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1lsg57scj | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the parabola $$(y-2)^2=x-1$$, the line $$x-2 y+4=0$$ and the positive coordinate axes is _________.</p> | [] | null | 5 | <p>Solving the equations</p>
<p>$$\begin{array}{r}
(y-2)^2=x-1 \text { and } x-2 y+4=0 \\
x=2(y-2)
\end{array}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxplac/f4bdc645-97cb-4588-be1e-e674c4cd7e92/51166c40-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxplad.png?format=png" data-orsr... | integer | jee-main-2024-online-30th-january-evening-shift | 4,957 |
luxweb0r | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in square units) of the region enclosed by the ellipse $$x^2+3 y^2=18$$ in the first quadrant below the line $$y=x$$ is</p> | [{"identifier": "A", "content": "$$\\sqrt{3} \\pi+1$$\n"}, {"identifier": "B", "content": "$$\\sqrt{3} \\pi$$\n"}, {"identifier": "C", "content": "$$\\sqrt{3} \\pi-\\frac{3}{4}$$\n"}, {"identifier": "D", "content": "$$\\sqrt{3} \\pi+\\frac{3}{4}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1i51n7/b27b05e5-86d3-44de-bbce-888961f304b0/435f7e30-0f40-11ef-a754-b58ce0eab988/file-1lw1i51n8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1i51n7/b27b05e5-86d3-44de-bbce-888961f304b0/435f7e30-0f40-11ef-a754-b58ce0eab988... | mcq | jee-main-2024-online-9th-april-evening-shift | 4,959 |
luy6z4gc | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The parabola $$y^2=4 x$$ divides the area of the circle $$x^2+y^2=5$$ in two parts. The area of the smaller part is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}+5 \\sin ^{-1}\\left(\\frac{2}{\\sqrt{5}}\\right)$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{3}+\\sqrt{5} \\sin ^{-1}\\left(\\frac{2}{\\sqrt{5}}\\right)$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{3}+5 \\sin ^{-1}\\left(\\frac{2}{\\sqrt{5}}\\right)$$\n"}, {"... | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw36wee0/c97c824c-882e-4206-aafc-00ef66493663/dfd02e70-102d-11ef-a47a-5f0284b5aece/file-1lw36wee1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw36wee0/c97c824c-882e-4206-aafc-00ef66493663/dfd02e70-102d-11ef-a47a-5f0284b5aece... | mcq | jee-main-2024-online-9th-april-morning-shift | 4,960 |
lv0vxbvp | maths | area-under-the-curves | area-bounded-between-the-curves | <p>One of the points of intersection of the curves $$y=1+3 x-2 x^2$$ and $$y=\frac{1}{x}$$ is $$\left(\frac{1}{2}, 2\right)$$. Let the area of the region enclosed by these curves be $$\frac{1}{24}(l \sqrt{5}+\mathrm{m})-\mathrm{n} \log _{\mathrm{e}}(1+\sqrt{5})$$, where $$l, \mathrm{~m}, \mathrm{n} \in \mathbf{N}$$. Th... | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "29"}, {"identifier": "C", "content": "31"}, {"identifier": "D", "content": "32"}] | ["A"] | null | <p>Solving curves $$y=1+3 x-2 x^2 ~\& ~y=\frac{1}{x}$$</p>
<p>$$\begin{aligned}
& 2 x^3-3 x^2-x+1=0 \\
& \Rightarrow \quad(2 x-1)\left(x^2-x-1\right)=0 \\
& \Rightarrow \quad x=\frac{1}{2}, x=\frac{1 \pm \sqrt{5}}{2}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/... | mcq | jee-main-2024-online-4th-april-morning-shift | 4,961 |
lv2er3kw | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in sq. units) of the region described by $$
\left\{(x, y): y^2 \leq 2 x \text {, and } y \geq 4 x-1\right\}
$$ is</p> | [{"identifier": "A", "content": "$$\\frac{9}{32}$$\n"}, {"identifier": "B", "content": "$$\\frac{11}{12}$$\n"}, {"identifier": "C", "content": "$$\\frac{8}{9}$$\n"}, {"identifier": "D", "content": "$$\\frac{11}{32}$$"}] | ["A"] | null | <p>$$\text { Area }=\int_\limits{-\frac{1}{2}}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhbnwbo/8e96ed5b-9608-42df-8bc6-dc691e8b1a15/7a1fa240-17f3-11ef-b996-7dd2f40f5f82/file-1lwhbnwbp.png?format=png" data-orsrc="https://app-content.cdn.exa... | mcq | jee-main-2024-online-4th-april-evening-shift | 4,962 |
lv3ve4fr | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region in the first quadrant inside the circle $$x^2+y^2=8$$ and outside the parabola $$y^2=2 x$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{2}-\\frac{1}{3}$$\n"}, {"identifier": "B", "content": "$$\\pi-\\frac{1}{3}$$\n"}, {"identifier": "C", "content": "$$\\pi-\\frac{2}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{2}-\\frac{2}{3}$$"}] | ["C"] | null | <p>We have, $$x^2+y^2=8$$ and $$y^2=2 x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4imme9/55025775-792e-443b-a53e-cd7d760a7686/861b5510-10e8-11ef-8bc0-edb751127fbd/file-1lw4immea.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4imme9/55025775-792e-443b-a53e-cd7... | mcq | jee-main-2024-online-8th-april-evening-shift | 4,963 |
lv5gs18q | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area of the region enclosed by the curve $$y=\min \{\sin x, \cos x\}$$ and the $$x$$ axis between $$x=-\pi$$ to $$x=\pi$$ be $$A$$. Then $$A^2$$ is equal to __________.</p> | [] | null | 16 | <p>$$y=f(x)=\min \{\sin x, \cos x\}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8v4grs/3efd55a7-6040-4108-bf47-c4327f08c8f1/86796680-134c-11ef-8bbe-1b4949638519/file-1lw8v4grt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8v4grs/3efd55a7-6040-4108-bf47-c4327f0... | integer | jee-main-2024-online-8th-april-morning-shift | 4,964 |
lv7v3qu3 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the parabolas $$y=x^2-5 x$$ and $$y=7 x-x^2$$ is ________.</p> | [] | null | 198 | <p>$$\begin{aligned}
y=x^2-5 x, y=7 x-x^2 & \Rightarrow \quad x^2-5 x=7 x-x^2 \\
& \Rightarrow \quad x=0, x=6
\end{aligned}$$</p>
<p>$$\text { Area }=\int_\limits0^6\left[\left(7 x-x^2\right)-(x-5 x)\right] d x$$</p>
<p>$$=\int_\limits0^6\left(12 x-2 x^2\right) d x=6 x-\left.\frac{2 x^3}{3}\right|_0 ^6$$</p>
<p>$$=216-... | integer | jee-main-2024-online-5th-april-morning-shift | 4,965 |
lv9s20di | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed between the curves $$y=x|x|$$ and $$y=x-|x|$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{8}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{4}{3}$$"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>$$y=x|x|$$ & $$y=x-|x|$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweeutim/25a3e531-aad5-453f-b0c9-bc32519834c1/9c0ef0e0-1659-11ef-b26b-2993e9de41b8/file-1lweeutin.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweeutim/25a3e531-aad5-453f-b0c9-bc32519834c1/... | mcq | jee-main-2024-online-5th-april-evening-shift | 4,966 |
lvb294kx | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0<\mathrm{a}<1\right\}$$ is $$\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7}$$ then the value of $$7 \mathrm{a}-3$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-$$1"}] | ["D"] | null | <p>$$\left\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0< a<1\right\}$$</p>
<p>$$ \Rightarrow \int\limits_1^2 {\left( {{1 \over x} - {a \over {{x^2}}}} \right)dx = \left| {\ln |x| + {a \over x}} \right|_1^2} $$</p>
<p>$$\begin{aligned}
& \left(\ln 2+\frac{a}{2}\right)-(\ln 1+a)=\ln 2-\frac{a}{2} \\
... | mcq | jee-main-2024-online-6th-april-evening-shift | 4,967 |
lvc57b84 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area of the region enclosed by the curves $$y=3 x, 2 y=27-3 x$$ and $$y=3 x-x \sqrt{x}$$ be $$A$$. Then $$10 A$$ is equal to</p> | [{"identifier": "A", "content": "172"}, {"identifier": "B", "content": "154"}, {"identifier": "C", "content": "162"}, {"identifier": "D", "content": "184"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwcyt7aa/3fbcb7c5-4c8c-4aaa-876a-5ac80c7dce41/145c0910-158e-11ef-bb3a-95d759b5a950/file-1lwcyt7ab.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwcyt7aa/3fbcb7c5-4c8c-4aaa-876a-5ac80c7dce41/145c0910-158e-11ef-bb3a-95d759b5a950... | mcq | jee-main-2024-online-6th-april-morning-shift | 4,968 |
cZPz7mFQ5ox2xoIa | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area enclosed between the curve $$y = {\log _e}\left( {x + e} \right)$$ and the coordinate axes is : | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$2$$"}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$4$$"}] | ["A"] | null | The graph of the curve $$y = {\log _e}\left( {x + e} \right)$$ is as shown in the fig.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267502/exam_images/ssxskaas7h8ixpoi326x.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Area Under The Curves Question 130 English Exp... | mcq | aieee-2005 | 4,969 |
TRI2HT1xJbgKZHyz | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let $$f(x)$$ be a non - negative continuous function such that the area bounded by the curve $$y=f(x),$$ $$x$$-axis and the ordinates $$x = {\pi \over 4}$$ and $$x = \beta > {\pi \over 4}$$ is $$\left( {\beta \sin \beta + {\pi \over 4}\cos \beta + \sqrt 2 \beta } \right).$$ Then $$f\left( {{\pi \over 2}} \ri... | [{"identifier": "A", "content": "$$\\left( {{\\pi \\over 4} + \\sqrt 2 - 1} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{\\pi \\over 4} - \\sqrt 2 + 1} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {1 - {\\pi \\over 4} - \\sqrt 2 } \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {1... | ["D"] | null | Given that
<br><br>$$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta $$
<br><br>Differentiating $$w.r.t$$ $$\beta $$
<br><br>$$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2 $$
<br><br>$$f\left( {{\pi \... | mcq | aieee-2005 | 4,970 |
55bGpDRnxxSxO83e | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let g(x) = cosx<sup>2</sup>, f(x) = $$\sqrt x $$ and $$\alpha ,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation 18x<sup>2</sup> - 9$$\pi $$x + $${\pi ^2}$$ = 0. Then the area (in sq. units) bounded by the curve
<br/>y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 ... | [{"identifier": "A", "content": "$${1 \\over 2}\\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 2}\\left( {\\sqrt 3 - 1} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 2}\\left( {\\sqrt 3 + 1} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {\\sqrt 3 - ... | ["B"] | null | Given quadratic equation,
<br><br>$$18{x^2} - 9\pi x + {\pi ^2} = 0$$
<br><br>$$ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$$
<br><br>$$ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$$
<br><br>$$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \r... | mcq | jee-main-2018-offline | 4,971 |
K7S3uSiqOCZZI0fYX9H7G | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area of the region
<br/><br/>A = {(x, y) : 0 $$ \le $$ y $$ \le $$x |x| + 1 and $$-$$1 $$ \le $$ x $$ \le $$1} in sq. units, is : | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265519/exam_images/gatckweagely4rxump82.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Area Under The Curves Question 110 English Explanation">
... | mcq | jee-main-2019-online-9th-january-evening-slot | 4,972 |
bauLwyzOpatnYqvKltGUV | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let S($$\alpha $$) = {(x, y) : y<sup>2</sup>
$$ \le $$ x, 0 $$ \le $$ x $$ \le $$ $$\alpha $$} and A($$\alpha $$)
is area of the region S($$\alpha $$). If for a $$\lambda $$, 0 < $$\lambda $$ < 4,
A($$\lambda $$) : A(4) = 2 : 5, then $$\lambda $$ equals | [{"identifier": "A", "content": "$$2{\\left( {{4 \\over {25}}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "B", "content": "$$2{\\left( {{2 \\over {5}}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "C", "content": "$$4{\\left( {{4 \\over {25}}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "D", "content": "$$4{\\left( ... | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267162/exam_images/qcblehdbr1lbbocet0ni.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263555/exam_images/rfv4vtuykmtleeqymrfe.webp"><img src="https://res.c... | mcq | jee-main-2019-online-8th-april-evening-slot | 4,974 |
aqpkRU5oUGyE2Sx1Fh1kmm2mc49 | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area bounded by the curve 4y<sup>2</sup> = x<sup>2</sup>(4 $$-$$ x)(x $$-$$ 2) is equal to : | [{"identifier": "A", "content": "$${\\pi \\over {16}}$$"}, {"identifier": "B", "content": "$${\\pi \\over {8}}$$"}, {"identifier": "C", "content": "$${3\\pi \\over {2}}$$"}, {"identifier": "D", "content": "$${3\\pi \\over {8}}$$"}] | ["C"] | null | Given,<br><br>4y<sup>2</sup> = x<sup>2</sup>(4 $$-$$ x)(x $$-$$ 2) ..... (1)<br><br>Here, Left hand side 4y<sup>2</sup> is always positive. So Right hand side should also be positive.<br><br>In x$$\in$$ [2, 4] Right hand side is positive.<br><br>By putting y = $$-$$y in equation (1), equation remains same. So, graph is... | mcq | jee-main-2021-online-18th-march-evening-shift | 4,976 |
1ktkeocfd | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = $${3 \over 2}$$ and the curve y = 1 + 4x $$-$$ x<sup>2</sup>, then 12 m is equal to _____________. | [] | null | 26 | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxvqwyud/3b08f599-748b-4dc3-95b3-f0ac0c10ea2e/861eb060-6af6-11ec-b350-33e20cd86462/file-1kxvqwyue.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxvqwyud/3b08f599-748b-4dc3-95b3-f0ac0c10ea2e/861eb060-6af6-11ec-b350-33e20cd8646... | integer | jee-main-2021-online-31st-august-evening-shift | 4,977 |
1l55hj2ov | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | <p>The area of the bounded region enclosed by the curve <br/><br/>$$y = 3 - \left| {x - {1 \over 2}} \right| - |x + 1|$$ and the x-axis is :</p> | [{"identifier": "A", "content": "$${9 \\over 4}$$"}, {"identifier": "B", "content": "$${45 \\over 16}$$"}, {"identifier": "C", "content": "$${27 \\over 8}$$"}, {"identifier": "D", "content": "$${63 \\over 16}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l99thyns/e3d7926f-e2f7-4b54-82b0-67049b426481/fa7cf380-4c79-11ed-b94d-45a8040c2a81/file-1l99thynt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l99thyns/e3d7926f-e2f7-4b54-82b0-67049b426481/fa7cf380-4c79-11ed-b94d-45a8040c2a81/fi... | mcq | jee-main-2022-online-28th-june-evening-shift | 4,978 |
5d6Ld5edVvT7r0SH | maths | binomial-theorem | binomial-theorem-for-any-index | The positive integer just greater than $${\left( {1 + 0.0001} \right)^{10000}}$$ is | [{"identifier": "A", "content": "4 "}, {"identifier": "B", "content": "5 "}, {"identifier": "C", "content": "2 "}, {"identifier": "D", "content": "3 "}] | ["D"] | null | $${\left( {1 + 0.0001} \right)^{10000}}$$ = $${\left( {1 + {1 \over {{{10}^4}}}} \right)^{10000}}$$
<br><br>= 1 + 10000$${ \times {1 \over {{{10}^4}}}}$$ + $${{10000\left( {9999} \right)} \over {2!}} \times {\left( {{1 \over {{{10}^4}}}} \right)^2}$$+......$$\infty $$
<br><br>< 1 + 1 + $${1 \over {2!}}$$ + $${1 \ov... | mcq | aieee-2002 | 4,979 |
1krzmg688 | maths | binomial-theorem | binomial-theorem-for-any-index | The lowest integer which is greater <br/><br/>than $${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$ is ______________. | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["A"] | null | Let $$P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$<br><br>Let $$x = {10^{100}}$$<br><br>$$ \Rightarrow P = {\left( {1 + {1 \over x}} \right)^x}$$<br><br>$$ \Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\,
2 \,}} \right. }}.{1 \over {{x^2}}} + ... | mcq | jee-main-2021-online-25th-july-evening-shift | 4,980 |
1lsgcji8h | maths | binomial-theorem | binomial-theorem-for-any-index | <p>$$\text { Number of integral terms in the expansion of }\left\{7^{\left(\frac{1}{2}\right)}+11^{\left(\frac{1}{6}\right)}\right\}^{824} \text { is equal to _________. }$$</p> | [] | null | 138 | <p>General term in expansion of $$\left((7)^{1 / 2}+(11)^{1 / 6}\right)^{824}$$ is $$\mathrm{t}_{\mathrm{r}+1}={ }^{824} \mathrm{C}_{\mathrm{r}}(7)^{\frac{824-\mathrm{r}}{2}}(11)^{\mathrm{r} / 6}$$</p>
<p>For integral term, $$r$$ must be multiple of 6.</p>
<p>Hence $$r=0,6,12, ....... 822$$</p> | integer | jee-main-2024-online-30th-january-morning-shift | 4,982 |
lv0vxd0y | maths | binomial-theorem | binomial-theorem-for-any-index | <p>The sum of all rational terms in the expansion of $$\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$$ is equal to :</p> | [{"identifier": "A", "content": "633"}, {"identifier": "B", "content": "6131"}, {"identifier": "C", "content": "3133"}, {"identifier": "D", "content": "931"}] | ["C"] | null | <p>$$\begin{aligned}
& T_{r+1}={ }^{15} \mathrm{C}_r\left(2^{1 / 5}\right)^{15-r}\left(5^{1 / 3}\right)^r \\
& ={ }^{15} C_r 5^{r / 3} 2^{\left(3-\frac{r}{5}\right)}
\end{aligned}$$</p>
<p>For rational terms,</p>
<p>$$\frac{r}{3}$$ and $$\frac{r}{5}$$ must be integer</p>
<p>3 and 5 divide $$r \Rightarrow 15$$ divides $... | mcq | jee-main-2024-online-4th-april-morning-shift | 4,983 |
PmsYoXcOZSnaCuzH | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The remainder left out when $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ is divided by 9 is : | [{"identifier": "A", "content": "2 "}, {"identifier": "B", "content": "7 "}, {"identifier": "C", "content": "8 "}, {"identifier": "D", "content": "0"}] | ["A"] | null | $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$
<br><br>= $${\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$$
<br><br>= $${\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}$$
<br><br> = $$\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)$$
<br> &nbs... | mcq | aieee-2009 | 4,984 |
njUyy7DYscXbuI3Jxazra | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If (27)<sup>999</sup> is divided by 7, then the remainder is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["D"] | null | We have,
<br><br>$${{{{\left( {27} \right)}^{999}}} \over 7}$$
<br><br>= $${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$$
<br><br>= $${{28\,\lambda - 1} \over 7}$$
<br><br>= $${{28\,\lambda - 7 + 7 - 1} \over \lambda }$$
<br><br>= $${{7\left( {4\lambda - 1} \right) + 6} \over 7}$$
<br><br>$$\therefore\,\,\,$$ Rema... | mcq | jee-main-2017-online-8th-april-morning-slot | 4,985 |
y6YAjngUyXiNUzKKBU7k9k2k5e343pn | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The greatest positive integer k, for which 49<sup>k</sup> + 1 is a factor of the sum <br/>49<sup>125</sup> + 49<sup>124</sup> + ..... + 49<sup>2</sup> + 49 + 1, is: | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "63"}, {"identifier": "D", "content": "65"}] | ["C"] | null | 1 + 49 + 49<sup>2</sup>
+ ..... + 49<sup>125</sup>
<br><br>sum of G.P. = $${{1.\left( {{{49}^{126}} - 1} \right)} \over {49 - 1}}$$
<br><br>= $${{\left( {{{49}^{63}} + 1} \right)\left( {{{49}^{63}} - 1} \right)} \over {48}}$$
<br><br>Also 49<sup>63</sup> - 1
<br><br>= (1 + 48)<sup>63</sup> - 1
<br><br>= [<sup>63</sup>... | mcq | jee-main-2020-online-7th-january-morning-slot | 4,986 |
aFW8brFlz4zwZsvzfZ1klt9oix2 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)<sup>2022</sup> is divided by 8 is __________. | [] | null | 1 | Let x = 4k + 3<br><br>(2020 + x)<sup>2022</sup><br><br>= (2020 + 4k + 3)<sup>2022</sup><br><br>= (4(505) + 4k + 3)<sup>2022</sup>
<br><br>= (4P + 3)<sup>2022</sup><br><br>= (4P + 4 $$-$$ 1)<sup>2022</sup><br><br>= (4A $$-$$ 1)<sup>2022</sup><br><br><sup>2022</sup>C<sub>0</sub>(4A)<sup>0</sup>($$-$$1)<sup>2022</sup> + <... | integer | jee-main-2021-online-25th-february-evening-slot | 4,987 |
ofQvn9MZjWjzpi3QoA1klt9p75z | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The total number of two digit numbers 'n', such that 3<sup>n</sup> + 7<sup>n</sup> is a multiple of 10, is __________. | [] | null | 45 | $$ \because $$ $${7^n} = {(10 - 3)^n} = 10k + {( - 3)^n}$$<br><br>$${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$$<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264628/exam_images/ukvxzlftpfikzghxsw5c.webp"><source media="(max-width: 500px)" srcset="https:... | integer | jee-main-2021-online-25th-february-evening-slot | 4,988 |
5UlwtPdmE6ETPIebwf1kmjbieu8 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If (2021)<sup>3762</sup> is divided by 17, then the remainder is __________. | [] | null | 4 | 2021 = 17m - 2
<br><br>(2021)<sup>3762</sup> = (17m $$-$$ 2)<sup>3762</sup> = multiple of 17 + 2<sup>3762</sup><br><br>= 17$$\lambda$$ + 2<sup>2</sup> (2<sup>4</sup>)<sup>940</sup><br><br>= 17$$\lambda$$ + 4 (17 $$-$$ 1)<sup>940</sup><br><br>= 17$$\lambda$$ + 4 (17$$\mu$$ + 1)<br><br>= 17k + 4; (k $$\in$$ I)<br><br>$$ ... | integer | jee-main-2021-online-17th-march-morning-shift | 4,989 |
1l5ai3wa2 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>If $${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$, then the remainder when K is divided by 6 is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>$${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$</p>
<p>$$ \Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}$$... | mcq | jee-main-2022-online-25th-june-morning-shift | 4,992 |
1l5bb3q9d | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder on dividing 1 + 3 + 3<sup>2</sup> + 3<sup>3</sup> + ..... + 3<sup>2021</sup> by 50 is _________.</p> | [] | null | 4 | <p>Given,</p>
<p>$$1 + 3 + {3^2} + {3^3} + \,\,.....\,\, + \,\,{3^{2021}}$$</p>
<p>$$ = {3^0} + {3^1} + {3^2} + {3^3} + \,\,....\,\, + \,\,{3^{2021}}$$</p>
<p>This is a G.P with common ratio = 3</p>
<p>$$\therefore$$ Sum $$ = {{1({3^{2022}} - 1)} \over {3 - 1}}$$</p>
<p>$$ = {{{3^{2022}} - 1} \over 2}$$</p>
<p>$$ = {{{... | integer | jee-main-2022-online-24th-june-evening-shift | 4,993 |
1l6f0uwiw | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$(11)^{1011}+(1011)^{11}$$ is divided by 9 is</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <p>$${\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)$$</p>
<p>For $${\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)$$</p>
<p>$${2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}... | mcq | jee-main-2022-online-25th-july-evening-shift | 4,995 |
1l6jb84eh | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$(2021)^{2022}+(2022)^{2021}$$ is divided by 7 is</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <p>$${(2021)^{2022}} + {(2022)^{2021}}$$</p>
<p>$$ = {(7k - 2)^{2022}} + {(7{k_1} - 1)^{2021}}$$</p>
<p>$$ = {\left[ {{{(7k - 2)}^3}} \right]^{674}} + {(7{k_1})^{2021}} - 2021{(7{k_1})^{2020}}\, + \,....\, - 1$$</p>
<p>$$ = {(7{k_2} - 1)^{674}} + (7m - 1)$$</p>
<p>$$ = (7n + 1) + (7m - 1) = 7(m + n)$$ (multiple of 7)</... | mcq | jee-main-2022-online-27th-july-morning-shift | 4,996 |
1ldoo7i54 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder, when $$19^{200}+23^{200}$$ is divided by 49 , is ___________.</p> | [] | null | 29 | $19^{200}+23^{200}$
<br/><br/>= $(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201}$
<br/><br/>Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
<br/><br/>$=49 \lambda+470$
<br/><br/>$=49(\lambda+9)+29$
<br/><br/>$$ \therefore $$ Remainder $=29$ | integer | jee-main-2023-online-1st-february-morning-shift | 4,998 |
ldr0sc3g | maths | binomial-theorem | divisibility-concept-and-remainder-concept | $50^{\text {th }}$ root of a number $x$ is 12 and $50^{\text {th }}$ root of another number $y$ is 18 . Then the remainder obtained on dividing $(x+y)$ by 25 is ____________. | [] | null | 23 | <p>Given $${x^{{1 \over {50}}}} = 12 \Rightarrow x = {12^{50}}$$</p>
<p>$${y^{{1 \over {50}}}} = 18 \Rightarrow y = {18^{50}}$$</p>
<p>$$12\equiv13$$ (Mod 25)</p>
<p>$$12^2\equiv19$$ (Mod 25)</p>
<p>$$12^3\equiv-3$$ (Mod 25)</p>
<p>$$12^9\equiv-2$$ (Mod 25)</p>
<p>$$12^{10}\equiv-1$$ (Mod 25)</p>
<p>$$12^{50}\equiv-1$$... | integer | jee-main-2023-online-30th-january-evening-shift | 5,000 |
1ldu60obo | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when (2023)$$^{2023}$$ is divided by 35 is __________.</p> | [] | null | 7 | $$
\begin{aligned}
& (2023)^{2023} \\\\
& =(2030-7)^{2023} \\\\
& =(35 \mathrm{~K}-7)^{2023} \\\\
& ={ }^{2023} \mathrm{C}_0(35 \mathrm{~K})^{2023}(-7)^0+{ }^{2023} \mathrm{C}_1(35 \mathrm{~K})^{2022}(-7)+ \\\\
& \ldots . .+\ldots \ldots .+{ }^{2023} \mathrm{C}_{2023}(-7)^{2023} \\\\
& =35 \mathrm{~N}-7^{2023} \\\\
& \... | integer | jee-main-2023-online-25th-january-evening-shift | 5,001 |
1lgvpgm7l | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Let the number $$(22)^{2022}+(2022)^{22}$$ leave the remainder $$\alpha$$ when divided by 3 and $$\beta$$ when divided by 7. Then $$\left(\alpha^{2}+\beta^{2}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "5"}] | ["D"] | null | We have, $(22)^{2022}+(2022)^{22}$
<br/><br/>As 2022 is completely divisible by 3
<br/><br/>So, $(2022)^{22}$ is also divisible by 3
<br/><br/>$(22)^{2022}=(21+1)^{2022}=(3 \times 7+1)^{2022}=7 m+1$
<br/><br/>$\Rightarrow(22)^{2022}$ leave a remainder 1 , when divisible by 3 .
<br/><br/>$\therefore(22)^{2022}+(2022)^{2... | mcq | jee-main-2023-online-10th-april-evening-shift | 5,004 |
1lgyl43d6 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>$$25^{190}-19^{190}-8^{190}+2^{190}$$ is divisible by :</p> | [{"identifier": "A", "content": "14 but not by 34"}, {"identifier": "B", "content": "neither 14 nor 34"}, {"identifier": "C", "content": "both 14 and 34"}, {"identifier": "D", "content": "34 but not by 14"}] | ["D"] | null | The given expression is divisible by 6 and 17 .
<br/><br/>Also, $25^{190}-8^{190}$ is not divisible by 7
<br/><br/>but $19^{190}-2^{190}$ is divisible by 7 ,
<br/><br/>So, $25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by 34 but not by 14 . | mcq | jee-main-2023-online-8th-april-evening-shift | 5,005 |
1lh00oqex | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The largest natural number $$n$$ such that $$3^{n}$$ divides $$66 !$$ is ___________.</p> | [] | null | 31 | We have,
<br/><br/>$$
\begin{aligned}
& {\left[\frac{66}{3}\right]=22} \\\\
& {\left[\frac{66}{3^2}\right]=7} \\\\
& {\left[\frac{66}{3^3}\right]=2}
\end{aligned}
$$
<br/><br/>Highest powers of 3 is greater than 66. So, their g.i.f. is always 0.
<br/><br/>$\therefore$ Required natural number $=22+7+2=31$ | integer | jee-main-2023-online-8th-april-morning-shift | 5,006 |
1lh2y2hd7 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Among the statements :</p>
<p>(S1) : $$2023^{2022}-1999^{2022}$$ is divisible by 8</p>
<p>(S2) : $$13(13)^{n}-12 n-13$$ is divisible by 144 for infinitely many $$n \in \mathbb{N}$$</p> | [{"identifier": "A", "content": "both (S1) and (S2) are incorrect"}, {"identifier": "B", "content": "only (S1) is correct"}, {"identifier": "C", "content": "only (S2) is correct"}, {"identifier": "D", "content": "both (S1) and (S2) are correct"}] | ["D"] | null | We have, $S_1$ : $(2023)^{2022}-(1999)^{2022}$
<br/><br/>$$
\begin{aligned}
& =(1999+24)^{2022}-(1999)^{2022}={ }^{2022} C_0(1999)^{2022}(24)^0 \\\\
& +{ }^{2022} C_1(1999)^{2021}(24)^1+{ }^{2022} C_2(1999)^{2020}(24)^2 \\\\
& +\ldots-(1999)^{2022} \\\\
& ={ }^{2022} C_1(1999)^{2021}(24)+{ }^{2022} C_2(1999)^{2022}(24)... | mcq | jee-main-2023-online-6th-april-evening-shift | 5,007 |
jaoe38c1lsfl8i9p | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Remainder when $$64^{32^{32}}$$ is divided by 9 is equal to ________.</p> | [] | null | 1 | <p>Let $$32^{32}=\mathrm{t}$$</p>
<p>$$\begin{aligned}
& 64^{32^{32}}=64^t=8^{2 t}=(9-1)^{2 t} \\
& =9 \mathrm{k}+1
\end{aligned}$$</p>
<p>Hence remainder $$=1$$</p> | integer | jee-main-2024-online-29th-january-evening-shift | 5,008 |
luy9clej | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$428^{2024}$$ is divided by 21 is __________.</p> | [] | null | 1 | <p>$$\begin{aligned}
& 428=21 \times 20+8 \\
\Rightarrow \quad & (428)^{2024} \equiv(20 \times 21+8)^{2024} \equiv 8^{2024}(\bmod 21) \\
& 8^2=21 \times 3+1 \\
& 8^{2024}=(21 \times 3+1)^{1012} \\
\Rightarrow \quad & 8^{2024} \equiv(21 \times 3+1)^{1012}(\bmod 21) \\
& \equiv 1^{2012}(\bmod 21) \\
& 428^{2024} \equiv 1... | integer | jee-main-2024-online-9th-april-morning-shift | 5,009 |
AuQk01qTdDzVQGQ4 | maths | binomial-theorem | general-term | The number of integral terms in the expansion of $${\left( {\sqrt 3 + \root 8 \of 5 } \right)^{256}}$$ is | [{"identifier": "A", "content": "35 "}, {"identifier": "B", "content": "32 "}, {"identifier": "C", "content": "33 "}, {"identifier": "D", "content": "34 "}] | ["C"] | null | General term = $${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$$
<br>= $${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$$
<br><br>When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.
<br... | mcq | aieee-2003 | 5,011 |
0TpOHs7adCRBwUIj | maths | binomial-theorem | general-term | The coefficient of $${x^n}$$ in expansion of $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$ is | [{"identifier": "A", "content": "$${\\left( { - 1} \\right)^{n - 1}}n$$ "}, {"identifier": "B", "content": "$${\\left( { - 1} \\right)^n}\\left( {1 - n} \\right)$$ "}, {"identifier": "C", "content": "$${\\left( { - 1} \\right)^{n - 1}}{\\left( {n - 1} \\right)^2}$$ "}, {"identifier": "D", "content": "$$\\left( {n - 1} ... | ["B"] | null | Given $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$
<br><br>= $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$
<br><br>General term of $${\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$$
<br><br>$$\therefore$$ Term containing $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ ... | mcq | aieee-2004 | 5,012 |
rTXLqH9tIJS0LPKw | maths | binomial-theorem | general-term | If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation | [{"identifier": "A", "content": "$$a - b = 1$$ "}, {"identifier": "B", "content": "$$a + b = 1$$"}, {"identifier": "C", "content": "$${a \\over b} = 1$$ "}, {"identifier": "D", "content": "$$ab = 1$$ "}] | ["D"] | null | General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is T<sub>r+1</sub>.
<br><br>T<sub>r+1</sub> = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$
<br><br>= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}... | mcq | aieee-2005 | 5,013 |
cpKuiseKzcfD4PcY | maths | binomial-theorem | general-term | If the coefficients of r<sup>th</sup>, (r+1)<sup>th</sup>, and (r + 2)<sup>th</sup> terms in the binomial expansion of $${{\rm{(1 + y )}}^m}$$ are in A.P., then m and r satisfy the equation | [{"identifier": "A", "content": "$${m^2} - m(4r - 1) + 4\\,{r^2} - 2 = 0$$ "}, {"identifier": "B", "content": "$${m^2} - m(4r + 1) + 4\\,{r^2} + 2 = 0$$ "}, {"identifier": "C", "content": "$${m^2} - m(4r + 1) + 4\\,{r^2} - 2 = 0$$ "}, {"identifier": "D", "content": "$${m^2} - m(4r - 1) + 4\\,{r^2} + 2 = 0$$ "}] | ["C"] | null | Let r = 2
<br><br>$$\therefore$$ 2nd, 3rd and 4th terms are in AP.
<br><br>2nd term = T<sub>2</sub> = $${}^m{C_1}.y$$
<br><br>Coefficient of T<sub>2</sub> = $${}^m{C_1}$$
<br><br>3rd term = T<sub>3</sub> = $${}^m{C_2}.{y^2}$$
<br><br>Coefficient of T<sub>3</sub> = $${}^m{C_2}$$
<br><br>4th term = T<sub>4</sub> = $${}^m... | mcq | aieee-2005 | 5,014 |
3Hpkd6AMmJo5duCX | maths | binomial-theorem | general-term | The term independent of $$x$$ in expansion of
<br/> $${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$ is | [{"identifier": "A", "content": "4 "}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "210"}, {"identifier": "D", "content": "310"}] | ["C"] | null | $${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \... | mcq | jee-main-2013-offline | 5,016 |
XA8U3jV98E5XEVrRGyjEj | maths | binomial-theorem | general-term | If the coefficients of x<sup>−2</sup> and x<sup>−4</sup> in the expansion of $${\left( {{x^{{1 \over 3}}} + {1 \over {2{x^{{1 \over 3}}}}}} \right)^{18}},\left( {x > 0} \right),$$ are m and n respectively, then $${m \over n}$$ is equal to : | [{"identifier": "A", "content": "182"}, {"identifier": "B", "content": "$${4 \\over 5}$$"}, {"identifier": "C", "content": "$${5 \\over 4}$$"}, {"identifier": "D", "content": "27"}] | ["A"] | null | T<sub>r+1</sub> = <sup>18</sup>C<sub>r</sub> $${\left( {{x^{{1 \over 3}}}} \right)^{18 - r}}$$ . $${\left( {{1 \over {2{x^{{1 \over 3}}}}}} \right)^r}$$
<br><br>= <sup>18</sup>C<sub>r</sub> $${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 -... | mcq | jee-main-2016-online-10th-april-morning-slot | 5,018 |
MegnCy5y6WH4JxsjLHqWr | maths | binomial-theorem | general-term | The coefficient of x<sup>−5</sup> in the binomial expansion of
<br/><br/>$${\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},$$ where x $$ \ne $$ 0, 1, is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$ 4"}, {"identifier": "D", "content": "$$-$$ 1"}] | ["A"] | null | $${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \... | mcq | jee-main-2017-online-9th-april-morning-slot | 5,019 |
SVFbx781jSVe3AbPjftul | maths | binomial-theorem | general-term | The total number of irrational terms in the binomial expansion of (7<sup>1/5</sup> – 3<sup>1/10</sup>)<sup>60</sup> is : | [{"identifier": "A", "content": "54 "}, {"identifier": "B", "content": "55"}, {"identifier": "C", "content": "49"}, {"identifier": "D", "content": "48"}] | ["A"] | null | General term T<sub>r+1</sub> = <sup>60</sup><sup></sup>C<sub>r</sub>, $${7^{{{60 - r} \over 5}}}{3^{{r \over {10}}}}$$
<br><br>$$ \therefore $$ for rational term, r = 0, 10, 20, 30, 40, 50, 60
<br><br>$$ \Rightarrow $$ no of rational terms = 7
<br><br>$$ \therefore $$ number of irration... | mcq | jee-main-2019-online-12th-january-evening-slot | 5,020 |
toC7gwkJsULkHFwpDn3rsa0w2w9jxb4rh40 | maths | binomial-theorem | general-term | The term independent of x in the expansion of
<br/>$$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$ is equal to : | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "- 108"}, {"identifier": "C", "content": "- 36"}, {"identifier": "D", "content": "- 72"}] | ["C"] | null | Given expression = $$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$
<br><br>= $${1 \over {60}}{\left( {2{x^3} - {3 \over {{x^2}}}} \right)^6} - {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$
<br><br>So its general term is
<br><br>T<sub>r +... | mcq | jee-main-2019-online-12th-april-evening-slot | 5,021 |
yO8G2rHmbJfQTsDYPk3rsa0w2w9jx23emrp | maths | binomial-theorem | general-term | The smallest natural number n, such that the coefficient of x in the expansion of $${\left( {{x^2} + {1 \over {{x^3}}}} \right)^n}$$ is <sup>n</sup>C<sub>23</sub>, is : | [{"identifier": "A", "content": "23"}, {"identifier": "B", "content": "58"}, {"identifier": "C", "content": "38"}, {"identifier": "D", "content": "35"}] | ["C"] | null | General term<br><br>
$${T_{r + 1}} = {}^n{C_r}{x^{2n - 2r}}.{x^{ - 3r}}$$<br><br>
$$ \therefore $$ $$2n - 5r = 1 \Rightarrow 2n = 5r + 1$$<br><br>
$$ \therefore $$ $$r = {{2n - 1} \over 5}$$<br><br>
$$ \Rightarrow $$ Coefficient of x = $${}^n{C_{\left( {{{2n - 1} \over 5}} \right)}} = {}^n{C_{23}}$$<br><br>
$$ \Rightar... | mcq | jee-main-2019-online-10th-april-evening-slot | 5,023 |
3TQ0nlDrlHmOpomuWHpzD | maths | binomial-theorem | general-term | If the fourth term in the binomial expansion of $${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$$
(x > 0) is 20 × 8<sup>7</sup>, then a value of
x is : | [{"identifier": "A", "content": "8<sup>\u20132</sup>"}, {"identifier": "B", "content": "8<sup>2</sup>"}, {"identifier": "C", "content": "8<sup>3</sup>"}, {"identifier": "D", "content": "8"}] | ["B"] | null | $${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$$
<br><br>Given T<sub>4</sub> = 20 × 8<sup>7</sup>
<br><br>$$ \Rightarrow $$ $${}^6{C_3}{\left( {{2 \over x}} \right)^3}{\left( {{x^{{{\log }_8}x}}} \right)^3}$$ = 20 × 8<sup>7</sup>
<br><br>$$ \Rightarrow $$ 20$$ \times $$$${8 \over {{x^3}}} \times {x^{3{{\log }_... | mcq | jee-main-2019-online-9th-april-morning-slot | 5,024 |
8ItKCdunQp0EBnpdGh1Tw | maths | binomial-theorem | general-term | A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $${\left( {{2^{1/3}} + {1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}}$$ is : | [{"identifier": "A", "content": "1 : 2(6)<sup>1/3</sup>"}, {"identifier": "B", "content": "1 : 4(6)<sup>1/3</sup>"}, {"identifier": "C", "content": "2(36)<sup>1/3</sup> : 1"}, {"identifier": "D", "content": "4(36)<sup>1/3</sup> : 1"}] | ["D"] | null | $${{{T_5}} \over {T_5^1}} = {{{}^{10}{C_4}{{\left( {{2^{1/3}}} \right)}^{10 - 4}}{{\left( {{1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)}^4}} \over {{}^{10}{C_4}{{\left( {{1 \over {2\left( {{3^{1/3}}} \right)}}} \right)}^{10 - 4}}{{\left( {{2^{1/3}}} \right)}^4}}} = 4.{\left( {36} \right)^{1/3}}$$ | mcq | jee-main-2019-online-12th-january-morning-slot | 5,026 |
xJn72qJEiHbMt3IWxBsj5 | maths | binomial-theorem | general-term | The positive value of $$\lambda $$ for which the co-efficient of x<sup>2</sup>
in the expression x<sup>2</sup> $${\left( {\sqrt x + {\lambda \over {{x^2}}}} \right)^{10}}$$ is 720, is - | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$2\\sqrt 2 $$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\sqrt 5 $$"}] | ["A"] | null | The general term in the expansion of the binomial expression $(a+b)^n$ is
<br/><br/>$$
T_{r+1}={ }^n C_r a^{n-r} b^r
$$
<br/><br/>Therefore, the general term in the expansion of the binomial expression <br/><br/>$x^2\left(\sqrt{x}+\frac{\lambda}{x^2}\right)^{10}$ is
<br/><br/>$$
\begin{aligned}
T_{r+1} & =x^2\left({ }... | mcq | jee-main-2019-online-10th-january-evening-slot | 5,027 |
p3tVtM7wc6QwB2WNUeyvG | maths | binomial-theorem | general-term | If the third term in the binomial expansion <br/>of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is - | [{"identifier": "A", "content": "$$2\\sqrt 2 $$"}, {"identifier": "B", "content": "$$4\\sqrt 2 $$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["D"] | null | $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$
<br><br>$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$
<br><br>$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$
<br><br>$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$
<br><br>$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$
<br><br>$$ \Rightarrow 2... | mcq | jee-main-2019-online-10th-january-morning-slot | 5,028 |
mfa9gcoLNQtl4I0BUejgy2xukg0d0dyo | maths | binomial-theorem | general-term | If the constant term in the binomial expansion
of <br/>$${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$$ is 405, then |k| equals : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["A"] | null | $${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$$
<br><br>r<sup>th</sup> term of the expansion,
<br><br>T<sub>r+1</sub> = <sup>10</sup>C<sub>r</sub>$${\left( {\sqrt x } \right)^{10 - r}}{\left( {{{ - k} \over {{x^2}}}} \right)^r}$$
<br><br>= <sup>10</sup>C<sub>r</sub>.$${x^{{{10 - r} \over 2}}}.{\left( { - k} \r... | mcq | jee-main-2020-online-6th-september-evening-slot | 5,030 |
57I50P6FMvU7vioAmejgy2xukfjjs70u | maths | binomial-theorem | general-term | The natural number m, for which the coefficient of x in the binomial expansion of<br/><br/>
$${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$$ is 1540, is ............. | [] | null | 13 | General term,
<br><br>$${T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}$$<br><br>$$ \because $$ $${}^{22}{C_3} = {}^{22}{C_{19}} = 1540$$<br><br>$$ \therefore $$ $$r = 3\,or\,19$$<br><br>$$22m - mr - 2r = 1$$<br><br>$$m = {{2r + 1} \over {22 - 5}}$$<b... | integer | jee-main-2020-online-5th-september-morning-slot | 5,031 |
9JtYhkWW4qfFVXr41hjgy2xukf443l60 | maths | binomial-theorem | general-term | If the term independent of x in the expansion of
<br/>$${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$$ is k, then 18 k is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "11"}] | ["C"] | null | General term,
<br><br>$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}{x^2}} \right]^{9 - r}}{\left( { - {1 \over {3x}}} \right)^r}$$<br><br>$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}} \right]^{9 - r}}{\left( { - {1 \over 3}} \right)^r}{x^{18 - 3r}}$$<br><br>For independent of x <br><br>18 $$ - $$ 3r = 0 $$ \Rightarrow... | mcq | jee-main-2020-online-3rd-september-evening-slot | 5,032 |
dPu5jf9EUPJ9Gi2nwLjgy2xukf0wsoo1 | maths | binomial-theorem | general-term | If the number of integral terms in the expansion
<br/>of (3<sup>1/2</sup> + 5<sup>1/8</sup>)<sup>n</sup> is exactly 33, then the least value
of n is : | [{"identifier": "A", "content": "264"}, {"identifier": "B", "content": "256"}, {"identifier": "C", "content": "128"}, {"identifier": "D", "content": "248"}] | ["B"] | null | General term of the expression,<br><br>$${T_{r + 1}} = {}^n{C_r}{\left( {{3^{{1 \over 2}}}} \right)^{n - r}}{\left( {{5^{{1 \over 8}}}} \right)^r}$$<br><br>$$ = {}^n{C_r}{\left( 3 \right)^{{{n - r} \over 2}}}{\left( 5 \right)^{{r \over 8}}}$$<br><br>We will get integral term when $${{n - r} \over 2}$$ and $${r \over 8}... | mcq | jee-main-2020-online-3rd-september-morning-slot | 5,033 |
teMqBnHZQ2oEeeo2dBjgy2xukewmb3vi | maths | binomial-theorem | general-term | Let
$$\alpha $$ > 0,
$$\beta $$ > 0 be such that
<br/>$$\alpha $$<sup>3</sup> + $$\beta $$<sup>2</sup> = 4. If the
maximum value of the term independent of x in
<br/>the binomial expansion of
$${\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}}$$
is 10k,
<br/>then k is equal to : | [{"identifier": "A", "content": "176"}, {"identifier": "B", "content": "336"}, {"identifier": "C", "content": "352"}, {"identifier": "D", "content": "84"}] | ["B"] | null | General term
<br><br>T<sub>r + 1</sub> = <sup>10</sup>C<sub>r</sub>$${\alpha ^{10 - r}}.{\left( x \right)^{{{10 - r} \over 9}}}.{\beta ^r}{\left( x \right)^{ - {r \over 6}}}$$
<br><br>= <sup>10</sup>C<sub>r</sub>$${\alpha ^{10 - r}}{\beta ^r}.{\left( x \right)^{{{10 - r} \over 9} - {r \over 6}}}$$
<br><br>If T<sub>r + ... | mcq | jee-main-2020-online-2nd-september-morning-slot | 5,034 |
ClPz4ye6HjzLfomsWE7k9k2k5khuqun | maths | binomial-theorem | general-term | In the expansion of $${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$$, if $${\ell _1}$$ is
the least value of the term independent of x
when $${\pi \over 8} \le \theta \le {\pi \over 4}$$ and $${\ell _2}$$ is the least value of the
term independent of x when $${\pi \over {16}} \le \th... | [{"identifier": "A", "content": "8 : 1"}, {"identifier": "B", "content": "16 : 1"}, {"identifier": "C", "content": "1 : 8"}, {"identifier": "D", "content": "1 : 16"}] | ["B"] | null | T<sub>r + 1</sub> = <sup>16</sup>C<sub>r</sub>$${\left( {{x \over {\cos \theta }}} \right)^{16 - r}}{\left( {{1 \over {x\sin \theta }}} \right)^r}$$
<br><br>= <sup>16</sup>C<sub>r</sub>$${\left( x \right)^{16 - 2r}} \times {1 \over {{{\left( {\cos \theta } \right)}^{16 - r}}{{\left( {\sin \theta } \right)}^r}}}$$
<br><... | mcq | jee-main-2020-online-9th-january-evening-slot | 5,035 |
whKykbRukRrJZDlsAZ7k9k2k5hjr4v7 | maths | binomial-theorem | general-term | If $$\alpha $$ and $$\beta $$ be the coefficients of x<sup>4</sup> and x<sup>2</sup>
respectively in the expansion of<br/>
$${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}$$, then | [{"identifier": "A", "content": "$$\\alpha + \\beta = 60$$"}, {"identifier": "B", "content": "$$\\alpha - \\beta = 60$$"}, {"identifier": "C", "content": "$$\\alpha + \\beta = -30$$"}, {"identifier": "D", "content": "$$\\alpha - \\beta = -132$$"}] | ["D"] | null | (x+a)<sup>n </sup>+ (x – a)<sup>n</sup>
= 2(T<sub>1</sub>
+ T<sub>3</sub>
+ T<sub>5</sub>
+.....)
<br><br>$${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}$$
<br><br>= 2[T<sub>1</sub>
+ T<sub>3</sub>
+ T<sub>5</sub>
+ T<sub>7</sub>
]
<br><br>= 2[<sup>6</sup>C<sub>0</sub>... | mcq | jee-main-2020-online-8th-january-evening-slot | 5,036 |
Qc54t3Qr25N0Yctsjd7k9k2k5fiux85 | maths | binomial-theorem | general-term | The coefficient of x<sup>7</sup>
in the expression
<br/>(1 + x)<sup>10</sup> + x(1 + x)<sup>9</sup>
+ x<sup>2</sup>(1 + x)<sup>8</sup>
+ ......+ x<sup>10</sup> is: | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "330"}, {"identifier": "C", "content": "420"}, {"identifier": "D", "content": "210"}] | ["B"] | null | (1 + x)<sup>10</sup> + x(1 + x)<sup>9</sup>
+ x<sup>2</sup>(1 + x)<sup>8</sup>
+ ......+ x<sup>10</sup>
<br><br>This is a G.P where
<br><br>First term, a = (1 + x)<sup>10</sup>
<br><br>common ratio, r = $${x \over {1 + x}}$$
<br><br>Number of terms = 11
<br><br>Sum of G.P
<br><br>= $${{{{\left( {1 + x} \right)}^{10}}... | mcq | jee-main-2020-online-7th-january-evening-slot | 5,037 |
7j0Ea0CxBrRDVTtcVm1klug2yrr | maths | binomial-theorem | general-term | The maximum value of the term independent of 't' in the expansion <br/>of $${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$$ where x$$\in$$(0, 1) is : | [{"identifier": "A", "content": "$${{10!} \\over {\\sqrt 3 {{(5!)}^2}}}$$"}, {"identifier": "B", "content": "$${{2.10!} \\over {3\\sqrt 3 {{(5!)}^2}}}$$"}, {"identifier": "C", "content": "$${{10!} \\over {3{{(5!)}^2}}}$$"}, {"identifier": "D", "content": "$${{2.10!} \\over {3{{(5!)}^2}}}$$"}] | ["B"] | null | $${T_{r + 1}} = {}^{10}{C_r}{(t{x^{1/5}})^{10 - r}}{\left[ {{{{{(1 - x)}^{1/10}}} \over t}} \right]^r}$$<br><br>$$ = {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}$$<br><br>$$ \Rightarrow 10 - 2r = 0 \Rightarrow r = 5$$
<br><br>$$ \therefore $$ $${T_6} = {}^{10}{C_5} \time... | mcq | jee-main-2021-online-26th-february-morning-slot | 5,038 |
jHrylvcW7IqT8Wci3y1kmhx5zo4 | maths | binomial-theorem | general-term | If n is the number of irrational terms in the <br/>expansion of $${\left( {{3^{1/4}} + {5^{1/8}}} \right)^{60}}$$, then (n $$-$$ 1) is divisible by : | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "26"}] | ["D"] | null | $${T_{r + 1}} = {}^{60}{C_r}{\left( {{3^{1/4}}} \right)^{60 - r}}{\left( {{5^{1/8}}} \right)^r}$$<br><br>rational if $${{60 - r} \over 4},{r \over 8}$$, both are whole numbers, $$r \in \{ 0,1,2,......60\} $$<br><br>$${{60 - r} \over 4} \in W \Rightarrow r \in \{ 0,4,8,....60\} $$<br><br>and $${r \over 8} \in W \Rightar... | mcq | jee-main-2021-online-16th-march-morning-shift | 5,039 |
H3IeTt4KAIUgRCZ0P31kmjb2rhf | maths | binomial-theorem | general-term | If the fourth term in the expansion of $${(x + {x^{{{\log }_2}x}})^7}$$ is 4480, then the value of x where x$$\in$$N is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["D"] | null | T<sub>4</sub> = $${}^7{C_3}{x^4}{({x^{{{\log }_2}x}})^3} = 4480$$<br><br>$$ \Rightarrow 35{x^4}{({x^{{{\log }_2}x}})^3} = 4480$$<br><br>$$ \Rightarrow {x^4}{({x^{{{\log }_2}x}})^3} = 128$$<br><br>take log w.r.t. base 2 we get,
<br><br>$$4{\log _2}x + 3{\log _2}({x^{{{\log }_2}x}}) = {\log _2}128$$<br><br>Let $${\log _2... | mcq | jee-main-2021-online-17th-march-morning-shift | 5,040 |
3EvGB4JDbb7CRrEJIt1kmknn38u | maths | binomial-theorem | general-term | Let the coefficients of third, fourth and fifth terms in the expansion of $${\left( {x + {a \over {{x^2}}}} \right)^n},x \ne 0$$, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ___________. | [] | null | 4 | $${T_{r + 1}} = {n_{C_r}}{x^{n - r}}.{\left( {{a \over {{x^2}}}} \right)^r}$$<br><br>$$ = {}^n{C_r}{a^r}{x^{n - 3r}}$$<br><br>$${T_3} = {}^n{C_2}{a^2}{x^{n - 6}}$$, $${T_4} = {}^n{C_3}{a^3}{x^{n - 9}}$$, $${T_5} = {}^n{C_4}{a^4}{x^{n - 12}}$$<br><br>Now, $${{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{... | integer | jee-main-2021-online-17th-march-evening-shift | 5,041 |
1krq0nwdb | maths | binomial-theorem | general-term | The number of rational terms in the binomial expansion of $${\left( {{4^{{1 \over 4}}} + {5^{{1 \over 6}}}} \right)^{120}}$$ is _______________. | [] | null | 21 | $${\left( {{4^{{1 \over 4}}} + {5^{{1 \over 6}}}} \right)^{120}}$$<br><br>$${T_{r + 1}} = {}^{120}{C_r}{({2^{1/2}})^{120 - r}}{(5)^{r/6}}$$<br><br>for rational terms r = 6$$\lambda$$ <br><br>0 $$\le$$ r $$\le$$ 120<br><br>So total no of terms are 21. | integer | jee-main-2021-online-20th-july-morning-shift | 5,043 |
1krubh7ej | maths | binomial-theorem | general-term | If the constant term, in binomial expansion of $${\left( {2{x^r} + {1 \over {{x^2}}}} \right)^{10}}$$ is 180, then r is equal to __________________. | [] | null | 8 | $${\left( {2{x^r} + {1 \over {{x^2}}}} \right)^{10}}$$<br><br>General term $$ = {}^{10}{C_R}{(2{x^2})^{10 - R}}{x^{ - 2R}}$$<br><br>$$ \Rightarrow {2^{10 - R}}{}^{10}{C_R} = 180$$ ....... (1)<br><br>& (10 $$-$$ R)r $$-$$ 2R = 0<br><br>$$r = {{2R} \over {10 - R}}$$<br><br>$$r = {{2(R - 10)} \over {10 - R}} + {{20} \... | integer | jee-main-2021-online-22th-july-evening-shift | 5,044 |
1krw3ge50 | maths | binomial-theorem | general-term | The term independent of 'x' in the expansion of <br/>$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$, where x $$\ne$$ 0, 1 is equal to ______________. | [] | null | 210 | $${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \... | integer | jee-main-2021-online-25th-july-morning-shift | 5,045 |
1krz59pdx | maths | binomial-theorem | general-term | The sum of all those terms which are rational numbers in the <br/><br/>expansion of (2<sup>1/3</sup> + 3<sup>1/4</sup>)<sup>12</sup> is : | [{"identifier": "A", "content": "89"}, {"identifier": "B", "content": "27"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "43"}] | ["D"] | null | $${T_{r + 1}} = {}^{12}{C_r}{\left( {{2^{1/3}}} \right)^r}.{\left( {{3^{1/4}}} \right)^{12 - 4}}$$<br><br>T<sub>r + 1</sub> will be rational number when r = 0, 3, 6, 9, 12 & r = 0, 4, 8, 12<br><br>$$\Rightarrow$$ r = 0, 12<br><br>T<sub>1</sub> + T<sub>13</sub> = 1 $$\times$$ 3<sup>3</sup> + 1 $$\times$$ 2<sup>4</su... | mcq | jee-main-2021-online-25th-july-evening-shift | 5,047 |
1krzrliu5 | maths | binomial-theorem | general-term | If the co-efficient of x<sup>7</sup> and x<sup>8</sup> in the expansion of $${\left( {2 + {x \over 3}} \right)^n}$$ are equal, then the value of n is equal to _____________. | [] | null | 55 | $${}^n{C_7}{2^{n - 7}}{1 \over {{3^7}}} = {}^n{C_8}{2^{n - 8}}{1 \over {{3^8}}}$$<br><br>$$\Rightarrow$$ n $$-$$ 7 = 48 $$\Rightarrow$$ n = 55 | integer | jee-main-2021-online-25th-july-evening-shift | 5,049 |
1ktisyc5x | maths | binomial-theorem | general-term | If $$\left( {{{{3^6}} \over {{4^4}}}} \right)k$$ is the term, independent of x, in the binomial expansion of $${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$$, then k is equal to ___________. | [] | null | 55 | $${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$$<br><br>$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{x \over 4}} \right)^{12 - r}}{\left( {{{12} \over {{x^2}}}} \right)^r}$$<br><br>$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^{12 - r}}{\left( {12} \right)^r}\,.\,{(x)^{1... | integer | jee-main-2021-online-31st-august-morning-shift | 5,051 |
1l54uczwv | maths | binomial-theorem | general-term | <p>Let the coefficients of x<sup>$$-$$1</sup> and x<sup>$$-$$3</sup> in the expansion of $${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}},x > 0$$, be m and n respectively. If r is a positive integer such that $$m{n^2} = {}^{15}{C_r}\,.\,{2^r}$$, then the value of r is equal to __________.... | [] | null | 5 | <p>Given, Binomial expansion</p>
<p>$${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}}$$</p>
<p>$$\therefore$$ General Term</p>
<p>$${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {2{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( { - {1 \over {{x^{{1 \over 5}}}}}} \right)^r}$$</p>
<p>$$ = {}^{15}{C_r... | integer | jee-main-2022-online-29th-june-evening-shift | 5,052 |
1l55h6lp9 | maths | binomial-theorem | general-term | <p>The term independent of x in the expansion of <br/><br/>$$(1 - {x^2} + 3{x^3}){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}},\,x \ne 0$$ is :</p> | [{"identifier": "A", "content": "$${7 \\over {40}}$$"}, {"identifier": "B", "content": "$${33 \\over {200}}$$"}, {"identifier": "C", "content": "$${39 \\over {200}}$$"}, {"identifier": "D", "content": "$${11 \\over {50}}$$"}] | ["B"] | null | <p>General term of Binomial expansion $${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$$ is</p>
<p>$${T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}$$</p>
<p>$$ = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left... | mcq | jee-main-2022-online-28th-june-evening-shift | 5,053 |
1l567u2dw | maths | binomial-theorem | general-term | <p>The number of positive integers k such that the constant term in the binomial expansion of $${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$$, x $$\ne$$ 0 is 2<sup>8</sup> . l, where l is an odd integer, is ______________.</p> | [] | null | 2 | <p>Given Binomial expression is</p>
<p>$${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$$</p>
<p>General term,</p>
<p>$${T_{r + 1}} = {}^{12}{C_r}{(2{x^3})^r}\,.\,{\left( {{3 \over {{x^k}}}} \right)^{12 - r}}$$</p>
<p>$$ = \left( {{}^{12}{C_r}\,.\,{2^r}\,.\,{3^{12 - r}}} \right)\,.\,{x^{3r - 12k + kr}}$$</p>
<p>For... | integer | jee-main-2022-online-28th-june-morning-shift | 5,054 |
1l57p2f15 | maths | binomial-theorem | general-term | <p>If the coefficient of x<sup>10</sup> in the binomial expansion of $${\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt 5 } \over {{x^{{1 \over 3}}}}}} \right)^{60}}$$ is $${5^k}\,.\,l$$, where l, k $$\in$$ N and l is co-prime to 5, then k is equal to _____________.</p> | [] | null | 5 | <p>Given Binomial Expansion</p>
<p>$$ = {\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt x } \over {{5^{{1 \over 3}}}}}} \right)^{60}}$$</p>
<p>$$\therefore$$ General term</p>
<p>$${T_{r + 1}} = {}^{60}{C_r}\,.\,{\left( {{{{x^{1/2}}} \over {{5^{1/4}}}}} \right)^{60 - r}}\,.\,{\left( {{{{5^{1/2}}} \over {{x^{1/... | integer | jee-main-2022-online-27th-june-morning-shift | 5,055 |
1l59jwnfk | maths | binomial-theorem | general-term | <p>The coefficient of x<sup>101</sup> in the expression $${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}} + \,\,.....\,\, + \,\,{x^{500}}$$, x > 0, is</p> | [{"identifier": "A", "content": "<sup>501</sup>C<sub>101</sub> (5)<sup>399</sup>"}, {"identifier": "B", "content": "<sup>501</sup>C<sub>101</sub> (5)<sup>400</sup>"}, {"identifier": "C", "content": "<sup>501</sup>C<sub>100</sub> (5)<sup>400</sup>"}, {"identifier": "D", "content": "<sup>500</sup>C<sub>101</sub> (5)<sup>... | ["A"] | null | <p>Given,</p>
<p>$${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}}\,\, + $$ ...... $${x^{500}}$$</p>
<p>This is a G.P. with first term $${(5 + x)^{500}}$$</p>
<p>Common ratio $$ = {{x{{(5 + x)}^{499}}} \over {{{(5 + x)}^{500}}}} = {x \over {5 + x}}$$ and 501 terms present.</p>
<p>$$\therefore$$ Sum $$ = {{{{(... | mcq | jee-main-2022-online-25th-june-evening-shift | 5,056 |
1l59l7q3l | maths | binomial-theorem | general-term | <p>If the sum of the co-efficient of all the positive even powers of x in the binomial expansion of $${\left( {2{x^3} + {3 \over x}} \right)^{10}}$$ is $${5^{10}} - \beta \,.\,{3^9}$$, then $$\beta$$ is equal to ____________.</p> | [] | null | 83 | <p>Given, Binomial Expansion</p>
<p>$${\left( {2{x^3} + {3 \over x}} \right)^{10}}$$</p>
<P>General term</p>
<p>$${T_{r + 1}} = {}^{10}{C_r}\,.\,{(2{x^3})^{10 - r}}\,.\,{\left( {{3 \over x}} \right)^r}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{2^{10 - r}}\,.\,{3^r}\,.\,{x^{30 - 3r}}\,.\,{x^{ - r}}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,... | integer | jee-main-2022-online-25th-june-evening-shift | 5,057 |
1l5vzfdaf | maths | binomial-theorem | general-term | <p>For two positive real numbers a and b such that $${1 \over {{a^2}}} + {1 \over {{b^3}}} = 4$$, then minimum value of the constant term in the expansion of $${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$$ is :</p> | [{"identifier": "A", "content": "$${{105} \\over 2}$$"}, {"identifier": "B", "content": "$${{105} \\over 4}$$"}, {"identifier": "C", "content": "$${{105} \\over 8}$$"}, {"identifier": "D", "content": "$${{105} \\over 16}$$"}] | ["C"] | null | <p>Given, Binomial expansion,</p>
<p>$${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$$</p>
<p>General term,</p>
<p>$${T_{r + 1}} = {}^{10}{C_r}\,.\,{\left( {a{x^{{1 \over 8}}}} \right)^{10 - r}}\,.\,{\left( {b{x^{ - {1 \over {12}}}}} \right)^r}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^... | mcq | jee-main-2022-online-30th-june-morning-shift | 5,058 |
1l6dx5rjl | maths | binomial-theorem | general-term | <p>If the maximum value of the term independent of $$t$$ in the expansion of $$\left(\mathrm{t}^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0$$, is $$\mathrm{K}$$, then $$8 \mathrm{~K}$$ is equal to ____________.</p> | [] | null | 6006 | <p>General term of $${\left( {{t^2}{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{15}}$$ is</p>
<p>$${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {{t^2}{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( {{{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^r}$$</p>
<p>$$ = {}^{15}{C_r}\,.\,{t^{30 - 2r}}\,.... | integer | jee-main-2022-online-25th-july-morning-shift | 5,059 |
1l6klfrzl | maths | binomial-theorem | general-term | <p>Let for the $$9^{\text {th }}$$ term in the binomial expansion of $$(3+6 x)^{\mathrm{n}}$$, in the increasing powers of $$6 x$$, to be the greatest for $$x=\frac{3}{2}$$, the least value of $$\mathrm{n}$$ is $$\mathrm{n}_{0}$$. If $$\mathrm{k}$$ is the ratio of the coefficient of $$x^{6}$$ to the coefficient of $$x^... | [] | null | 24 | <p>$${(3 + 6x)^n} = {3^n}{(1 + 2x)^n}$$</p>
<p>If T<sub>9</sub> is numerically greatest term</p>
<p>$$\therefore$$ $${T_8} \le {T_9} \le {T_{10}}$$</p>
<p>$${}^n{C_7}{3^{n - 7}}{(6x)^7} \le {}^n{C_8}{3^{n - 8}}{(6x)^8} \ge {}^n{C_9}{3^{n - 9}}{(6x)^9}$$</p>
<p>$$ \Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {... | integer | jee-main-2022-online-27th-july-evening-shift | 5,060 |
1ldo784qa | maths | binomial-theorem | general-term | <p>Let the sixth term in the binomial expansion of $${\left( {\sqrt {{2^{{{\log }_2}\left( {10 - {3^x}} \right)}}} + \root 5 \of {{2^{(x - 2){{\log }_2}3}}} } \right)^m}$$ in the increasing powers of $$2^{(x-2) \log _{2} 3}$$, be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion ... | [] | null | 4 | ${ }^m C_1,{ }^m C_2,{ }^m C_3$ are first, third and fifth term of $A P$
<br/><br/>$$
\begin{aligned}
\therefore \quad & a={ }^m C_1 \\\\
& a+2 d={ }^m C_2 \\\\
& a+4 d={ }^m C_3 \\\\
\therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\
\Rightarrow & m=7 \text { or } m=2 \\\\
\because & m=2 \text { is not possible } \\\\
\th... | integer | jee-main-2023-online-1st-february-evening-shift | 5,062 |
1ldo7goz9 | maths | binomial-theorem | general-term | <p>If the term without $$x$$ in the expansion of $$\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22}$$ is 7315 , then $$|\alpha|$$ is equal to ___________.</p> | [] | null | 1 | Given expansion $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}$
<br/><br/>$$
T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r
$$
<br/><br/>For constant term
<br/><br/>$$
\begin{aligned}
& \frac{44-2 r}{3}-3 r=0 \\\\
& \Rightarrow r=4
\end{aligned}
$$
<br/><br/>Now ${ }^{22... | integer | jee-main-2023-online-1st-february-evening-shift | 5,063 |
ldoaj02i | maths | binomial-theorem | general-term | The coefficient of $x^{-6}$, in the
<br/><br/>expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$, is | [] | null | 5040 | Coeff of $x^{-6}$ in $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$
<br/><br/>$$
\begin{aligned}
T_{r+1} & ={ }^{9} C_{r}\left(\frac{4 x}{5}\right)^{9-r}\left(\frac{5}{2 x^{2}}\right)^{r} \\\\
9-3 r & =-6 \\\\
r & =5
\end{aligned}
$$
<br/><br/>Coeff of $x^{-6}={ }^{9} C_{5}\left(\frac{4}{5}\right)^{4}\left(\frac{... | integer | jee-main-2023-online-31st-january-evening-shift | 5,064 |
ldoavd66 | maths | binomial-theorem | general-term | If the constant term in the binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}$ is $-84$ and the coefficient of $x^{-3 l}$ is
$2^{\alpha} \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to ________. | [] | null | 98 | Given binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}$
<br/><br/>$$
\begin{aligned}
T_{r+1} & ={ }^{9} C_{r}\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^{l}}\right)^{r} \\
& ={ }^{9} C_{r} x^{\frac{45-5 r}{2}-lr} \cdot 2^{r-9} \cdot 4^{r} \cdot(-1)^{r}
\end{aligned}
$... | integer | jee-main-2023-online-31st-january-evening-shift | 5,065 |
1ldptjmpy | maths | binomial-theorem | general-term | <p>Let $$\alpha>0$$, be the smallest number such that the expansion of $$\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}$$ has a term $$\beta x^{-\alpha}, \beta \in \mathbb{N}$$. Then $$\alpha$$ is equal to ___________.</p> | [] | null | 2 | $\mathrm{T}_{\mathrm{r}+1}={ }^{30} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{30-\mathrm{r}}\left(\frac{2}{\mathrm{x}^{3}}\right)^{\mathrm{r}}$
<br/><br/>$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
<br/><br/>$\frac{60-11 \mathrm{r}}{3}<0
$
<br/... | integer | jee-main-2023-online-31st-january-morning-shift | 5,066 |
1ldr72ghg | maths | binomial-theorem | general-term | <p>If the coefficient of $$x^{15}$$ in the expansion of $$\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15}$$ is equal to the coefficient of $$x^{-15}$$ in the expansion of $$\left(a x^{1 / 3}-\frac{1}{b x^{3}}\right)^{15}$$, where $$a$$ and $$b$$ are positive real numbers, then for each such ordered p... | [{"identifier": "A", "content": "a = 3b"}, {"identifier": "B", "content": "ab = 1"}, {"identifier": "C", "content": "ab = 3"}, {"identifier": "D", "content": "a = b"}] | ["B"] | null | <p>For $$\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)$$</p>
<p>$${T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}$$</p>
<p>$$\therefore$$ $${x^{15}} \to 3(15 - r) - {r \over 3} = 15$$</p>
<p>$$ \Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9$$</p>
<p>Simila... | mcq | jee-main-2023-online-30th-january-morning-shift | 5,067 |
1ldswq8el | maths | binomial-theorem | general-term | <p>Let the coefficients of three consecutive terms in the binomial expansion of $$(1+2x)^n$$ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.</p> | [] | null | 1120 | $\mathrm{t}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2 \mathrm{x})^{\mathrm{r}}$
<br/><br/>
$$
\begin{aligned}
& \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\
& \Rightarrow \frac{\frac{n !}{(r-1) ... | integer | jee-main-2023-online-29th-january-morning-shift | 5,068 |
1ldswsczz | maths | binomial-theorem | general-term | <p>If the co-efficient of $$x^9$$ in $${\left( {\alpha {x^3} + {1 \over {\beta x}}} \right)^{11}}$$ and the co-efficient of $$x^{-9}$$ in $${\left( {\alpha x - {1 \over {\beta {x^3}}}} \right)^{11}}$$ are equal, then $$(\alpha\beta)^2$$ is equal to ___________.</p> | [] | null | 1 | Coefficient of $\mathrm{x}^{9}$ in $\left(\alpha x^{3}+\frac{1}{\beta x}\right)={ }^{11} C_{6} \cdot \frac{\alpha^{5}}{\beta^{6}}$
<br/><br/>
$\because$ Both are equal
<br/><br/>
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
<br/><br/>
$\Rightarrow... | integer | jee-main-2023-online-29th-january-morning-shift | 5,069 |
1ldv2styk | maths | binomial-theorem | general-term | <p>The constant term in the expansion of $${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$$ is ___________.</p> | [] | null | 1080 | Constant term in the expansion of
<br/><br/>
$$
\begin{aligned}
& \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5} \\\\
& \frac{1}{x^{35}}\left(2 x^{8}+1+3 x^{9}\right)^{5} \\\\
& \frac{1}{x^{35}}\left(1+x^{8}(3 x+2)\right)^{5}
\end{aligned}
$$
<br/><br/>
Term independent of $x=$ coefficient of $x^{35}$ in
<br/><br/>
$$
\b... | integer | jee-main-2023-online-25th-january-morning-shift | 5,070 |
1ldwxmgwd | maths | binomial-theorem | general-term | <p>Let the sum of the coefficients of the first three terms in the expansion of $${\left( {x - {3 \over {{x^2}}}} \right)^n},x \ne 0.~n \in \mathbb{N}$$, be 376. Then the coefficient of $$x^4$$ is __________.</p> | [] | null | 405 | $S=1-3 n+\frac{9 n(n-1)}{2}=376$
<br/><br/>
$$
\begin{aligned}
& 3 n^{2}-5 n-250=0 \\\\
& n=10, \frac{-25}{3} \text { (Rejected) } \\\\
& T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\
& ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\
& ={ }^{10} C_{r} x^{10-3 r}(-3)^{r}
\end{aligned}
$$
<br/><br/>
H... | integer | jee-main-2023-online-24th-january-evening-shift | 5,071 |
1lgowbuor | maths | binomial-theorem | general-term | <p>The coefficient of $$x^{5}$$ in the expansion of $$\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{26}{3}$$"}, {"identifier": "B", "content": "$$\\frac{80}{9}$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["B"] | null | Given, $$\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$$
<br/><br/>General term,
<br/><br/>$$
\begin{aligned}
& T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r} \\\\
& \therefore 15-5 \mathrm{r}=5 \\\\
& \therefore \mathrm{r}=2 \\\\
& T_3=10\left(\frac... | mcq | jee-main-2023-online-13th-april-evening-shift | 5,072 |
1lgq11k4d | maths | binomial-theorem | general-term | <p>Let $$\alpha$$ be the constant term in the binomial expansion of $$\left(\sqrt{x}-\frac{6}{x^{\frac{3}{2}}}\right)^{n}, n \leq 15$$. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $$x^{-n}$$ is $$\lambda \alpha$$, then $$\lambda$$ is equal to _____________.</p> | [] | null | 36 | Given expression $(\sqrt{x}-\frac{6}{x^{3/2}})^n$. Here, $a = \sqrt{x}$ and $b = -\frac{6}{x^{3/2}}$.
<br/><br/>The $r$-th term of the binomial expansion of $(a+b)^n$ is given by
<br/><br/>$T_{r} = {}^n{C_r}a^{n-r}b^{r}$.
<br/><br/>Substitute $a$ and $b$ in this formula, we get:
<br/><br/>$T_{r} = {}^n{C_r}(\sqrt{... | integer | jee-main-2023-online-13th-april-morning-shift | 5,073 |
1lgsubacw | maths | binomial-theorem | general-term | <p>The sum of the coefficients of three consecutive terms in the binomial expansion of $$(1+\mathrm{x})^{\mathrm{n}+2}$$, which are in the ratio $$1: 3: 5$$, is equal to :</p> | [{"identifier": "A", "content": "63"}, {"identifier": "B", "content": "92"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "41"}] | ["A"] | null | The problem asks for the sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$, which are in the ratio 1 : 3 : 5.
<br/><br/>Given that the ratios of the coefficients are 1:3:5, we let the terms be $T_r$, $T_{r+1}$, and $T_{r+2}$. The coefficients of these terms are ${ }^{n+2} C... | mcq | jee-main-2023-online-11th-april-evening-shift | 5,074 |
1lgvpl71w | maths | binomial-theorem | general-term | <p>If the coefficients of $$x$$ and $$x^{2}$$ in $$(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$$ are 4 and $$-$$5 respectively, then $$2 p+3 q$$ is equal to :</p> | [{"identifier": "A", "content": "66"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "69"}, {"identifier": "D", "content": "63"}] | ["D"] | null | We have, coefficient of $x$ in $(1+x)^p(1-x)^q=4$ and
<br/><br/>coefficient of $x^2$ in $(1+x)^p(1-x)^q=-5$
<br/><br/>$$
\begin{aligned}
& (1+x)^p(1-x)^q \\\\
& =\left(1+p x+\frac{p(p-1)}{2} x^2+\ldots\right)\left(1-q x+\frac{q(q-1)}{2} x^2+\ldots\right) \\\\
& =1+(p-q) x+\left(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q\ri... | mcq | jee-main-2023-online-10th-april-evening-shift | 5,078 |
1lgxt15ef | maths | binomial-theorem | general-term | <p>If the coefficient of $${x^7}$$ in $${\left( {ax - {1 \over {b{x^2}}}} \right)^{13}}$$ and the coefficient of $${x^{ - 5}}$$ in $${\left( {ax + {1 \over {b{x^2}}}} \right)^{13}}$$ are equal, then $${a^4}{b^4}$$ is equal to :</p> | [{"identifier": "A", "content": "22"}, {"identifier": "B", "content": "33"}, {"identifier": "C", "content": "44"}, {"identifier": "D", "content": "11"}] | ["A"] | null | The given expression is $\left(a x-\frac{1}{b x^2}\right)^{13}$
<br/><br/>So,
<br/><br/>$$
\begin{aligned}
T_{r+1} & ={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r \\\\
& ={ }^{13} C_r(a)^{13-r} x^{13-r-2 r}(-1 / b)^r \\\\
& ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r}
\end{aligned}
$$
<br/><... | mcq | jee-main-2023-online-10th-april-morning-shift | 5,079 |
1lgyliytd | maths | binomial-theorem | general-term | <p>The absolute difference of the coefficients of $$x^{10}$$ and $$x^{7}$$ in the expansion of $$\left(2 x^{2}+\frac{1}{2 x}\right)^{11}$$ is equal to :</p> | [{"identifier": "A", "content": "$$11^{3}-11$$"}, {"identifier": "B", "content": "$$13^{3}-13$$"}, {"identifier": "C", "content": "$$12^{3}-12$$"}, {"identifier": "D", "content": "$$10^{3}-10$$"}] | ["C"] | null | General term of $\left(2 x^2+\frac{1}{2 x}\right)^{11}$ is :
<br/><br/>$$
\begin{aligned}
& \mathrm{T}_{\mathrm{r}+1}={ }^{11} \mathrm{C}_r\left(2 x^2\right){ }^{11-r}\left(\frac{1}{2 x}\right)^r \\\\
& ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-2 r} 2^{-r} x^{-r} \\\\
& ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-3 r}
\end{align... | mcq | jee-main-2023-online-8th-april-evening-shift | 5,080 |
1lh00ko4t | maths | binomial-theorem | general-term | <p>Let $$[t]$$ denote the greatest integer $$\leq t$$. If the constant term in the expansion of $$\left(3 x^{2}-\frac{1}{2 x^{5}}\right)^{7}$$ is $$\alpha$$, then $$[\alpha]$$ is equal to ___________.</p> | [] | null | 1275 | Let $\mathrm{T}_{r+1}$ be the constant term.
<br/><br/>$$
\mathrm{T}_{r+1}={ }^7 \mathrm{C}_r\left(3 x^2\right)^{7-r}\left(\frac{-1}{2 x^5}\right)^r
$$
<br/><br/>For constant term, power of $x$ should be zero.
<br/><br/>$$
\begin{aligned}
& \text { i.e., } 14-2 r-5 r=0 \\\\
& \Rightarrow 14=7 r \Rightarrow r=2
\end{ali... | integer | jee-main-2023-online-8th-april-morning-shift | 5,081 |
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