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1l6dwd0i9 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the locus of the centre $$(\alpha, \beta), \beta>0$$, of the circle which touches the circle $$x^{2}+(y-1)^{2}=1$$ externally and also touches the $$x$$-axis be $$\mathrm{L}$$. Then the area bounded by $$\mathrm{L}$$ and the line $$y=4$$ is:</p> | [{"identifier": "A", "content": "$$\n\\frac{32 \\sqrt{2}}{3}\n$$"}, {"identifier": "B", "content": "$$\n\\frac{40 \\sqrt{2}}{3}\n$$"}, {"identifier": "C", "content": "$$\\frac{64}{3}$$"}, {"identifier": "D", "content": "$$\n\\frac{32}{3}\n$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97rzcly/4fa8a86d-de28-44c0-be52-7dde80194483/7a763360-4b5a-11ed-bfde-e1cb3fafe700/file-1l97rzclz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97rzcly/4fa8a86d-de28-44c0-be52-7dde80194483/7a763360-4b5a-11ed-bfde-e1cb3fafe700/fi... | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6f3sq57 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $$4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0$$ at the point ($$-$$2, 3) be A. Then 8A is equal to ______________.</p> | [] | null | 170 | $$
\begin{aligned}
& 4 x^3-3 x y^2+6 x^2-5 x y-8 y^2+9 x+14=0 \text { at } P(-2,3) \\\\
& 12 x^2-3\left(y^2+2 y x y^{\prime}\right)+12 x-5\left(x y^{\prime}+y\right)-16 y y^{\prime} + 9=0 \\\\
& 48-3\left(9-12 y^{\prime}\right)-24-5\left(-2 y^{\prime}+3\right)-48 y^{\prime}+9 =0 \\\\
& y^{\prime}=-9 /... | integer | jee-main-2022-online-25th-july-evening-shift |
1l6ggjssj | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y<sup>a</sup> is $${{364} \over 3}$$, is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "9"}] | ["B"] | null | <p>$$\mathrm{a}$$ is a odd natural number and</p>
<p>$$\left| {\int\limits_1^3 {{y^a}dy} } \right| = {{364} \over 3}$$</p>
<p>$$ \Rightarrow \left| {{1 \over {a + 1}}\left( {{y^{a + 1}}} \right)_1^3} \right| = {{364} \over 3}$$</p>
<p>$$ \Rightarrow {{{3^{a + 1}} - 1} \over {a + 1}} = \, \pm \,{{364} \over 3}$$</p>
<p>... | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6hzgs0d | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area bounded by the curves $$y=\left|x^{2}-1\right|$$ and $$y=1$$ is</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}(\\sqrt{2}+1)$$"}, {"identifier": "B", "content": "$$\\frac{4}{3}(\\sqrt{2}-1)$$"}, {"identifier": "C", "content": "$$2(\\sqrt{2}-1)$$"}, {"identifier": "D", "content": "$$\\frac{8}{3}(\\sqrt{2}-1)$$"}] | ["D"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nnzsdz/1978812c-886a-41da-a99e-ba9650919e28/fdc43870-2c7e-11ed-a18d-5933e4fde865/file-1l7nnzse0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nnzsdz/1978812c-886a-41da-a99e-ba9650919e28/fdc43870-2c7e-11ed-a18d-5933e4fde86... | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6jbsxxf | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the smaller region enclosed by the curves $$y^{2}=8 x+4$$ and $$x^{2}+y^{2}+4 \sqrt{3} x-4=0$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{3}(2-12 \\sqrt{3}+8 \\pi)$$"}, {"identifier": "B", "content": "$$\\frac{1}{3}(2-12 \\sqrt{3}+6 \\pi)$$"}, {"identifier": "C", "content": "$$\\frac{1}{3}(4-12 \\sqrt{3}+8 \\pi)$$"}, {"identifier": "D", "content": "$$\\frac{1}{3}(4-12 \\sqrt{3}+6 \\pi)$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ps543w/90436956-c0bb-4a4d-b417-ec2e224aeb75/c7ae54c0-2da8-11ed-8542-f96181a425b5/file-1l7ps543x.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ps543w/90436956-c0bb-4a4d-b417-ec2e224aeb75/c7ae54c0-2da8-11ed-8542-f96181a425b5... | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6kjv7su | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by $$y \leq 4 x^{2}, x^{2} \leq 9 y$$ and $$y \leq 4$$, is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{40}{3}$$"}, {"identifier": "B", "content": "$$\\frac{56}{3}$$"}, {"identifier": "C", "content": "$$\\frac{112}{3}$$"}, {"identifier": "D", "content": "$$\\frac{80}{3}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7q9zzgg/929f9d9b-01c4-4a4d-8457-02912d2b4071/9d5df400-2dee-11ed-a744-1fb8f3709cfa/file-1l7q9zzgh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7q9zzgg/929f9d9b-01c4-4a4d-8457-02912d2b4071/9d5df400-2dee-11ed-a744-1fb8f3709cfa... | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6kk0129 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Consider a curve $$y=y(x)$$ in the first quadrant as shown in the figure. Let the area $$\mathrm{A}_{1}$$ is twice the area $$\mathrm{A}_{2}$$. Then the normal to the curve perpendicular to the line $$2 x-12 y=15$$ does NOT pass through the point.</p>
<p><img src="data:image/png;base64,UklGRioRAABXRUJQVlA4IB4RAABQ2g... | [{"identifier": "A", "content": "(6, 21)"}, {"identifier": "B", "content": "(8, 9)"}, {"identifier": "C", "content": "(10, $$-$$4)"}, {"identifier": "D", "content": "(12, $$-$$15)"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qa2cbj/f16ef20a-d0a5-47f2-a99e-77b83471fd40/deec2e00-2dee-11ed-a744-1fb8f3709cfa/file-1l7qa2cbk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qa2cbj/f16ef20a-d0a5-47f2-a99e-77b83471fd40/deec2e00-2dee-11ed-a744-1fb8f3709cfa... | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nmba5c | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed by the curves $$y=\log _{e}\left(x+\mathrm{e}^{2}\right), x=\log _{e}\left(\frac{2}{y}\right)$$ and $$x=\log _{\mathrm{e}} 2$$, above the line $$y=1$$ is:</p> | [{"identifier": "A", "content": "$$2+\\mathrm{e}-\\log _{\\mathrm{e}} 2$$"}, {"identifier": "B", "content": "$$1+e-\\log _{e} 2$$"}, {"identifier": "C", "content": "$$e-\\log _{e} 2$$"}, {"identifier": "D", "content": "$$1+\\log _{e} 2$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97u3d0g/088f2998-8bd4-4720-8b55-fa1efe186501/bc448a00-4b62-11ed-80b9-4154b7faa509/file-1l97u3d0h.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97u3d0g/088f2998-8bd4-4720-8b55-fa1efe186501/bc448a00-4b62-11ed-80b9-4154b7faa509/fi... | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6p2kie8 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region</p>
<p>$$\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{5}{2} \\sin ^{-1}\\left(\\frac{3}{5}\\right)-\\frac{1}{2}$$"}, {"identifier": "B", "content": "$$\\frac{5 \\pi}{4}-\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$\\frac{3 \\pi}{4}+\\frac{3}{2}$$"}, {"identifier": "D", "content": "$$\\frac{5 \\pi}{4}-\\frac{1}{2}$$"}] | ["D"] | null | <p>$$A = \int\limits_{ - 1}^1 {\left( {\sqrt {5 - {x^2}} - (1 - x)} \right)dx + \int\limits_1^2 {\left( {\sqrt {5 - {x^2}} - (x - 1)} \right)dx} } $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssldo3/57b8d8ee-cd1f-4f78-8605-15da9040ba0e/e5f60c30-2f50-11ed-85dd-19dc023e9ad1/file-1l7ssldo... | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo65ln2 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region given by $$\{ (x,y):xy \le 8,1 \le y \le {x^2}\} $$ is :</p> | [{"identifier": "A", "content": "$$16{\\log _e}2 - {{14} \\over 3}$$"}, {"identifier": "B", "content": "$$8{\\log _e}2 - {{13} \\over 3}$$"}, {"identifier": "C", "content": "$$16{\\log _e}2 + {7 \\over 3}$$"}, {"identifier": "D", "content": "$$8{\\log _e}2 + {7 \\over 6}$$"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5fqsjf/90c628e8-42f9-49e3-ac52-f0f944e1d31e/2287dcb0-ad0e-11ed-a86d-8dfe0389db88/file-1le5fqsjg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5fqsjf/90c628e8-42f9-49e3-ac52-f0f944e1d31e/2287dcb0-ad0e-11ed-a86d-8dfe0389db88/fi... | mcq | jee-main-2023-online-1st-february-evening-shift |
ldoah9qp | maths | area-under-the-curves | area-bounded-between-the-curves | Let the area of the region
<br/><br/>$\left\{(x, y):|2 x-1| \leq y \leq\left|x^{2}-x\right|, 0 \leq x \leq 1\right\}$ be $\mathrm{A}$.
<br/><br/>Then $(6 \mathrm{~A}+11)^{2}$ is equal to | [] | null | 125 | For $B$,
<br><br>$$
\begin{aligned}
& x-x^{2}=2 x-1 \\\\
& x^{2}+x-1=0 \\\\
& x=\frac{-1+\sqrt{5}}{2}
\end{aligned}
$$
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leeiutst/47959be2-54bf-4c50-8327-bce6176bbe32/5bf957e0-b20d-11ed-9c9a-c3bd979bc154/file-1leeiutsu.png?format=png" data-ors... | integer | jee-main-2023-online-31st-january-evening-shift |
1ldoocgjv | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$A$$ be the area bounded by the curve $$y=x|x-3|$$, the $$x$$-axis and the ordinates $$x=-1$$ and $$x=2$$. Then $$12 A$$ is equal to ____________.</p> | [] | null | 62 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le9xvce9/cce9188c-fc73-4ab5-a87d-33da42a11b53/28073f10-af88-11ed-bd02-1d2a2a7b6687/file-1le9xvcea.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le9xvce9/cce9188c-fc73-4ab5-a87d-33da42a11b53/28073f10-af88-11ed-bd02-1d2a2a7b6687/fi... | integer | jee-main-2023-online-1st-february-morning-shift |
1ldptrjv7 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let for $$x \in \mathbb{R}$$,</p>
<p>$$
f(x)=\frac{x+|x|}{2} \text { and } g(x)=\left\{\begin{array}{cc}
x, & x<0 \\
x^{2}, & x \geq 0
\end{array}\right. \text {. }
$$</p>
<p>Then area bounded by the curve $$y=(f \circ g)(x)$$ and the lines $$y=0,2 y-x=15$$ is equal to __________.</p> | [] | null | 72 | $f(x)=\frac{x+|x|}{2}=\left[\begin{array}{ll}x & x \geq 0 \\ 0 & x<0\end{array}\right.$
<br><br>$g(x)=\left[\begin{array}{cc}x^{2} & x \geq 0 \\ x & x<0\end{array}\right.$
<br><br>$f(x)=f[g(x)]=\left[\begin{array}{cl}g(x), & g(x) \geq 0 \\ 0, & g(x)<0\end{array}\right.$
<br><br>$\ope... | integer | jee-main-2023-online-31st-january-morning-shift |
ldqwiizg | maths | area-under-the-curves | area-bounded-between-the-curves | Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-p x+\frac{5}{4} p=0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq(x-q)^2, 0 \leq x \leq q\right\}$ is : | [{"identifier": "A", "content": "$\\frac{125}{3}$"}, {"identifier": "B", "content": "243"}, {"identifier": "C", "content": "164"}, {"identifier": "D", "content": "25"}] | ["B"] | null | <p>Given equation : $$4{x^2} - 4px + 5p = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leolev6m/9d2f1ed0-3ff3-4daa-b4d0-ae63d5dd71a4/31fa37e0-b797-11ed-b103-ed967fad3dff/file-1leolev6n.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leolev6m/9d2f1ed0-3ff3-4daa-b4d0... | mcq | jee-main-2023-online-30th-january-evening-shift |
ldr01u60 | maths | area-under-the-curves | area-bounded-between-the-curves | Let $A$ be the area of the region
<br/><br/>$\left\{(x, y): y \geq x^2, y \geq(1-x)^2, y \leq 2 x(1-x)\right\}$.
<br/><br/>Then $540 \mathrm{~A}$ is equal to : | [] | null | 25 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leol5gd1/d39e5f85-8fa2-457b-8f2b-aa98ad9bb499/2c3ee450-b796-11ed-b103-ed967fad3dff/file-1leol5gd2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leol5gd1/d39e5f85-8fa2-457b-8f2b-aa98ad9bb499/2c3ee450-b796-11ed-b103-ed967fad3dff... | integer | jee-main-2023-online-30th-january-evening-shift |
1ldr7y2pf | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$\alpha$$ be the area of the larger region bounded by the curve $$y^{2}=8 x$$ and the lines $$y=x$$ and $$x=2$$, which lies in the first quadrant. Then the value of $$3 \alpha$$ is equal to ___________.</p> | [] | null | 22 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leq0fubs/f2abef4a-3124-4350-839e-3e7bfc1aa857/beb62c80-b85e-11ed-9fed-b1659a6c339b/file-1leq0fubt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leq0fubs/f2abef4a-3124-4350-839e-3e7bfc1aa857/beb62c80-b85e-11ed-9fed-b1659a6c339b... | integer | jee-main-2023-online-30th-january-morning-shift |
1ldsf7k2m | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$A = \left\{ {(x,y):\left| {\cos x - \sin x} \right| \le y \le \sin x,0 \le x \le {\pi \over 2}} \right\}$$ is</p> | [{"identifier": "A", "content": "$$\\sqrt 5 + 2\\sqrt 2 - 4.5$$"}, {"identifier": "B", "content": "$$1 - {3 \\over {\\sqrt 2 }} + {4 \\over {\\sqrt 5 }}$$"}, {"identifier": "C", "content": "$$\\sqrt 5 - 2\\sqrt 2 + 1$$"}, {"identifier": "D", "content": "$${3 \\over {\\sqrt 5 }} - {3 \\over {\\sqrt 2 }} + 1$$"}] | ["C"] | null | <p>$$
|\cos x-\sin x| \leq y \leq \sin x
$$
<br><br>Intersection point of $\cos x-\sin x=\sin x$
<br><br>$$
\Rightarrow \tan x=\frac{1}{2}
$$
<br><br>Let $\psi=\tan ^{-1} \frac{1}{2}$
<br><br>So, $\tan \psi=\frac{1}{2}, \sin \psi=\frac{1}{\sqrt{5}}, \cos \psi=\frac{2}{\sqrt{5}}$</p>
<p><img src="https://app-content.cdn... | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsv2h4c | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$\Delta$$ be the area of the region $$\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\}$$. Then $${1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$2\\sqrt 3 - {1 \\over 3}$$"}, {"identifier": "B", "content": "$$2\\sqrt 3 - {2 \\over 3}$$"}, {"identifier": "C", "content": "$$\\sqrt 3 - {4 \\over 3}$$"}, {"identifier": "D", "content": "$$\\sqrt 3 - {2 \\over 3}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leksols2/a3491772-b199-41ab-9761-0d39a13abce7/7e1c5220-b580-11ed-b843-fd540edb80bf/file-1leksols3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leksols2/a3491772-b199-41ab-9761-0d39a13abce7/7e1c5220-b580-11ed-b843-fd540edb80bf/fi... | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldsvuk9z | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$[x]$$ denote the greatest integer $$\le x$$. Consider the function $$f(x) = \max \left\{ {{x^2},1 + [x]} \right\}$$. Then the value of the integral $$\int\limits_0^2 {f(x)dx} $$ is</p> | [{"identifier": "A", "content": "$${{5 + 4\\sqrt 2 } \\over 3}$$"}, {"identifier": "B", "content": "$${{4 + 5\\sqrt 2 } \\over 3}$$"}, {"identifier": "C", "content": "$${{8 + 4\\sqrt 2 } \\over 3}$$"}, {"identifier": "D", "content": "$${{1 + 5\\sqrt 2 } \\over 3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ldt0ucjn/f3408cd8-1181-40fe-aecc-1f2a69b3243c/74ae7e20-a63a-11ed-a341-61f046f8f7d7/file-1ldt0ucjo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ldt0ucjn/f3408cd8-1181-40fe-aecc-1f2a69b3243c/74ae7e20-a63a-11ed-a341-61f046f8f7d7... | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldsvyln7 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}$$ and<br/><br/> $$
B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. }
$$.</p>
<p>Then the ratio of the area of A to the area of B is</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{\\pi+1}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi-1}{\\pi+1}$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{\\pi-1}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi+1}{\\pi-1}$$"}] | ["B"] | null | <p>$y^{2}+(x-1)^{2}=4$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lekro7yb/473e481f-1c3a-4e67-a338-a22cacc1490d/8a4d0930-b57c-11ed-9ad9-3b1cedbe69d8/file-1lekro7yc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lekro7yb/473e481f-1c3a-4e67-a338-a22cacc1490d/8a4d0930-... | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldv36uu9 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area enclosed by the parabolas $$\mathrm{P_1:2y=5x^2}$$ and $$\mathrm{P_2:x^2-y+6=0}$$ is equal to the area enclosed by $$\mathrm{P_1}$$ and $$\mathrm{y=\alpha x,\alpha > 0}$$, then $$\alpha^3$$ is equal to ____________.</p> | [] | null | 600 | $x^{2}+6=\frac{5}{2} x^{2} \Rightarrow x=\pm 2$
<br/><br/>
$$
\begin{aligned}
& \text { Area between } P_{1} \text { and } P_{2} \quad \text { [Say } \left.A_{1}\right] \\\\
& =\int\limits_{-2}^{2}\left(x^{2}+6\right)-\frac{5}{2} x^{2} d x \\\\
& =2 \int\limits_{0}^{2}\left(6-\frac{3}{2} x^{2}\right) d x=2\left[6 x-\f... | integer | jee-main-2023-online-25th-january-morning-shift |
1ldwxskv5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region bounded by the curves $$y^2-2y=-x,x+y=0$$ is A, then 8 A is equal to __________</p> | [] | null | 36 | Area enclosed by
<br><br>
$$
\begin{aligned}
& y^{2}-2 y=-x \\\\
& x+y=0
\end{aligned}
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5i5l5v/1c86c1a0-48dd-4e1e-9be0-6401a21b507d/903a5b30-ad17-11ed-8a8c-4d67f5492755/file-1le5i5l5w.png?format=png" data-orsrc="https://app-content.cdn... | integer | jee-main-2023-online-24th-january-evening-shift |
1ldyb9d7r | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed by the curves $${y^2} + 4x = 4$$ and $$y - 2x = 2$$ is :</p> | [{"identifier": "A", "content": "$${{22} \\over 3}$$"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "$${{23} \\over 3}$$"}, {"identifier": "D", "content": "$${{25} \\over 3}$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le1yf9t4/b18e1d13-83c3-4532-94a6-9b8efbcd5fcf/0d044770-ab24-11ed-b3ea-6525e5fb1f1e/file-1le1yf9t5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le1yf9t4/b18e1d13-83c3-4532-94a6-9b8efbcd5fcf/0d044770-ab24-11ed-b3ea-6525e5fb1f1e/fi... | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnz10hf | maths | area-under-the-curves | area-bounded-between-the-curves | If the area bounded by the curve $2 y^{2}=3 x$, lines $x+y=3, y=0$ and outside the circle $(x-3)^{2}+y^{2}=2$ is $\mathrm{A}$, then $4(\pi+4 A)$ is equal to ____________. | [] | null | 42 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrn6cu8/fb34c18c-fa58-4f6a-a195-c84349a81234/e04c5800-e0dc-11ed-9ecd-e999028462e7/file-1lgrn6cu9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrn6cu8/fb34c18c-fa58-4f6a-a195-c84349a81234/e04c5800-e0dc-11ed-9ecd-e999028462e7/fi... | integer | jee-main-2023-online-15th-april-morning-shift |
1lgoxjwx4 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$\left\{(x, y): x^{2} \leq y \leq\left|x^{2}-4\right|, y \geq 1\right\}$$ is</p> | [{"identifier": "A", "content": "$$\\frac{4}{3}(4 \\sqrt{2}+1)$$"}, {"identifier": "B", "content": "$$\\frac{3}{4}(4 \\sqrt{2}+1)$$"}, {"identifier": "C", "content": "$$\\frac{4}{3}(4 \\sqrt{2}-1)$$"}, {"identifier": "D", "content": "$$\\frac{3}{4}(4 \\sqrt{2}-1)$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh2jl21d/1eaa8b9f-d025-4054-8bc8-91786b0e9ad1/73b7e010-e6db-11ed-948f-4b963ec65c15/file-1lh2jl21e.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh2jl21d/1eaa8b9f-d025-4054-8bc8-91786b0e9ad1/73b7e010-e6db-11ed-948f-4b963ec65c15/fi... | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpxipld | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the curve
$$f(x)=\max \{\sin x, \cos x\},-\pi \leq x \leq \pi$$ and the $$x$$-axis is</p> | [{"identifier": "A", "content": "$$2 \\sqrt{2}(\\sqrt{2}+1)$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$2(\\sqrt{2}+1)$$"}, {"identifier": "D", "content": "$$4(\\sqrt{2})$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh3urauz/df812b2b-41d3-412b-9523-6e99a63e5a07/ee7558b0-e793-11ed-9b1f-65ec4fb7911b/file-1lh3urav0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh3urauz/df812b2b-41d3-412b-9523-6e99a63e5a07/ee7558b0-e793-11ed-9b1f-65ec4fb7911b/fi... | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgrgc23k | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the curve $$y=x^{3}$$ and its tangent at the point $$(-1,-1)$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{23}{4}$$"}, {"identifier": "B", "content": "$$\\frac{19}{4}$$"}, {"identifier": "C", "content": "$$\\frac{27}{4}$$"}, {"identifier": "D", "content": "$$\\frac{31}{4}$$<br/><br/>"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1li62ldz6/e3e1ec7b-7f4a-4711-90a5-23ea1380cf4d/5bb8a710-fc98-11ed-bb91-ad55392bec91/file-1li62ldz7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1li62ldz6/e3e1ec7b-7f4a-4711-90a5-23ea1380cf4d/5bb8a710-fc98-11ed-bb91-ad55392bec91/fi... | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgswa4aj | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If A is the area in the first quadrant enclosed by the curve $$\mathrm{C: 2 x^{2}-y+1=0}$$, the tangent to $$\mathrm{C}$$ at the point $$(1,3)$$ and the line $$\mathrm{x}+\mathrm{y}=1$$, then the value of $$60 \mathrm{~A}$$ is _________.</p> | [] | null | 16 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lie8i8gh/9671b86e-ee06-496d-abb6-609a695b7ace/c0004c10-0115-11ee-9f57-5de63d0f488b/file-1lie8i8gi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lie8i8gh/9671b86e-ee06-496d-abb6-609a695b7ace/c0004c10-0115-11ee-9f57-5de63d0f488b/fi... | integer | jee-main-2023-online-11th-april-evening-shift |
1lguvhkpl | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Area of the region $$\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\}$$ is</p> | [{"identifier": "A", "content": "$$2 \\pi+\\frac{16}{3}$$"}, {"identifier": "B", "content": "$$\\pi-\\frac{8}{3}$$"}, {"identifier": "C", "content": "$$\\pi+\\frac{8}{3}$$"}, {"identifier": "D", "content": "$$2 \\pi-\\frac{16}{3}$$"}] | ["D"] | null | We have,
<br><br>$$
x^2+(y-2)^2 \leq 2^2 \text { and } x^2 \geq 2 y
$$
<br><br>On solving the given equation of parabola and circle, we get
<br><br>$$
\begin{aligned}
2 y+(y-2)^2 & =4 \\\\
\Rightarrow y =0 \text { or } 2
\end{aligned}
$$
<br><br>If $y=0$, then $x=0$ and
<br><br>If $y=2$, then $x= \pm 2$
<br><br>$\t... | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvqkz5x | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\}$$ is $$\mathrm{A}$$, then $$6 \mathrm{A}+16 \sqrt{2}$$ is equal to __________.</p> | [] | null | 27 | $$
\text { We have, }\left\{(x, y):\left|x^2-2\right| \leq y \leq x\right\}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnk7a0xn/c3e34f25-2aa9-4803-b342-54bfe7409450/bb60e0b0-675a-11ee-a06a-699a057b80c4/file-6y3zli1lnk7a0xo.png?format=png" data-orsrc="https://app-content.cdn.examgo... | integer | jee-main-2023-online-10th-april-evening-shift |
1lgxw9wy2 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$y = p(x)$$ be the parabola passing through the points $$( - 1,0),(0,1)$$ and $$(1,0)$$. If the area of the region $$\{ (x,y):{(x + 1)^2} + {(y - 1)^2} \le 1,y \le p(x)\} $$ is A, then $$12(\pi - 4A)$$ is equal to ___________.</p> | [] | null | 16 | Let, $y=p(x)$ be the parabola passing through the points $(-1,0)(0,1)(1,0)$.
<br><br>Now, to find the area of the region
<br><br>$$
\left\{(x, y) ;(x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\right\}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnda9au3/d6943506-1c0a-4dc0-87a7-f279a28202cc/e... | integer | jee-main-2023-online-10th-april-morning-shift |
1lgyom5cr | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area enclosed by the lines $$x+y=2, \mathrm{y}=0, x=0$$ and the curve $$f(x)=\min \left\{x^{2}+\frac{3}{4}, 1+[x]\right\}$$ where $$[x]$$
denotes the greatest integer $$\leq x$$, be $$\mathrm{A}$$. Then the value of $$12 \mathrm{~A}$$ is _____________.</p> | [] | null | 17 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyue4ph/cd17397d-b57f-4fb7-a489-8f746007324f/e1368850-5b9b-11ee-83da-a3f80d422da4/file-6y3zli1lmyue4pi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lmyue4ph/cd17397d-b57f-4fb7-a489-8f746007324f/e1368850-5b9b-11ee-83... | integer | jee-main-2023-online-8th-april-evening-shift |
1lgzyidm2 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$\left\{(x, y): x^{2} \leq y \leq 8-x^{2}, y \leq 7\right\}$$ is :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "21"}] | ["C"] | null | The given curves are
<br><br>$x^2 \leq y, y \leq 8-x^2 ; y \leq 7$
<br><br>On solving, we get $x^2=8-x^2$
<br><br>$$
\begin{aligned}
& \Rightarrow x^2=4 \\\\
& \Rightarrow x= \pm 2
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljg19fl1/2d19d263-28f7-4e45-975b-... | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh23quiq | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$S=\left\{(x, y): 2 y-y^{2} \leq x^{2} \leq 2 y, x \geq y\right\}$$ is equal to $$\frac{n+2}{n+1}-\frac{\pi}{n-1}$$, then the natural number $$n$$ is equal to ___________.</p> | [] | null | 5 | Given region,
<br><br>$$
S=\left\{(x, y): 2 y-y^2 \leq x^2 \leq 2 y, x \geq y\right\}
$$
<br><br>Here, we have three curves
<br><br>$$
\begin{aligned}
&2 y-y^2 =x^2 ..........(i)\\\\
&x^2 =2 y ..........(2)\\\\
&\text {and}~~ x = y ...........(3)
\end{aligned}
$$
<br><br><img src="https://app-content.cd... | integer | jee-main-2023-online-6th-april-morning-shift |
1lh2xt7m5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area bounded by the curves $$y=|x-1|+|x-2|$$ and $$y=3$$ is equal to :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "3"}] | ["B"] | null | Given equation of curve $y=|x-1|+|x-2|$ and $y=3$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo8c5sm6/ed381416-4cb6-4a32-b174-452035f058cf/a2f26cd0-74a0-11ee-8723-4d48ce392782/file-6y3zli1lo8c5sm7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lo8c5sm6/... | mcq | jee-main-2023-online-6th-april-evening-shift |
lsan0p30 | maths | area-under-the-curves | area-bounded-between-the-curves | Three points $\mathrm{O}(0,0), \mathrm{P}\left(\mathrm{a}, \mathrm{a}^2\right), \mathrm{Q}\left(-\mathrm{b}, \mathrm{b}^2\right), \mathrm{a}>0, \mathrm{~b}>0$, are on the parabola $y=x^2$. Let $\mathrm{S}_1$ be the area of the region bounded by the line $\mathrm{PQ}$ and the parabola, and $\mathrm{S}_2$ be the ar... | [] | null | 7 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsogvt13/49cb1ddc-aaf9-4143-8cc4-b6d4940befaa/82ba6c70-ccb0-11ee-aa98-13f456b8f7af/file-6y3zli1lsogvt14.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsogvt13/49cb1ddc-aaf9-4143-8cc4-b6d4940befaa/82ba6c70-ccb0-11ee-aa... | integer | jee-main-2024-online-1st-february-evening-shift |
lsan1ar0 | maths | area-under-the-curves | area-bounded-between-the-curves | The sum of squares of all possible values of $k$, for which area of the region bounded by the parabolas $2 y^2=\mathrm{k} x$ and $\mathrm{ky}^2=2(y-x)$ is maximum, is equal to : | [] | null | 8 | Given $k y^2=2(y-x)$ .........(i)
<br/><br/>$$
2 y^2=k x
$$ .........(ii)
<br/><br/>Point of intersection of (i) and (ii)
<br/><br/>$$
\begin{aligned}
& k y^2=2\left(y-\frac{2 y^2}{k}\right) \\\\
& \Rightarrow y=0, k y=2\left(1-\frac{2 y}{k}\right)
\end{aligned}
$$
<br/><br/>$\begin{aligned} & k y+\frac{4 y}{k}=2 \\\\... | integer | jee-main-2024-online-1st-february-evening-shift |
lsaonhys | maths | area-under-the-curves | area-bounded-between-the-curves | The area enclosed by the curves $x y+4 y=16$ and $x+y=6$ is equal to : | [{"identifier": "A", "content": "$28-30 \\log _{\\mathrm{e}} 2$"}, {"identifier": "B", "content": "$30-28 \\log _{\\mathrm{e}} 2$"}, {"identifier": "C", "content": "$30-32 \\log _{\\mathrm{e}} 2$"}, {"identifier": "D", "content": "$32-30 \\log _{\\mathrm{e}} 2$"}] | ["C"] | null | <p>To find the enclosed area between the two curves $ x y+4 y=16 $ and $ x+y=6 $, we need to determine the region of intersection and integrate the difference of the functions over the interval where they intersect.</p>
<p>First, let's solve the equations simultaneously to find the points of intersection.</p>
<p>The ... | mcq | jee-main-2024-online-1st-february-morning-shift |
lsblh7jy | maths | area-under-the-curves | area-bounded-between-the-curves | Let the area of the region $\left\{(x, y): x-2 y+4 \geqslant 0, x+2 y^2 \geqslant 0, x+4 y^2 \leq 8, y \geqslant 0\right\}$ be $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime numbers. Then $\mathrm{m}+\mathrm{n}$ is equal to _____________. | [] | null | 119 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1g6xtj/f303b300-000e-4c93-be80-f4117fb2f938/ffa44560-d3d3-11ee-a874-bd0f61840084/file-1lt1g6xtk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1g6xtj/f303b300-000e-4c93-be80-f4117fb2f938/ffa44560-d3d3-11ee-a874-bd0f61840084... | integer | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscojzx5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\left\{(x, y): 0 \leq y \leq \min \left\{2 x, 6 x-x^2\right\}\right\}$$ is $$\mathrm{A}$$, then $$12 \mathrm{~A}$$ is equal to ________.</p> | [] | null | 304 | <p>We have</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1wj8og/d813b308-3b51-4315-93b6-2504383585b1/e7ce7b00-d413-11ee-b9d5-0585032231f0/file-1lt1wj8oh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1wj8og/d813b308-3b51-4315-93b6-2504383585b1/e7ce7b00-d413-11ee-b9... | integer | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4h6fi | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the parabolas $$y=4 x-x^2$$ and $$3 y=(x-4)^2$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{32}{9}$$\n"}, {"identifier": "B", "content": "$$\\frac{14}{3}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwhl5f/5e904568-fdcb-45b9-b0dc-a84be5bf8837/9e21fa30-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwhl5g.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwhl5f/5e904568-fdcb-45b9-b0dc-a84be5bf8837/9e21fa30-ca2d-11ee... | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse5f37p | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$\left\{(x, y): y^2 \leq 4 x, x<4, \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0, x \neq 3\right\}$$ is</p> | [{"identifier": "A", "content": "$$\\frac{32}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{16}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{8}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{64}{3}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& y^2 \leq 4 x, x<4 \\
& \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0 \\
& \text { Case - I}: y>0 \\
& \frac{x(x-1)(x-2)}{(x-3)(x-4)}>0 \\
& x \in(0,1) \cup(2,3) \\
& \text { Case }- \text { II : y<0 } \\
& \frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \in(1,2) \cup(3,4)
\... | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf0hs1k | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in sq. units) of the part of the circle $$x^2+y^2=169$$ which is below the line $$5 x-y=13$$ is $$\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$$, where $$\alpha, \beta$$ are coprime numbers. Then $$\alpha+\beta$$ is equal to __________.</p> | [] | null | 171 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt30icn5/b76421b2-6e9f-4b5a-94e8-92add4114de7/3c39af10-d4b0-11ee-8384-811001421c41/file-1lt30icn6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt30icn5/b76421b2-6e9f-4b5a-94e8-92add4114de7/3c39af10-d4b0-11ee-8384-811001421c41... | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsf0no1u | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the points of intersection of two distinct conics $$x^2+y^2=4 b$$ and $$\frac{x^2}{16}+\frac{y^2}{b^2}=1$$ lie on the curve $$y^2=3 x^2$$, then $$3 \sqrt{3}$$ times the area of the rectangle formed by the intersection points is _________.</p> | [] | null | 432 | <p>Putting $$y^2=3 x^2$$ in both the conics</p>
<p>We get $$x^2=b$$ and $$\frac{b}{16}+\frac{3}{b}=1$$</p>
<p>$$\Rightarrow \mathrm{b}=4,12 \quad(\mathrm{b}=4$$ is rejected because curves coincide)</p>
<p>$$\therefore \mathrm{b}=12$$</p>
<p>Hence points of intersection are</p>
<p>$$( \pm \sqrt{12}, \pm 6) \Rightarrow \... | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfl0gg5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area of the region $$\left\{(x, y): 0 \leq x \leq 3,0 \leq y \leq \min \left\{x^2+2,2 x+2\right\}\right\}$$ be A. Then $$12 \mathrm{~A}$$ is equal to __________.</p> | [] | null | 164 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8vkvl/77732020-1aec-484b-950e-aed3c54f53c2/8d586e10-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8vkvm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8vkvl/77732020-1aec-484b-950e-aed3c54f53c2/8d586e10-ce37-11ee... | integer | jee-main-2024-online-29th-january-evening-shift |
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 |
1lsgb2phy | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in square units) of the region bounded by the parabola $$y^2=4(x-2)$$ and the line $$y=2 x-8$$, is :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnpz3w/f37f9151-5383-482a-a0a5-619fb7183b76/d1c573c0-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnpz3x.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnpz3w/f37f9151-5383-482a-a0a5-619fb7183b76/d1c573c0-cde4-11ee-a0... | mcq | jee-main-2024-online-30th-january-morning-shift |
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 |
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 |
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 |
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 |
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 |
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 |
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>$$=216-... | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s20di | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed between the curves $$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|$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweeutim/25a3e531-aad5-453f-b0c9-bc32519834c1/9c0ef0e0-1659-11ef-b26b-2993e9de41b8/file-1lweeutin.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweeutim/25a3e531-aad5-453f-b0c9-bc32519834c1/... | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294kx | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\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 |
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 |
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 |
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>$$f\left( {{\pi \... | mcq | aieee-2005 |
55bGpDRnxxSxO83e | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let g(x) = cosx<sup>2</sup>, f(x) = $$\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>$$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \r... | mcq | jee-main-2018-offline |
K7S3uSiqOCZZI0fYX9H7G | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area of the region
<br/><br/>A = {(x, y) : 0 $$ \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}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265519/exam_images/gatckweagely4rxump82.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Area Under The Curves Question 110 English Explanation">
... | mcq | jee-main-2019-online-9th-january-evening-slot |
6Oe5Nkd1BXdoY6FgrjK1p | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area (in sq. units) of the region<br/>
A = { (x, y) $$ \in $$ R × R| 0 $$ \le $$ x $$ \le $$ 3, 0 $$ \le $$ y $$ \le $$ 4,
y $$ \le $$ x<sup>2</sup> + 3x} is : | [{"identifier": "A", "content": "$${{59} \\over 6}$$"}, {"identifier": "B", "content": "$${{26} \\over 3}$$"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "$${{53} \\over 6}$$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264987/exam_images/bujwhpy9e56yw8c0s7ne.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267785/exam_images/zjv3gmgivbhrqjdtzjj2.webp"><img src="https://res.c... | mcq | jee-main-2019-online-8th-april-morning-slot |
bauLwyzOpatnYqvKltGUV | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let S($$\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 |
nVJQQ0RIj7eHctBMc91klux59dv | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let A<sub>1</sub> be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A<sub>2</sub> be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = $${\pi \over 2}$$ in the first quadrant. Then, | [{"identifier": "A", "content": "$${A_1}:{A_2} = 1:\\sqrt 2 $$ and $${A_1} + {A_2} = 1$$"}, {"identifier": "B", "content": "$${A_1} = {A_2}$$ and $${A_1} + {A_2} = \\sqrt 2 $$"}, {"identifier": "C", "content": "$$2{A_1} = {A_2}$$ and $${A_1} + {A_2} = 1 + \\sqrt 2 $$"}, {"identifier": "D", "content": "$${A_1}:{A_2} = 1... | ["A"] | null | $${A_1} + {A_2} = \int\limits_0^{\pi /2} {\cos x.\,dx = \left. {\sin x} \right|_0^{\pi /2}} = 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264467/exam_images/ar9w3rtouhitqkvqu61e.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (... | mcq | jee-main-2021-online-26th-february-evening-slot |
aqpkRU5oUGyE2Sx1Fh1kmm2mc49 | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area bounded by the curve 4y<sup>2</sup> = x<sup>2</sup>(4 $$-$$ x)(x $$-$$ 2) is equal to : | [{"identifier": "A", "content": "$${\\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$$ [2, 4] Right hand side is positive.<br><br>By putting y = $$-$$y in equation (1), equation remains same. So, graph is... | mcq | jee-main-2021-online-18th-march-evening-shift |
1ktkeocfd | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = $${3 \over 2}$$ and the curve y = 1 + 4x $$-$$ x<sup>2</sup>, then 12 m is equal to _____________. | [] | null | 26 | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxvqwyud/3b08f599-748b-4dc3-95b3-f0ac0c10ea2e/861eb060-6af6-11ec-b350-33e20cd86462/file-1kxvqwyue.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxvqwyud/3b08f599-748b-4dc3-95b3-f0ac0c10ea2e/861eb060-6af6-11ec-b350-33e20cd8646... | integer | jee-main-2021-online-31st-august-evening-shift |
1l55hj2ov | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | <p>The area of the bounded region enclosed by the curve <br/><br/>$$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}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l99thyns/e3d7926f-e2f7-4b54-82b0-67049b426481/fa7cf380-4c79-11ed-b94d-45a8040c2a81/file-1l99thynt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l99thyns/e3d7926f-e2f7-4b54-82b0-67049b426481/fa7cf380-4c79-11ed-b94d-45a8040c2a81/fi... | mcq | jee-main-2022-online-28th-june-evening-shift |
5d6Ld5edVvT7r0SH | maths | binomial-theorem | binomial-theorem-for-any-index | The positive integer just greater than $${\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 |
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>$$ \Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\,
2 \,}} \right. }}.{1 \over {{x^2}}} + ... | mcq | jee-main-2021-online-25th-july-evening-shift |
lsamwj6p | maths | binomial-theorem | binomial-theorem-for-any-index | Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively<br/><br/> in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}$ is : | [{"identifier": "A", "content": "$\\frac{1}{9}$"}, {"identifier": "B", "content": "$\\frac{1}{4}$"}, {"identifier": "C", "content": "$\\frac{4}{9}$"}, {"identifier": "D", "content": "$\\frac{9}{4}$"}] | ["D"] | null | $\begin{aligned} & \mathrm{t}_7={ }^{18} \mathrm{C}_6\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^6={ }^{18} \mathrm{C}_6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^6} \\\\ & \mathrm{t}_{13}={ }^{18} \mathrm{C}_{12}\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^6\left(\... | mcq | jee-main-2024-online-1st-february-evening-shift |
1lsgcji8h | maths | binomial-theorem | binomial-theorem-for-any-index | <p>$$\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$$</p> | integer | jee-main-2024-online-30th-january-morning-shift |
lv0vxd0y | maths | binomial-theorem | binomial-theorem-for-any-index | <p>The sum of all rational terms in the expansion of $$\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$$ divides $... | mcq | jee-main-2024-online-4th-april-morning-shift |
PmsYoXcOZSnaCuzH | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The remainder left out when $${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)$$
<br> &nbs... | mcq | aieee-2009 |
njUyy7DYscXbuI3Jxazra | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If (27)<sup>999</sup> is divided by 7, then the remainder is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["D"] | null | We have,
<br><br>$${{{{\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\,\,\,$$ Rema... | mcq | jee-main-2017-online-8th-april-morning-slot |
y6YAjngUyXiNUzKKBU7k9k2k5e343pn | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The greatest positive integer k, for which 49<sup>k</sup> + 1 is a factor of the sum <br/>49<sup>125</sup> + 49<sup>124</sup> + ..... + 49<sup>2</sup> + 49 + 1, is: | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "63"}, {"identifier": "D", "content": "65"}] | ["C"] | null | 1 + 49 + 49<sup>2</sup>
+ ..... + 49<sup>125</sup>
<br><br>sum of G.P. = $${{1.\left( {{{49}^{126}} - 1} \right)} \over {49 - 1}}$$
<br><br>= $${{\left( {{{49}^{63}} + 1} \right)\left( {{{49}^{63}} - 1} \right)} \over {48}}$$
<br><br>Also 49<sup>63</sup> - 1
<br><br>= (1 + 48)<sup>63</sup> - 1
<br><br>= [<sup>63</sup>... | mcq | jee-main-2020-online-7th-january-morning-slot |
aFW8brFlz4zwZsvzfZ1klt9oix2 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)<sup>2022</sup> is divided by 8 is __________. | [] | null | 1 | Let x = 4k + 3<br><br>(2020 + x)<sup>2022</sup><br><br>= (2020 + 4k + 3)<sup>2022</sup><br><br>= (4(505) + 4k + 3)<sup>2022</sup>
<br><br>= (4P + 3)<sup>2022</sup><br><br>= (4P + 4 $$-$$ 1)<sup>2022</sup><br><br>= (4A $$-$$ 1)<sup>2022</sup><br><br><sup>2022</sup>C<sub>0</sub>(4A)<sup>0</sup>($$-$$1)<sup>2022</sup> + <... | integer | jee-main-2021-online-25th-february-evening-slot |
ofQvn9MZjWjzpi3QoA1klt9p75z | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The total number of two digit numbers 'n', such that 3<sup>n</sup> + 7<sup>n</sup> is a multiple of 10, is __________. | [] | null | 45 | $$ \because $$ $${7^n} = {(10 - 3)^n} = 10k + {( - 3)^n}$$<br><br>$${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$$<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264628/exam_images/ukvxzlftpfikzghxsw5c.webp"><source media="(max-width: 500px)" srcset="https:... | integer | jee-main-2021-online-25th-february-evening-slot |
5UlwtPdmE6ETPIebwf1kmjbieu8 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If (2021)<sup>3762</sup> is divided by 17, then the remainder is __________. | [] | null | 4 | 2021 = 17m - 2
<br><br>(2021)<sup>3762</sup> = (17m $$-$$ 2)<sup>3762</sup> = multiple of 17 + 2<sup>3762</sup><br><br>= 17$$\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 |
1ktgq16qr | maths | binomial-theorem | divisibility-concept-and-remainder-concept | 3 $$\times$$ 7<sup>22</sup> + 2 $$\times$$ 10<sup>22</sup> $$-$$ 44 when divided by 18 leaves the remainder __________. | [] | null | 15 | 3(1 + 6)<sup>22</sup> + 2 . (1 + 9)<sup>22</sup> $$-$$ 44 = (3 + 2 $$-$$ 44) = 18 . I<br><br>= $$-$$ 39 + 18 . I<br><br>= (54 $$-$$ 39) + 18(I $$-$$ 3)<br><br>= 15 + 18I<sub>1</sub><br><br>$$\Rightarrow$$ Remainder = 15 | integer | jee-main-2021-online-27th-august-evening-shift |
1l587i0nd | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when (2021)<sup>2023</sup> is divided by 7 is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["C"] | null | (2021)<sup>2023</sup></p>
<p>= (2016 + 5)<sup>2023</sup> [here 2016 is divisible by 7]</p>
<p>= <sup>2023</sup>C<sub>0</sub> (2016)<sup>2023</sup> + .......... + <sup>2023</sup>C<sub>2022</sub> (2016) (5)<sup>2022</sup> + <sup>2023</sup>C<sub>2023</sub> (5)<sup>2023</sup></p>
<p>= 2016 [<sup>2023</sup>C<sub>0</sub> . (... | mcq | jee-main-2022-online-26th-june-morning-shift |
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 |
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>$$ = {{{... | integer | jee-main-2022-online-24th-june-evening-shift |
1l5c0s15h | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when 3<sup>2022</sup> is divided by 5 is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>$${3^{2022}}$$</p>
<p>$$ = {({3^2})^{1011}}$$</p>
<p>$$ = {(9)^{1011}}$$</p>
<p>$$ = {(10 - 1)^{1011}}$$</p>
<p>$$ = {}^{1011}{C_0}{(10)^{1011}} + \,\,.....\,\, + \,\,{}^{1011}{C_{1010}}\,.\,{(10)^1} - {}^{1011}{C_{1011}}$$</p>
<p>$$ = 10\left[ {{}^{1011}{C_0}{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}... | mcq | jee-main-2022-online-24th-june-morning-shift |
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>$${2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}... | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6jb84eh | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$(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)$$ (multiple of 7)</... | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6m5vk3w | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$7^{2022}+3^{2022}$$ is divided by 5 is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["C"] | null | $$
\begin{aligned}
& 7^{2022}+3^{2022} \\\\
& =\left(7^2\right)^{1011}+\left(3^2\right)^{1011} \\\\
&=(50-1)^{1011}+(10-1)^{1011} \\\\
&= (50^{1011}-1011.50^{1010}+\ldots-1) \\\\
& + (10^{1011}-1011.10^{1010}+\ldots . .-1) \\\\
&= 5 m-1+5 n-1=5(m+n)-2 \\\\
&= 5(m+n)-5+3=5(m+n-1)+3 \\\\
&= 5 k+3 \\\\
& \therefore \text ... | mcq | jee-main-2022-online-28th-july-morning-shift |
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 |
1ldptern1 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder on dividing $$5^{99}$$ by 11 is ____________.</p> | [] | null | 9 | $5^{99}=5^{4} .5^{95}$
<br/><br/>$=625\left[5^{5}\right]^{19}$
<br/><br/>$=625[3125]^{19}$
<br/><br/>$=625[3124+1]^{19}$
<br/><br/>$=625[11 \mathrm{k} \times 19+1]$
<br/><br/>$=625 \times 11 \mathrm{k} \times 19+625$
<br/><br/>$=11 \mathrm{k}_{1}+616+9$
<br/><br/>$=11\left(\mathrm{k}_{2}\right)+9$
<br/><br/>Rem... | integer | jee-main-2023-online-31st-january-morning-shift |
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 |
1ldu60obo | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when (2023)$$^{2023}$$ is divided by 35 is __________.</p> | [] | null | 7 | $$
\begin{aligned}
& (2023)^{2023} \\\\
& =(2030-7)^{2023} \\\\
& =(35 \mathrm{~K}-7)^{2023} \\\\
& ={ }^{2023} \mathrm{C}_0(35 \mathrm{~K})^{2023}(-7)^0+{ }^{2023} \mathrm{C}_1(35 \mathrm{~K})^{2022}(-7)+ \\\\
& \ldots . .+\ldots \ldots .+{ }^{2023} \mathrm{C}_{2023}(-7)^{2023} \\\\
& =35 \mathrm{~N}-7^{2023} \\\\
& \... | integer | jee-main-2023-online-25th-january-evening-shift |
1lgoy2e8y | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder, when $$7^{103}$$ is divided by 17, is __________</p> | [] | null | 12 | $7^{103}=7 \times 7^{102}$
<br/><br/>$$
\begin{aligned}
& =7 \times(49)^{51} \\\\
& =7 \times(51-2)^{51}
\end{aligned}
$$
<br/><br/>Remainder = $7 \times(-2)^{51}$
<br/><br/>$$
\begin{aligned}
& =-7\left(2^3 \cdot(16)^{12}\right) \\\\
& =-56(17-1)^{12}
\end{aligned}
$$
<br/><br/>Remainder $=-56 \times(-1)^{12}=-56+68=1... | integer | jee-main-2023-online-13th-april-evening-shift |
1lgpy5uw8 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Fractional part of the number $$\frac{4^{2022}}{15}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{8}{15}$$"}, {"identifier": "B", "content": "$$\\frac{4}{15}$$"}, {"identifier": "C", "content": "$$\\frac{1}{15}$$"}, {"identifier": "D", "content": "$$\\frac{14}{15}$$"}] | ["C"] | null | $$
\begin{aligned}
& \left\{\frac{4^{2022}}{15}\right\}=\left\{\frac{2^{4044}}{15}\right\} \\\\
& =\left\{\frac{(1+15)^{1011}}{15}\right\} \\\\
& =\frac{1}{15}
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgvpgm7l | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Let the number $$(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)$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "5"}] | ["D"] | null | We have, $(22)^{2022}+(2022)^{22}$
<br/><br/>As 2022 is completely divisible by 3
<br/><br/>So, $(2022)^{22}$ is also divisible by 3
<br/><br/>$(22)^{2022}=(21+1)^{2022}=(3 \times 7+1)^{2022}=7 m+1$
<br/><br/>$\Rightarrow(22)^{2022}$ leave a remainder 1 , when divisible by 3 .
<br/><br/>$\therefore(22)^{2022}+(2022)^{2... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgyl43d6 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>$$25^{190}-19^{190}-8^{190}+2^{190}$$ is divisible by :</p> | [{"identifier": "A", "content": "14 but not by 34"}, {"identifier": "B", "content": "neither 14 nor 34"}, {"identifier": "C", "content": "both 14 and 34"}, {"identifier": "D", "content": "34 but not by 14"}] | ["D"] | null | The given expression is divisible by 6 and 17 .
<br/><br/>Also, $25^{190}-8^{190}$ is not divisible by 7
<br/><br/>but $19^{190}-2^{190}$ is divisible by 7 ,
<br/><br/>So, $25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by 34 but not by 14 . | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh00oqex | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The largest natural number $$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}
$$
<br/><br/>Highest powers of 3 is greater than 66. So, their g.i.f. is always 0.
<br/><br/>$\therefore$ Required natural number $=22+7+2=31$ | integer | jee-main-2023-online-8th-april-morning-shift |
1lh2y2hd7 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Among the statements :</p>
<p>(S1) : $$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 |
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 |
luy9clej | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$428^{2024}$$ is divided by 21 is __________.</p> | [] | null | 1 | <p>$$\begin{aligned}
& 428=21 \times 20+8 \\
\Rightarrow \quad & (428)^{2024} \equiv(20 \times 21+8)^{2024} \equiv 8^{2024}(\bmod 21) \\
& 8^2=21 \times 3+1 \\
& 8^{2024}=(21 \times 3+1)^{1012} \\
\Rightarrow \quad & 8^{2024} \equiv(21 \times 3+1)^{1012}(\bmod 21) \\
& \equiv 1^{2012}(\bmod 21) \\
& 428^{2024} \equiv 1... | integer | jee-main-2024-online-9th-april-morning-shift |
v11RSlXUNBj0iZzn | maths | binomial-theorem | general-term | The coefficients of $${x^p}$$ and $${x^q}$$ in the expansion of $${\left( {1 + x} \right)^{p + q}}$$ are | [{"identifier": "A", "content": "equal "}, {"identifier": "B", "content": "equal with opposite signs "}, {"identifier": "C", "content": "reciprocals of each other "}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | Here in this expansion $${\left( {1 + x} \right)^{p + q}}$$
<br><br>The general term = $${T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}$$
<br><br>$$\therefore$$ $${x^p}$$ will be present in the term = $${}^{p + q}{C_p}.{\left( x \right)^p}$$
<br><br>So coefficient of $${x^p}$$ = $${}^{p + q}{C_p}$$
<br><br>And $${x... | mcq | aieee-2002 |
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 |
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