A) 1100
B) 1250
C) 1050
D) 5250
Correct Answer: C
Solution :
\[p(t)=1000+\frac{1000\,t}{100+{{t}^{2}}}\] .... (i) On differentiating both side with respect to t, \[p\,(t)=0=\frac{(1000+{{t}^{2}})\,(1000)-1000\,\,t(2t)}{{{(100+{{t}^{2}})}^{2}}}\] \[=1000\frac{(100-{{t}^{2}})}{{{(100+{{t}^{2}})}^{2}}}\] ?. (ii) put p(t) = 0 for maxima or minima \[\Rightarrow \,\,100-{{t}^{2}}=0\] \[\Rightarrow \] \[t=\pm \,\,10\] Now, again differentiatmg equation (ii) w.r. to x \[p\,(t)=1000\] \[\left[ \frac{{{(100+{{t}^{2}})}^{2}}(-2t)-100-{{t}^{2}})2(100+{{t}^{2}})\,2t}{{{(100+{{t}^{2}})}^{4}}} \right]\] \[=1000\,\,t\frac{\left[ 100+{{t}^{2}})\,(-2)-100-{{t}^{2}})\,(4) \right]}{{{(100+{{t}^{4}})}^{2}}}\] \[=-1000\,t\frac{[200+2\,{{t}^{2}}]}{{{(100+{{t}^{2}})}^{4}}}\] At \[t=10\] \[p\,(t)<0\] \[\therefore \] The maximum value is \[p(10)=1000+\frac{10000}{100+100}\] \[=1000+\frac{10000}{200}=1000+50\] \[=1050\]
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