InterviewSolution
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InterviewSolution
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Let `f(theta)=(1)/(tan^9 theta){(1+tantheta)^10+(2+tantheta)^10+....+(20+tantheta)^10}-20tan theta` The left hand limit of `f(theta)" as "theta to (pi)/(2)`,isA. `1900`B. `2000`C. `2100`D. `2200` |
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Answer» Correct Answer - C Let `xtan theta`. Then, `f(tan^-1x)=((1+x)^10+(2x+x)^10+.....+ (20+x)^10-20x^10)/(x^9)` and `xtooo "as" theta to (pi^-)/(2)` `lim_(theta to(pi^-)/(2))f(theta) =lim_(xtooo) f(tan^-1x)` ` =lim_(xt0 oo) ((1+x)^10+(2+x)^10+....+(20+x)^10-20x^10)/(x^9)` ` =lim_(xto oo) (sum_(r=1)^(20)(r+x)^10-x^10)/(x^9)` ` =lim_(xtooo) sum_(r=1)^(20) (.^10C_(0) r^10+.^10C_(1) r^9+.....+.^10C_(9))/ (x^9)` ` =sum_(r=1)^(20)lim_(xtooo) {(.^10C_(0) r^10)/(x^9)+(.^10C_(1) r^9)/(r^8)+.....+.^10C_(9) r}` ` =sum_(r=1)^(20).^10C_(0)r=10 (sum_(r=1)^(20)r)=10xx(20xx21)/(2)=2100` |
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