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JEE Advanced · Mathematics · 2013

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1
Match List I with List II and select the correct answer using the code given below the lists: List I: P. Volume of the parallelepiped determined by vectors a⃗,b⃗,c⃗\vec a,\vec b,\vec c is 22. Then the volume of the parallelepiped determined by vectors 2(a⃗×b⃗), 3(b⃗×c⃗), (c⃗×a⃗)2(\vec a\times\vec b),\ 3(\vec b\times\vec c),\ (\vec c\times\vec a) is Q. Volume of the parallelepiped determined by vectors a⃗,b⃗,c⃗\vec a,\vec b,\vec c is 55. Then the volume of the parallelepiped determined by vectors 3(a⃗+b⃗), (b⃗+c⃗), 2(c⃗+a⃗)3(\vec a+\vec b),\ (\vec b+\vec c),\ 2(\vec c+\vec a) is R. Area of a triangle with adjacent sides determined by vectors a⃗\vec a and b⃗\vec b is 2020. Then the area of the triangle with adjacent sides determined by vectors (2a⃗+3b⃗)(2\vec a+3\vec b) and (a⃗−b⃗)(\vec a-\vec b) is S. Area of a parallelogram with adjacent sides determined by vectors a⃗\vec a and b⃗\vec b is 3030. Then the area of the parallelogram with adjacent sides determined by vectors (a⃗+b⃗)(\vec a+\vec b) and a⃗\vec a is List II: 1. 100100 2. 3030 3. 2424 4. 6060 Codes:
Not attempted
A
(A) P→4, Q→2, R→3, S→1P\to4,\ Q\to2,\ R\to3,\ S\to1
B
(B) P→2, Q→3, R→1, S→4P\to2,\ Q\to3,\ R\to1,\ S\to4
C
(C) P→3, Q→4, R→1, S→2P\to3,\ Q\to4,\ R\to1,\ S\to2
D
(D) P→1, Q→4, R→3, S→2P\to1,\ Q\to4,\ R\to3,\ S\to2
2
A box B1B_1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_3 contains 3 white balls, 4 red balls and 5 black balls. If 1 ball is drawn from each of the boxes B1B_1, B2B_2 and B3B_3, the probability that all 3 drawn balls are of the same colour is
Not attempted
A
82648\dfrac{82}{648}
B
90648\dfrac{90}{648}
C
558648\dfrac{558}{648}
D
566648\dfrac{566}{648}
3
A box B1B_1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_3 contains 3 white balls, 4 red balls and 5 black balls. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2B_2 is
Not attempted
A
116181\dfrac{116}{181}
B
126181\dfrac{126}{181}
C
65181\dfrac{65}{181}
D
55181\dfrac{55}{181}
4
The function f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣f(x)=2|x|+|x+2|-\big||x+2|-2|x|\big| has a local minimum or a local maximum at x=x=
Not attempted
A
−2-2
B
−23-\dfrac{2}{3}
C
22
D
23\dfrac{2}{3}
5
Let f:[0,1]→Rf:[0,1]\to\mathbb{R} (the set of all real numbers) be a function. Suppose the function ff is twice differentiable, f(0)=f(1)=0f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥ex, x∈[0,1]f''(x)-2f'(x)+f(x)\geq e^x,\ x\in[0,1]. Which of the following is true for 0<x<10<x<1?
Not attempted
A
0<f(x)<∞0<f(x)<\infty
B
−12<f(x)<12-\dfrac12<f(x)<\dfrac12
C
−14<f(x)<1-\dfrac14<f(x)<1
D
−∞<f(x)<0-\infty<f(x)<0
6
Let f:[0,1]→Rf:[0,1]\to\mathbb{R} (the set of all real numbers) be a function. Suppose the function ff is twice differentiable, f(0)=f(1)=0f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥ex, x∈[0,1]f''(x)-2f'(x)+f(x)\geq e^x,\ x\in[0,1]. If the function e−xf(x)e^{-x}f(x) assumes its minimum in the interval [0,1][0,1] at x=14x=\dfrac14, which of the following is true?
Not attempted
A
f′(x)<f(x), 14<x<34f'(x)<f(x),\ \dfrac14<x<\dfrac34
B
f′(x)>f(x), 0<x<14f'(x)>f(x),\ 0<x<\dfrac14
C
f′(x)<f(x), 0<x<14f'(x)<f(x),\ 0<x<\dfrac14
D
f′(x)<f(x), 34<x<1f'(x)<f(x),\ \dfrac34<x<1
7
A line L:y=mx+3L: y=mx+3 meets the yy-axis at E(0,3)E(0,3) and the arc of the parabola y2=16x, 0≤y≤6y^2=16x,\ 0\le y\le6 at the point F(x0,y0)F(x_0,y_0). The tangent to the parabola at F(x0,y0)F(x_0,y_0) intersects the yy-axis at G(0,y1)G(0,y_1). The slope mm of the line LL is chosen such that the area of the triangle EFGEFG has a local maximum. Match List I with List II and select the correct answer using the code given below the lists: List I: P. m=m= Q. Maximum area of △EFG\triangle EFG is R. y0=y_0= S. y1=y_1= List II: 1. 12\dfrac12 2. 44 3. 22 4. 11 Codes:
Not attempted
A
(A) P→4, Q→1, R→2, S→3P\to4,\ Q\to1,\ R\to2,\ S\to3
B
(B) P→3, Q→4, R→1, S→2P\to3,\ Q\to4,\ R\to1,\ S\to2
C
(C) P→1, Q→3, R→2, S→4P\to1,\ Q\to3,\ R\to2,\ S\to4
D
(D) P→1, Q→3, R→4, S→2P\to1,\ Q\to3,\ R\to4,\ S\to2
8
Let ω\omega be a complex cube root of unity with ω≠1\omega\neq1 and P=[pij]P=[p_{ij}] be an n×nn\times n matrix with pij=ωi+jp_{ij}=\omega^{i+j}. Then P2≠0P^2\neq0, when n=n=
Not attempted
A
5757
B
5555
C
5858
D
5656
9
Let w=3+i2w=\dfrac{\sqrt3+i}{2} and P={wn:n=1,2,3,…}P=\{w^n: n=1,2,3,\ldots\}. Further H1={z∈C:Re⁡z>12}H_1=\left\{z\in\mathbb{C}: \operatorname{Re} z>\dfrac12\right\} and H2={z∈C:Re⁡z<−12}H_2=\left\{z\in\mathbb{C}: \operatorname{Re} z<-\dfrac12\right\}, where C\mathbb{C} is the set of all complex numbers. If z1∈P∩H1z_1\in P\cap H_1, z2∈P∩H2z_2\in P\cap H_2 and OO represents the origin, then ∠z1Oz2=\angle z_1Oz_2=
Not attempted
A
π2\dfrac{\pi}{2}
B
π6\dfrac{\pi}{6}
C
2π3\dfrac{2\pi}{3}
D
5π6\dfrac{5\pi}{6}
10
If 3x=4x−13^x=4^{x-1}, then x=x=
Not attempted
A
2log⁡322log⁡32−1\dfrac{2\log_3 2}{2\log_3 2-1}
B
22−log⁡23\dfrac{2}{2-\log_2 3}
C
11−log⁡43\dfrac{1}{1-\log_4 3}
D
2log⁡232log⁡23−1\dfrac{2\log_2 3}{2\log_2 3-1}
11
Let S=S1∩S2∩S3S=S_1\cap S_2\cap S_3, where S1={z∈C:∣z∣<4}S_1=\{z\in\mathbb{C}: |z|<4\}, S2={z∈C:Im⁡[z−1+3 i1−3 i]>0}S_2=\left\{z\in\mathbb{C}: \operatorname{Im}\left[\dfrac{z-1+\sqrt3\,i}{1-\sqrt3\,i}\right]>0\right\} and S3={z∈C:Re⁡ z>0}S_3=\{z\in\mathbb{C}: \operatorname{Re}\,z>0\}. Area of S=S=
Not attempted
A
10π3\dfrac{10\pi}{3}
B
20π3\dfrac{20\pi}{3}
C
16π3\dfrac{16\pi}{3}
D
32π3\dfrac{32\pi}{3}
12
Let S=S1∩S2∩S3S=S_1\cap S_2\cap S_3, where S1={z∈C:∣z∣<4}S_1=\{z\in\mathbb{C}: |z|<4\}, S2={z∈C:Im⁡[z−1+3 i1−3 i]>0}S_2=\left\{z\in\mathbb{C}: \operatorname{Im}\left[\dfrac{z-1+\sqrt3\,i}{1-\sqrt3\,i}\right]>0\right\} and S3={z∈C:Re⁡ z>0}S_3=\{z\in\mathbb{C}: \operatorname{Re}\,z>0\}. min⁡z∈S∣1−3i−z∣=\displaystyle\min_{z\in S}|1-3i-z|=
Not attempted
A
2−32\dfrac{2-\sqrt3}{2}
B
2+32\dfrac{2+\sqrt3}{2}
C
3−32\dfrac{3-\sqrt3}{2}
D
3+32\dfrac{3+\sqrt3}{2}
13
In a triangle PQRPQR, PP is the largest angle and cos⁡P=13\cos P=\dfrac13. Further the incircle of the triangle touches the sides PQPQ, QRQR and RPRP at NN, LL and MM respectively, such that the lengths of PNPN, QLQL, RMRM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
Not attempted
A
1616
B
1818
C
2424
D
2222
14
Match List I with List II and select the correct answer using the code given below the lists: List I: P. (1y2(cos⁡(tan⁡−1y)+ysin⁡(tan⁡−1y)cot⁡(sin⁡−1y)+tan⁡(sin⁡−1y))2+y4)1/2\left(\dfrac1{y^2}\left(\dfrac{\cos(\tan^{-1}y)+y\sin(\tan^{-1}y)}{\cot(\sin^{-1}y)+\tan(\sin^{-1}y)}\right)^2+y^4\right)^{1/2} takes value Q. If cos⁡x+cos⁡y+cos⁡z=0=sin⁡x+sin⁡y+sin⁡z\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z then a possible value of cos⁡x−y2\cos\dfrac{x-y}{2} is R. If cos⁡(π4−x)cos⁡2x+sin⁡xsin⁡2xsec⁡x=cos⁡xsin⁡2xsec⁡x+cos⁡(π4+x)cos⁡2x\cos\left(\dfrac{\pi}{4}-x\right)\cos2x+\sin x\sin2x\sec x=\cos x\sin2x\sec x+\cos\left(\dfrac{\pi}{4}+x\right)\cos2x then a possible value of sec⁡x\sec x is S. If cot⁡(sin⁡−11−x2)=sin⁡(tan⁡−1(x6))\cot\left(\sin^{-1}\sqrt{1-x^2}\right)=\sin\left(\tan^{-1}(x\sqrt6)\right), x≠0x\neq0, then a possible value of xx is List II: 1. 1253\dfrac12\sqrt{\dfrac53} 2. 2\sqrt2 3. 12\dfrac12 4. 11 Codes:
Not attempted
A
(A) P→4, Q→3, R→1, S→2P\to4,\ Q\to3,\ R\to1,\ S\to2
B
(B) P→4, Q→3, R→2, S→1P\to4,\ Q\to3,\ R\to2,\ S\to1
C
(C) P→3, Q→4, R→2, S→1P\to3,\ Q\to4,\ R\to2,\ S\to1
D
(D) P→3, Q→4, R→1, S→2P\to3,\ Q\to4,\ R\to1,\ S\to2
15
Two lines L1:x=5, y3−α=z−2L_1: x=5,\ \dfrac{y}{3-\alpha}=\dfrac{z}{-2} and L2:x=α, y−1=z2−αL_2: x=\alpha,\ \dfrac{y}{-1}=\dfrac{z}{2-\alpha} are coplanar. Then α\alpha can take value(s)
Not attempted
A
11
B
22
C
33
D
44
16
Consider the lines L1:x−12=y−1=z+31L_1: \dfrac{x-1}{2}=\dfrac{y}{-1}=\dfrac{z+3}{1}, L2:x−41=y+31=z+32L_2: \dfrac{x-4}{1}=\dfrac{y+3}{1}=\dfrac{z+3}{2} and the planes P1:7x+y+2z=3P_1: 7x+y+2z=3, P2:3x+5y−6z=4P_2: 3x+5y-6z=4. Let ax+by+cz=dax+by+cz=d be the equation of the plane passing through the point of intersection of lines L1L_1 and L2L_2, and perpendicular to planes P1P_1 and P2P_2. Match List I with List II and select the correct answer using the code given below the lists: List I: P. a=a= Q. b=b= R. c=c= S. d=d= List II: 1. 1313 2. −3-3 3. 11 4. −2-2 Codes:
Not attempted
A
(A) P→3, Q→2, R→4, S→1P\to3,\ Q\to2,\ R\to4,\ S\to1
B
(B) P→1, Q→3, R→4, S→2P\to1,\ Q\to3,\ R\to4,\ S\to2
C
(C) P→3, Q→2, R→1, S→4P\to3,\ Q\to2,\ R\to1,\ S\to4
D
(D) P→2, Q→4, R→1, S→3P\to2,\ Q\to4,\ R\to1,\ S\to3
17
Circle(s) touching the xx-axis at a distance 33 from the origin and having an intercept of length 272\sqrt7 on the yy-axis is (are)
Not attempted
A
x2+y2−6x+8y+9=0x^2+y^2-6x+8y+9=0
B
x2+y2−6x+7y+9=0x^2+y^2-6x+7y+9=0
C
x2+y2−6x−8y+9=0x^2+y^2-6x-8y+9=0
D
x2+y2−6x−7y+9=0x^2+y^2-6x-7y+9=0
18
Let PQPQ be a focal chord of the parabola y2=4axy^2=4ax. The tangents to the parabola at PP and QQ meet at a point lying on the line y=2x+a, a>0y=2x+a,\ a>0. Length of chord PQPQ is
Not attempted
A
7a7a
B
5a5a
C
2a2a
D
3a3a
19
Let PQPQ be a focal chord of the parabola y2=4axy^2=4ax. The tangents to the parabola at PP and QQ meet at a point lying on the line y=2x+a, a>0y=2x+a,\ a>0. If chord PQPQ subtends an angle θ\theta at the vertex of y2=4axy^2=4ax, then tan⁡θ=\tan\theta=
Not attempted
A
237\dfrac23\sqrt7
B
−237-\dfrac23\sqrt7
C
235\dfrac23\sqrt5
D
−235-\dfrac23\sqrt5
20
For a∈Ra \in \mathbb{R} (the set of all real numbers), a≠−1a \neq -1, lim⁡n→∞(1a+2a+⋯+na)(n+1)a−1[(na+1)+(na+2)+⋯+(na+n)]=160\displaystyle\lim_{n\to\infty}\frac{\left(1^{a}+2^{a}+\cdots+n^{a}\right)}{(n+1)^{a-1}\left[(na+1)+(na+2)+\cdots+(na+n)\right]}=\frac{1}{60}. Then a=a=
Not attempted
A
55
B
77
C
−152-\dfrac{15}{2}
D
−172-\dfrac{17}{2}
21
Consider the set of eight vectors V={ai^+bj^+ck^:a,b,c∈{−1,1}}V=\{a\hat i+b\hat j+c\hat k : a,b,c\in\{-1,1\}\}. Three non-coplanar vectors can be chosen from VV in 2p2^p ways. Then pp is
Not attempted
Your answer: —
Correct answer: 5
22
A vertical line passing through the point (h,0)(h,0) intersects the ellipse x24+y23=1\dfrac{x^2}{4}+\dfrac{y^2}{3}=1 at the points PP and QQ. Let the tangents to the ellipse at PP and QQ meet at the point RR. If Δ(h)=\Delta(h)= area of the triangle PQRPQR, Δ1=max⁡1/2≤h≤1Δ(h)\Delta_1=\displaystyle\max_{1/2\le h\le1}\Delta(h) and Δ2=min⁡1/2≤h≤1Δ(h)\Delta_2=\displaystyle\min_{1/2\le h\le1}\Delta(h), then 85Δ1−8Δ2=\dfrac{8}{\sqrt5}\Delta_1-8\Delta_2=
Not attempted
Your answer: —
Correct answer: 9
23
Of the three independent events E1,E2E_1,E_2 and E3E_3, the probability that only E1E_1 occurs is α\alpha, only E2E_2 occurs is β\beta and only E3E_3 occurs is γ\gamma. Let the probability pp that none of the events E1,E2E_1,E_2 or E3E_3 occurs satisfy the equations (α−2β)p=αβ(\alpha-2\beta)p=\alpha\beta and (β−3γ)p=2βγ(\beta-3\gamma)p=2\beta\gamma. All the given probabilities are assumed to lie in the interval (0,1)(0,1). Then Probability of occurrence of E1Probability of occurrence of E3=\dfrac{\text{Probability of occurrence of }E_1}{\text{Probability of occurrence of }E_3}=
Not attempted
Your answer: —
Correct answer: 6
24
A pack contains nn cards numbered from 11 to nn. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 12241224. If the smaller of the numbers on the removed cards is kk, then k−20=k-20=
Not attempted
Your answer: —
Correct answer: 5
25
The coefficients of three consecutive terms of (1+x)n+5(1+x)^{n+5} are in the ratio 5:10:145:10:14. Then n=n=
Not attempted
Your answer: —
Correct answer: 6
26
A line ll passing through the origin is perpendicular to the lines l1:(3+t)i^+(−1+2t)j^+(4+2t)k^, −∞<t<∞l_1:(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k,\ -\infty<t<\infty and l2:(3+2s)i^+(3+2s)j^+(2+s)k^, −∞<s<∞l_2:(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k,\ -\infty<s<\infty. Then, the coordinate(s) of the point(s) on l2l_2 at a distance of 17\sqrt{17} from the point of intersection of ll and l1l_1 is (are)
Not attempted
A
(73,73,53)\left(\dfrac73,\dfrac73,\dfrac53\right)
B
(−1,−1,0)(-1,-1,0)
C
(1,1,1)(1,1,1)
D
(79,79,89)\left(\dfrac79,\dfrac79,\dfrac89\right)
27
Let f(x)=xsin⁡πxf(x)=x\sin\pi x, x>0x>0. Then for all natural numbers nn, f′(x)f'(x) vanishes at
Not attempted
A
a unique point in the interval (n,n+12)\left(n,n+\dfrac12\right)
B
a unique point in the interval (n+12,n+1)\left(n+\dfrac12,n+1\right)
C
a unique point in the interval (n,n+1)(n,n+1)
D
two points in the interval (n,n+1)(n,n+1)
28
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:158:15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are
Not attempted
A
24
B
32
C
45
D
60
29
For 3×33\times3 matrices MM and NN, which of the following statement(s) is (are) NOT correct?
Not attempted
A
NTMNN^TMN is symmetric or skew symmetric, according as MM is symmetric or skew symmetric
B
MN−NMMN-NM is skew symmetric for all symmetric matrices MM and NN
C
MNMN is symmetric for all symmetric matrices MM and NN
D
(adj M)(adj N)=adj(MN)(\text{adj}\,M)(\text{adj}\,N)=\text{adj}(MN) for all invertible matrices MM and NN
30
Let Sn=∑k=14n(−1)k(k+1)2k2S_n=\displaystyle\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2. Then SnS_n can take value(s)
Not attempted
A
1056
B
1088
C
1120
D
1332
31
Let PR⃗=3i^+j^−2k^\vec{PR}=3\hat{i}+\hat{j}-2\hat{k} and SQ⃗=i^−3j^−4k^\vec{SQ}=\hat{i}-3\hat{j}-4\hat{k} determine diagonals of a parallelogram PQRSPQRS and PT⃗=i^+2j^+3k^\vec{PT}=\hat{i}+2\hat{j}+3\hat{k} be another vector. Then the volume of the parallelepiped determined by the vectors PT⃗,PQ⃗\vec{PT},\vec{PQ} and PS⃗\vec{PS} is
Not attempted
A
5
B
20
C
10
D
30
32
Perpendiculars are drawn from points on the line x+22=y+1−1=z3\dfrac{x+2}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3} to the plane x+y+z=3x+y+z=3. The feet of perpendiculars lie on the line
Not attempted
A
x5=y−18=z−2−13\dfrac{x}{5}=\dfrac{y-1}{8}=\dfrac{z-2}{-13}
B
x2=y−13=z−2−5\dfrac{x}{2}=\dfrac{y-1}{3}=\dfrac{z-2}{-5}
C
x4=y−13=z−2−7\dfrac{x}{4}=\dfrac{y-1}{3}=\dfrac{z-2}{-7}
D
x2=y−1−7=z−25\dfrac{x}{2}=\dfrac{y-1}{-7}=\dfrac{z-2}{5}
33
For a>b>c>0a>b>c>0, the distance between (1,1)(1,1) and the point of intersection of the lines ax+by+c=0ax+by+c=0 and bx+ay+c=0bx+ay+c=0 is less than 222\sqrt2. Then
Not attempted
A
a+b−c>0a+b-c>0
B
a−b+c<0a-b+c<0
C
a−b+c>0a-b+c>0
D
a+b−c<0a+b-c<0
34
The value of cot⁡(∑n=123cot⁡−1(1+∑k=1n2k))\cot\left(\displaystyle\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right) is
Not attempted
A
2325\dfrac{23}{25}
B
2523\dfrac{25}{23}
C
2324\dfrac{23}{24}
D
2423\dfrac{24}{23}
35
A curve passes through the point (1,π6)\left(1,\dfrac{\pi}{6}\right). Let the slope of the curve at each point (x,y)(x,y) be yx+sec⁡(yx)\dfrac{y}{x}+\sec\left(\dfrac{y}{x}\right), x>0x>0. Then the equation of the curve is
Not attempted
A
sin⁡(yx)=log⁡x+12\sin\left(\dfrac{y}{x}\right)=\log x+\dfrac12
B
cosec(yx)=log⁡x+2\text{cosec}\left(\dfrac{y}{x}\right)=\log x+2
C
sec⁡(2yx)=log⁡x+2\sec\left(\dfrac{2y}{x}\right)=\log x+2
D
cos⁡(2yx)=log⁡x+12\cos\left(\dfrac{2y}{x}\right)=\log x+\dfrac12
36
The area enclosed by the curves y=sin⁡x+cos⁡xy=\sin x+\cos x and y=∣cos⁡x−sin⁡x∣y=|\cos x-\sin x| over the interval [0,π2]\left[0,\dfrac{\pi}{2}\right] is
Not attempted
A
4(2−1)4(\sqrt2-1)
B
22(2−1)2\sqrt2(\sqrt2-1)
C
2(2+1)2(\sqrt2+1)
D
22(2+1)2\sqrt2(\sqrt2+1)
37
The number of points in (−∞,∞)(-\infty,\infty), for which x2−xsin⁡x−cos⁡x=0x^2 - x\sin x - \cos x = 0, is
Not attempted
A
6
B
4
C
2
D
0
38
Let f:[12,1]→Rf:\left[\dfrac12,1\right]\to\mathbb{R} (the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x)<2f(x)f'(x) < 2f(x) and f(12)=1f\left(\dfrac12\right)=1. Then the value of ∫1/21f(x) dx\displaystyle\int_{1/2}^{1} f(x)\,dx lies in the interval
Not attempted
A
(2e−1,2e)(2e-1,2e)
B
(e−1,2e−1)(e-1,2e-1)
C
(e−12,e−1)\left(\dfrac{e-1}{2},e-1\right)
D
(0,e−12)\left(0,\dfrac{e-1}{2}\right)
39
Four persons independently solve a certain problem correctly with probabilities 12,34,14,18\dfrac{1}{2},\dfrac{3}{4},\dfrac{1}{4},\dfrac{1}{8}. Then the probability that the problem is solved correctly by at least one of them is
Not attempted
A
235256\dfrac{235}{256}
B
21256\dfrac{21}{256}
C
3256\dfrac{3}{256}
D
253256\dfrac{253}{256}
40
Let complex numbers α\alpha and 1α\dfrac{1}{\alpha} lie on circles (x−x0)2+(y−y0)2=r2(x-x_0)^2+(y-y_0)^2=r^2 and (x−x0)2+(y−y0)2=4r2(x-x_0)^2+(y-y_0)^2=4r^2 respectively. If z0=x0+iy0z_0=x_0+iy_0 satisfies 2∣z0∣2=r2+22|z_0|^2=r^2+2, then ∣α∣=|\alpha|=
Not attempted
A
12\dfrac{1}{\sqrt2}
B
12\dfrac{1}{2}
C
17\dfrac{1}{\sqrt7}
D
13\dfrac{1}{3}
Write both circle conditions as ∣α−z0∣2=r2|\alpha-z_0|^2=r^2 and ∣1/α−z0∣2=4r2|1/\alpha - z_0|^2=4r^2, expand using Re\text{Re}, and exploit rotational symmetry: since rotating α\alpha and z0z_0 by the same angle preserves both circle equations and ∣z0∣|z_0|, you may choose z0z_0 to be real without loss of generality.
Deconstruction\textbf{Deconstruction}: We need ∣α∣|\alpha| given two circle-membership conditions on α\alpha and 1/α1/\alpha, linked through z0z_0 by 2∣z0∣2=r2+22|z_0|^2=r^2+2. Strategy\textbf{Strategy}: Expand ∣α−z0∣2=r2|\alpha-z_0|^2=r^2 and ∣1/α−z0∣2=4r2|1/\alpha-z_0|^2=4r^2 in terms of Re(αzˉ0)\text{Re}(\alpha\bar z_0) and Re(αz0)\text{Re}(\alpha z_0). The whole configuration is invariant under simultaneously rotating α\alpha and z0z_0 by the same angle (this preserves both circle equations and ∣z0∣|z_0|, hence ∣α∣|\alpha| is unaffected), so we may choose z0z_0 real, which forces Re(αzˉ0)=Re(αz0)\text{Re}(\alpha\bar z_0)=\text{Re}(\alpha z_0) and collapses the system to one variable. Execution\textbf{Execution}: Write α=p+iq\alpha=p+iq. Circle 1 gives ∣α∣2−2Re(αzˉ0)+∣z0∣2=r2|\alpha|^2-2\text{Re}(\alpha\bar z_0)+|z_0|^2=r^2. Circle 2 (on 1/α1/\alpha) gives 1∣α∣2−2Re(αz0)∣α∣2+∣z0∣2=4r2\dfrac{1}{|\alpha|^2}-\dfrac{2\text{Re}(\alpha z_0)}{|\alpha|^2}+|z_0|^2=4r^2. - Choose z0=x0z_0=x_0 real (WLOG by rotation). Then Re(αzˉ0)=Re(αz0)=px0=:A\text{Re}(\alpha\bar z_0)=\text{Re}(\alpha z_0)=p x_0=:A. - Let s=∣α∣2s=|\alpha|^2. Eq.(1): A=s+x02−r2A=s+x_0^2-r^2. - Eq.(2): 1−As=4r2−x02⇒1−A=s(4r2−x02)\dfrac{1-A}{s}=4r^2-x_0^2 \Rightarrow 1-A=s(4r^2-x_0^2). - Substituting AA: 1−x02+r2=s(1−x02+4r2)1-x_0^2+r^2=s(1-x_0^2+4r^2), so s=1−x02+r21−x02+4r2s=\dfrac{1-x_0^2+r^2}{1-x_0^2+4r^2}. - Given constraint (with z0z_0 real, ∣z0∣2=x02|z_0|^2=x_0^2): 2x02=r2+2⇒x02=r2+222x_0^2=r^2+2 \Rightarrow x_0^2=\dfrac{r^2+2}{2}. - Then 1−x02+r2=1−r2+22+r2=r221-x_0^2+r^2 = 1-\dfrac{r^2+2}{2}+r^2=\dfrac{r^2}{2}, and 1−x02+4r2=1−r2+22+4r2=7r221-x_0^2+4r^2=1-\dfrac{r^2+2}{2}+4r^2=\dfrac{7r^2}{2}. - So s=r2/27r2/2=17s=\dfrac{r^2/2}{7r^2/2}=\dfrac{1}{7}, giving ∣α∣=17|\alpha|=\dfrac{1}{\sqrt7}. Verification\textbf{Verification}: The result is independent of rr (it cancels), consistent with the problem expecting a numeric answer purely from the given relation 2∣z0∣2=r2+22|z_0|^2=r^2+2 — a strong consistency check that the algebra is dimensionally sound. Final Answer\textbf{Final Answer}: (C) 17\dfrac{1}{\sqrt7}