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1. If f(x),g(x),h(x) are continuous for all x belongs to 'R'. and h(x)=Lt n→∞ f(x)+x 2n g(x)/1+x 2n , f(1)/g(1 )=K , f(-1)/g(-1)= L then k+L=? 2. A and B are two fixed points on fixed circle and 'p' is a moving point on the above circle,angle APB= 60 , AB =8, range of PA * PB is ? 3. The population of a country increases by 2% per year.In 100 years population increases by 'p' times then [p]=? ([.] denotes greatest integer function)

1.   If f(x),g(x),h(x) are continuous for all x belongs to 'R'. and


h(x)=Lt n→∞ f(x)+x2ng(x)/1+x2n , f(1)/g(1)=K, f(-1)/g(-1)=L then k+L=?

2. A and B are two fixed points on fixed circle and 'p' is a moving    point on the above circle,angleAPB=60 , AB=8, range of PA*PB is ?

3. The population of a country increases by 2% per year.In 100 years population  increases by 'p' times then [p]=?  ([.] denotes greatest integer function)




 







Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
3.
Population increases by 2% every year. Let ‘p’ be the population at any time & p0be the initial population.
Given:
\frac{dp}{dt} = .02p
\int \frac{dp}{p} = \int .02dt
\int_{p_{0}}^{p} \frac{dp}{p} = \int_{0}^{100} .02dt
(lnp)_{p_{0}}^{p} = (.02t)_{0}^{100}
ln\frac{p}{p_{0}} = (.02.100) = 2
p = p_{0}e^{2}
Increase in the population:
p - p_{0} = p_{0} e^{2} -p_{0} = p_{0} (e^{2}-1)
[e^{2}-1] = [2.718^{2}-1] = [6.3875] = 6
3.
What is the range. Its not clear. Please rewrite this part again.
1.
h(x) = \lim_{n\rightarrow \infty }(f(x) + \frac{x^{2n}.g(x)}{1+x^{2n}})
Since functions h(x), f(x) & g(x) are continuous in R. There limit should exist at -1 & 1.
Limit at x =1
LHL = RHL
\lim_{x\rightarrow 1^{-}}h(x) = \lim_{x\rightarrow 1^{+}}h(x)
LHL = \lim_{n\rightarrow \infty }\lim_{h\rightarrow 0}(f(1-h)+\frac{g(1-h).(1-h)^{2n}}{1+(1-h)^{2n}})
LHL = f(1)
RHL = \lim_{n\rightarrow \infty }\lim_{h\rightarrow 0}(f(1+h)+\frac{g(1+h).(1+h)^{2n}}{1+(1+h)^{2n}})
RHL = f(1) + g(1)
LHL =RHL
f(1) = f(1)+g(1)
\Rightarrow g(1) = 0
Limit at x = -1
LHL = \lim_{n\rightarrow \infty }\lim_{h\rightarrow 0}(f(-1-h)+\frac{g(-1-h).(-1-h)^{2n}}{1+(-1-h)^{2n}})
LHL = \lim_{n\rightarrow \infty }\lim_{h\rightarrow 0}(f(-1-h)+\frac{g(-1-h).(1+h)^{2n}}{1+(1+h)^{2n}})
LHL = f(-1) + g(-1)
RHL = \lim_{n\rightarrow \infty }\lim_{h\rightarrow 0}(f(-1+h)+\frac{g(-1+h).(-1+h)^{2n}}{1+(-1+h)^{2n}})
RHL = \lim_{n\rightarrow \infty }\lim_{h\rightarrow 0}(f(-1+h)+\frac{g(-1+h).(1-h)^{2n}}{1+(1-h)^{2n}})
RHL = f(-1)
LHL=RHL
\Rightarrow g(-1) = 0
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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