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1l55j8k9n | maths | complex-numbers | conjugate-of-complex-numbers | <p>Sum of squares of modulus of all the complex numbers z satisfying $$\overline z = i{z^2} + {z^2} - z$$ is equal to ___________.</p> | [] | null | 2 | Let $z=x+i y$
<br/><br/>
So $2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)$
<br/><br/>
$\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and
<br/><br/>
$$
x^{2}-y^{2}+2 x y=0\quad\dots(ii)
$$
<br/><br/>
From (i) and (ii) we get
<br/><br/>
$$
x=0 \text { or } y=-\frac{1}{2}
$$
<br/><br/>
When $x=0$ we get $y=0$
<br/><br/>
Whe... | integer | jee-main-2022-online-28th-june-evening-shift | 5,455 |
1l6jennth | maths | complex-numbers | conjugate-of-complex-numbers | <p>Let $$S=\left\{z \in \mathbb{C}: z^{2}+\bar{z}=0\right\}$$. Then $$\sum\limits_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z))$$ is equal to ______________.</p> | [] | null | 0 | <p>$$\because$$ $${z^2} + \overline z = 0$$</p>
<p>Let $$z = x + iy$$</p>
<p>$$\therefore$$ $${x^2} - {y^2} + 2ixy + x - iy = 0$$</p>
<p>$$({x^2} - {y^2} + x) + i(2xy - y) = 0$$</p>
<p>$$\therefore$$ $${x^2} + {y^2} = 0$$ and $$(2x - 1)y = 0$$</p>
<p>if $$x = \, + \,{1 \over 2}$$ then $$y = \, \pm \,{{\sqrt 3 } \over ... | integer | jee-main-2022-online-27th-july-morning-shift | 5,458 |
1l6p0k7zx | maths | complex-numbers | conjugate-of-complex-numbers | <p>If $$z=2+3 i$$, then $$z^{5}+(\bar{z})^{5}$$ is equal to :</p> | [{"identifier": "A", "content": "244"}, {"identifier": "B", "content": "224"}, {"identifier": "C", "content": "245"}, {"identifier": "D", "content": "265"}] | ["A"] | null | <p>$$z = (2 + 3i)$$</p>
<p>$$ \Rightarrow {z^5} = (2 + 3i){\left( {{{(2 + 3i)}^2}} \right)^2}$$</p>
<p>$$ = (2 + 3i){( - 5 + 12i)^2}$$</p>
<p>$$ = (2 + 3i)( - 119 - 120i)$$</p>
<p>$$ = - 238 - 240i - 357i + 360$$</p>
<p>$$ = 122 - 597i$$</p>
<p>$${\overline z ^5} = 122 + 597i$$</p>
<p>$${z^5} + {\overline z ^5} = 244$... | mcq | jee-main-2022-online-29th-july-morning-shift | 5,459 |
lgnyjdfm | maths | complex-numbers | conjugate-of-complex-numbers | If the set $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\right\}$ is equal to<br/><br/> the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to : | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "27"}, {"identifier": "C", "content": "42"}, {"identifier": "D", "content": "30"}] | ["D"] | null | Let $z_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)$<br/><br/>
Let $\mathrm{z}=3+\mathrm{iy}$<br/><br/> $\bar{z}=3-i y$<br/><br/>
$$
\begin{aligned}
& z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)} \\\\
& =\frac{9+y^2+i(2 y)}{8-8 i y} \\\\
& =\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)} \\\\
& \op... | mcq | jee-main-2023-online-15th-april-morning-shift | 5,460 |
1lgow71ev | maths | complex-numbers | conjugate-of-complex-numbers | <p>Let $$S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}$$. Then $$\sum_\limits{z \in \mathrm{S}}|z|^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{7}{2}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\frac{5}{2}$$"}] | ["B"] | null | Let $z=x+i y$
<br/><br/>$$
\begin{aligned}
&\bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right) \\\\
&\Rightarrow x-i y=i\left(x^2-y^2+2 i x y+x\right) \\\\
& x-i y=i\left(x^2-y^2+x\right)-2 x y \\\\
&x=-2 x y \Rightarrow x(2 y+1)=0 \\\\
&\Rightarrow x=0, y=\frac{-1}{2} ........(i)\\\\
&-y=x^2-y^2+x ...........(ii)
... | mcq | jee-main-2023-online-13th-april-evening-shift | 5,461 |
1lgsugbnn | maths | complex-numbers | conjugate-of-complex-numbers | <p>For $$a \in \mathbb{C}$$, let $$\mathrm{A}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)\}$$ and $$\mathrm{B}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}$$. Then among the two statements :</p>
<p>(S1): If $$\operatorname{Re}(a), \operatornam... | [{"identifier": "A", "content": "both are false"}, {"identifier": "B", "content": "only (S1) is true"}, {"identifier": "C", "content": "only (S2) is true"}, {"identifier": "D", "content": "both are true"}] | ["A"] | null | We are given that $a \in \mathbb{C}$ and $z \in \mathbb{C}$.
<br/><br/>Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \in \mathbb{R}$
<br/><br/>We are also given two sets A and B defined as follows :
<br/><br/>- A is the set of all complex numbers $z$ for which the real part of $(a + \overline{z}... | mcq | jee-main-2023-online-11th-april-evening-shift | 5,462 |
rPNBPY8xjQoyu1Iq | maths | complex-numbers | cube-roots-of-unity | If $$z = x - iy$$ and $${z^{{1 \over 3}}} = p + iq$$, then
<br/><br/>$${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$ is equal to : | [{"identifier": "A", "content": "- 2"}, {"identifier": "B", "content": "- 1"}, {"identifier": "C", "content": "2 "}, {"identifier": "D", "content": "1"}] | ["A"] | null | Given $${z^{{1 \over 3}}} = p + iq$$
<br><br>$$ \Rightarrow $$ z = (p + iq)<sup>3</sup>
<br><br>= p<sup>3</sup> + (iq)<sup>3</sup> +3p(iq)(p + iq)
<br><br>= p<sup>3</sup> - iq<sup>3</sup> +3ip<sup>2</sup>q - 3pq<sup>2</sup>
<br><br>= p(p<sup>2</sup> - 3q<sup>2</sup>) - iq(q<sup>2</sup> - 3p<sup>2</sup>)
<br><br>Given t... | mcq | aieee-2004 | 5,463 |
b3Gq6JEgA0LBu57T | maths | complex-numbers | cube-roots-of-unity | If the cube roots of unity are 1, $$\omega \,,\,{\omega ^2}$$ then the roots of the equation $${(x - 1)^3}$$ + 8 = 0, are : | [{"identifier": "A", "content": "$$ - 1, - 1 + 2\\,\\,\\omega , - 1 - 2\\,\\,{\\omega ^2}$$ "}, {"identifier": "B", "content": "$$ - 1, - 1, - 1$$ "}, {"identifier": "C", "content": "$$ - 1,1 - 2\\omega ,1 - 2{\\omega ^2}$$ "}, {"identifier": "D", "content": "$$ - 1,1 + 2\\omega ,1 + 2{\\omega ^2}$$ "}] | ["C"] | null | $${\left( {x - 1} \right)^3} + 8 = 0$$
<br><br>$$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$
<br><br>$$ \Rightarrow x - 1 = - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$
<br><br>or $$\,\,\,x = - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,... | mcq | aieee-2005 | 5,464 |
15Ymvfe9vqnI6Ico | maths | complex-numbers | cube-roots-of-unity | If $$\alpha ,\beta \in C$$ are the distinct roots of the equation
<br/>x<sup>2</sup> - x + 1 = 0, then $${\alpha ^{101}} + {\beta ^{107}}$$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["D"] | null | Given equation,
<br><br>x<sup>2</sup> $$-$$ x + 1 = 0
<br><br>Roots of this equation
<br><br>x = $${{1 \pm \sqrt 3 i} \over 2}$$
<br><br>$$\therefore\,\,\,$$ $$ \propto \, = \,{{1 + \sqrt 3 \,i} \over 2}$$
<br><br>and $$\beta = \,{{1 - \sqrt 3 \,i} \over 2}$$
<br><br>We know;
<br><br>$$\omega = {{ - 1 + \sqrt 3 \,i... | mcq | jee-main-2018-offline | 5,466 |
5MdWZd7KqONKCwZmNrjgy2xukfagwe44 | maths | complex-numbers | cube-roots-of-unity | If a and b are real numbers such that<br/> $${\left( {2 + \alpha } \right)^4} = a + b\alpha $$ <br/>where $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$ then a + b is
equal to : | [{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "57"}] | ["B"] | null | $$\alpha = \omega $$ as given $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$
<br><br>$$ \Rightarrow {(2 + \omega )^4} = a + b\omega \,({\omega ^3} = 1)$$<br><br>$$ \Rightarrow {2^4} + {4.2^3}\omega + {6.2^2}{\omega ^3} + 4.2.\,{\omega ^3} + {\omega ^4} = a + b\omega $$<br><br>$$ \Rightarrow 16 + 32\omega + 24{\omega ^2}... | mcq | jee-main-2020-online-4th-september-evening-slot | 5,467 |
ZDaEWjLQBunSbbEG | maths | complex-numbers | de-moivre's-theorem | If $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$ then : | [{"identifier": "A", "content": "x = 2n + 1, where n is any positive integer"}, {"identifier": "B", "content": "x = 4n , where n is any positive integer"}, {"identifier": "C", "content": "x = 2n, where n is any positive integer\n"}, {"identifier": "D", "content": "x = 4n + 1, where n is any positive integer.\n"}] | ["B"] | null | $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$
<br><br>$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$
<br><br>$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$
<br>... | mcq | aieee-2003 | 5,468 |
xAltR1yM4WVfFWDJ7KLTZ | maths | complex-numbers | de-moivre's-theorem | Let z<sub>0</sub> be a root of the quadratic equation, x<sup>2</sup> + x + 1 = 0, If z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$, then arg z is equal to : | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "0"}] | ["A"] | null | 1 + x + x<sup>2</sup> = 0
<br><br>x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$
<br><br>z<sub>0</sub> = w, w<sup>2</sup>
<br><br>Now
<br><br>z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$
<br><br>z = 3 + 6iw<sup>81</sup> $$-$$ 3iw<sup>93</sup> (... | mcq | jee-main-2019-online-9th-january-evening-slot | 5,470 |
241XVRmztPvAOIbOcO5fL | maths | complex-numbers | de-moivre's-theorem | Let $$z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.$$ If R(z) and 1(z) respectively denote the real and imaginary parts of z, then : | [{"identifier": "A", "content": "R(z) = $$-$$ \n3"}, {"identifier": "B", "content": "R(z) < 0 and I(z) > 0"}, {"identifier": "C", "content": "I(z) = 0"}, {"identifier": "D", "content": "R(z) > 0 and I(z) > 0"}] | ["C"] | null | $$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$$
<br><br>$$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$$
<br><br>$$ = {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$$
<br><br>$$ = \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {... | mcq | jee-main-2019-online-10th-january-evening-slot | 5,471 |
GpTWf3x6bsc54rlvWQCvy | maths | complex-numbers | de-moivre's-theorem | Let $${\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\,$$ where x and y are real numbers, then y $$-$$ x equals : | [{"identifier": "A", "content": "$$-$$ 85 "}, {"identifier": "B", "content": "85"}, {"identifier": "C", "content": "$$-$$ 91 "}, {"identifier": "D", "content": "91"}] | ["D"] | null | $${\left( { - 2 - {i \over 3}} \right)^3} = - {{{{\left( {6 + i} \right)}^3}} \over {27}}$$
<br><br>$$ = {{ - 198 - 107i} \over {27}} = {{x + iy} \over {27}}$$
<br><br>Hence, $$y - x = 198 - 107 = 91$$ | mcq | jee-main-2019-online-11th-january-morning-slot | 5,472 |
3MeWrbrvU6nSZ6rQ0zoop | maths | complex-numbers | de-moivre's-theorem | If $$z = {{\sqrt 3 } \over 2} + {i \over 2}\left( {i = \sqrt { - 1} } \right)$$,
<br/><br/>then (1 + iz + z<sup>5</sup> + iz<sup>8</sup>)<sup>9</sup> is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "\u20131"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "(-1 + 2i)<sup>9</sup>"}] | ["B"] | null | $$z = {{\sqrt 3 } \over 2} + {i \over 2}$$
<br><br>$$ \Rightarrow $$ z = $$\cos {\pi \over 6}$$ + i $$\sin {\pi \over 6}$$
<br><br>$$ \Rightarrow $$ z = $${e^{i{\pi \over 6}}}$$
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267414/exam_images/mjophfx1aur2l... | mcq | jee-main-2019-online-8th-april-evening-slot | 5,473 |
jZ7ETjEkx74JNPhvsujgy2xukfqax1w7 | maths | complex-numbers | de-moivre's-theorem | The value of $${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}$$ is : | [{"identifier": "A", "content": "\u20132<sup>15</sup>i"}, {"identifier": "B", "content": "\u20132<sup>15</sup>"}, {"identifier": "C", "content": "2<sup>15</sup>i"}, {"identifier": "D", "content": "6<sup>5</sup>"}] | ["A"] | null | $${\left( {{{ - 1 + i\sqrt 3 } \over {1 - i}}} \right)^{30}}$$
<br><br>= $${\left( {{{2\omega } \over {1 - i}}} \right)^{30}}$$
<br><br>= $${{{2^{30}}.{\omega ^{30}}} \over {{{\left( {{{\left( {1 - i} \right)}^2}} \right)}^{15}}}}$$
<br><br>= $${{{2^{30}}.1} \over {{{\left( {1 + {i^{^2}} - 2i} \right)}^{15}}}}$$
<br><b... | mcq | jee-main-2020-online-5th-september-evening-slot | 5,475 |
ZH8p2PoFt0hgYa1xTs1klrms6yw | maths | complex-numbers | de-moivre's-theorem | Let $$i = \sqrt { - 1} $$. If $${{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 - i)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 + i)}^{24}}}} = k$$, and $$n = [|k|]$$ be the greatest integral part of | k |. Then $$\sum\limits_{j = 0}^{n + 5} {{{(j + 5)}^2} - \sum\limits_{j = 0}^{n + 5} ... | [] | null | 310 | $${(1 + i)^2} = 1 + {i^2} + 2i = 1 - 1 + 2i = 2i$$<br><br>$${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$$<br><br>We know,<br><br>$$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $$<br><br>$$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $$<br><br>and $$ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$$<br><br>$$ \R... | integer | jee-main-2021-online-24th-february-evening-slot | 5,476 |
1ktd1lmo6 | maths | complex-numbers | de-moivre's-theorem | If $${\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)$$, then p and q are roots of the equation : | [{"identifier": "A", "content": "$${x^2} - \\left( {\\sqrt 3 - 1} \\right)x - \\sqrt 3 = 0$$"}, {"identifier": "B", "content": "$${x^2} + \\left( {\\sqrt 3 + 1} \\right)x + \\sqrt 3 = 0$$"}, {"identifier": "C", "content": "$${x^2} + \\left( {\\sqrt 3 - 1} \\right)x - \\sqrt 3 = 0$$"}, {"identifier": "D", "content... | ["A"] | null | $${\left( {2{e^{i\pi /6}}} \right)^{100}} = {2^{99}}(p + iq)$$<br><br>$${2^{100}}\left( {\cos {{50\pi } \over 3} + i\sin {{50\pi } \over 3}} \right) = {2^{99}}(p + iq)$$<br><br>$$p + iq = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)$$<br><br>p = $$-$$1, q = $$\sqrt 3 $$<br><br>$${x^2} - (\sqrt 3 ... | mcq | jee-main-2021-online-26th-august-evening-shift | 5,477 |
1ktd50jva | maths | complex-numbers | de-moivre's-theorem | The least positive integer n such that $${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}},i = \sqrt { - 1} $$ is a positive integer, is ___________. | [] | null | 6 | $${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}} = {{{{(2i)}^n}} \over {{{( - 2i)}^{{{n - 2} \over 2}}}}}$$<br><br>$$ = {{{{(2i)}^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}} = {{{2^{{{n + 2} \over 2}}};{i^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}}$$<br><br>This is positive integer for n... | integer | jee-main-2021-online-26th-august-evening-shift | 5,478 |
ldo8iriz | maths | complex-numbers | de-moivre's-theorem | The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to : | [{"identifier": "A", "content": "$\\cos \\frac{\\pi}{12}-i \\sin \\frac{\\pi}{12}$"}, {"identifier": "B", "content": "$\\sqrt{2}\\left(\\cos \\frac{\\pi}{12}+i \\sin \\frac{\\pi}{12}\\right)$"}, {"identifier": "C", "content": "$\\sqrt{2} i\\left(\\cos \\frac{5 \\pi}{12}-i \\sin \\frac{5 \\pi}{12}\\right)$"}, {"identifi... | ["D"] | null | $\mathrm{Z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}=\frac{\mathrm{i}-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}}$
<br/><br/>$=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}} \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}} \mathrm{i}}{\frac{1}{2}-\sqrt{3 / 2} \mathrm{i}}=\frac{\sqrt{3}... | mcq | jee-main-2023-online-31st-january-evening-shift | 5,479 |
1ldwx26gd | maths | complex-numbers | de-moivre's-theorem | <p>The value of $${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$$ is</p> | [{"identifier": "A", "content": "$$ - {1 \\over 2}\\left( {1 - i\\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$ - {1 \\over 2}\\left( {\\sqrt 3 - i} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 2}\\left( {1 - i\\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {\\sq... | ["B"] | null | $z=\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^{3}$
<br/><br/>
$$
\begin{aligned}
& 1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}=1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18} \\\\
& =1+2 \cos ^{2} \frac{5 \pi}{36}-1+2 i \sin \frac{5 \pi}{36} \cos \... | mcq | jee-main-2023-online-24th-january-evening-shift | 5,480 |
de5FYCyGl5T5tbGQ | maths | complex-numbers | modulus-of-complex-numbers | If $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$, then $$z$$ lies on : | [{"identifier": "A", "content": "an ellipse"}, {"identifier": "B", "content": "a circle"}, {"identifier": "C", "content": "a straight line"}, {"identifier": "D", "content": "a parabola"}] | ["C"] | null | Given $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$
<br><br>$$\therefore$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$$
<br><br>$$ \Rightarrow $$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$$
<b... | mcq | aieee-2005 | 5,483 |
iSzQWpXnIqltJoVX | maths | complex-numbers | modulus-of-complex-numbers | A complex number z is said to be unimodular if $$\,\left| z \right| = 1$$. Suppose $${z_1}$$ and $${z_2}$$ are complex numbers such that $${{{z_1} - 2{z_2}} \over {2 - {z_1}\overline {{z_2}} }}$$ is unimodular and $${z_2}$$ is not unimodular. Then the point $${z_1}$$ lies on a : | [{"identifier": "A", "content": "circle of radius 2."}, {"identifier": "B", "content": "circle of radius $${\\sqrt 2 }$$."}, {"identifier": "C", "content": "straight line parallel to x-axis"}, {"identifier": "D", "content": "straight line parallel to y-axis."}] | ["A"] | null | $$\left| {{{{z_1} - 2{z_2}} \over {2 - {z_1}{{\overline z }_2}}}} \right| = 1 \Rightarrow {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2}$$
<br><br>$$ \Rightarrow \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,... | mcq | jee-main-2015-offline | 5,486 |
kVD6ruGEq9JlGdhtNQrqq | maths | complex-numbers | modulus-of-complex-numbers | If |z $$-$$ 3 + 2i| $$ \le $$ 4 then the difference between the greatest value and the least value of |z| is : | [{"identifier": "A", "content": "$$2\\sqrt {13} $$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "4 + $$\\sqrt {13} $$"}, {"identifier": "D", "content": "$$\\sqrt {13} $$"}] | ["A"] | null | $$\left| {z - \left( {3 - 2i} \right)} \right| \le 4$$ represents a circle whose center is (3, $$-$$2) and radius = 4.
<br><br>$$\left| z \right|$$ = $$\left| z -0\right|$$ represents the distance of point 'z' from origin (0, 0)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266491/... | mcq | jee-main-2018-online-15th-april-evening-slot | 5,487 |
l79mOqWrm4oKN8xh1VHfS | maths | complex-numbers | modulus-of-complex-numbers | Let z<sub>1</sub> and z<sub>2</sub> be any two non-zero complex numbers such that $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.$$ If $$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$ then : | [{"identifier": "A", "content": "$${\\rm I}m\\left( z \\right) = 0$$"}, {"identifier": "B", "content": "$$\\left| z \\right| = \\sqrt {{17 \\over 2}} $$"}, {"identifier": "C", "content": "$$\\left| z \\right| =$$ $${1 \\over 2}\\sqrt {9 + 16{{\\cos }^2}\\theta } $$"}, {"identifier": "D", "content": "Re(z) $$=$$ 0"}] | ["C"] | null | Given, $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$$
<br><br>$$ \Rightarrow $$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$
<br><br>$$ \Rightarrow $$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$
<br><br>As we know, for any co... | mcq | jee-main-2019-online-10th-january-morning-slot | 5,488 |
LrLuK5pVFXqE5Ymzxn9Vn | maths | complex-numbers | modulus-of-complex-numbers | Let z<sub>1</sub> and z<sub>2</sub> be two complex numbers satisfying | z<sub>1</sub> | = 9 and | z<sub>2</sub> – 3 – 4i | = 4. Then the minimum value of
| z<sub>1</sub> – z<sub>2</sub> | is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\sqrt 2 $$"}] | ["A"] | null | $$\left| {{z_1}} \right| = 9,\,\,\left| {{z_2} - \left( {3 + 4i} \right)} \right| = 4$$
<br><br>$${C_1},(0,0)$$ radius r<sub>1</sub> = 9
<br><br>C<sub>2</sub> (3, 4), radius r<sub>2</sub> = 4
<br><br>C<sub>1</sub>C<sub>2</sub> = $$\left| {{r_1} - {r_2}} \right| = 5$$
<br><br>$$ \therefore $$ Circle touches i... | mcq | jee-main-2019-online-12th-january-evening-slot | 5,490 |
lOLmjWCdgMiaiJ2BWW7k9k2k5khn9k5 | maths | complex-numbers | modulus-of-complex-numbers | If z be a complex number satisfying
|Re(z)| + |Im(z)| = 4, then |z| cannot be : | [{"identifier": "A", "content": "$$\\sqrt {10} $$"}, {"identifier": "B", "content": "$$\\sqrt {7} $$"}, {"identifier": "C", "content": "$$\\sqrt {{{17} \\over 2}} $$"}, {"identifier": "D", "content": "$$\\sqrt {8} $$"}] | ["B"] | null | Let z = x + iy
<br><br>given that |Re(z)| + |Im(z)| = 4
<br><br>$$ \therefore $$ |x| + |y| = 4
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264539/exam_images/tkloovrsmfpi6fjklioa.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th... | mcq | jee-main-2020-online-9th-january-evening-slot | 5,492 |
peRrOzS3nUH7gT3DYS1klri4bcj | maths | complex-numbers | modulus-of-complex-numbers | If the least and the largest real values of a, for which the <br/>equation z + $$\alpha $$|z – 1| + 2i = 0
(z $$ \in $$ C and i = $$\sqrt { - 1} $$) has a solution, are p and q respectively; then 4(p<sup>2</sup> + q<sup>2</sup>) is equal to __________. | [] | null | 10 | $$x + iy + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0$$<br><br>$$ \therefore $$ y + 2 = 0 and $$x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$$<br><br>y = $$-$$2 & $${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$$<br><br>$${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^... | integer | jee-main-2021-online-24th-february-morning-slot | 5,493 |
K1j14MVoLV2InNAjlE1kmhwyz0n | maths | complex-numbers | modulus-of-complex-numbers | Let a complex number z, |z| $$\ne$$ 1, <br/><br/>satisfy $${\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2$$. Then, the largest value of |z| is equal to ____________. | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}] | ["D"] | null | $${{|z| + 11} \over {{{(|z| - 1)}^2}}} \ge {1 \over 2}$$<br><br>$$2|z| + 22 \ge {(|z| - 1)^2}$$<br><br>$$2|z| + 22 \ge \,|z{|^2} - 2|z| + 1$$<br><br>$$|z{|^2} - 4|z| - 21 \le 0$$<br><br>$$(|z| - 7)(|z| + 3) \le 0$$<br><br>$$ \Rightarrow \,|z| \le 7$$<br><br>$$ \therefore $$ $$|z{|_{\max }} = 7$$ | mcq | jee-main-2021-online-16th-march-morning-shift | 5,494 |
ye1emd881Y8Z3i9nuW1kmiwzsy1 | maths | complex-numbers | modulus-of-complex-numbers | The least value of |z| where z is complex number which satisfies the inequality $$\exp \left( {{{(|z| + 3)(|z| - 1)} \over {||z| + 1|}}{{\log }_e}2} \right) \ge {\log _{\sqrt 2 }}|5\sqrt 7 + 9i|,i = \sqrt { - 1} $$, is equal to : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\sqrt 5 $$"}] | ["B"] | null | Let | z | = t, t $$ \ge $$ 0<br><br>$${e^{{{(t + 3)(t - 1)} \over {t + 1}}{{\log }_e}2}} \ge {\log _{\sqrt 2 }}16 = 8$$ ($$ \because $$ t + 1 > 0)<br><br>$${2^{{{(t + 3)(t - 1)} \over {t + 1}}}} \ge {2^3}$$<br><br>$${{(t + 3)(t - 1)} \over {t + 1}} \ge 3$$<br><br>$${t^2} + 2t - 3 \ge 3t + 3$$<br><br>$${t^2} - t - 6 ... | mcq | jee-main-2021-online-16th-march-evening-shift | 5,495 |
1ktk9aapu | maths | complex-numbers | modulus-of-complex-numbers | If z is a complex number such that $${{z - i} \over {z - 1}}$$ is purely imaginary, then the minimum value of | z $$-$$ (3 + 3i) | is : | [{"identifier": "A", "content": "$$2\\sqrt 2 - 1$$"}, {"identifier": "B", "content": "$$3\\sqrt 2 $$"}, {"identifier": "C", "content": "$$6\\sqrt 2 $$"}, {"identifier": "D", "content": "$$2\\sqrt 2 $$"}] | ["D"] | null | $${{z - i} \over {z - 1}}$$ is purely imaginary number<br><br>Let $$z = x + iy$$<br><br>$$\therefore$$ $${{x + i(y - 1)} \over {(x - 1) + i(y)}} \times {{(x - 1) - iy} \over {(x - 1) - iy}}$$<br><br>$$ \Rightarrow {{x(x - 1) + y(y - 1) + i( - y - x + 1)} \over {{{(x - 1)}^2} + {y^2}}}$$ is purely imaginary number<br><b... | mcq | jee-main-2021-online-31st-august-evening-shift | 5,496 |
1l5bay6xr | maths | complex-numbers | modulus-of-complex-numbers | <p>Let S = {z $$\in$$ C : |z $$-$$ 3| $$\le$$ 1 and z(4 + 3i) + $$\overline z $$(4 $$-$$ 3i) $$\le$$ 24}. If $$\alpha$$ + i$$\beta$$ is the point in S which is closest to 4i, then 25($$\alpha$$ + $$\beta$$) is equal to ___________.</p> | [] | null | 80 | <p>Here $$|z - 3| < 1$$</p>
<p>$$ \Rightarrow {(x - 3)^2} + {y^2} < 1$$</p>
<p>and $$z = (4 + 3i) + \overline z (4 - 3i) \le 24$$</p>
<p>$$ \Rightarrow 4x - 3y \le 12$$</p>
<p>$$\tan \theta = {4 \over 3}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5v605na/b9bf1f33-cb41-4f86-92e8-80... | integer | jee-main-2022-online-24th-june-evening-shift | 5,498 |
1l5c0riod | maths | complex-numbers | modulus-of-complex-numbers | <p>Let $$A = \{ z \in C:1 \le |z - (1 + i)| \le 2\} $$</p>
<p>and $$B = \{ z \in A:|z - (1 - i)| = 1\} $$. Then, B :</p> | [{"identifier": "A", "content": "is an empty set"}, {"identifier": "B", "content": "contains exactly two elements"}, {"identifier": "C", "content": "contains exactly three elements"}, {"identifier": "D", "content": "is an infinite set"}] | ["D"] | null | <p>Let, $$z = x + iy$$</p>
<p>Given, $$1 \le \left| {z - (1 + i)} \right| \le 2$$</p>
<p>$$ \Rightarrow 1 \le \left| {x + iy - 1 - i} \right| \le 2$$</p>
<p>$$ \Rightarrow 1 \le \left| {(x - 1) + i(y - 1)} \right| \le 2$$</p>
<p>$$ \Rightarrow 1 \le \sqrt {{{(x - 1)}^2} + {{(y - 1)}^2}} \le 2$$</p>
<p>It represent two... | mcq | jee-main-2022-online-24th-june-morning-shift | 5,499 |
1l6duxx3p | maths | complex-numbers | modulus-of-complex-numbers | <p>For $$\mathrm{n} \in \mathbf{N}$$, let $$\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\}$$ and $$\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}$$. Then the number of elements in the set $$\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\rig... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["D"] | null | $S_{n}=\left\{z \in C:|z-3+2 i|=\frac{n}{4}\right\}$ represents a circle with centre $C_{1}(3,-2)$ and radius $r_{1}=\frac{n}{4}$
<br/><br/>
Similarly $T_{n}$ represents circle with centre $C_{2}(2,-3)$ and radius $r_{2}=\frac{1}{n}$
<br/><br/>
$\text { As } S_{n} \cap T_{n}=\phi$<br/><br/>
$$
\begin{array}{llr}
C_{1} ... | mcq | jee-main-2022-online-25th-july-morning-shift | 5,500 |
1l6f0l8mz | maths | complex-numbers | modulus-of-complex-numbers | <p>For $$z \in \mathbb{C}$$ if the minimum value of $$(|z-3 \sqrt{2}|+|z-p \sqrt{2} i|)$$ is $$5 \sqrt{2}$$, then a value Question: of $$p$$ is _____________.</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\frac{7}{2}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7bug80c/85c5d49d-7e9c-43b0-b928-429394532f1d/d5c2fcc0-25fe-11ed-b97b-f3de74335fe6/file-1l7bug80d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7bug80c/85c5d49d-7e9c-43b0-b928-429394532f1d/d5c2fcc0-25fe-11ed-b97b-f3de74335fe6... | mcq | jee-main-2022-online-25th-july-evening-shift | 5,501 |
1l6jb3png | maths | complex-numbers | modulus-of-complex-numbers | <p>Let the minimum value $$v_{0}$$ of $$v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}$$ is attained at $${ }{z}=z_{0}$$. Then $$\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "1000"}, {"identifier": "B", "content": "1024"}, {"identifier": "C", "content": "1105"}, {"identifier": "D", "content": "1196"}] | ["A"] | null | <p>Let $$z = x + iy$$</p>
<p>$$v = {x^2} + {y^2} + {(x - 3)^2} + {y^2} + {x^2} + {(y - 6)^2}$$</p>
<p>$$ = (3{x^2} - 6x + 9) + (3{y^2} - 12y + 36)$$</p>
<p>$$ = 3({x^2} + {y^2} - 2x - 4y + 15)$$</p>
<p>$$ = 3[{(x - 1)^2} + {(y - 2)^2} + 10]$$</p>
<p>$${v_{\min }}$$ at $$z = 1 + 2i = {z_0}$$ and $${v_0} = 30$$</p>
<p>so... | mcq | jee-main-2022-online-27th-july-morning-shift | 5,503 |
1l6m60fcu | maths | complex-numbers | modulus-of-complex-numbers | <p>Let $$S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}$$ and $$S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}$$. Then, for $$z_{1} \in S_{1}$$ and $$z_{2} \in S_{2}$$, the least value of $$\left|z_{2}-z_{1}\right|$$ is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "$$\\frac{5}{2}$$"}] | ["C"] | null | <p>$$\because$$ $${\left| {{Z_2} + |{Z_2} - 1|} \right|^2} = {\left| {{Z_2} - |{Z_2} + 1|} \right|^2}$$</p>
<p>$$ \Rightarrow \left( {{Z_2} + |{Z_2} - 1|} \right)\left( {{{\overline Z }_2} + |{Z_2} - 1|} \right) = \left( {{Z_2} - |{Z_2} + 1|} \right)\left( {{{\overline Z }_2} - |{Z_2} + 1|} \right)$$</p>
<p>$$ \Rightar... | mcq | jee-main-2022-online-28th-july-morning-shift | 5,504 |
1l6rdnyvc | maths | complex-numbers | modulus-of-complex-numbers | <p>If $$z \neq 0$$ be a complex number such that $$\left|z-\frac{1}{z}\right|=2$$, then the maximum value of $$|z|$$ is :</p> | [{"identifier": "A", "content": "$$\\sqrt{2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\sqrt{2}-1$$"}, {"identifier": "D", "content": "$$\\sqrt{2}+1$$"}] | ["D"] | null | <p>We know,</p>
<p>$$\left| {|{z_1}| - |{z_2}|} \right| \le \left| {{z_1} + {z_2}} \right| \le |{z_1}| + |{z_2}|$$</p>
<p>$$\therefore$$ $$\left| {|z| - {1 \over {|z|}}} \right| \le \left| {z - {1 \over z}} \right|$$</p>
<p>$$ \Rightarrow \left| {|z| - {1 \over {|z|}}} \right| \le 2$$ [Given $$\left| {z - {1 \over z}} ... | mcq | jee-main-2022-online-29th-july-evening-shift | 5,505 |
1ldu5cviu | maths | complex-numbers | modulus-of-complex-numbers | <p>Let $$z$$ be a complex number such that $$\left| {{{z - 2i} \over {z + i}}} \right| = 2,z \ne - i$$. Then $$z$$ lies on the circle of radius 2 and centre :</p> | [{"identifier": "A", "content": "(0, $$-$$2)"}, {"identifier": "B", "content": "(0, 0)"}, {"identifier": "C", "content": "(0, 2)"}, {"identifier": "D", "content": "(2, 0)"}] | ["A"] | null | $\left|\frac{z-2 i}{z+i}\right|=2$
<br/><br/>
$\Rightarrow (z-2 i)(\bar{z}+2 i)=4(z+i)(\bar{z}-i)$
<br/><br/>
$\Rightarrow z \bar{z}+2 i z-2 i \bar{z}+4=4(z \bar{z}-z i+\overline{z i}+1)$
<br/><br/>
$\Rightarrow 3 z \bar{z}-6 i z+6 i \bar{z}=0$
<br/><br/>
$\Rightarrow z \bar{z}-2 i z+2 i \bar{z}=0$
<br/><br/>
$\theref... | mcq | jee-main-2023-online-25th-january-evening-shift | 5,506 |
lsam3k7w | maths | complex-numbers | modulus-of-complex-numbers | If $z$ is a complex number such that $|z| \leqslant 1$, then the minimum value of $\left|z+\frac{1}{2}(3+4 i)\right|$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$\\frac{5}{2}$"}, {"identifier": "C", "content": "$\\frac{3}{2}$"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <p>To find the minimum value of $\left|z+\frac{1}{2}(3+4i)\right|$, where $z$ is a complex number with $|z| \leqslant 1$, we can think of this geometrically as the distance from any point inside or on the boundary of the unit circle in the complex plane to the fixed point $\frac{1}{2}(3+4i)$.</p>
<p>To make this more ... | mcq | jee-main-2024-online-1st-february-evening-shift | 5,508 |
luy9clkm | maths | complex-numbers | modulus-of-complex-numbers | <p>The sum of the square of the modulus of the elements in the set $$\{z=\mathrm{a}+\mathrm{ib}: \mathrm{a}, \mathrm{b} \in \mathbf{Z}, z \in \mathbf{C},|z-1| \leq 1,|z-5| \leq|z-5 \mathrm{i}|\}$$ is __________.</p> | [] | null | 9 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw3in2th/8eab885e-7362-426e-8f81-a5a9157aded7/ca24c350-105b-11ef-9330-f7b57aaea8e9/file-1lw3in2ti.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw3in2th/8eab885e-7362-426e-8f81-a5a9157aded7/ca24c350-105b-11ef-9330-f7b57aaea8e9... | integer | jee-main-2024-online-9th-april-morning-shift | 5,509 |
lv7v3k13 | maths | complex-numbers | modulus-of-complex-numbers | <p>Consider the following two statements :</p>
<p>Statement I: For any two non-zero complex numbers $$z_1, z_2,(|z_1|+|z_2|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \text {, and }$$</p>
<p>Statement II : If $$x, y, z$$ are three distinct... | [{"identifier": "A", "content": "both Statement I and Statement II are incorrect.\n"}, {"identifier": "B", "content": "Statement I is correct but Statement II is incorrect.\n"}, {"identifier": "C", "content": "Statement I is incorrect but Statement II is correct.\n"}, {"identifier": "D", "content": "both Statement I an... | ["B"] | null | <p>$$\begin{aligned}
& \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}=\lambda \\
& \Rightarrow a^2=\lambda^2|(y-z)|^2 \\
& b^2=\lambda^2|(z-x)|^2 \\
& c^2=\lambda^2|(x-y)|^2 \\
& \frac{a^2(\overline{y-z})}{(y-z)(y-z)}=\frac{a^2(\bar{y}-\bar{z})}{|y-z|^2}=\frac{a^2(\bar{y}-\bar{z})}{\frac{a^2}{\lambda^2}}=\lambda^2(\ba... | mcq | jee-main-2024-online-5th-april-morning-shift | 5,510 |
lvb2948r | maths | complex-numbers | modulus-of-complex-numbers | <p>If $$z_1, z_2$$ are two distinct complex number such that $$\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$$, then</p> | [{"identifier": "A", "content": "either $$z_1$$ lies on a circle of radius $$\\frac{1}{2}$$ or $$z_2$$ lies on a circle of radius 1.\n"}, {"identifier": "B", "content": "$$z_1$$ lies on a circle of radius $$\\frac{1}{2}$$ and $$z_2$$ lies on a circle of radius 1.\n"}, {"identifier": "C", "content": "either $$z_1$$ lies... | ["C"] | null | <p>$$\begin{aligned}
& \left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2 \\
& \left|z_1-2 z_2\right|=\left|1-2 z_1 \bar{z}_2\right| \\
& \Rightarrow\left(z_1-2 z_2\right)\left(\bar{z}_1-2 \bar{z}_2\right)=\left(1-2 z_1 \bar{z}_2\right)\left(1-2 \bar{z}_1 z_2\right) \\
& \Rightarrow\left|z_1\right|^2+4\left|z_2... | mcq | jee-main-2024-online-6th-april-evening-shift | 5,511 |
Ppy3N7ZDNZ9Ix0Av | maths | complex-numbers | nth-roots-of-unity | The value of $$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\,\,\cos {{2k\pi } \over {11}}} \right)} $$ is : | [{"identifier": "A", "content": "i"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "- 1"}, {"identifier": "D", "content": "- i"}] | ["D"] | null | $$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)} $$
<br><br>$$ = i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)} $$
<br><br>$$ = i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\lim... | mcq | aieee-2006 | 5,512 |
1dXMQ34Gy45CNn7DdBZAT | maths | complex-numbers | nth-roots-of-unity | Let $$\alpha $$ and $$\beta $$ be two roots of the equation x<sup>2</sup> + 2x + 2 = 0 , then $$\alpha ^{15}$$ + $$\beta ^{15}$$ is equal to : | [{"identifier": "A", "content": "-256"}, {"identifier": "B", "content": "512"}, {"identifier": "C", "content": "-512"}, {"identifier": "D", "content": "256"}] | ["A"] | null | Given equation,
<br><br>x<sup>2</sup> + 2x + 2 = 0
<br><br>$$ \therefore $$ x = $${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$$
<br><br>x = $$-$$ 1 $$ \pm $$ i
<br><br>$$ \therefore $$ $$\alpha $$ = $$-$$ 1 + i
<br><br>and $$\beta $$ = $$-$$ 1 $$-$$ i
<br><br><u>Note </u> :
<br><br>x + iy = r ... | mcq | jee-main-2019-online-9th-january-morning-slot | 5,513 |
noAYQK9TnN7SiKwnAIFy2 | maths | complex-numbers | nth-roots-of-unity | If $$\alpha $$ and $$\beta $$ be the roots of the equation
x<sup>2</sup> – 2x + 2 = 0, then the least value of n for which $${\left( {{\alpha \over \beta }} \right)^n} = 1$$ is :
| [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["C"] | null | x<sup>2</sup> – 2x + 2 = 0
<br><br>$$ \therefore $$ x = $${{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i$$
<br><br>Now, $${\alpha \over \beta } = {{1 + i} \over {1 - i}} = {{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}} = i$$
<br><br>or $${\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( {1 - i} \right)}^2}}... | mcq | jee-main-2019-online-8th-april-morning-slot | 5,514 |
DLDIcZtF5DiyreYTGnjgy2xukf0zyu08 | maths | complex-numbers | nth-roots-of-unity | If $${\left( {{{1 + i} \over {1 - i}}} \right)^{{m \over 2}}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{{n \over 3}}} = 1$$, (m, n
$$ \in $$ N) then the
greatest common divisor of the least values of
m and n is _______ . | [] | null | 4 | $${\left( {{{1 + i} \over {1 - i}}} \right)^{m/2}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{n/3}} = 1$$<br><br>$$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$$<br><br>$$ \Rightarrow {(i)^{m/2}} = {( - i)^{... | integer | jee-main-2020-online-3rd-september-morning-slot | 5,515 |
1ktbgurkn | maths | complex-numbers | nth-roots-of-unity | Let $$z = {{1 - i\sqrt 3 } \over 2}$$, $$i = \sqrt { - 1} $$. Then the value of $$21 + {\left( {z + {1 \over z}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^3} + .... + {\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$$ is ______________. | [] | null | 13 | $$z = {{1 - \sqrt {3i} } \over 2} = {e^{ - i{\pi \over 3}}}$$<br><br>$${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$$<br><br>$$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right)... | integer | jee-main-2021-online-26th-august-morning-shift | 5,516 |
lv5gsy4m | maths | complex-numbers | nth-roots-of-unity | <p>If the set $$R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}$$ has $$m$$ elements and $$\sum_\limits{n=1}^m\left(1-i^{n !}\right)=x+i y$$, where $$i=\sqrt{-1}$$, then the value of $$m+x+y$$ is</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "5"}] | ["A"] | null | <p>$$R=\{(a, b): a+5 b=42\}$$</p>
<p>Then $$R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}$$</p>
<p>$$\begin{aligned}
& \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \\
& \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \\
& =5-i \\
& \therefore x=5, y... | mcq | jee-main-2024-online-8th-april-morning-shift | 5,518 |
jO4bSQV4DP64o4Ji8Ljgy2xukezaaujn | maths | complex-numbers | square-root-of-a-complex-number | The imaginary part of
<br/>$${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$$ can be : | [{"identifier": "A", "content": "-2$$\\sqrt 6 $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$\\sqrt 6 $$"}, {"identifier": "D", "content": "-$$\\sqrt 6 $$"}] | ["A"] | null | $$3 + 2\sqrt { - 54} $$<br><br>
$$ = 9 - 6 + 2\sqrt { - 54} $$<br><br>
$$ = 9 + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$$<br><br>
$$ = {3^2} + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$$<br><br>
$$ = {\left( {3 + \sqrt 6 i} \right)^2}$$<br><br>
Similarly, $$\left( {3 - 2\sqrt { - 54} } \right) = {\left( {3 ... | mcq | jee-main-2020-online-2nd-september-evening-slot | 5,519 |
JQ9MUck61xQ6DdPr | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}}$$ is | [{"identifier": "A", "content": "$${1 \\over {p + 1}}$$"}, {"identifier": "B", "content": "$${1 \\over {1 - p}}$$"}, {"identifier": "C", "content": "$${1 \\over p} - {1 \\over {p - 1}}$$"}, {"identifier": "D", "content": "$${1 \\over {p + 2}}$$"}] | ["A"] | null | We have $$\mathop {\lim }\limits_{x \to \infty } {{{1^p} + {2^p} + .... + {n^p}} \over {{n^{p + 1}}}};$$
<br><br>$$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{{{r^p}} \over {{n^p}.n}}} = \int\limits_0^1 {{x^p}} dx$$
<br><br>$$ = {\left[ {{{{x^{p + 1}}} \over {p + 1}}} \right]_0} = {1 \over {p + 1}}$... | mcq | aieee-2002 | 5,520 |
wXfQ6ecjBBTNFejK | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}$$ - $$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}$$ | [{"identifier": "A", "content": "$${1 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 30}$$"}, {"identifier": "C", "content": "zero"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["A"] | null | The given expression can be written as
<br><br>$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\sum\limits_{r = 1}^n {\left( {{r \over n}} \right)} ^4} - \mathop {\lim }\limits_{n \to \infty } {1 \over n}.\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\left( {{r \over n}} \right)^3}$$
<br><br>$$ = \int\limits_... | mcq | aieee-2003 | 5,521 |
ze5mgsSsHsysqEx4 | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{e^{{r \over n}}}} $$ is | [{"identifier": "A", "content": "$$e+1$$ "}, {"identifier": "B", "content": "$$e-1$$ "}, {"identifier": "C", "content": "$$1-e$$"}, {"identifier": "D", "content": "$$e$$"}] | ["B"] | null | $$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} {e^{{r \over n}}}\,\,$$
<br><br>$$\left[ {} \right.$$ Using definite integrals as limit of sum $$\left. {} \right]$$
<br><br>$$ = \int\limits_\theta ^1 {{e^x}} dx = e - 1$$ | mcq | aieee-2004 | 5,522 |
zQuI65ExuF62vE2u | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right]$$
<br/>equals | [{"identifier": "A", "content": "$${1 \\over 2}\\sec 1$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$cosec 1"}, {"identifier": "C", "content": "tan 1"}, {"identifier": "D", "content": "$${1 \\over 2}$$tan 1"}] | ["D"] | null | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}} + {3 \over {{n^2}}}se{c^2}{9 \over {{n^2}}} + ... + {1 \over n}{{\sec }^2}1} \right]$$
<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } {r \over {{n^2}}}{\sec ^2}{{{r^2}} \ov... | mcq | aieee-2005 | 5,523 |
Dk9DSWeqwOHhPHwxzNN0b | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{x \to \infty } \left( {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + {n \over {{n^2} + {3^2}}} + ..... + {1 \over {5n}}} \right)$$ is equal to : | [{"identifier": "A", "content": "tan<sup>\u20131 </sup>(2)"}, {"identifier": "B", "content": "tan<sup>\u20131 </sup>(3)"}, {"identifier": "C", "content": "$${\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{n \over {{n^2} + {r^2}}}} $$
<br><br>$$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{1 \over {n\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}} = \int\limits_0^2 {{{dx} \over {1 + {x^2}}}} } = {\tan ^{ - 1}}2$$ | mcq | jee-main-2019-online-12th-january-evening-slot | 5,526 |
hSHUIycDWkH38RPecf3rsa0w2w9jwy2hm13 | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } \left( {{{{{(n + 1)}^{1/3}}} \over {{n^{4/3}}}} + {{{{(n + 2)}^{1/3}}} \over {{n^{4/3}}}} + ....... + {{{{(2n)}^{1/3}}} \over {{n^{4/3}}}}} \right)$$
<br/>is equal to : | [{"identifier": "A", "content": "$${4 \\over 3}{\\left( 2 \\right)^{3/4}}$$"}, {"identifier": "B", "content": "$${3 \\over 4}{\\left( 2 \\right)^{4/3}} - {3 \\over 4}$$"}, {"identifier": "C", "content": "$${4 \\over 3}{\\left( 2 \\right)^{4/3}}$$"}, {"identifier": "D", "content": "$${3 \\over 4}{\\left( 2 \\right)^{4/3... | ["B"] | null | $$\mathop {\lim }\limits_{n \to \infty } {{{{(n + 1)}^{{1 \over 3}}} + {{(n + 2)}^{{1 \over 3}}} + ..... + {{\left( {n + n} \right)}^{{1 \over 3}}}} \over {n{{(n)}^{{1 \over 3}}}}}$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{{(n + r)}^{{1 \over 3}}}} \over {n.{n^{{1 \over 3... | mcq | jee-main-2019-online-10th-april-morning-slot | 5,527 |
KPNhsqLgDWKQZvRCQ31klt9lykx | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]$$ is equal to : | [{"identifier": "A", "content": "$${{1 \\over 2}}$$"}, {"identifier": "B", "content": "$${{1 \\over 3}}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${{1 \\over 4}}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ... + {n \over {{{(2n - 1)}^2}}}} \right]$$
<br><br>$$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{\left( {n + 0} \right)}^2}}} + {n \over {{{\left( {n + 1} \right)}^2}}} + {n \over {{{\l... | mcq | jee-main-2021-online-25th-february-evening-slot | 5,528 |
gaG80xeroFIPmrjcDJ1kmhzhgz7 | maths | definite-integration | definite-integral-as-a-limit-of-sum | Let f : (0, 2) $$ \to $$ R be defined as f(x) = log<sub>2</sub>$$\left( {1 + \tan \left( {{{\pi x} \over 4}} \right)} \right)$$. Then, $$\mathop {\lim }\limits_{n \to \infty } {2 \over n}\left( {f\left( {{1 \over n}} \right) + f\left( {{2 \over n}} \right) + ... + f(1)} \right)$$ is equal to ___________. | [] | null | 1 | $$E = 2\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} f\left( {{r \over n}} \right)$$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {{\pi x} \over 4}} \right)dx} $$ ..... (i)<br><br>replacing x $$ \to $$ 1 $$-$$ x<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {\... | integer | jee-main-2021-online-16th-march-morning-shift | 5,529 |
1ks04wdm1 | maths | definite-integration | definite-integral-as-a-limit-of-sum | The value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{(2j - 1) + 8n} \over {(2j - 1) + 4n}}} $$ is equal to : | [{"identifier": "A", "content": "$$5 + {\\log _e}\\left( {{3 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$2 - {\\log _e}\\left( {{2 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$3 + 2{\\log _e}\\left( {{2 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$1 + 2{\\log _e}\\left( {{3 \\... | ["D"] | null | $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{\left( {{{2j} \over n} - {1 \over n} + 8} \right)} \over {\left( {{{2j} \over n} - {1 \over n} + 4} \right)}}} $$<br><br>$$\int\limits_0^1 {{{2x + 8} \over {2x + 4}}dx = \int\limits_0^1 {dx + \int\limits_0^1 {{4 \over {2x + 4}}} dx} } $$<br><b... | mcq | jee-main-2021-online-27th-july-morning-shift | 5,530 |
1ktbg2i67 | maths | definite-integration | definite-integral-as-a-limit-of-sum | The value of <br/><br/>$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}} $$ is : | [{"identifier": "A", "content": "$${1 \\over 2}{\\tan ^{ - 1}}(2)$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\tan ^{ - 1}}(4)$$"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}(4)$$"}, {"identifier": "D", "content": "$${1 \\over 4}{\\tan ^{ - 1}}(4)$$"}] | ["B"] | null | $$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}} $$<br><br>$$ \Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx} $$<br><br>$$ \Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}... | mcq | jee-main-2021-online-26th-august-morning-shift | 5,531 |
1l5b84rpc | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + {{{n^2}} \over {({n^2} + 9)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 8} + {1 \\over 4}{\\log _e}2$$"}, {"identifier": "B", "content": "$${\\pi \\over 4} + {1 \\over 8}{\\log _e}2$$"}, {"identifier": "C", "content": "$${\\pi \\over 4} - {1 \\over 8}{\\log _e}2$$"}, {"identifier": "D", "content": "$${\\pi \\over 8} + {\\log _e}\\sqrt 2 $... | ["A"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{n^2}} \over {({n^2} + {r^2})(n + r)}}} $$</... | mcq | jee-main-2022-online-24th-june-evening-shift | 5,533 |
1l6dxdfr3 | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$
\begin{aligned}
&\text { If } \lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)] \\
&=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^{k}+2^{k}+3^{k}+\ldots+n^{k}\right]
\end{aligned}$$,
then the integral value of $$\mathrm{k}$$ is equal to _____... | [] | null | 5 | $\lim\limits_{n \rightarrow \infty}\left(\frac{n+1}{n}\right)^{k-1} \frac{1}{n} \sum_{r=1}^{n}\left(k+\frac{r}{n}\right)=33 \lim\limits_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\left(\frac{r}{n}\right)^{k}$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \int_{0}^{1}(k+x) d x=33 \int_{0}^{1} x^{k} d x \\\\
&\Righta... | integer | jee-main-2022-online-25th-july-morning-shift | 5,535 |
1l6f0ysr5 | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$\mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,...\,\, + \,\,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-$$2"}] | ["C"] | null | <p>$$I = \mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,.....\,\, + \,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$$</p>
<p>Let $${2^n} = t$$ and i... | mcq | jee-main-2022-online-25th-july-evening-shift | 5,536 |
1l6ghx5h6 | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>If $$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$ and $$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} $$, $$x \in (0,1)$$, then :</p> | [{"identifier": "A", "content": "$$2\\sqrt 2 f\\left( {{a \\over 2}} \\right) = f'\\left( {{a \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$f\\left( {{a \\over 2}} \\right)f'\\left( {{a \\over 2}} \\right) = \\sqrt 2 $$"}, {"identifier": "C", "content": "$$\\sqrt 2 f\\left( {{a \\over 2}} \\right) = f'\\le... | ["C"] | null | <p>$$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{2 \over {1 + {{\left( {{k \over n}} \right)}^2}}}} $$</p>
<p>$$a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = 2{{\tan }^{ - 1}}x\i... | mcq | jee-main-2022-online-26th-july-morning-shift | 5,537 |
1ldom3jmb | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {1 + n}} + {1 \over {2 + n}} + {1 \over {3 + n}}\, + \,...\, + \,{1 \over {2n}}} \right]$$ is equal to</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${\\log _e}2$$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{2 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$${\\log _e}\\left( {{3 \\over 2}} \\right)$$"}] | ["B"] | null | $$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\
& =\int_0^1 \frac{d x... | mcq | jee-main-2023-online-1st-february-morning-shift | 5,538 |
ldqvscjt | maths | definite-integration | definite-integral-as-a-limit-of-sum | $\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$\\frac{19}{3}$"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "12"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}} $$</p>
<p>$$ = \int_0^1 {3{{(2 + x)}^2}\,dx} $$</p>
<p>$$ = \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1$$</p>
<p>$$ = {3^3} - {2^3} = 19$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 5,539 |
1lh2zlowl | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>Let $$f(x)=\frac{x}{\left(1+x^{n}\right)^{\frac{1}{n}}}, x \in \mathbb{R}-\{-1\}, n \in \mathbb{N}, n > 2$$.</p>
<p>If $$f^{n}(x)=\left(f \circ f \circ f \ldots .\right.$$. upto $$n$$ times) $$(x)$$, then
<br/><br/>$$\lim _\limits{n \rightarrow \infty} \int_\limits{0}^{1} x^{n-2}\left(f^{n}(x)\right) d x$$ is eq... | [] | null | 0 | $$
\begin{aligned}
& \text { We have, } f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}} \\\\
& \therefore f(f(x))=\frac{f(x)}{\left(1+\left[f(x)^n\right]^{1 / n}\right.}\\\\
&=\frac{\frac{x}{\left(1+x^n\right)^{1 / n}}}{\left(1+\frac{x^n}{1+x^n}\right)^{1 / n}}\\\\
&=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\\\
& f(f(f(x)))=\... | integer | jee-main-2023-online-6th-april-evening-shift | 5,541 |
luy9clal | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>Let $$\lim _\limits{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right.$$ $$\left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)$$ be $$\frac{\pi}{k... | [] | null | 32 | <p>$$\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\ldots \frac{n}{\sqrt{n^4+n^4}}\right) \\
& -\lim _{n \rightarrow \infty}\left(\frac{2 n}{\left(n^2+1\right)\left(\sqrt{n^4+1}\right)}\right)+\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+1}}+\cdots \frac{2 n \cdot n^2}{\l... | integer | jee-main-2024-online-9th-april-morning-shift | 5,543 |
I6qJxSWERiZM8zTy | maths | definite-integration | newton-lebnitz-rule-of-differentiation | The value of $$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over xsinx}$$ is | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{d \over {dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over {{d \over x}\left( {x\sin x} \right)}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{{{\sec }^2}{x^2}.2x} \over {\sin \,x + x\,\cos \,x}}$$ (by $$L'$$ Hospital rule)
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{2{{\sec... | mcq | aieee-2003 | 5,544 |
pJMN8tkqVsHXcQwI | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $$f:R \to R$$ be a differentiable function having $$f\left( 2 \right) = 6$$,
<br/>$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals : | [{"identifier": "A", "content": "$$24$$ "}, {"identifier": "B", "content": "$$36$$ "}, {"identifier": "C", "content": "$$12$$ "}, {"identifier": "D", "content": "$$18$$ "}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$$
<br/><br/>This limit resembles a derivative because the fraction has the form $0/0$ as $x \to 2$ since both... | mcq | aieee-2005 | 5,545 |
PrYwpygyiAmYysm5OI6VH | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f be a differentiable function from
<br/><br/><b>R</b> to <b>R</b> such that $$\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$$ <br/><br/>for all $$x,y \in $$ <b>R</b>.
<br/><br/>If $$f\left( 0 \right) = 1$$
<br/><br/>then $$\int\limits_0^1 {{f^2}} ... | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "0"}] | ["A"] | null | $$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$$
<br><br>$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$
<br><br>$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2... | mcq | jee-main-2019-online-9th-january-evening-slot | 5,547 |
M4m67SR6ySoaeO4fudc40 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If $$\int\limits_0^x \, $$f(t) dt = x<sup>2</sup> + $$\int\limits_x^1 \, $$ t<sup>2</sup>f(t) dt then f '$$\left( {{1 \over 2}} \right)$$ is - | [{"identifier": "A", "content": "$${{18} \\over {25}}$$"}, {"identifier": "B", "content": "$${{6} \\over {25}}$$"}, {"identifier": "C", "content": "$${{24} \\over {25}}$$"}, {"identifier": "D", "content": "$${{4} \\over {5}}$$"}] | ["C"] | null | $$\int\limits_0^x \, $$f(t) dt = x<sup>2</sup> + $$\int\limits_x^1 \, $$ t<sup>2</sup>f(t) dt f '$$\left( {{1 \over 2}} \right)$$ = ?
<br><br>Differentiate w.r.t. 'x'
<br><br>f(x) = 2x + 0 $$-$$ x<sup>2</sup> f(x)
<br><br>f(x) = $${{2x} \over {1 + {x^2}}}$$ $... | mcq | jee-main-2019-online-10th-january-evening-slot | 5,548 |
7eehTp46kzo6JgEura18hoxe66ijvwu5o7l | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If f : R $$ \to $$ R is a differentiable function and
f(2) = 6,<br/> then $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$ is :- | [{"identifier": "A", "content": "2f'(2)"}, {"identifier": "B", "content": "24f'(2)"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "12f'(2)"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$
<br><br>This is $${0 \over 0}$$ form so we use L – Hospital Rule.
<br><br>= $$\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}$$
<br><br>= $${2f\left( 2 \right).f'\lef... | mcq | jee-main-2019-online-9th-april-evening-slot | 5,549 |
rLxHSZ2R8C1omj8xb43rsa0w2w9jx6g9d50 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f : R $$ \to $$ R be a continuously differentiable function such that f(2) = 6 and f'(2) = $${1 \over {48}}$$. If $$\int\limits_6^{f\left( x \right)} {4{t^3}} dt$$ = (x - 2)g(x), then $$\mathop {\lim }\limits_{x \to 2} g\left( x \right)$$ is equal to : | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "24"}] | ["A"] | null | Given $$\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)$$
<br><br>$$ \Rightarrow g(x) = $$ $${{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$$<br><br>
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 2} g(x) = $$$$\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x... | mcq | jee-main-2019-online-12th-april-morning-slot | 5,550 |
Cj07jRGhIXsCZBtHfG7k9k2k5hjthqb | maths | definite-integration | newton-lebnitz-rule-of-differentiation | $$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$ is equal to | [{"identifier": "A", "content": "$$ - {1 \\over 5}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 10}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$ {1 \\over 10}$$"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$
<br><br>This is in $${0 \over 0}$$ form.
<br><br>So apply newton leibniz rule
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$$ = 0 | mcq | jee-main-2020-online-8th-january-evening-slot | 5,551 |
ccqUxwJrneLUb7B9yj7k9k2k5kh9wos | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let a function ƒ : [0, 5] $$ \to $$ R be continuous,
ƒ(1) = 3 and F be defined as :<br/><br/>
$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$ <br/><br/>Then for the function F, the point x = 1 is : | [{"identifier": "A", "content": "a point of inflection."}, {"identifier": "B", "content": "a point of local maxima."}, {"identifier": "C", "content": "a point of local minima."}, {"identifier": "D", "content": "not a critical point."}] | ["C"] | null | $$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$
<br><br>$$ \Rightarrow $$ F'(x) = x<sup>2</sup>g(x) = x<sup>2</sup>$$\int\limits_1^t {f(u)du} $$
<br><br>$$ \therefore $$ F'(1) = (1)(0) = 0
<br><br>Now, F''(x) = 2xg(x) + x<sup>2</sup>g'(x)
<br><br>F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3
<br><br>[ As g'(t) = f(t); g'(1) = f'(1)... | mcq | jee-main-2020-online-9th-january-evening-slot | 5,552 |
eWTJ6SlzQ17mZgakyQjgy2xukfw0k3gs | maths | definite-integration | newton-lebnitz-rule-of-differentiation | $$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$ | [{"identifier": "A", "content": "is equal to 0"}, {"identifier": "B", "content": "is equal to $${1 \\over 2}$$"}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "is equal to $$ - {1 \\over 2}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$$$\left( {{0 \over 0}} \right)$$
<br><br>Apply L Hospital Rule
<br><br>= $$\mathop {\lim }\limits_{x \to 1} {{2\left( {x - 1}... | mcq | jee-main-2020-online-6th-september-morning-slot | 5,553 |
O2ZShb2HjKklQdRLCL1klrfaiok | maths | definite-integration | newton-lebnitz-rule-of-differentiation | $$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {15}}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {0^ + }} {{\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \over {{x^3}}}$$<br><br>This is in $${0 \over 0}$$ form, so use L' Hospital rule<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{d \over {dx}}\left( {\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \right)} \over {{d \over {dx}}\le... | mcq | jee-main-2021-online-24th-february-morning-slot | 5,554 |
U00iEUsehx8dZ22WFB1kluwqi02 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ be a differentiable function for all x$$\in$$R. Then f(x) equals : | [{"identifier": "A", "content": "$${e^{({e^{x - 1}})}}$$"}, {"identifier": "B", "content": "$$2{e^{{e^x}}} - 1$$"}, {"identifier": "C", "content": "$$2{e^{{e^x} - 1}} - 1$$"}, {"identifier": "D", "content": "$${e^{{e^x}}} - 1$$"}] | ["C"] | null | $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ .... (1)<br><br>Differentiating both sides w.r.t. x<br><br>$$f'(x) = {e^x}.f(x) + {e^x}$$ (Using Newton L:eibnitz Theorem)<br><br>$$ \Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}$$<br><br>Integrating w.r.t. x<br><br>$$\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx}... | mcq | jee-main-2021-online-26th-february-evening-slot | 5,555 |
kS6QZFldzuh02L2VvE1kmhzjbbd | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If the normal to the curve y(x) = $$\int\limits_0^x {(2{t^2} - 15t + 10)dt} $$ at a point (a, b) is parallel to the line x + 3y = $$-$$5, a > 1, then the value of | a + 6b | is equal to ___________. | [] | null | 406 | Normal to the curve at point P(a, b) is parallel to the line x + 3y = $$-$$5.
<br><br>m<sub>normal</sub> = $$ - {1 \over 3}$$
<br><br>$$ \therefore $$ m<sub>tangent</sub> = 3 = $${{dy} \over {dx}}$$
<br><br>Given y(x) = $$\int\limits_0^x {(2{t^2} - 15t + 10)dt} $$
<br><br>$$ \Rightarrow $$ y'(x) = (2x<sup>2</sup> $$-$$... | integer | jee-main-2021-online-16th-march-morning-shift | 5,556 |
1kryf7cd2 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f : (a, b) $$\to$$ R be twice differentiable function such that $$f(x) = \int_a^x {g(t)dt} $$ for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least : | [{"identifier": "A", "content": "twelve roots in (a, b)"}, {"identifier": "B", "content": "five roots in (a, b)"}, {"identifier": "C", "content": "seven roots in (a, b)"}, {"identifier": "D", "content": "three roots in (a, b)"}] | ["C"] | null | $$f(x) = \int_a^x {g(t)dt} $$
<br><br>$$ \Rightarrow $$ f′(x) = g(x)
<br><br>$$ \Rightarrow $$ f′'(x) = g'(x)
<br><br>Given, g(x).g'(x) = 0
<br><br>$$ \Rightarrow $$ f′(x).f′'(x) = 0
<br><br>Also given f(x) has exactly 5 roots.
<br><br>So from Rolle's theorem we can say,
<br><br>f′(x) has 4 roots and f′'(x) has 3 roots... | mcq | jee-main-2021-online-27th-july-evening-shift | 5,558 |
1ks0cs08h | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $$F:[3,5] \to R$$ be a twice differentiable function on (3, 5) such that <br/><br/>$$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$. If $$F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$$, then $$\alpha$$ + $$\beta$$ is equal to _______________. | [] | null | 16 | $$F(3) = 0$$<br><br>$${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$<br><br>$${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$$<br><br>$$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$$<br><br>$${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$$<br><br>$$y{e^{\i... | integer | jee-main-2021-online-27th-july-morning-shift | 5,559 |
1ktiotbdz | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } $$, $$0 \le x \le 1$$ and f(0) = 0, then $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} $$ : | [{"identifier": "A", "content": "equals 0"}, {"identifier": "B", "content": "equals 1"}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "equals $${1 \\over 2}$$"}] | ["D"] | null | $$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } ,\,0 \le x \le 1$$<br><br>differentiating both the sides<br><br>$$\sqrt {1 - {{(f'(x))}^2}} = f(x)$$<br><br>$$ \Rightarrow 1 - {(f'(x))^2} = {f^2}(x)$$<br><br>$${{f'(x)} \over {\sqrt {1 - {f^2}(x)} }} = 1$$<br><br>$${\sin ^{ - 1}}f(x) = x + C$$<br><br>$$\b... | mcq | jee-main-2021-online-31st-august-morning-shift | 5,560 |
1ktnzk3or | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f : R $$\to$$ R be a continuous function. Then $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$ is equal to : | [{"identifier": "A", "content": "f (2)"}, {"identifier": "B", "content": "2f (2)"}, {"identifier": "C", "content": "2f $$\\left( {\\sqrt 2 } \\right)$$"}, {"identifier": "D", "content": "4f (2)"}] | ["B"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}.{{\left[ {f({{\sec }^2}x).2\sec x.\sec x\tan x} \right]} \over {2x}}$$<br><br>= $$\mathop {\lim }\l... | mcq | jee-main-2021-online-1st-september-evening-shift | 5,561 |
1l56qx530 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>If m and n respectively are the number of local maximum and local minimum points of the function $$f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $$, then the ordered pair (m, n) is equal to</p> | [{"identifier": "A", "content": "(3, 2)"}, {"identifier": "B", "content": "(2, 3)"}, {"identifier": "C", "content": "(2, 2)"}, {"identifier": "D", "content": "(3, 4)"}] | ["B"] | null | <p>$$f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $$</p>
<p>$$f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0$$</p>
<p>$$x = 0$$, or $$({x^2} - 4)({x^2} - 1) = 0$$</p>
<p>$$x = 0,$$ $$x = \pm 2,\, \pm 1$$</p>
<p>Now, $$f'(x) = {{2x(x + 1)(x - 1)(x + 2)(x - 2)} \over {\le... | mcq | jee-main-2022-online-27th-june-evening-shift | 5,562 |
1l56qytz0 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let f be a differentiable function in $$\left( {0,{\pi \over 2}} \right)$$. If $$\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x} $$, then $${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$6 - 9\\sqrt 2 $$"}, {"identifier": "B", "content": "$$6 - {9 \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${9 \\over 2} - 6\\sqrt 2 $$"}, {"identifier": "D", "content": "$${9 \\over {\\sqrt 2 }} - 6$$"}] | ["B"] | null | <p>$$\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x} $$</p>
<p>$$ \Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x$$</p>
<p>$$ \Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x$$</p>
<p>$$ \Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x$$</p>
<p><img src="https://app-c... | mcq | jee-main-2022-online-27th-june-evening-shift | 5,563 |
1l6f3imiz | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$f$$ be a twice differentiable function on $$\mathbb{R}$$. If $$f^{\prime}(0)=4$$ and $$f(x) + \int\limits_0^x {(x - t)f'(t)dt = \left( {{e^{2x}} + {e^{ - 2x}}} \right)\cos 2x + {2 \over a}x} $$, then $$(2 a+1)^{5}\, a^{2}$$ is equal to _______________.</p> | [] | null | 8 | $$
\begin{aligned}
\because f(x)+\int_0^x(x-t) f^{\prime}(t) d t & =\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2 x}{a} ~~...(i)
\end{aligned}
$$<br/><br/>
Here $f(0)=2 \hspace{0.5cm} ...(ii)$<br/><br/>
On differentiating equation (i) w.r.t. $x$ we get :<br/><br/>
$$
\begin{aligned}
& f^{\prime}(x)+\int_0^x f^{\prime... | integer | jee-main-2022-online-25th-july-evening-shift | 5,564 |
1l6klsu23 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let f be a differentiable function satisfying $$f(x)=\frac{2}{\sqrt{3}} \int\limits_{0}^{\sqrt{3}} f\left(\frac{\lambda^{2} x}{3}\right) \mathrm{d} \lambda, x>0$$ and $$f(1)=\sqrt{3}$$. If $$y=f(x)$$ passes through the point $$(\alpha, 6)$$, then $$\alpha$$ is equal to _____________.</p> | [] | null | 12 | <p>$$\because$$ $$f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0} $$</p>
<p>On differentiating both sides w.r.t., x, we get</p>
<p>$$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \righ... | integer | jee-main-2022-online-27th-july-evening-shift | 5,565 |
1l6m6dywt | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>The minimum value of the twice differentiable function $$f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$$, $$x \in \mathbf{R}$$, is :</p> | [{"identifier": "A", "content": "$$-\\frac{2}{\\sqrt{\\mathrm{e}}}$$"}, {"identifier": "B", "content": "$$-2 \\sqrt{\\mathrm{e}}$$"}, {"identifier": "C", "content": "$$-\\sqrt{\\mathrm{e}}$$"}, {"identifier": "D", "content": "$$\\frac{2}{\\sqrt{\\mathrm{e}}}$$"}] | ["A"] | null | <p>$$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$</p>
<p>$$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$</p>
<p>$${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} $$</p>
<p>Differentiate on both side</p>
<p>$${e^{ - x}}f'(x) + ( - f(x){e^{ - x}}... | mcq | jee-main-2022-online-28th-july-morning-shift | 5,566 |
ldo9n0jy | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If $\phi(x)=\frac{1}{\sqrt{x}} \int\limits_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, x>0$,
<br/><br/>then $\emptyset^{\prime}\left(\frac{\pi}{4}\right)$ is equal to : | [{"identifier": "A", "content": "$\\frac{4}{6+\\sqrt{\\pi}}$"}, {"identifier": "B", "content": "$\\frac{4}{6-\\sqrt{\\pi}}$"}, {"identifier": "C", "content": "$\\frac{8}{\\sqrt{\\pi}}$"}, {"identifier": "D", "content": "$\\frac{8}{6+\\sqrt{\\pi}}$"}] | ["D"] | null | $\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$
<br/><br/>$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$
<br/><br/>$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right)... | mcq | jee-main-2023-online-31st-january-evening-shift | 5,567 |
1ldwxrkys | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$f$$ be $$a$$ differentiable function defined on $$\left[ {0,{\pi \over 2}} \right]$$ such that $$f(x) > 0$$ and $$f(x) + \int_0^x {f(t)\sqrt {1 - {{({{\log }_e}f(t))}^2}} dt = e,\forall x \in \left[ {0,{\pi \over 2}} \right]}$$. Then $$\left( {6{{\log }_e}f\left( {{\pi \over 6}} \right)} \right)^2$$ is e... | [] | null | 27 | $f(x)+\int_{0}^{x} f(t) \sqrt{1-\left(\log _{e} f(t)\right)^{2}} d t=e\quad...(1)$
<br/><br/>
So, $f(0)=e$
<br/><br/>
Now differentiate w.r. to $x$
<br/><br/>
$$
\begin{gathered}
f^{\prime}(x)+f(x) \sqrt{1-\left(\log _{e} f(x)^{2}\right.}=0 \\\\
\frac{f^{\prime}(x)}{f(x) \sqrt{1-\left(\log _{e} f(x)\right)^{2}}}=-1
\en... | integer | jee-main-2023-online-24th-january-evening-shift | 5,569 |
lsan5y2d | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $f:(0, \infty) \rightarrow \mathbf{R}$ and $\mathrm{F}(x)=\int\limits_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}$. If $\mathrm{F}\left(x^2\right)=x^4+x^5$, then $\sum\limits_{\mathrm{r}=1}^{12} f\left(\mathrm{r}^2\right)$ is equal to ____________. | [] | null | 219 | $F(x)=\int\limits_0^x t \cdot f(t) d t$
<br/><br/>$\begin{aligned} & F'(x)=x f(x) \\\\ & F\left(x^2\right)=x^4+x^5, \quad \text { let } x^2=t \\\\ & F(t)=t^2+t^{5 / 2} \\\\ & F^{\prime}(t)=2 t+5 / 2 t^{3 / 2} \\\\ & t \cdot f(t)=2 t+5 / 2 t^{3 / 2} \\\\ & f(t)=2+5 / 2 r^{1 / 2}\end{aligned}$
<br/><br/>$\begin{aligned} ... | integer | jee-main-2024-online-1st-february-evening-shift | 5,571 |
jaoe38c1lsf0dr2z | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{3 \\pi^2}{4}$$\n"}, {"identifier": "B", "content": "$$\\frac{3 \\pi^2}{8}$$\n"}, {"identifier": "C", "content": "$$\\frac{3 \\pi}{4}$$\n"}, {"identifier": "D", "content": "$$\\frac{3 \\pi}{8}$$"}] | ["B"] | null | <p>Using L'hospital rule</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\
& =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \\
& =\fra... | mcq | jee-main-2024-online-29th-january-morning-shift | 5,573 |
jaoe38c1lsfl2jrf | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let the slope of the line $$45 x+5 y+3=0$$ be $$27 r_1+\frac{9 r_2}{2}$$ for some $$r_1, r_2 \in \mathbb{R}$$. Then $$\lim _\limits{x \rightarrow 3}\left(\int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} d t\right)$$ is equal to _________.</p> | [] | null | 12 | <p>According to the question,</p>
<p>$$\begin{aligned}
& 27 r_1+\frac{9 r_2}{2}=-9 \\
& \lim _\limits{x \rightarrow 3} \frac{\int_\limits3^x 8 t^2 d t}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} \\
& =\lim _\limits{x \rightarrow 3} \frac{8 x^2}{\frac{3 r_2^2}{2}-2 r_2 x-3 r_1 x^2-3} \text { (using LH' Rule) } \\
& =\frac{7... | integer | jee-main-2024-online-29th-january-evening-shift | 5,574 |
luxweovm | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{3 \\pi^2}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{9 \\pi^2}{8}$$\n"}, {"identifier": "C", "content": "$$\\frac{5 \\pi^2}{9}$$\n"}, {"identifier": "D", "content": "$$\\frac{11 \\pi^2}{10}$$"}] | ["B"] | null | <p>$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$</p>
<p>Using Newton Leibniz theorem</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{\s... | mcq | jee-main-2024-online-9th-april-evening-shift | 5,576 |
lv2er436 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$f(x)=\int_0^x\left(t+\sin \left(1-e^t\right)\right) d t, x \in \mathbb{R}$$. Then, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{6}$$\n"}, {"identifier": "B", "content": "$$-\\frac{1}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "D", "content": "$$-\\frac{2}{3}$$"}] | ["B"] | null | <p>Given $$f(x)=\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t$$</p>
<p>Now, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text { form }\right)$$</p>
<p>$$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3} \\
& =\lim _{x \rig... | mcq | jee-main-2024-online-4th-april-evening-shift | 5,577 |
hqF2I8KjE6uVCfhw | maths | definite-integration | properties-of-definite-integration | $${I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} $$ then $$\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]$$ equals | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$\\infty $$ "}, {"identifier": "D", "content": "zero"}] | ["B"] | null | $${I_n} + {I_{n + 2}}$$
<br><br>$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx$$
<br><br>$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx$$
<br><br>$$ = \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}$$
<br><br>$$ = {{1 - 0} \over {n + 1}} = {1 \over {n + 1... | mcq | aieee-2002 | 5,578 |
fOTnK3XKAyiYtvBR | maths | definite-integration | properties-of-definite-integration | $$\int\limits_0^{10\pi } {\left| {\sin x} \right|dx} $$ is | [{"identifier": "A", "content": "$$20$$"}, {"identifier": "B", "content": "$$8$$"}, {"identifier": "C", "content": "$$10$$"}, {"identifier": "D", "content": "$$18$$"}] | ["A"] | null | $$I = \int\limits_0^{10\pi } {\left| {\sin x} \right|} dx$$
<br><br>$$ = 10\int\limits_0^\pi {\left| {\sin x\,} \right|} \,dx$$
<br><br>$$ = 10\int\limits_0^\pi {\sin \,x\,dx} $$
<br><br>$$\left[ \, \right.$$ as $$\left| {\sin x} \right|$$ is periodic with period $$\pi $$
<br><br>and $$sin$$ $$x > 0$$ if $$0 < ... | mcq | aieee-2002 | 5,579 |
O2Ykyj1nWTzQNYTw | maths | definite-integration | properties-of-definite-integration | $$\int\limits_0^2 {\left[ {{x^2}} \right]dx} $$ is | [{"identifier": "A", "content": "$$2 - \\sqrt 2 $$ "}, {"identifier": "B", "content": "$$2 + \\sqrt 2 $$"}, {"identifier": "C", "content": "$$\\,\\sqrt 2 - 1$$ "}, {"identifier": "D", "content": "$$ - \\sqrt 2 - \\sqrt 3 + 5$$ "}] | ["D"] | null | $$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx$$
<br... | mcq | aieee-2002 | 5,580 |
e87VKbcXQlihzLai | maths | definite-integration | properties-of-definite-integration | $$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx$$ is | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 4}$$ "}, {"identifier": "B", "content": "$${{\\pi ^2}}$$ "}, {"identifier": "C", "content": "zero "}, {"identifier": "D", "content": "$${\\pi \\over 2}$$ "}] | ["B"] | null | $$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx$$
<br><br>$$ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx$$
<br><br>$$ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;$$
<br><br>$$\l... | mcq | aieee-2002 | 5,581 |
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