \right)\ge 5\]
D) \[f\left( 6 \right)\le 5\]
Correct Answer: B
Solution :
Since, f?(x) is differentiable \[=\frac{10}{100}\times 6+\frac{10}{100}\times 4\] \[=0.6+0.4=1\] By Lagrange?s mean value theorem, \[\frac{\Delta R}{R}=\frac{1}{10}\] \[{{r}_{i}}=\frac{1}{\beta }\sqrt{\frac{2mv}{q}}\] \[\frac{r\alpha }{{{r}_{p}}}=\sqrt{\frac{{{m}_{\alpha }}}{{{m}_{p}}}}\sqrt{\frac{{{q}_{p}}}{{{q}_{\alpha }}}}\] (given) \[\frac{{{r}_{\alpha }}}{10}=\sqrt{\frac{4}{2}}\] \[{{r}_{\alpha }}=10\sqrt{2}cm\] \[R=\frac{V}{l}\] [\[{{\left( \frac{\Delta k}{k} \right)}_{retained}}={{\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}={{\left( \frac{A-1}{A+1} \right)}^{2}}\] f(1) = - 2] \[{{\left( \frac{A-1}{A+1} \right)}^{2}}E.\] \[\tan \delta '=\frac{\tan \delta }{\cos \theta }=\frac{\tan {{45}^{\circ }}}{\cos {{30}^{\circ }}}\] \[\tan \delta =\frac{1}{\sqrt{3}/2}\] \[=\frac{2}{\sqrt{3}}\]
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