01Easy210×since 2002Q3285limx→∞(nn2+12+nn2+22+nn2+32+.....+15n)\mathop {\lim }\limits_{x \to \infty } \left( {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + {n \over {{n^2} + {3^2}}} + ..... + {1 \over {5n}}} \right)x→∞lim(n2+12n+n2+22n+n2+32n+.....+5n1) is equal to :
02Easy210×since 2002Q3398The value of ∑n=1100∫n−1nex−[x]dx\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} }n=1∑100n−1∫nex−[x]dx, where [ x ] is the greatest integer ≤\le≤ x, is :A100eB100(e −-− 1)C100(1 + e)D100(1 −-− e)Check answerSkip
03Easy210×since 2002Q3279limn→∞1p+2p+3p+.....+npnp+1\mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}}n→∞limnp+11p+2p+3p+.....+np isA1p+1{1 \over {p + 1}}p+11B11−p{1 \over {1 - p}}1−p1C1p−1p−1{1 \over p} - {1 \over {p - 1}}p1−p−11D1p+2{1 \over {p + 2}}p+21Check answerSkip
04Medium210×since 2002Q3280limn→∞1+24+34+....+n4n5\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}n→∞limn51+24+34+....+n4 - limn→∞1+23+33+....+n3n5\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}n→∞limn51+23+33+....+n3A15{1 \over 5}51B130{1 \over 30}301CzeroD14{1 \over 4}41Check answerSkip
05Easy210×since 2002Q3281Limn→∞∑r=1n1nern\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{e^{{r \over n}}}}n→∞Limr=1∑nn1enr isAe+1e+1e+1Be−1e-1e−1C1−e1-e1−eDeeeCheck answerSkip
06Medium210×since 2002Q3282limn→∞[1n2sec21n2+2n2sec24n2....+1nsec21]\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right]n→∞lim[n21sec2n21+n22sec2n24....+n1sec21] equalsA12sec1{1 \over 2}\sec 121sec1B12{1 \over 2}21cosec 1Ctan 1D12{1 \over 2}21tan 1Check answerSkip
07Medium210×since 2002Q3283limn→∞((n+1)(n+2)...3nn2n)1n\mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n}}}n→∞lim(n2n(n+1)(n+2)...3n)n1 is equal to:A9e2{9 \over {{e^2}}}e29B3 log 3−23\,\log \,3 - 23log3−2C18e4{{18} \over {{e^4}}}e418D27e2{{27} \over {{e^2}}}e227Check answerSkip
08Hard210×since 2002Q3284If limn→∞ 1a+2a+......+na(n+1)a−1[(na+1)+(na+2)+.....+(na+n)]=160\mathop {\lim }\limits_{n \to \infty } \,\,{{{1^a} + {2^a} + ...... + {n^a}} \over {{{(n + 1)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]}} = {1 \over {60}}n→∞lim(n+1)a−1[(na+1)+(na+2)+.....+(na+n)]1a+2a+......+na=601 for some positive real number a, then a is equal to :A7B8C152{{15} \over 2}215D172{{17} \over 2}217Check answerSkip
09Medium210×since 2002Q3286limn→∞((n+1)1/3n4/3+(n+2)1/3n4/3+.......+(2n)1/3n4/3)\mathop {\lim }\limits_{n \to \infty } \left( {{{{{(n + 1)}^{1/3}}} \over {{n^{4/3}}}} + {{{{(n + 2)}^{1/3}}} \over {{n^{4/3}}}} + ....... + {{{{(2n)}^{1/3}}} \over {{n^{4/3}}}}} \right)n→∞lim(n4/3(n+1)1/3+n4/3(n+2)1/3+.......+n4/3(2n)1/3) is equal to :A43(2)3/4{4 \over 3}{\left( 2 \right)^{3/4}}34(2)3/4B34(2)4/3−34{3 \over 4}{\left( 2 \right)^{4/3}} - {3 \over 4}43(2)4/3−43C43(2)4/3{4 \over 3}{\left( 2 \right)^{4/3}}34(2)4/3D34(2)4/3−43{3 \over 4}{\left( 2 \right)^{4/3}} - {4 \over 3}43(2)4/3−34Check answerSkip
10Medium210×since 2002Q3287limn→∞[1n+n(n+1)2+n(n+2)2+........+n(2n+1)2]\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]n→∞lim[n1+(n+1)2n+(n+2)2n+........+(2n+1)2n] is equal to :A12{{1 \over 2}}21B13{{1 \over 3}}31C1D14{{1 \over 4}}41Check answerSkip
11Easy210×since 2002Q3288The value of limn→∞1n∑j=1n(2j−1)+8n(2j−1)+4n\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{(2j - 1) + 8n} \over {(2j - 1) + 4n}}}n→∞limn1j=1∑n(2j−1)+4n(2j−1)+8n is equal to :A5+loge(32)5 + {\log _e}\left( {{3 \over 2}} \right)5+loge(23)B2−loge(23)2 - {\log _e}\left( {{2 \over 3}} \right)2−loge(32)C3+2loge(23)3 + 2{\log _e}\left( {{2 \over 3}} \right)3+2loge(32)D1+2loge(32)1 + 2{\log _e}\left( {{3 \over 2}} \right)1+2loge(23)Check answerSkip
12Easy210×since 2002Q3289The value of limn→∞1n∑r=02n−1n2n2+4r2\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}}n→∞limn1r=0∑2n−1n2+4r2n2 is :A12tan−1(2){1 \over 2}{\tan ^{ - 1}}(2)21tan−1(2)B12tan−1(4){1 \over 2}{\tan ^{ - 1}}(4)21tan−1(4)Ctan−1(4){\tan ^{ - 1}}(4)tan−1(4)D14tan−1(4){1 \over 4}{\tan ^{ - 1}}(4)41tan−1(4)Check answerSkip
13Medium210×since 2002Q3290If Un=(1+1n2)(1+22n2)2.....(1+n2n2)n{U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^nUn=(1+n21)(1+n222)2.....(1+n2n2)n, then limn→∞(Un)−4n2\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}n→∞lim(Un)n2−4 is equal to :Ae216{{{e^2}} \over {16}}16e2B4e{4 \over e}e4C16e2{{16} \over {{e^2}}}e216D4e2{4 \over {{e^2}}}e24Check answerSkip
14Medium210×since 2002Q3291limn→∞(n2(n2+1)(n+1)+n2(n2+4)(n+2)+n2(n2+9)(n+3)+ .... + n2(n2+n2)(n+n))\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + {{{n^2}} \over {({n^2} + 9)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)n→∞lim((n2+1)(n+1)n2+(n2+4)(n+2)n2+(n2+9)(n+3)n2+....+(n2+n2)(n+n)n2) is equal to :Aπ8+14loge2{\pi \over 8} + {1 \over 4}{\log _e}28π+41loge2Bπ4+18loge2{\pi \over 4} + {1 \over 8}{\log _e}24π+81loge2Cπ4−18loge2{\pi \over 4} - {1 \over 8}{\log _e}24π−81loge2Dπ8+loge2{\pi \over 8} + {\log _e}\sqrt 28π+loge2Check answerSkip
15Medium210×since 2002Q3292limn→∞∑r=1nr2r2−7rn+6n2\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}}n→∞limr=1∑n2r2−7rn+6n2r is equal to :Aloge(32){\log _e}\left( {{{\sqrt 3 } \over 2}} \right)loge(23)Bloge(334){\log _e}\left( {{{3\sqrt 3 } \over 4}} \right)loge(433)Cloge(274){\log _e}\left( {{{27} \over 4}} \right)loge(427)Dloge(43){\log _e}\left( {{4 \over 3}} \right)loge(34)Check answerSkip
16Medium210×since 2002Q3293limn→∞12n(11−12n+11−22n+11−32n+ ... + 11−2n−12n)\mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,...\,\, + \,\,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)n→∞lim2n11−2n11+1−2n21+1−2n31+...+1−2n2n−11 is equal toA12\frac{1}{2}21B1C2D−-−2Check answerSkip
17Medium210×since 2002Q3294If a=limn→∞∑k=1n2nn2+k2a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}}a=n→∞limk=1∑nn2+k22n and f(x)=1−cosx1+cosxf(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}}f(x)=1+cosx1−cosx, x∈(0,1)x \in (0,1)x∈(0,1), then :A22f(a2)=f′(a2)2\sqrt 2 f\left( {{a \over 2}} \right) = f'\left( {{a \over 2}} \right)22f(2a)=f′(2a)Bf(a2)f′(a2)=2f\left( {{a \over 2}} \right)f'\left( {{a \over 2}} \right) = \sqrt 2f(2a)f′(2a)=2C2f(a2)=f′(a2)\sqrt 2 f\left( {{a \over 2}} \right) = f'\left( {{a \over 2}} \right)2f(2a)=f′(2a)Df(a2)=2f′(a2)f\left( {{a \over 2}} \right) = \sqrt 2 f'\left( {{a \over 2}} \right)f(2a)=2f′(2a)Check answerSkip
18Easy210×since 2002Q3295limn→∞[11+n+12+n+13+n + ... + 12n]\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {1 + n}} + {1 \over {2 + n}} + {1 \over {3 + n}}\, + \,...\, + \,{1 \over {2n}}} \right]n→∞lim[1+n1+2+n1+3+n1+...+2n1] is equal toA0Bloge2{\log _e}2loge2Cloge(23){\log _e}\left( {{2 \over 3}} \right)loge(32)Dloge(32){\log _e}\left( {{3 \over 2}} \right)loge(23)Check answerSkip
19Easy210×since 2002Q3296limn→∞3n{4+(2+1n)2+(2+2n)2+…+(3−1n)2}\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}n→∞limn3{4+(2+n1)2+(2+n2)2+…+(3−n1)2} is equal to :A0B193\frac{19}{3}319C19D12Check answerSkip
20Medium210×since 2002Q3297Among (S1): \lim_\limits{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)=1 (S2) : \lim_\limits{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}AOnly (S1) is trueBBoth (S1) and (S2) are trueCBoth (S1) and (S2) are falseDOnly (S2) is trueCheck answerSkip