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KooM1QkVr1JhZisA3F1kluga4m4 | maths | sequences-and-series | summation-of-series | The sum of the infinite series <br/>$$1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ......$$ is equal to : | [{"identifier": "A", "content": "$${9 \\over 4}$$"}, {"identifier": "B", "content": "$${13 \\over 4}$$"}, {"identifier": "C", "content": "$${15 \\over 4}$$"}, {"identifier": "D", "content": "$${11 \\over 4}$$"}] | ["B"] | null | $$S = 1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + ....$$<br><br>$${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {7 \over {{3^3}}} + {{12} \over {{3^4}}} + ....$$<br><br>$$2S = 1 + {1 \over 3} + {5 \over {{3^2}}} + {5 \over {{3^3}}} + {5 \over {{3^4}}} + ....$$ + up to infinite ... | mcq | jee-main-2021-online-26th-february-morning-slot |
ZBMT4NvKr5BSRHqBGT1kluxngbh | maths | sequences-and-series | summation-of-series | The sum of the series <br/><br/>$$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $$ is equal to : | [{"identifier": "A", "content": "$${{41} \\over 8}e + {{19} \\over 8}{e^{ - 1}} - 10$$"}, {"identifier": "B", "content": "$${{41} \\over 8}e - {{19} \\over 8}{e^{ - 1}} - 10$$"}, {"identifier": "C", "content": "$${{41} \\over 8}e + {{19} \\over 8}{e^{ - 1}} + 10$$"}, {"identifier": "D", "content": "$$ - {{41} \\over 8}... | ["B"] | null | $$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $$<br><br>Put 2n + 1 = r, where r = 3, 5, 7, .......<br><br>$$ \Rightarrow n = {{r - 1} \over 2}$$<br><br>$${{{n^2} + 6n + 10} \over {(2n + 1)!}} = {{{{\left( {{{r - 1} \over 2}} \right)}^2} + 3r - 3 + 10} \over {r!}} $$
<br><br>$$= {{{r^2} + 10r + 2... | mcq | jee-main-2021-online-26th-february-evening-slot |
zd4RTaPsITd0LFIGzU1kmiz8sd6 | maths | sequences-and-series | summation-of-series | S<sub>n</sub>(x) = log<sub>a<sup>1/2</sup></sub>x + log<sub>a<sup>1/3</sup></sub>x + log<sub>a<sup>1/6</sup></sub>x + log<sub>a<sup>1/11</sup></sub>x + log<sub>a<sup>1/18</sup></sub>x + log<sub>a<sup>1/27</sup></sub>x + ...... up to n-terms, where a > 1. If S<sub>24</sub>(x) = 1093 and S<sub>12</sub>(2x) = 265, then... | [] | null | 16 | $${S_n}(x) = {\log _a}{x^2} + {\log _a}{x^3} + {\log _a}{x^6} + {\log _a}{x^{11}}$$<br><br>$${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$$<br><br>$${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$$<br><br>$${S_r} = 2 + 3 + 6 + 11$$
<br><br>$$ \therefore $$ T<sub>n</sub> = 2 + (1 + 3 + 5 +..... | integer | jee-main-2021-online-16th-march-evening-shift |
qu0cXEoBuHthtOcLsa1kmlisv2k | maths | sequences-and-series | summation-of-series | If $$\alpha$$, $$\beta$$ are natural numbers such that <br/>100<sup>$$\alpha$$</sup> $$-$$ 199$$\beta$$ = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through ($$\alpha$$, $$\beta$$) and origin is : | [{"identifier": "A", "content": "540"}, {"identifier": "B", "content": "550"}, {"identifier": "C", "content": "530"}, {"identifier": "D", "content": "510"}] | ["B"] | null | RHS = $$\sum\limits_{r = 0}^{99} {(100 - r)(100 + r)} $$<br><br>$$ = {(100)^3} - {{99 \times 100 \times 199} \over 6} = {(100)^3} - (1650)199$$<br><br>LHS = (100)<sup>$$\alpha$$</sup> $$-$$ (199)<sup>$$\beta$$</sup><br><br>So, $$\alpha$$ = 3, $$\beta$$ = 1650<br><br>Slope = tan$$\theta$$ = $${\beta \over \alpha }$$<br... | mcq | jee-main-2021-online-18th-march-morning-shift |
XHol4qKySWK9HAFu5P1kmlix67q | maths | sequences-and-series | summation-of-series | $${1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}}$$ is equal to | [{"identifier": "A", "content": "$${{101} \\over {404}}$$"}, {"identifier": "B", "content": "$${{25} \\over {101}}$$"}, {"identifier": "C", "content": "$${{101} \\over {408}}$$"}, {"identifier": "D", "content": "$${{99} \\over {400}}$$"}] | ["B"] | null | $$S = \sum\limits_{r = 1}^{100} {{1 \over {{{(2n + 1)}^2} - 1}}} $$<br><br>$$ = \sum\limits_{r = 1}^{100} {{1 \over {(2n + 1 + 1)(2n + 1 - 1)}}} $$<br><br>$$ = \sum\limits_{r = 1}^{100} {{1 \over {2n(2n + 2)}}} $$<br><br>$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {{1 \over {n(n + 1)}}} $$<br><br>$$ = {1 \over 4}\sum\lim... | mcq | jee-main-2021-online-18th-march-morning-shift |
1krruwb6f | maths | sequences-and-series | summation-of-series | If sum of the first 21 terms of the series $${\log _{{9^{1/2}}}}x + {\log _{{9^{1/3}}}}x + {\log _{{9^{1/4}}}}x + .......$$, where x > 0 is 504, then x is equal to | [{"identifier": "A", "content": "243"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "81"}] | ["D"] | null | $$s = 2{\log _9}x + 3{\log _9}x + ....... + 22{\log _9}x$$<br><br>$$s = {\log _9}x(2 + 3 + ..... + 22)$$<br><br>$$s = {\log _9}x\left\{ {{{21} \over 2}(2 + 22)} \right\}$$<br><br>Given, $$252{\log _9}x = 504$$<br><br>$$ \Rightarrow {\log _9}x = 2 \Rightarrow x = 81$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1krrx00sq | maths | sequences-and-series | summation-of-series | For k $$\in$$ N, let $${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $$, where $$\alpha > 0$$. Then the value of $$100{\left( {{{{A_{14}} + {A_{15}}} \over {{A_{13}}}}} \right)^2}$$ is equal to _____________. | [] | null | 9 | $${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $$<br><br>$${A_{14}} = {1 \over {( - 14)( - 13)......( - 1)(1).......(6)}} = {1 \over {14!.6!}}$$<br><br>$${A_{15}} = {1 \over {( - 15)( - 14)......( - 1)(1).......(5)}} = {1 \over {15!.5!}}... | integer | jee-main-2021-online-20th-july-evening-shift |
1krrx9mvg | maths | sequences-and-series | summation-of-series | Let $$\left\{ {{a_n}} \right\}_{n = 1}^\infty $$ be a sequence such that a<sub>1</sub> = 1, a<sub>2</sub> = 1 and $${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$$ for all n $$\ge$$ 1. Then the value of $$47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}} $$ is equal to ______________. | [] | null | 7 | $${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$$, let $$\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} = P$$<br><br>Divide by 8<sup>n</sup> we get<br><br>$${{{a_{n + 2}}} \over {{8^n}}} = {{2{a_{n + 1}}} \over {{8^n}}} + {{{a_n}} \over {{8^n}}}$$<br><br>$$ \Rightarrow 64{{{a_{n + 2}}} \over {{8^{n + 2}}}} = {{16{a_{n + 1... | integer | jee-main-2021-online-20th-july-evening-shift |
1krw23dgt | maths | sequences-and-series | summation-of-series | If the value of<br/><br/> $${\left( {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....upto\,\infty } \right)^{{{\log }_{(0.25)}}\left( {{1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....upto\,\infty } \right)}}$$<br/><br/> is $$l$$, then $$l$$<sup>2</sup> is equal to _______________. | [] | null | 3 | $$l = {\left( {\underbrace {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}}}_S + ....} \right)^{{{\log }_{0.25}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ...} \right)}}$$<br><br>$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....$$<br><br>$${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + ... | integer | jee-main-2021-online-25th-july-morning-shift |
1ktbcep0a | maths | sequences-and-series | summation-of-series | The sum of the series <br/><br/>$${1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$ when x = 2 is : | [{"identifier": "A", "content": "$$1 + {{{2^{101}}} \\over {{4^{101}} - 1}}$$"}, {"identifier": "B", "content": "$$1 + {{{2^{100}}} \\over {{4^{101}} - 1}}$$"}, {"identifier": "C", "content": "$$1 - {{{2^{100}}} \\over {{4^{100}} - 1}}$$"}, {"identifier": "D", "content": "$$1 - {{{2^{101}}} \\over {{2^{400}} - 1}}$$"}] | ["D"] | null | $$S = {1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$$<br><br>$$S + {1 \over {1 - x}} = {1 \over {1 - x}} + {1 \over {x + 1}} + ...... = {2 \over {1 - {x^2}}} + {2 \over {1 + {x^2}}} + ....$$<br><br>$$S + {1 \over {1 - x}} = {{{2^{101}}} \ove... | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktehkmgu | maths | sequences-and-series | summation-of-series | If 0 < x < 1, then $${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$$, is equal to : | [{"identifier": "A", "content": "$$x\\left( {{{1 + x} \\over {1 - x}}} \\right) + {\\log _e}(1 - x)$$"}, {"identifier": "B", "content": "$$x\\left( {{{1 - x} \\over {1 + x}}} \\right) + {\\log _e}(1 - x)$$"}, {"identifier": "C", "content": "$${{1 - x} \\over {1 + x}} + {\\log _e}(1 - x)$$"}, {"identifier": "D", "conten... | ["A"] | null | Let $$t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty $$<br><br>$$ = \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty $$<br><br>$$ = 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}... | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktehsn1f | maths | sequences-and-series | summation-of-series | If for x, y $$\in$$ R, x > 0, y = log<sub>10</sub>x + log<sub>10</sub>x<sup>1/3</sup> + log<sub>10</sub>x<sup>1/9</sup> + ...... upto $$\infty$$ terms <br/><br/>and $${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$$, then the ordered pair (x, y) is equal to : | [{"identifier": "A", "content": "(10<sup>6</sup>, 6)"}, {"identifier": "B", "content": "(10<sup>4</sup>, 6)"}, {"identifier": "C", "content": "(10<sup>2</sup>, 3)"}, {"identifier": "D", "content": "(10<sup>6</sup>, 9)"}] | ["D"] | null | $${{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}$$<br><br>$$ \Rightarrow {\log _{10}}x = 6 \Rightarrow x = {10^6}$$<br><br>Now, <br><br>$$y = ({\log _{10}}x) + \left( {{{\log }_{10}}{x^{{1 \over 3}}}} \right) + \left( {{{\log }_{10}}{x^{{1 \over 9}}}} \right) + ....\infty $$<b... | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktg3am8n | maths | sequences-and-series | summation-of-series | If 0 < x < 1 and $$y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$$, then the value of e<sup>1 + y</sup> at $$x = {1 \over 2}$$ is : | [{"identifier": "A", "content": "$${1 \\over 2}{e^2}$$"}, {"identifier": "B", "content": "2e"}, {"identifier": "C", "content": "$${1 \\over 2}\\sqrt e $$"}, {"identifier": "D", "content": "2e<sup>2</sup>"}] | ["A"] | null | $$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$$<br><br>$$ = ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)$$<br><br>$$ = {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ..... | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktioa9ro | maths | sequences-and-series | summation-of-series | The sum of 10 terms of the series
<br/><br/>$${3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....$$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${{120} \\over {121}}$$"}, {"identifier": "C", "content": "$${{99} \\over {100}}$$"}, {"identifier": "D", "content": "$${{143} \\over {144}}$$"}] | ["B"] | null | $$S = {{{2^2} - {1^2}} \over {{1^2} \times {2^2}}} + {{{3^2} - {2^2}} \over {{2^2} \times {3^2}}} + {{{4^2} - {3^2}} \over {{3^2} \times {4^2}}} + ...$$<br><br>$$ = \left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right] + \left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] + \left[ {{1 \over {{3^2}}} - {1 \over {{4^... | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktkeexuz | maths | sequences-and-series | summation-of-series | If $$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$$, then 160 S is equal to ________. | [] | null | 305 | $$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$$<br><br>$${1 \over 5}S = {7 \over 5} + {9 \over {{5^3}}} + {{13} \over {{5^4}}} + ....$$<br><br>On subtracting<br><br>$${4 \over 5}S = {7 \over 5} + {2 \over {{5^2}}} + {4 \over {{5^3}}} + {6 \over {{5^4}}} + ....$$<br><br>$$S =... | integer | jee-main-2021-online-31st-august-evening-shift |
1kto7yln2 | maths | sequences-and-series | summation-of-series | Let S<sub>n</sub> = 1 . (n $$-$$ 1) + 2 . (n $$-$$ 2) + 3 . (n $$-$$ 3) + ..... + (n $$-$$ 1) . 1, n $$\ge$$ 4.<br/><br/>The sum $$\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $$ is equal to : | [{"identifier": "A", "content": "$${{e - 1} \\over 3}$$"}, {"identifier": "B", "content": "$${{e - 2} \\over 6}$$"}, {"identifier": "C", "content": "$${e \\over 3}$$"}, {"identifier": "D", "content": "$${e \\over 6}$$"}] | ["A"] | null | Let T<sub>r</sub> = r(n $$-$$ r)<br><br>T<sub>r</sub> = nr $$-$$ r<sup>2</sup><br><br>$$ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {{T_r} = \sum\limits_{r = 1}^n {(nr - {r^2})} } $$<br><br>$${S_n} = {{n\,.\,(n)(n + 1)} \over 2} - {{n(n + 1)(2n + 1)} \over 6}$$<br><br>$${S_n} = {{n(n - 1)(n + 1)} \over 6}$$<br><br>Now, ... | mcq | jee-main-2021-online-1st-september-evening-shift |
1l545a1mq | maths | sequences-and-series | summation-of-series | <p>Let $$\{ {a_n}\} _{n = 0}^\infty $$ be a sequence such that $${a_0} = {a_1} = 0$$ and $${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$$ for all n $$\ge$$ 0. Then, $$\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}}} $$ is equal to:</p> | [{"identifier": "A", "content": "$${6 \\over {343}}$$"}, {"identifier": "B", "content": "$${7 \\over {216}}$$"}, {"identifier": "C", "content": "$${8 \\over {343}}$$"}, {"identifier": "D", "content": "$${{49} \\over {216}}$$"}] | ["B"] | null | <p>$${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$$ & $${a_0} = {a_1} = 0$$</p>
<p>$${a_2} = 2{a_1} - {a_0} + 1 = 1$$</p>
<p>$${a_3} = 2{a_2} - {a_1} + 1 = 3$$</p>
<p>$${a_4} = 2{a_3} - {a_2} + 1 = 6$$</p>
<p>$${a_5} = 2{a_4} - {a_3} + 1 = 10$$</p>
<p>$$\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}} = {{{a_2}} \over {{... | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54apl39 | maths | sequences-and-series | summation-of-series | <p>The sum of the infinite series $$1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + {{35} \over {{6^4}}} + {{51} \over {{6^5}}} + {{70} \over {{6^6}}} + \,\,.....$$ is equal to :</p> | [{"identifier": "A", "content": "$${{425} \\over {216}}$$"}, {"identifier": "B", "content": "$${{429} \\over {216}}$$"}, {"identifier": "C", "content": "$${{288} \\over {125}}$$"}, {"identifier": "D", "content": "$${{280} \\over {125}}$$"}] | ["C"] | null | <p>$$S = 1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + $$ ...... (1)</p>
<p>$${1 \over 6}S = {1 \over 6} + {5 \over {{6^2}}} + {{12} \over {{6^3}}} + $$ ...... (2)</p>
<p>$$S - {1 \over 6}S = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $$ ........</p>
<p>$$ \Rightarrow {{5S} \over 6} ... | mcq | jee-main-2022-online-29th-june-evening-shift |
1l56q6r0z | maths | sequences-and-series | summation-of-series | <p>Let $$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + \,.....$$. Then 4S is equal to</p> | [{"identifier": "A", "content": "$${\\left( {{7 \\over 3}} \\right)^2}$$"}, {"identifier": "B", "content": "$${{{7^3}} \\over {{3^2}}}$$"}, {"identifier": "C", "content": "$${\\left( {{7 \\over 3}} \\right)^3}$$"}, {"identifier": "D", "content": "$${{{7^2}} \\over {{3^3}}}$$"}] | ["C"] | null | <p>$$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + $$ ..... ...... (i)</p>
<p>$${1 \over 7}S = {2 \over 7} + {6 \over {{7^2}}} + {{12} \over {{7^3}}} + {{20} \over {{7^4}}} + $$ .... ....... (ii)</p>
<p>(i) - (ii)</p>
<p>$${6 \over 7}S = 2 + {4 \over 7} + {6 \over {{7^2}}} +... | mcq | jee-main-2022-online-27th-june-evening-shift |
1l57p5mqd | maths | sequences-and-series | summation-of-series | <p>If the sum of the first ten terms of the series</p>
<p>$${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$$</p>
<p>is $${m \over n}$$, where m and n are co-prime numbers, then m + n is equal to ______________.</p> | [] | null | 276 | <p>$${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$$</p>
<p>$$ = {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$$</p>
<p>$$ = {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$$</p>
<p>$${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $$</... | integer | jee-main-2022-online-27th-june-morning-shift |
1l58f35j3 | maths | sequences-and-series | summation-of-series | <p>If $$A = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$ and $$B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$, then $${A \over B}$$ is equal to :</p> | [{"identifier": "A", "content": "$${{11} \\over 9}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$$${{11} \\over 9}$$"}, {"identifier": "D", "content": "$$-$$$${{11} \\over 3}$$"}] | ["C"] | null | <p>$$A = \sum\limits_{n = 1}^\infty {{1 \over {{{(3 + {{( - 1)}^n})}^n}}}} $$ and $$B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{(3 + {{( - 1)}^n})}^n}}}} $$</p>
<p>$$A = {1 \over 2} + {1 \over {{4^2}}} + {1 \over {{2^3}}} + {1 \over {{4^4}}} + $$ ........</p>
<p>$$B = {{ - 1} \over 2} + {1 \over {{4^2}}}... | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59jzej6 | maths | sequences-and-series | summation-of-series | <p>The sum 1 + 2 . 3 + 3 . 3<sup>2</sup> + ......... + 10 . 3<sup>9</sup> is equal to :</p> | [{"identifier": "A", "content": "$${{2\\,.\\,{3^{12}} + 10} \\over 4}$$"}, {"identifier": "B", "content": "$${{19\\,.\\,{3^{10}} + 1} \\over 4}$$"}, {"identifier": "C", "content": "$$5\\,.\\,{3^{10}} - 2$$"}, {"identifier": "D", "content": "$${{9\\,.\\,{3^{10}} + 1} \\over 2}$$"}] | ["B"] | null | <p>Let $$S = 1\,.\,{3^0} + 2\,.\,{3^1} + 3\,.\,{3^2} + \,\,......\,\, + \,\,10\,.\,{3^9}$$</p>
<p>$$3S = 1\,.\,{3^1} + 2\,.\,{3^2} + \,\,..........\,\, + \,\,10\,.\,{3^{10}}$$</p>
<p>___________________________________________________________</p>
<p>$$ - 2S = (1\,.\,{3^0} + 1\,.\,{3^1} + 1\,.\,{3^2} + \,\,........\,\, ... | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5ajtlod | maths | sequences-and-series | summation-of-series | <p>For a natural number n, let $${\alpha _n} = {19^n} - {12^n}$$. Then, the value of $${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}}$$ is ___________.</p> | [] | null | 4 | <p>$${\alpha _n} = {19^n} - {12^n}$$</p>
<p>Let equation of roots 12 & 19 i.e.</p>
<p>$${x^2} - 31x + 228 = 0$$</p>
<p>$$ \Rightarrow (31 - x) = {{228} \over x}$$ (where x can be 19 or 12)</p>
<p>$$\therefore$$ $${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}} = {{31({{19}^9} - {{12}^9}) - ({{19}^{10}} - {{12}... | integer | jee-main-2022-online-25th-june-morning-shift |
1l5ajy4fx | maths | sequences-and-series | summation-of-series | <p>The greatest integer less than or equal to the sum of first 100 terms of the sequence $${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$$ ...... is equal to ___________.</p> | [] | null | 98 | <p>$$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, + $$ ....</p>
<p>$$ = \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)} $$</p>
<p>$$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$$</p>
<p>$$ = 98 + 2{\left( {{2 \ove... | integer | jee-main-2022-online-25th-june-morning-shift |
1l5vzhxr8 | maths | sequences-and-series | summation-of-series | <p>The value of $$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}} + \,\,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,.....\,\, + \,\,11}}$$ is equal to:</p> | [{"identifier": "A", "content": "$${{20} \\over {11}}$$"}, {"identifier": "B", "content": "$${{11} \\over {6}}$$"}, {"identifier": "C", "content": "$${{241} \\over {132}}$$"}, {"identifier": "D", "content": "$${{21} \\over {11}}$$"}] | ["B"] | null | <p>Given,</p>
<p>$$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}}\, + \,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,....\,\, + \,11}}$$</p>
<p>General term,</p>
<p>$${T_n} = {1 \over {1 + 2 + 3\, + \,\,....\,\, + \,\,n}}$$</p>
<p>$$ = {1 \over {{{n(n + 1)} \over 2}}}$$</p>
<p>$$ = {2 \over {n(n + 1)}}$$</p>
<p>$$ = 2\left[... | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6dx9fyy | maths | sequences-and-series | summation-of-series | <p>Let $$a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$$ and $$b_{n}=a_{n}+b_{n-1}$$ for every <br/><br/>natural number $$n \geqslant 2$$. Then $$\sum\limits_{n = 1}^{15} {{a_n}.{b_n}} $$ is equal to ___________.</p> | [] | null | 27560 | <p>Given,</p>
<p>$${a_n} = {a_{n - 1}} + 2$$</p>
<p>$$ \Rightarrow {a_n} - {a_{n - 1}} = 2$$</p>
<p>$$\therefore$$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.</p>
<p>Also given $${a_1} = 1$$</p>
<p>$$\therefore$$ Series is = 1, 3, 5, 7 ......</p>
<p>$$... | integer | jee-main-2022-online-25th-july-morning-shift |
1l6f0wdoi | maths | sequences-and-series | summation-of-series | <p>The sum $$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}} $$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{7}{87}$$"}, {"identifier": "B", "content": "$$\\frac{7}{29}$$"}, {"identifier": "C", "content": "$$\\frac{14}{87}$$"}, {"identifier": "D", "content": "$$\\frac{21}{29}$$"}] | ["B"] | null | <p>$$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}} = {3 \over 4}\sum\limits_{n = 1}^{21} {{1 \over {4n - 1}} - {1 \over {4n + 3}}} } $$</p>
<p>$$ = {3 \over 4}\left[ {\left( {{1 \over 3} - {1 \over 7}} \right) + \left( {{1 \over 7} - {1 \over {11}}} \right) + \,\,....\,\, + \,\,\left( {{1 \over {83}} - {1 \ove... | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6gjc5oa | maths | sequences-and-series | summation-of-series | <p>The series of positive multiples of 3 is divided into sets : $$\{3\},\{6,9,12\},\{15,18,21,24,27\}, \ldots$$ Then the sum of the elements in the $$11^{\text {th }}$$ set is equal to ____________.</p> | [] | null | 6993 | <p>Given series</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbx3j2/7964fa43-5af9-4fc1-96c1-0a4102aef97d/c56543e0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbx3j3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbx3j2/7964fa43-5af9-4fc1-96c1-0a4102aef97d/c56543e0-2c4f-1... | integer | jee-main-2022-online-26th-july-morning-shift |
1l6hzs0vf | maths | sequences-and-series | summation-of-series | <p>If $$\sum\limits_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$$, where m and n are co-prime, then $$m+n$$ is equal to _____________.</p> | [] | null | 166 | <p>$$\sum\limits_{k = 1}^{10} {{k \over {{k^4} + {k^2} + 1}}} $$</p>
<p>$$ = {1 \over 2}\left[ {\sum\limits_{k = 1}^{10} {\left( {{1 \over {{k^2} - k + 1}} - {1 \over {{k^2} + k + 1}}} \right)} } \right.$$</p>
<p>$$ = {1 \over 2}\left[ {1 - {1 \over 3} + {1 \over 3} - {1 \over 7} + {1 \over 7} - {1 \over {13}}\, + \,..... | integer | jee-main-2022-online-26th-july-evening-shift |
1l6klic2q | maths | sequences-and-series | summation-of-series | <p>$$
\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\cdots+
\frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$$ is equal to _____________.</p> | [] | null | 120 | <p>$${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{2n(2{n^2} + 3n + 1) -... | integer | jee-main-2022-online-27th-july-evening-shift |
1l6m6bitt | maths | sequences-and-series | summation-of-series | <p>Consider the sequence $$a_{1}, a_{2}, a_{3}, \ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$$ for $$\mathrm{n}=1,2,3, \ldots .$$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\ri... | [{"identifier": "A", "content": "$$-$$30"}, {"identifier": "B", "content": "$$-$$31"}, {"identifier": "C", "content": "$$-$$60"}, {"identifier": "D", "content": "$$-$$61"}] | ["C"] | null | <p>$${a_{n + 2}} = {2 \over {{a_{n + 1}}}} + {a_n}$$</p>
<p>$$ \Rightarrow {a_n}{a_{n + 1}} + 1 = {a_{n + 1}}{a_{n + 2}} - 1$$</p>
<p>$$ \Rightarrow {a_{n + 2}}{a_{n + 1}} - {a_n}\,.\,{a_{n + 1}} = 2$$</p>
<p>For</p>
<p>$$\matrix{
{n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \cr
{n = 2} & {{a_4}{a_3} - {a_3}{a_2} = ... | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6npcsi3 | maths | sequences-and-series | summation-of-series | $${6 \over {{3^{12}}}} + {{10} \over {{3^{11}}}} + {{20} \over {{3^{10}}}} + {{40} \over {{3^9}}} + \,\,...\,\, + \,\,{{10240} \over 3} = {2^n}\,.\,m$$, where m is odd, then m . n is equal to ____________. | [] | null | 12 | <p>$${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right... | integer | jee-main-2022-online-28th-july-evening-shift |
1l6p3pptf | maths | sequences-and-series | summation-of-series | <p>If $$\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102}=\frac{\mathrm{k}}{101}$$, then 34 k is equal to _________.</p> | [] | null | 286 | <p>$$S = {1 \over {2 \times 3 \times 4}} + {1 \over {3 \times 4 \times 5}} + {1 \over {4 \times 5 \times 6}}\, + \,....\, + \,{1 \over {100 \times 101 \times 102}}$$</p>
<p>$$ = {1 \over {(3 - 1)\,.\,1}}\left[ {{1 \over {2 \times 3}} - {1 \over {101 \times 102}}} \right]$$</p>
<p>$$ = {1 \over 2}\left( {{1 \over 6} - {... | integer | jee-main-2022-online-29th-july-morning-shift |
1l6re3sg0 | maths | sequences-and-series | summation-of-series | <p>$$
\begin{aligned}
&\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\
&a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 .
\end{aligned}
$$</p>
<p>Then $$a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$$ is equal to</p> | [{"identifier": "A", "content": "483"}, {"identifier": "B", "content": "528"}, {"identifier": "C", "content": "575"}, {"identifier": "D", "content": "624"}] | ["B"] | null | <p>Given,</p>
<p>$${a_0} = {a_1} = 0$$</p>
<p>and $${a_{n + 2}} = 3{a_{n + 1}} - 2{a_n} + 1$$</p>
<p>For $$n = 0,\,{a_2} = 3{a_1} - 2{a_0} + 1$$</p>
<p>$$ = 3\,.\,0 - 2\,.\,0 + 1$$</p>
<p>$$ = 1$$</p>
<p>For $$n = 1,\,{a_3} = 3{a_2} - 2{a_1} + 1$$</p>
<p>$$ = 3\,.\,1 - 2\,.\,0 + 1$$</p>
<p>$$ = 4$$</p>
<p>For $$n = 2,\... | mcq | jee-main-2022-online-29th-july-evening-shift |
1ldo67p4f | maths | sequences-and-series | summation-of-series | <p>The sum $$\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}} $$ is equal to :</p> | [{"identifier": "A", "content": "$${{11e} \\over 2} + {7 \\over {2e}}$$"}, {"identifier": "B", "content": "$${{13e} \\over 4} + {5 \\over {4e}} - 4$$"}, {"identifier": "C", "content": "$${{11e} \\over 2} + {7 \\over {2e}} - 4$$"}, {"identifier": "D", "content": "$${{13e} \\over 4} + {5 \\over {4e}}$$"}] | ["B"] | null | $\begin{aligned} & \sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !} \\\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{2 n(2 n-1)+8 n+8}{(2 n) !} \\\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\\\ & e=1+1+\frac{1}{2 !}+\frac{1}{3 ... | mcq | jee-main-2023-online-1st-february-evening-shift |
ldoawens | maths | sequences-and-series | summation-of-series | The sum $1^{2}-2 \cdot 3^{2}+3 \cdot 5^{2}-4 \cdot 7^{2}+5 \cdot 9^{2}-\ldots+15 \cdot 29^{2}$ is _________. | [] | null | 6952 | $S=1^{2}-2.3^{2}+3.5^{2}-4.7^{2}+\ldots \ldots+15.29^{2}$
<br/><br/>Separating odd placed and even placed terms we get
<br/><br/>$$
\begin{aligned}
& \mathrm{S}=\left(1.1^2+3.5^2+\ldots .15 .(29)^2\right)-\left(2.3^2+4.7^2\right. \\
& +\ldots .+14 .(27)^2
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\sum_{r=1}^{8}(... | integer | jee-main-2023-online-31st-january-evening-shift |
1ldom86ia | maths | sequences-and-series | summation-of-series | <p>The sum of 10 terms of the series</p>
<p>$${1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....$$ is</p> | [{"identifier": "A", "content": "$${{58} \\over {111}}$$"}, {"identifier": "B", "content": "$${{56} \\over {111}}$$"}, {"identifier": "C", "content": "$${{55} \\over {111}}$$"}, {"identifier": "D", "content": "$${{59} \\over {111}}$$"}] | ["C"] | null | $$
\begin{aligned}
& T_n=\frac{n}{1+n^2+n^4} \\\\
& =\frac{n}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\right)} \\\\
& =\frac{1}{2}\left[\frac{\left(\mathrm{n}^2+\mathrm{n}+1\right)-\left(\mathrm{n}^2-\mathrm{n}+1\right)}{\left(\mathrm{n}^2-\mathrm{n}+1\right)\left(\mathrm{n}^2+\mathrm{n}+1\... | mcq | jee-main-2023-online-1st-february-morning-shift |
ldqzs9si | maths | sequences-and-series | summation-of-series | The $8^{\text {th }}$ common term of the series
<br/><br/>$$
\begin{aligned}
& S_1=3+7+11+15+19+\ldots . . \\\\
& S_2=1+6+11+16+21+\ldots . .
\end{aligned}
$$
<br/><br/>is : | [] | null | 151 | <p>First common term is 11</p>
<p>Common difference of series of common terms is LCM (4, 5) = 20</p>
<p>$$a_8=a+7d$$</p>
<p>$$=11+7\times20=151$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldr5km4h | maths | sequences-and-series | summation-of-series | <p>If $${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$$, then $${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$$ is equal to :</p> | [{"identifier": "A", "content": "$${{51} \\over {144}}$$"}, {"identifier": "B", "content": "$${{49} \\over {138}}$$"}, {"identifier": "C", "content": "$${{50} \\over {141}}$$"}, {"identifier": "D", "content": "$${{52} \\over {147}}$$"}] | ["C"] | null | <p>$$\sum\limits_{i = 1}^{25} {{a_i} = \sum {{{ - 2} \over {4{n^2} - 16n + 15}} = \sum {{{ - 2} \over {(2n - 5)(2n - 3)}}} } } $$</p>
<p>$$ = \sum\limits_{i = 1}^{25} {\left( {{1 \over {2n - 3}} - {1 \over {2n - 5}}} \right)} $$</p>
<p>$$ = \left[ {\left( {{1 \over { - 1}} - {1 \over { - 3}}} \right) + \left( {{1 \over... | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldr7tnf5 | maths | sequences-and-series | summation-of-series | <p>Let $$\sum_\limits{n=0}^{\infty} \frac{\mathrm{n}^{3}((2 \mathrm{n}) !)+(2 \mathrm{n}-1)(\mathrm{n} !)}{(\mathrm{n} !)((2 \mathrm{n}) !)}=\mathrm{ae}+\frac{\mathrm{b}}{\mathrm{e}}+\mathrm{c}$$, where $$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}$$ and $$e=\sum_\limits{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n}... | [] | null | 26 | <p>$$\sum\limits_{n = 0}^\infty {{{{n^3}(2n!) + (2n - 1)(n!)} \over {n!\,.\,(2n)!}}} $$</p>
<p>$$ = \sum\limits_{n = 0}^\infty {{{{n^3}} \over {n!}} + {{2n - 1} \over {2n!}}} $$</p>
<p>$$ = \sum\limits_{n = 0}^\infty {{3 \over {(n - 2)!}} + {1 \over {(n - 3)!}} + {1 \over {(n - 1)!}} + {1 \over {(2n - 1)!}} - {1 \ov... | integer | jee-main-2023-online-30th-january-morning-shift |
1ldsg00ct | maths | sequences-and-series | summation-of-series | <p>Let $$a_1=b_1=1$$ and $${a_n} = {a_{n - 1}} + (n - 1),{b_n} = {b_{n - 1}} + {a_{n - 1}},\forall n \ge 2$$. If $$S = \sum\limits_{n = 1}^{10} {{{{b_n}} \over {{2^n}}}} $$ and $$T = \sum\limits_{n = 1}^8 {{n \over {{2^{n - 1}}}}} $$, then $${2^7}(2S - T)$$ is equal to ____________.</p> | [] | null | 461 | <p>$$\because$$ $${a_n} = {a_{n - 1}} + (n - 1)$$ and $${a_1} = {b_1} = 1$$</p>
<p>$${b_n} = {b_{n - 1}} + {a_{n - 1}}$$</p>
<p>$$\therefore$$ $${b_{n + 1}} = 2{b_n} - {b_{n - 1}} + n - 1$$</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;bord... | integer | jee-main-2023-online-29th-january-evening-shift |
1ldwxkndi | maths | sequences-and-series | summation-of-series | <p>If $${{{1^3} + {2^3} + {3^3}\, + \,...\,up\,to\,n\,terms} \over {1\,.\,3 + 2\,.\,5 + 3\,.\,7\, + \,...\,up\,to\,n\,terms}} = {9 \over 5}$$, then the value of $$n$$ is</p> | [] | null | 5 | Given $\frac{1^{3}+2^{3}+3^{3}+\ldots \text { up to } n \text { terms }}{1.3+2.5+3.7+\ldots \text { up to } n \text { terms }}=\frac{9}{5}$
<br/><br/>
Now
<br/><br/>
Let $S=1.3+2.5+3.7+\ldots$
<br/><br/>
$$
\begin{aligned}
& T_{n}=n \cdot(2 n+1) \\\\
& \therefore S=\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\\\
& \Rig... | integer | jee-main-2023-online-24th-january-evening-shift |
lgnyx1yy | maths | sequences-and-series | summation-of-series | If the sum of the series
<br/><br/>$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^{2}}-\frac{1}{2 \cdot 3}+\frac{1}{3^{2}}\right)+\left(\frac{1}{2^{3}}-\frac{1}{2^{2} \cdot 3}+\frac{1}{2 \cdot 3^{2}}-\frac{1}{3^{3}}\right)+$
<br/><br/>$\left(\frac{1}{2^{4}}-\frac{1}{2^{3} \cdot 3}+\frac{1}{2^{2} \cdot 3^{2}}-... | [] | null | 7 | We can rewrite the given series as follows :
<br/><br/>$$S = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right) + \left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right) + \left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right) + \ldots$$
<br/><br/>Th... | integer | jee-main-2023-online-15th-april-morning-shift |
1lgq0vrkc | maths | sequences-and-series | summation-of-series | <p>The sum to $$20$$ terms of the series $$2 \cdot 2^{2}-3^{2}+2 \cdot 4^{2}-5^{2}+2 \cdot 6^{2}-\ldots \ldots$$ is equal to __________.</p> | [] | null | 1310 | $$
\begin{aligned}
& \sum_{r=1}^{10}\left(2 \cdot(2 r)^2-(2 r+1)^2\right) \\\\
& =\sum_{r=1}^{10}\left(8 r^2-4 r^2-4 r-1\right) \\\\
& =\sum_{r=1}^{10}\left(4 r^2-4 r-1\right) \\\\
& =\frac{4 \cdot 10 \cdot 11 \cdot 21}{6}-4 \frac{10 \cdot 11}{2}-10 \\\\
& =44 \cdot 35-220-10 \\\\
& =1540-230=1310
\end{aligned}
$$ | integer | jee-main-2023-online-13th-april-morning-shift |
1lgreevjx | maths | sequences-and-series | summation-of-series | <p>Let $$< a_{\mathrm{n}} > $$ be a sequence such that $$a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$$. If $$28 \sum_\limits{k=1}^{10} \frac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}$$, where $$\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}$$ are the first $$\mathrm{m}$$ prime numbers,... | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["C"] | null | Given the sum of the first n terms, $S_n = \frac{n^2+3n}{(n+1)(n+2)}$, we can find the n<sup>th</sup> term $a_n$ as the difference between the sum of the first n terms and the sum of the first n-1 terms :
<br/><br/>So,
$$a_n = S_n - S_{n-1}$$
<br/><br/>Solving, we get :
<br/><br/>$$a_n = \frac{n^2+3n}{(n+1)(n+2)} ... | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgsw5qty | maths | sequences-and-series | summation-of-series | <p>For $$k \in \mathbb{N}$$, if the sum of the series $$1+\frac{4}{k}+\frac{8}{k^{2}}+\frac{13}{k^{3}}+\frac{19}{k^{4}}+\ldots$$ is 10 , then the value of $$k$$ is _________.</p> | [] | null | 2 | From the given series :
<br/><br/>$$10 = 1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$$
<br/><br/>We isolate the 1 to get :
<br/><br/>$$9 = \frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$$ .......(1)
<br/><br/>Divide each term in the equation by $k$ :
<br/><br/>$$\frac{9}{k} = \fra... | integer | jee-main-2023-online-11th-april-evening-shift |
1lguwu480 | maths | sequences-and-series | summation-of-series | <p>Let $$S=109+\frac{108}{5}+\frac{107}{5^{2}}+\ldots .+\frac{2}{5^{107}}+\frac{1}{5^{108}}$$. Then the value of $$\left(16 S-(25)^{-54}\right)$$ is equal to ___________.</p> | [] | null | 2175 | We have, $S=109+\frac{108}{5}+\frac{107}{5^2}+\ldots+\frac{2}{5^{107}}+\frac{1}{5^{108}}$ ...........(i) <br/><br/>$\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2}+\frac{107}{5^3}+\ldots .+\frac{2}{5^{108}}+\frac{1}{5^{109}}$ .............(ii)
<br/><br/>On subtracting Eq. (ii) from Eq. (i), we get
<br/><br/>$$
\begin{aligned... | integer | jee-main-2023-online-11th-april-morning-shift |
1lgvpd2gn | maths | sequences-and-series | summation-of-series | <p>If $$\mathrm{S}_{n}=4+11+21+34+50+\ldots$$ to $$n$$ terms, then $$\frac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_{9}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "227"}, {"identifier": "B", "content": "226"}, {"identifier": "C", "content": "220"}, {"identifier": "D", "content": "223"}] | ["D"] | null | Given that
<br/><br/>$$
\begin{aligned}
& \mathrm{S}_n=4+11+21+24+50+\ldots+\mathrm{T}_n \\\\
& \mathrm{~S}_n=~~~~~~~~4+11+21+34++\mathrm{T}_{n-1}+\mathrm{T}_n \\
& -\quad-\quad-\quad-\quad-\quad-\quad- \\
& \hline 0=4+7+10+13+16+\ldots\left(\mathrm{T}_n-\mathrm{T}_{n-2}\right)-\mathrm{T}_n
\end{aligned}
$$
<br/><br/>$... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxwg676 | maths | sequences-and-series | summation-of-series | <p>The sum of all those terms, of the arithmetic progression 3, 8, 13, ...., 373, which are not divisible by 3, is equal to ____________.</p> | [] | null | 9525 | The sum of all those terms, of the arithmetic progression $3,8,13 \ldots, 373$.
<br/><br/>Which are not divisible by 3 is
<br/><br/>$$
\begin{aligned}
& =(3+8+13+18+\ldots+373) -(3+18+33+\ldots+363) \\\\
& =\frac{75}{2}(3+373)-\frac{25}{2}(3+363) \\\\
& =\frac{75}{2} \times 376-\frac{25}{2} \times 366 \\\\
& =75 \times... | integer | jee-main-2023-online-10th-april-morning-shift |
1lgyl2od3 | maths | sequences-and-series | summation-of-series | <p>Let $$\mathrm{a}_{\mathrm{n}}$$ be the $$\mathrm{n}^{\text {th }}$$ term of the series $$5+8+14+23+35+50+\ldots$$ and $$\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$$. Then $$\mathrm{S}_{30}-a_{40}$$ is equal to :</p> | [{"identifier": "A", "content": "11280"}, {"identifier": "B", "content": "11290"}, {"identifier": "C", "content": "11310"}, {"identifier": "D", "content": "11260"}] | ["B"] | null | Let $\mathrm{S}_n=5+8+14+23+\ldots .+a_n$
<br/><br/>and $\mathrm{S}_n=0+5+8+14+\ldots .+a_n$
<br/><br/>On subtracting, we get
<br/><br/>$$
\begin{aligned}
& 0=5+3+6 \ldots-a_n \\\\
& \Rightarrow a_n=5+3+6+9+\ldots(n-1) \text { terms } \\\\
& =5+\left[\frac{(n-1)}{2}(6+(n-2) 3)\right]
\end{aligned}
$$
<br/><br/>$$
\be... | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh200ul0 | maths | sequences-and-series | summation-of-series | <p>The sum of the first $$20$$ terms of the series $$5+11+19+29+41+\ldots$$ is :</p> | [{"identifier": "A", "content": "3420"}, {"identifier": "B", "content": "3450"}, {"identifier": "C", "content": "3250"}, {"identifier": "D", "content": "3520"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{S}_n=5+11+19+29+41+\ldots .+\mathrm{T}_n \\\\
& \mathrm{S}_n=~~~~~~~~ 5+11+19+29+\ldots .+\mathrm{T}_{n-1}+\mathrm{T}_n \\\\
& \hline 0=5+6+8+10+12+\ldots . . \mathrm{T}_n
\end{aligned}
$$
<br/><br/>$$
\begin{array}{rlrl}
&0 =5+\frac{n-1}{2}[2 \times 6+(n-2)(2)]-T_n \\\\
&\Rightarrow T_n... | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2xwwfd | maths | sequences-and-series | summation-of-series | <p> If $$\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1$$ and $$1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$$ then $$m^{2}-n^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "220"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "240"}, {"identifier": "D", "content": "180"}] | ["C"] | null | Given $\operatorname{gcd}(m, n)=1$ and
<br/><br/>$$
\begin{aligned}
& \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\
& =1012 m^2 n \\\\
& \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\
& =1012 m^2 n \\\\
& \Rightarrow(1-2)(1+2)+(3-4)(3+4)+\ldots .(2021-2022) \\
& (2021+2022)+(202... | mcq | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lse57f85 | maths | sequences-and-series | summation-of-series | <p>The sum of the series $$\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots$$ up to 10 -terms is</p> | [{"identifier": "A", "content": "$$\\frac{45}{109}$$\n"}, {"identifier": "B", "content": "$$-\\frac{55}{109}$$\n"}, {"identifier": "C", "content": "$$\\frac{55}{109}$$\n"}, {"identifier": "D", "content": "$$-\\frac{45}{109}$$"}] | ["B"] | null | <p>General term of the sequence,</p>
<p>$$\begin{aligned}
& \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4} \\
& \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2} \\
& \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} \\
... | mcq | jee-main-2024-online-31st-january-morning-shift |
1lsg5cobe | maths | sequences-and-series | summation-of-series | <p>Let $$S_n$$ be the sum to $$n$$-terms of an arithmetic progression $$3,7,11$$,
If $$40<\left(\frac{6}{n(n+1)} \sum_\limits{k=1}^n S_k\right)<42$$, then $$n$$ equals ________.</p> | [] | null | 9 | <p>$$\begin{aligned}
& \mathrm{S}_{\mathrm{n}}= 3+7+11+\ldots \ldots \mathrm{n} \text { terms } \\
&=\frac{\mathrm{n}}{2}(6+(\mathrm{n}-1) 4)=3 \mathrm{n}+2 \mathrm{n}^2-2 \mathrm{n} \\
&=2 \mathrm{n}^2+\mathrm{n} \\
& \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}=2 \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K... | integer | jee-main-2024-online-30th-january-evening-shift |
1lsgcmznt | maths | sequences-and-series | summation-of-series | <p>Let $$\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots$$ upto 10 terms and $$\beta=\sum_\limits{n=1}^{10} n^4$$. If $$4 \alpha-\beta=55 k+40$$, then $$\mathrm{k}$$ is equal to __________.</p> | [] | null | 353 | <p>$$\begin{gathered}
\alpha=1^2+4^2+8^2 \ldots . \\
t_n=a^2+b n+c
\end{gathered}$$</p>
<p>$$\begin{aligned}
& 1=a+b+c \\
& 4=4 a+2 b+c \\
& 8=9 a+3 b+c
\end{aligned}$$</p>
<p>On solving we get, $$\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1$$</p>
<p>$$\begin{aligned}
& \alpha=\sum_{n=1}^{10}\left(\fr... | integer | jee-main-2024-online-30th-january-morning-shift |
luxwdj8i | maths | sequences-and-series | summation-of-series | <p>If $$\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots . .+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots \ldots+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$$, then $$\alpha$$ is equal to ___________.</p> | [] | null | 1011 | <p>$$\begin{aligned}
& \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\
& \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\
& \quad \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) ... | integer | jee-main-2024-online-9th-april-evening-shift |
luy6z4j2 | maths | sequences-and-series | summation-of-series | <p>If the sum of the series $$\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}$$ is equal to 5, then $$50 \mathrm{~d}$$ is equal to :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "20"}] | ["A"] | null | <p>$$\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5$$</p>
<p>Multiply and divide by $$d$$</p>
<p>$$\begin{aligned}
& \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\
& \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{... | mcq | jee-main-2024-online-9th-april-morning-shift |
lv2erzk5 | maths | sequences-and-series | summation-of-series | <p>The value of $$\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$$ is</p> | [{"identifier": "A", "content": "$$\\frac{305}{301}$$\n"}, {"identifier": "B", "content": "$$\\frac{306}{305}$$\n"}, {"identifier": "C", "content": "$$\\frac{32}{31}$$\n"}, {"identifier": "D", "content": "$$\\frac{31}{30}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n(n+1)^2}{\sum_\limits{n=1}^{100} n^2(n+1)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n^3+2 n^2+n}... | mcq | jee-main-2024-online-4th-april-evening-shift |
lv5gt51f | maths | sequences-and-series | summation-of-series | <p>Let the positive integers be written in the form :</p>
<p><img src="data:image/png;base64,UklGRvQEAABXRUJQVlA4IOgEAABwYwCdASoAA6UBP4HA3mc2MK4noHToSsAwCWlu4XdhG/P58iPxBcuiJs2bNmzZs2bNj6YykDSTJkyZMmTJkyZMmTJkyZMmTJkyZDQzmJcnECBAgQIECBAgQIECBAgQIECBAgJSFLZI8ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx48ePHjx4... | [] | null | 103 | <p>To solve this problem, we need to determine in which row the number $5310$ appears when positive integers are arranged in rows such that the $k^\text{th}$ row contains exactly $k$ numbers.</p>
<h3><strong>Understanding the Pattern</strong></h3>
<p><p><strong>First row ($k = 1$)</strong>: Contains 1 number.</p></p>
... | integer | jee-main-2024-online-8th-april-morning-shift |
lv5gt1x2 | maths | sequences-and-series | summation-of-series | <p>Let $$\alpha=\sum_\limits{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$$ and $$\beta=\left(\sum_\limits{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$$. If $$140<\frac{2 \alpha}{\beta}<281$$, then the value of $$n$$ is _________.</p> | [] | null | 5 | <p>$$\begin{aligned}
\alpha= & \sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r \\
& =4 \sum_{r=0}^n r^2{ }^n C_r+2 \sum_{r=0}^n r \cdot{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\
& =4 n(n+1) 2^{n-2}+2 \cdot n \cdot 2^{n-1}+2^n \\
& =2^n(n(n+1)+n+1)=2^n(n+1)^2 \\
& \beta=\sum_{r=0}^n\left(\frac{{ }^n C_r}{r+1}\right)+\left(\frac{1... | integer | jee-main-2024-online-8th-april-morning-shift |
lv7v4g04 | maths | sequences-and-series | summation-of-series | <p>If $$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$$ and $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=\mathrm{n}$$, then the point $$(\mathrm{m}, \mathrm{n})$$ lies on the line</p> | [{"identifier": "A", "content": "$$11(x-1)-100 y=0$$\n"}, {"identifier": "B", "content": "$$11 x-100 y=0$$\n"}, {"identifier": "C", "content": "$$11(x-1)-100(y-2)=0$$\n"}, {"identifier": "D", "content": "$$11(x-2)-100(y-1)=0$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\
& \text { and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\
& =\frac... | mcq | jee-main-2024-online-5th-april-morning-shift |
lvb2953r | maths | sequences-and-series | summation-of-series | <p>If $$\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x \neq 0$$, and $$(60)^2 \mathrm{~S}(60)=\mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b}$$, where $$a, b \in N$$, then $$(a+b)$$ equal to _________.</p> | [] | null | 3660 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwawri93/7135ddd1-f250-4342-8c29-b0037fcb56d9/81a82a80-146c-11ef-a0c7-b1f23fa7cdc4/file-1lwawri94.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwawri93/7135ddd1-f250-4342-8c29-b0037fcb56d9/81a82a80-146c-11ef-a0c7-b1f23fa7cdc4... | integer | jee-main-2024-online-6th-april-evening-shift |
WSabZw5gL3VAFWgcOz1qpqahkk8e9ppyl | maths | sets-and-relations | number-of-sets-and-relations | Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y $$ \subseteq $$ X, Z $$ \subseteq $$ X and Y $$ \cap $$ Z is empty, is : | [{"identifier": "A", "content": "3<sup>5</sup>"}, {"identifier": "B", "content": "2<sup>5</sup>"}, {"identifier": "C", "content": "5<sup>3</sup>"}, {"identifier": "D", "content": "5<sup>2</sup>"}] | ["A"] | null | For any element x<sub>i</sub> present in X, 4 cases arises while making subsets Y and Z.
<br><br><b>Case- 1 :</b> x<sub>i</sub> $$ \in $$ Y, x<sub>i</sub> $$ \in $$ Z $$ \Rightarrow $$ Y $$ \cap $$ Z $$ \ne $$ $$\phi $$
<br><br><b>Case- 2 :</b> x<sub>i</sub> $$ \in $$ Y, x<sub>i</sub> $$ \notin $$ Z $$ \Rightarrow $$ Y... | mcq | aieee-2012 |
l91h0jo8 | maths | sets-and-relations | number-of-sets-and-relations | Let A and B be two sets containing four and
two elements respectively. Then, the number
of subsets of the set A $\times$ B , each having atleast
three elements are | [{"identifier": "A", "content": "219"}, {"identifier": "B", "content": "256"}, {"identifier": "C", "content": "275"}, {"identifier": "D", "content": "510"}] | ["A"] | null | Given,<br/><br/>
$$
\begin{aligned}
&n(A)=4, n(B) =2 \\\\
&\Rightarrow n(A \times B) =8
\end{aligned}
$$<br/><br/>
Total number of subsets of set $(A \times B)=2^8$<br/><br/>
Number of subsets of set $A \times B$ having no element (i.e. $\phi)=1$<br/><br/>
Number of subsets of set $A \times B$ having one element $={ }^... | mcq | jee-main-2015-offline |
ITWzRryGcqebgm3ACxWNT | maths | sets-and-relations | number-of-sets-and-relations | Let P = {$$\theta $$ : sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 \,\cos \theta $$} <br/><br/>and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then | [{"identifier": "A", "content": "P $$ \\subset $$ Q and Q $$-$$ P $$ \\ne $$ $$\\phi $$"}, {"identifier": "B", "content": "Q $$ \\not\\subset $$ P "}, {"identifier": "C", "content": "P $$ \\not\\subset $$ Q"}, {"identifier": "D", "content": "P = Q"}] | ["D"] | null | Given,
<br><br>sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 $$cos$$\theta $$
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $$\left( {\sqrt 2 + 1} \right)$$cos$$\theta $$
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $${{2 - 1} \over {\sqrt 2 - 1}}$$cos$$\theta $$
<br><br>$$ \Rightarrow... | mcq | jee-main-2016-online-10th-april-morning-slot |
DFFepIdax5IrxuOX | maths | sets-and-relations | number-of-sets-and-relations | Two sets A and B are as under :
<br/><br/>A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};
<br/><br/>B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)<sup>2</sup> + 9(b - 5)<sup>2</sup> $$ \le $$ 36 };
<br/><br/>Then | [{"identifier": "A", "content": "neither A $$ \\subset $$ B nor B $$ \\subset $$ A"}, {"identifier": "B", "content": "B $$ \\subset $$ A"}, {"identifier": "C", "content": "A $$ \\subset $$ B"}, {"identifier": "D", "content": "A $$ \\cap $$ B = $$\\phi $$ ( an empty set )"}] | ["C"] | null | Given,
<br><br>$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$
<br><br>Let $$a - 6 = x$$ and $$b - 5 = y$$
<br><br>$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$
<br><br>$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$
<br><br>This is a equation of ellipse.
<br><br>This el... | mcq | jee-main-2018-offline |
WCJJppkqk0JTLAItniIrv | maths | sets-and-relations | number-of-sets-and-relations | Let S = {1, 2, 3, … , 100}. The number of non-empty subsets A of S such that the product of elements in A is even is : | [{"identifier": "A", "content": "2<sup>50</sup> \u2013 1 "}, {"identifier": "B", "content": "2<sup>50</sup> (2<sup>50</sup> $$-$$ 1)"}, {"identifier": "C", "content": "2<sup>100</sup> $$-$$ 1"}, {"identifier": "D", "content": "2<sup>50</sup> + 1"}] | ["B"] | null | S = {1,2,3, . . . .100}
<br><br>= Total non empty subsets-subsets with product of element is odd
<br><br>= 2<sup>100</sup> $$-$$ 1 $$-$$ 1[(2<sup>50</sup> $$-$$ 1)]
<br><br>= 2<sup>100</sup> $$-$$ 2<sup>50</sup>
<br><br>= 2<sup>50</sup> (2<sup>50</sup> $$-$$ 1) | mcq | jee-main-2019-online-12th-january-morning-slot |
qjItDZ6yIHBa8CspKAyI4 | maths | sets-and-relations | number-of-sets-and-relations | Let Z be the set of integers.
<br/>If A = {x $$ \in $$ Z : 2<sup>(x + 2) (x<sup>2</sup> $$-$$ 5x + 6)</sup> = 1} and
<br/>B = {x $$ \in $$ Z : $$-$$ 3 < 2x $$-$$ 1 < 9},
<br/>then the number of subsets of the set A $$ \times $$ B, is | [{"identifier": "A", "content": "2<sup>12</sup>"}, {"identifier": "B", "content": "2<sup>18</sup>"}, {"identifier": "C", "content": "2<sup>10</sup>"}, {"identifier": "D", "content": "2<sup>15</sup>"}] | ["D"] | null | A ={x $$ \in $$ z : 2<sup>(x+2)(x<sup>2</sup> $$-$$ 5x + 6) </sup> = 1}
<br><br>2<sup>(x+2)(x<sup>2</sup> $$-$$ 5x + 6)</sup> = 2<sup>0</sup> $$ \Rightarrow $$ x = $$-$$ 2, 2, 3
<br><br>A = {$$-$$2, 2, 3}
<br><br>B = {x $$\varepsilon $$ Z : $$-$$ < 2x $$-$$ 1 < 9}
<br><br>B = {0, 1, 2, 3, 4}
<br><br>A $$ \times $... | mcq | jee-main-2019-online-12th-january-evening-slot |
mW5EIn2EI2rmTjurp27k9k2k5foidxf | maths | sets-and-relations | number-of-sets-and-relations | Let X = {n $$ \in $$ N : 1 $$ \le $$ n $$ \le $$ 50}. If
<br/>A = {n $$ \in $$ X: n is a multiple of 2} and
<br/>B = {n $$ \in $$ X: n is a multiple of 7}, then the number of elements in the smallest subset of X
containing both A and B is ________. | [] | null | 29 | X = {1, 2, 3, 4, …, 50}
<br><br>A = {2, 4, 6, 8, …, 50} = 25 elements
<br><br>B = {7, 14, 21, 28, 35, 42, 49} = 7 elements
<br><br>Here n(A$$ \cup $$B) = n(A) + n(B) – n(A$$ \cap $$B)
<br><br>= 25 + 7 – 3 = 29 | integer | jee-main-2020-online-7th-january-evening-slot |
uh9zBvgdVXVFTfGA1mjgy2xukewrzn2p | maths | sets-and-relations | number-of-sets-and-relations | If R = {(x, y) : x, y
$$ \in $$ Z, x<sup>2</sup> + 3y<sup>2</sup>
$$ \le $$ 8} is a relation
on the set of integers Z, then the domain of R<sup>–1</sup> is : | [{"identifier": "A", "content": "{0, 1} "}, {"identifier": "B", "content": "{\u20132, \u20131, 1, 2}"}, {"identifier": "C", "content": "{\u20131, 0, 1}"}, {"identifier": "D", "content": "{\u20132, \u20131, 0, 1, 2}"}] | ["C"] | null | Given R = {(x, y) : x, y
$$ \in $$ Z, x<sup>2</sup> + 3y<sup>2</sup>
$$ \le $$ 8}
<br><br>So R = {(0,1), (0,–1), (1,0), (–1,0), (1,1), (1,-1)
<br>(-1,1), (-1,-1), (2,0), (-2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}
<br><br>$$ \Rightarrow $$ R : { -2, -1, 0, 1, 2} $$ \to $$ {-1, 0, 1}
<br><br>$$ \therefore $$ R<sup>... | mcq | jee-main-2020-online-2nd-september-morning-slot |
B2HRGK7dEczQF7C94Njgy2xukf0qgrp2 | maths | sets-and-relations | number-of-sets-and-relations | Consider the two sets :
<br/>A = {m $$ \in $$ R : both the roots of<br/> x<sup>2</sup>
– (m + 1)x + m + 4 = 0 are real}
<br/>and B = [–3, 5).
<br/>Which of the following is not true? | [{"identifier": "A", "content": "A $$ \\cap $$ B = {\u20133}"}, {"identifier": "B", "content": "B \u2013 A = (\u20133, 5)"}, {"identifier": "C", "content": "A $$ \\cup $$ B = R"}, {"identifier": "D", "content": "A - B = ($$ - $$$$ \\propto $$, $$ - $$3) $$ \\cup $$ (5, $$ \\propto $$)"}] | ["D"] | null | As roots are real so, $$D \ge 0$$<br><br>$${(m + 1)^2} - 4(m + 4) \ge 0$$<br><br>$$ \Rightarrow {m^2} - 2m - 15 \ge 0$$<br><br>$$ \Rightarrow $$ $$(m - 5)(m + 3) \ge 0$$<br><br>$$m\, \in \,$$($$ - $$$$ \propto $$, $$ - $$3] $$ \cup $$ [5, $$ \propto $$)<br><br>$$A= ( - $$$$ \propto $$, $$ - $$3] $$ \cup $$ [5, $$ \prop... | mcq | jee-main-2020-online-3rd-september-morning-slot |
1p4U7IKhG25B9jE6nSjgy2xukfakf6tf | maths | sets-and-relations | number-of-sets-and-relations | Let $$\mathop \cup \limits_{i = 1}^{50} {X_i} = \mathop \cup \limits_{i = 1}^n {Y_i} = T$$ where each X<sub>i</sub> contains 10 elements and each Y<sub>i</sub> contains 5 elements. If each element of the set T is an element of exactly 20 of sets X<sub>i</sub>’s and exactly 6 of sets Y<sub>i</sub>’s, then n is equal ... | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "45"}] | ["A"] | null | $$\mathop \cup \limits_{i = 1}^{50} {X_i} = $$ X<sub>1</sub>, X<sub>2</sub>,....., X<sub>50</sub> = 50 sets. Given each sets having 10 elements.
<br><br>So total elements = 50 $$ \times $$ 10
<br><br>$$\mathop \cup \limits_{i = 1}^n {Y_i} =$$ $$ Y<sub>1</sub>, Y<sub>2</sub>,....., Y<sub>n</sub> = n sets. Given each s... | mcq | jee-main-2020-online-4th-september-evening-slot |
3iz5nbJtSww5lL4Sg9jgy2xukfxgtuou | maths | sets-and-relations | number-of-sets-and-relations | Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more
than the total number of subsets of B, then the value of m.n is ______. | [] | null | 28 | Number of subsets of A = 2<sup>m</sup>
<br><br>Number of subsets of B = 2<sup>n</sup>
<br><br>Given = 2<sup>m</sup> – 2<sup>n</sup>
= 112
<br><br>$$ \therefore $$ m = 7, n = 4 (2<sup>7</sup> – 2<sup>4</sup>
= 112)
<br><br>$$ \therefore $$ m $$ \times $$ n = 7 $$ \times $$ 4 = 28 | integer | jee-main-2020-online-6th-september-morning-slot |
iQP8FXc5rG0uqHHQlB1klrjjed5 | maths | sets-and-relations | number-of-sets-and-relations | Let A = {n $$ \in $$ N: n is a 3-digit number}<br/><br>
B = {9k + 2: k $$ \in $$ N}
<br/><br>and C = {9k + $$l$$: k $$ \in $$ N} for some $$l ( 0 < l < 9)$$<br/><br>
If the sum of all the elements of the set A $$ \cap $$ (B $$ \cup $$ C) is 274 $$ \times $$ 400, then $$l$$ is equal to ________.</br></br><... | [] | null | 5 | In this problem, we're dealing with 3-digit numbers in set $A$, and subsets $B$ and $C$ which represent numbers of specific forms.
<br/><br/>1. First, we consider the numbers of the form $9k + 2$ (Set $B$) within the 3-digit range, which starts at 101 and ends at 992.
<br/><br/>2. We calculate the sum of these number... | integer | jee-main-2021-online-24th-february-morning-slot |
ElJhwaKkT1JdIO9sfl1kmhw02h6 | maths | sets-and-relations | number-of-sets-and-relations | The number of elements in the set {x $$\in$$ R : (|x| $$-$$ 3) |x + 4| = 6} is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["B"] | null | <b>Case 1 :</b><br><br>x $$ \le $$ $$-$$4<br><br>($$-$$x $$-$$ 3)($$-$$x $$-$$ 4) = 6<br><br>$$ \Rightarrow $$ (x + 3)(x + 4) = 6<br><br>$$ \Rightarrow $$ x<sup>2</sup> + 7x + 6 = 0<br><br>$$ \Rightarrow $$ x = $$-$$1 or $$-$$6<br><br>but x $$ \le $$ $$-$$4<br><br>x = $$-$$6<br><br><b>Case 2 :</b><br><br>x $$\in$$ ($$-... | mcq | jee-main-2021-online-16th-march-morning-shift |
Cy64jNuK5iXgalKmT51kmiwrnp6 | maths | sets-and-relations | number-of-sets-and-relations | Let A = {2, 3, 4, 5, ....., 30} and '$$ \simeq $$' be an equivalence relation on A $$\times$$ A, defined by (a, b) $$ \simeq $$ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "7"}] | ["D"] | null | ad = bc<br><br>(a, b) R (4, 3) $$ \Rightarrow $$ 3a = 4b<br><br>a = $${4 \over 3}$$b<br><br>b must be multiple of 3<br><br>b = {3, 6, 9 ..... 30}<br><br>(a, b) = {(4, 3), (8, 16), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)}<br><br>$$ \Rightarrow $$ 7 ordered pair | mcq | jee-main-2021-online-16th-march-evening-shift |
1krygqgjj | maths | sets-and-relations | number-of-sets-and-relations | Let A = {n $$\in$$ N | n<sup>2</sup> $$\le$$ n + 10,000}, B = {3k + 1 | k$$\in$$ N} an dC = {2k | k$$\in$$N}, then the sum of all the elements of the set A $$\cap$$(B $$-$$ C) is equal to _____________. | [] | null | 832 | B $$-$$ C $$ \equiv $$ {7, 13, 19, ......, 97, .......}<br><br>Now, n<sup>2</sup> $$-$$ n $$\le$$ 100 $$\times$$ 100<br><br>$$\Rightarrow$$ n(n $$-$$ 1) $$\le$$ 100 $$\times$$ 100<br><br>$$\Rightarrow$$ A = {1, 2, ......., 100}.<br><br>So, A$$\cap$$(B $$-$$ C) = {7, 13, 19, ......., 97}<br><br>Hence, sum = $${{16} \ove... | integer | jee-main-2021-online-27th-july-evening-shift |
1ktep0ng3 | maths | sets-and-relations | number-of-sets-and-relations | If A = {x $$\in$$ R : |x $$-$$ 2| > 1}, <br/>B = {x $$\in$$ R : $$\sqrt {{x^2} - 3} $$ > 1}, <br/>C = {x $$\in$$ R : |x $$-$$ 4| $$\ge$$ 2} and Z is the set of all integers, then the number of subsets of the <br/>set (A $$\cap$$ B $$\cap$$ C)<sup>c</sup> $$\cap$$ Z is ________________. | [] | null | 256 | A = ($$-$$$$\infty$$, 1) $$\cup$$ (3, $$\infty$$)<br><br>B = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ (2, $$\infty$$)<br><br>C = ($$-$$$$\infty$$, 2] $$\cup$$ [6, $$\infty$$)<br><br>So, A $$\cap$$ B $$\cap$$ C = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ [6, $$\infty$$)<br><br>z $$\cap$$ (A $$\cap$$ B $$\cap$$ C)' = {$$-$$2, $$-$$1,... | integer | jee-main-2021-online-27th-august-morning-shift |
1l567giit | maths | sets-and-relations | number-of-sets-and-relations | <p>Let R<sub>1</sub> and R<sub>2</sub> be relations on the set {1, 2, ......., 50} such that</p>
<p>R<sub>1</sub> = {(p, p<sup>n</sup>) : p is a prime and n $$\ge$$ 0 is an integer} and</p>
<p>R<sub>2</sub> = {(p, p<sup>n</sup>) : p is a prime and n = 0 or 1}.</p>
<p>Then, the number of elements in R<sub>1</sub> $$-$$ ... | [] | null | 8 | Given, ${R}_1=\left\{\left(p, p^n\right): p\right.$ is a Prime and $n \geq 0$ is an integer $\}$
<br/><br/>and, set $A=\{1,2,3 \ldots \ldots .50\}$
<br/><br/>$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47
<br/><br/>$\therefore$ We can calculate no. of elements in $\mathrm{... | integer | jee-main-2022-online-28th-june-morning-shift |
1l58apr54 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let A = {n $$\in$$ N : H.C.F. (n, 45) = 1} and</p>
<p>Let B = {2k : k $$\in$$ {1, 2, ......., 100}}. Then the sum of all the elements of A $$\cap$$ B is ____________.</p> | [] | null | 5264 | <p>Sum of all elements of A $$\cap$$ B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]</p>
<p>$$ = 2\left[ {{{100 \times 101} \over 2} - 3\left( {{{33 \times 34} \over 2}} \right) - 5\left( {{{20 \times 21} \over 2}} \right) + 15\left( {{{6 \times 7} \over 2}} \right)} \right]$$</p>
<p>$... | integer | jee-main-2022-online-26th-june-morning-shift |
1l58aqy4b | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\min \,\{ i,j\} } } $$ and $$B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \,\{ i,j\} } } $$. Then A + B is equal to _____________.</p> | [] | null | 1100 | <p>$$\sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\{ i,\,j\} } } $$</p>
<p>= {1, 1} {1, 2} {1, 3} ..... {1, 10}</p>
<p>{2, 1} {2, 2} {2, 3} ..... {2, 10}</p>
<p>{3, 1} {3, 2} {3, 3} ..... {3, 10}</p>
<p>$$ \vdots $$</p>
<p>{10, 1} {10, 2} {10, 3} ..... {10, 10}</p>
<p>Now, $$A = \sum\limits_{i = 1}^{10} {\sum\li... | integer | jee-main-2022-online-26th-june-morning-shift |
1l5bb3057 | maths | sets-and-relations | number-of-sets-and-relations | <p>The sum of all the elements of the set $$\{ \alpha \in \{ 1,2,.....,100\} :HCF(\alpha ,24) = 1\} $$ is __________.</p> | [] | null | 1633 | <p>The numbers upto 24 which gives g.c.d. with 24 equals to 1 are 1, 5, 7, 11, 13, 17, 19 and 23.</p>
<p>Sum of these numbers = 96</p>
<p>There are four such blocks and a number 97 is there upto 100.</p>
<p>$$\therefore$$ Complete sum</p>
<p>= 96 + (24 $$\times$$ 8 + 96) + (48 $$\times$$ 8 + 96) + (72 $$\times$$ 8 + 96... | integer | jee-main-2022-online-24th-june-evening-shift |
1l6f35cin | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3,4,5,6,7\}$$. Define $$B=\{T \subseteq A$$ : either $$1 \notin T$$ or $$2 \in T\}$$ and $$C=\{T \subseteq A: T$$ the sum of all the elements of $$T$$ is a prime number $$\}$$. Then the number of elements in the set $$B \cup C$$ is ________________.</p> | [] | null | 107 | <p>$$\because$$ $$(B \cup C)' = B'\, \cap C'$$</p>
<p>B' is a set containing sub sets of A containing element 1 and not containing 2.</p>
<p>And C' is a set containing subsets of A whose sum of elements is not prime.</p>
<p>So, we need to calculate number of subsets of {3, 4, 5, 6, 7} whose sum of elements plus 1 is co... | integer | jee-main-2022-online-25th-july-evening-shift |
1l6hzials | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3,4,5,6,7\}$$ and $$B=\{3,6,7,9\}$$. Then the number of elements in the set $$\{C \subseteq A: C \cap B \neq \phi\}$$ is ___________.</p> | [] | null | 112 | <p>As C $$\cap$$ B $$\ne$$ $$\phi$$, c must be not be formed by {1, 2, 4, 5}</p>
<p>$$\therefore$$ Number of subsets of A = 2<sup>7</sup> = 128</p>
<p>and number of subsets formed by {1, 2, 4, 5} = 16</p>
<p>$$\therefore$$ Required no. of subsets = 2<sup>7</sup> $$-$$ 2<sup>4</sup> = 128 $$-$$ 16 = 112</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6p0j53w | maths | sets-and-relations | number-of-sets-and-relations | <p>Let R be a relation from the set $$\{1,2,3, \ldots, 60\}$$ to itself such that $$R=\{(a, b): b=p q$$, where $$p, q \geqslant 3$$ are prime numbers}. Then, the number of elements in R is :</p> | [{"identifier": "A", "content": "600"}, {"identifier": "B", "content": "660"}, {"identifier": "C", "content": "540"}, {"identifier": "D", "content": "720"}] | ["B"] | null | <p>We have a set S = {1, 2, 3, ..., 60}, and a relation R defined on the set S. An element (a, b) belongs to the relation R if and only if b can be expressed as the product of two prime numbers p and q, where both p and q are greater than or equal to 3.</p>
<p>In terms of number theory, prime numbers are integers great... | mcq | jee-main-2022-online-29th-july-morning-shift |
1l6p3rlat | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$S=\{4,6,9\}$$ and $$T=\{9,10,11, \ldots, 1000\}$$. If $$A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in \mathbf{N}, a_{1}, a_{2}, a_{3}, \ldots, a_{k}\right.$$ $$\epsilon S\}$$, then the sum of all the elements in the set $$T-A$$ is equal to __________.</p> | [] | null | 11 | <p>Here $$S = \{ 4,6,9\} $$</p>
<p>And $$T = \{ 9,10,11,\,\,......,\,\,1000\} $$.</p>
<p>We have to find all numbers in the form of $$4x + 6y + 9z$$, where $$x,y,z \in \{ 0,1,2,\,......\} $$.</p>
<p>If a and b are coprime number then the least number from which all the number more than or equal to it can be express as ... | integer | jee-main-2022-online-29th-july-morning-shift |
1ldv2tpr1 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____________.</p> | [] | null | 43 | Elements of the type $3 \mathrm{k}=3$<br/><br/>
Elements of the type $3 \mathrm{k}+1=1,7,9$<br/><br/>
Elements of the type $3 \mathrm{k}+2=2,5,11$<br/><br/>
Subsets containing one element $S_1=1$<br/><br/>
Subsets containing two elements<br/><br/>
$$
S_2={ }^3 C_1 \times{ }^3 C_1=9
$$<br/><br/>
Subsets containing three... | integer | jee-main-2023-online-25th-january-morning-shift |
lgnzb2hr | maths | sets-and-relations | number-of-sets-and-relations | The number of elements in the set <br/><br/>$\left\{n \in \mathbb{N}: 10 \leq n \leq 100\right.$ and $3^{n}-3$ is a multiple of 7$\}$ is ___________. | [] | null | 15 | To determine the number of elements in the given set, we need to find how many natural numbers $n$ between $10$ and $100$ (inclusive) satisfy the condition that $3^n - 3$ is a multiple of $7$.
<br/><br/>Recall that for any integers $a$ and $b$, $a$ is a multiple of $b$ if there exists an integer $k$ such that $a = bk$... | integer | jee-main-2023-online-15th-april-morning-shift |
lgnzfh40 | maths | sets-and-relations | number-of-sets-and-relations | Let $A=\{1,2,3,4\}$ and $\mathrm{R}$ be a relation on the set $A \times A$ defined by <br/><br/>$R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$. Then the number of elements in $\mathrm{R}$ is ____________. | [] | null | 6 | $$
2a + 3b = 4c + 5d
$$
<br/><br/>
Given A = {1, 2, 3, 4}, the maximum value of $2a + 3b$ is 20, when (a, b) = (4, 4), and the minimum value of $4c + 5d$ is 9, when (c, d) = (1, 1). Therefore, the possible values for $2a + 3b = 4c + 5d$ are 9, 13, 14, 17, 18, and 19.
<br/><br/>
Now, let's find the combinations of (a, b... | integer | jee-main-2023-online-15th-april-morning-shift |
1lgsumb17 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$\mathrm{A}=\{1,3,4,6,9\}$$ and $$\mathrm{B}=\{2,4,5,8,10\}$$. Let $$\mathrm{R}$$ be a relation defined on $$\mathrm{A} \times \mathrm{B}$$ such that $$\mathrm{R}=\left\{\left(\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)\right): a_{1} \leq b_{2}\right.$$ and $$\left.b_{1} \leq a_{2}\right\}$$. Then the n... | [{"identifier": "A", "content": "180"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "52"}, {"identifier": "D", "content": "160"}] | ["D"] | null | Given that the sets are $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$, for the relation $\mathrm{R}$ on the set $A \times B$, we need to find the combinations of pairs that satisfy the conditions $a_1 \leq b_2$ and $b_1 \leq a_2$.
<br/><br/>We find the number of combinations by considering the possible values ... | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgvqdfo3 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$\mathrm{A}=\{2,3,4\}$$ and $$\mathrm{B}=\{8,9,12\}$$. Then the number of elements in the relation
$$\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right.$$ divides $$\mathrm{b}_{2}$$ and $$\mathrm{a}_{2}$$ divides $$\le... | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "12"}] | ["C"] | null |
<p>Given sets :
<br/>$ A = {2,3,4} $
<br/>$ B = {8,9,12} $</p>
<p>We want to find the number of elements of the form $( (a_1, b_1), (a_2, b_2) )$ such that :</p>
<ol>
<li>$ a_1 $ divides $ b_2 $</li>
<li>$ a_2 $ divides $ b_1 $</li>
</ol>
<p>For the first condition :
<br/>$ a_1 $ divides $ b_2 $
<br/>Given $ a_1 \in A... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxwachw | maths | sets-and-relations | number-of-sets-and-relations | <p>The number of elements in the set $$\{ n \in Z:|{n^2} - 10n + 19| < 6\} $$ is _________.</p> | [] | null | 6 | Given, $\left|n^2-10 n+19\right|<6$
<br/><br/>$\Rightarrow-6 < n^2-10 n+19 < 6$
<br/><br/>Take, $-6 < n^2-10 n+19$ and $n^2-10 n+19 < 6$
<br/><br/>$$
\begin{array}{ll}
\Rightarrow n^2-10 n+25 > 0 & \text { and }\quad n^2-10 n+13 < 0 \\\\
\Rightarrow(n-5)^2 > 0 & \text { and } n=\frac{10 \pm \sqrt{100-52}}{2}<0
\end{a... | integer | jee-main-2023-online-10th-april-morning-shift |
1lh23u6jh | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$\mathrm{A}=\{1,2,3,4, \ldots ., 10\}$$ and $$\mathrm{B}=\{0,1,2,3,4\}$$. The number of elements in the relation $$R=\left\{(a, b) \in A \times A: 2(a-b)^{2}+3(a-b) \in B\right\}$$ is ___________.</p> | [] | null | 18 | <p>Given sets :
<br/><br/>A={1,2,3,4, ............,10}
<br/><br/> B={0,1,2,3,4}
<br/><br/>We are looking for pairs $(a,b) \in A \times A$ such that :
<br/><br/>$ 2(a-b)^2 + 3(a-b) \in B $</p>
<p>Let's break down the relation :</p>
<p><strong>Case 1 :</strong> $ a-b = 0 $
<br/><br/>$ 2(a-b)^2 + 3(a-b) = 0 $
<br/><br... | integer | jee-main-2023-online-6th-april-morning-shift |
lsaqc1qh | maths | sets-and-relations | number-of-sets-and-relations | Let $A=\{1,2,3, \ldots, 20\}$. Let $R_1$ and $R_2$ two relation on $A$ such that
<br/><br/>$R_1=\{(a, b): b$ is divisible by $a\}$
<br/><br/>$R_2=\{(a, b): a$ is an integral multiple of $b\}$.
<br/><br/>Then, number of elements in $R_1-R_2$ is equal to _____________. | [] | null | 46 | <p>To determine the number of elements in $R_1 - R_2$, let's first articulate the meaning of both relations on set $A = \{1, 2, 3, \ldots, 20\}$:</p><p>$R_1$ includes pairs $(a, b)$ where $b$ is divisible by $a$. This includes pairs like $(1,1), (1,2), \ldots, (1,20)$ for $1$; similar series for $2$ up to $(2,20)$ (exc... | integer | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lscn4dk2 | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A$$ and $$B$$ be two finite sets with $$m$$ and $$n$$ elements respectively. The total number of subsets of the set $$A$$ is 56 more than the total number of subsets of $$B$$. Then the distance of the point $$P(m, n)$$ from the point $$Q(-2,-3)$$ is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["B"] | null | <p>$$\begin{aligned}
& 2^{\mathrm{m}}-2^{\mathrm{n}}=56 \\
& 2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 \\
& 2^{\mathrm{n}}=2^3 \text { and } 2^{\mathrm{m}-\mathrm{n}}-1=7 \\
& \Rightarrow \mathrm{n}=3 \text { and } 2^{\mathrm{m}-\mathrm{n}}=8 \\
& \Rightarrow \mathrm{n}=3 \text { and } \mathrm... | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd55mch | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3, \ldots \ldots \ldots \ldots, 100\}$$. Let $$R$$ be a relation on $$\mathrm{A}$$ defined by $$(x, y) \in R$$ if and only if $$2 x=3 y$$. Let $$R_1$$ be a symmetric relation on $$A$$ such that $$R \subset R_1$$ and the number of elements in $$R_1$$ is $$\mathrm{n}$$. Then, the minimum value of $$\math... | [] | null | 66 | <p>$$\begin{aligned}
& \mathrm{R}=\{(3,2),(6,4),(9,6),(12,8), \ldots \ldots \ldots .(99,66)\} \\
& \mathrm{n}(\mathrm{R})=33 \\
& \therefore 66
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse5zdng | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{1,2,3,4\}$$ and $$R=\{(1,2),(2,3),(1,4)\}$$ be a relation on $$\mathrm{A}$$. Let $$\mathrm{S}$$ be the equivalence relation on $$\mathrm{A}$$ such that $$R \subset S$$ and the number of elements in $$\mathrm{S}$$ is $$\mathrm{n}$$. Then, the minimum value of $$n$$ is __________.</p> | [] | null | 16 | $$
\begin{aligned}
& A=\{1,2,3,4\} \\\\
& R=\{(1,2),(2,3),(1,4)\}
\end{aligned}
$$
<br/><br/>$S$ is equivalence
for $R < S$ and reflexive
<br/><br/>$$
\{(1,1),(2,2),(3,3),(4,4)\}
$$
<br/><br/>for symmetric
<br/><br/>$$
\{(2,1),(4,1),(3,2)\}
$$
<br/><br/>for transitive
<br/><br/>$$
\{(1,3),(3,1),(4,2),(2,4)\}
$$
<br/><... | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsfkhne6 | maths | sets-and-relations | number-of-sets-and-relations | <p>If R is the smallest equivalence relation on the set $$\{1,2,3,4\}$$ such that $$\{(1,2),(1,3)\} \subset \mathrm{R}$$, then the number of elements in $$\mathrm{R}$$ is __________.</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "8"}] | ["B"] | null | <p>Given set $$\{1,2,3,4\}$$</p>
Minimum order pairs are</p>
<p>$$(1,1),(2,2),(3,3),(4,4),(3,1),(2,1),(2,3),(3,2),(1,3),(1,2)$$</p>
<p>Thus no. of elements $$=10$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg59rbo | maths | sets-and-relations | number-of-sets-and-relations | <p>The number of symmetric relations defined on the set $$\{1,2,3,4\}$$ which are not reflexive is _________.</p> | [] | null | 960 | <p>To find the number of symmetric relations on the set $$\{1,2,3,4\}$$ that are not reflexive, we first calculate the total number of symmetric relations and then subtract the count of those that are both symmetric and reflexive.</p><p>A symmetric relation involves pairs where if a pair (x, y) is in the relation, then... | integer | jee-main-2024-online-30th-january-evening-shift |
luy9cl8m | maths | sets-and-relations | number-of-sets-and-relations | <p>Let $$A=\{2,3,6,7\}$$ and $$B=\{4,5,6,8\}$$. Let $$R$$ be a relation defined on $$A \times B$$ by $$(a_1, b_1) R(a_2, b_2)$$ if and only if $$a_1+a_2=b_1+b_2$$. Then the number of elements in $$R$$ is __________.</p> | [] | null | 25 | <p>To find the number of elements in the relation $$R$$ defined on $$A \times B$$, we need to determine all pairs $$((a_1, b_1), (a_2, b_2))$$ such that $$a_1 + a_2 = b_1 + b_2$$, where $$a_1, a_2 \in A$$ and $$b_1, b_2 \in B$$.</p>
<p>First, consider all possible sums of pairs from set $$A$$ and set $$B$$.</p>
<p>Po... | integer | jee-main-2024-online-9th-april-morning-shift |
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