INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
Question – 1 : What is the inverse of the matrix A=(cosθsinθ0−sinθcosθ0001) :
Answer – (A) : (cosθ−sinθ0sinθcosθ0001)
Answer – (B) : (cosθ0−sinθ010sinθ0cosθ)
Answer – (C) : (1000cosθ−sinθ0sinθcosθ)
Answer – (D) : (cosθsinθ0−sinθcosθ0001)
View Answer
Correct Answer : (A)
Explanation : We know that A−1=adj(A)|A| …..(i) Now,|A|=cosθ(cosθ)−sinθ(−sinθ)=cos2θ+sin2θ=1Now, AdjA=(cosθ−sinθ0sinθcosθ0001)Putting the value in (i), ∴A−1=(cosθ−sinθ0sinθcosθ0001)
Question – 2 : The modulus-amplitude form of 3+i, where i=−1 is:
Answer – (A) : 2(cosπ6+isinπ6)
Answer – (B) : 2(cosπ6+isinπ6)
Answer – (C) : 4(cosπ3+isinπ3)
Answer – (D) : 4(cosπ6+isinπ6)
View Answer
Correct Answer : (B)
Explanation : Modulus amplitude form of (3+i) is:z=r(cosθ+isinθ)⇒ rcosθ=3,rsinθ=1⇒r2(cos2θ+sin2θ)=4⇒r2=4⇒r=22cosθ=3 and 2sinθ=1⇒cosθ=32 and sinθ=12⇒θ=π6 and θ=π6∴z=2(cosπ6+isinπ6)
Question – 3 : If the sum of the infinite GP is 43 and its first term is 34 then its common ratio is:
Answer – (A) : 716
Answer – (B) : 916
Answer – (C) : 19
Answer – (D) : 79
View Answer
Correct Answer : (A)
Explanation : S∞=a1−r where a ‘ be the first term and r be the common ratio of GP.∴43=341−r⇒1−r=3443⇒r=1−916⇒r=716
Question – 4 : How many four-digit numbers divisible by 10 can be formed using 1,5,0,6,7 without repetition of digits?
Answer – (A) : 24
Answer – (B) : 36
Answer – (C) : 44
Answer – (D) : 64
View Answer
Correct Answer : (A)
Explanation : Four-digit numbers divisible by10 using 1,5,0,6,7. Divisibility by 10 → no. should end with zero After fixing ‘zero’ at unit place, we are left with 4 numbers. After placing any number at tens place we are left with ‘3’ numbers. After placing any number at tens and hundreds place. Total =4×3×2=24
Question – 5 : In the expansion of (1+x)43 coefficients of (2r+1)th and (r+2)th terms are equal, then what is the value of r : (≠1)
Answer – (A) : 14
Answer – (B) : 14
Answer – (C) : 21
Answer – (D) : 22
View Answer
Correct Answer : (B)
Explanation : Given as:(1+x)43As we know, the coefficient of the r term in the expansion of (1+x)n is nCr−1 . Therefore, the coefficients of the (2r+1) and (r+2) terms in the given expansion are 43C2r+1−1 and 43Cr+2−1For these coefficients to be equal, we must have,⇒ 43C2r+1−1=43Cr+2−1⇒ 43C2r=43Cr+1⇒ 2r=r+1 or 2r+r+1=43[ since, nCr=nCs⇒r=s (or) r+s=n]⇒ 2r−r=1 or 3r+1=43⇒ 3r=43−1 r=1⇒ r=1 or 3r=42⇒ r=423⇒ r=1 or r=14∴r=14 [since, value 1 gives the same term]
Question – 6 : The product of r consecutive positive integers, divided by r! is:
Answer – (A) : A positive integer
Answer – (B) : Equal to r
Answer – (C) : A positive integer
Answer – (D) : None of these
View Answer
Correct Answer : (C)
Explanation : Consider r consecutive positive intergers,n+1,……,n+r−1,n+rThen, (n+1)…(n+r)r!=n!(n+1)……(n+r)n!r!=n+rCrWhich is a positive integer.
Question – 7 : If the cardinality of a set A is 4 and that of a set B is 3 , then what is the cardinality of the set AΔB ?
Answer – (A) : Cannot be determined
Answer – (B) : 5
Answer – (C) : 7
Answer – (D) : Cannot be determined
View Answer
Correct Answer : (D)
Explanation : Since sets A and B are not known, the cardinality of the set AΔB can not be determined.
Question – 8 : ∫dxez+e−z equals:
Answer – (A) : tan−1ex+c
Answer – (B) : log(ex−e−x)+c
Answer – (C) : tan−1ex+c
Answer – (D) : tan−1e−x+c
View Answer
Correct Answer : (C)
Explanation : I=∫exex+e−xdxI=∫exe2x+1dxLet, ex=t⇒exdx=dt=∫1t2+1dt=tan−1t+c∫exex+e−xdx =tan−1ex+c
Question – 9 : If the ratio AM to GM of two positive numbers a and b is 5:3 , then a:b is equal to:
Answer – (A) : 09:01
Answer – (B) : 02:09
Answer – (C) : 09:01
Answer – (D) : 05:03
View Answer
Correct Answer : (C)
Explanation : AMGM=53a+b2ab=53a+b2ab=53ab+12ab=53 [divide numerator & denominator by b ] Let, ab=y⇒ y+12y=53⇒ 3y+3=10ySquaring both sides,⇒ 9y2+9+18y=100y⇒ 9y2−82y+9=0⇒ 9y2−81y−y+9=0⇒ 9y(y−9)−1(y−9)=0⇒ (y−9)(9y−1)=0⇒ y=19 or y=9i.e. ab=19 or ab=91
Question – 10 : If A + B + C = π , then what is sin (A + B) + sin C equal to?
Answer – (A) : 2 sin C
Answer – (B) : 2 sin C
Answer – (C) : cos C – sin C
Answer – (D) : None of the above
View Answer
Correct Answer : (B)
Explanation : As we know, sin (π – θ) = sin θ Given that: A + B + C = π ⇒ A + B = π – C Now, sin (A + B) = sin (π -C) Now put the value in given equation, sin (A + B) + sin C = sin (π – C) + sin C = sin C + sin C (∵ sin (π – θ) = sin θ) = 2 sin C
Question – 11 : Which of these is a second-order differential equation?
Answer – (A) : y′y′′+y=sinx
Answer – (B) : y′y′′+y=sinx
Answer – (C) : y′′′y′′+y=0
Answer – (D) : None of these
View Answer
Correct Answer : (B)
Explanation : The order of a differential equation is the order of the highest order derivative. So, y′′y′+y=sinx has order 2 .
Question – 12 : What is i1000+i1001+i1002+i1003 equal to: (where i=−1 )
Answer – (A) : 0
Answer – (B) : i
Answer – (C) : −i
Answer – (D) : 1
View Answer
Correct Answer : (A)
Explanation : i1000+i1001+i1002+i1003=i1000+i1000⋅i+i1002+i1002⋅i=(i2)500+(i2)500⋅i+(i2)501+(i2)501⋅i=(−1)500+(−1)500⋅i+(−1)501+(−1)501⋅i=1+i−1−i=0
Question – 13 : What is the principal argument of (−1−i) , (where i=−1 )?
Answer – (A) : −3π4
Answer – (B) : −π4
Answer – (C) : −3π4
Answer – (D) : 3π4
View Answer
Correct Answer : (C)
Explanation : Given, Z=−1−iZ=r(cosθ+isinθ)Here, r is modulus and θ is argument.rcosθ=−1 and rsinθ=−1⇒r2=2⇒r=2Undefined control sequence therefore⇒cosθ=−122sinθ=−1⇒sinθ=−−12Here both cosθ and sinθ are negative.So, θ lies in 3 rd quadrant.Argument =−3π4
Question – 14 : Find the equation of the curve with differential equation (1+y2)dx=xydy, and passing through (1,0) :
Answer – (A) : x2−y2=1
Answer – (B) : 4×2−y2=4
Answer – (C) : x2+y2=1
Answer – (D) : 4×2+y2=4
View Answer
Correct Answer : (A)
Explanation : Given,(1+y2)dx=xydy⇒2xdx=2y1+y2dy… (i) Let t=1+y2⇒dt=2ydy… ..(ii) integrating (i) and (ii), Using ⇒∫2xdx=∫dtt+logcTherefore, logx2=logct⇒x2=c(1+y2) …..(i) Put (x,y)=(1,0)⇒1=c(1+0)⇒c=1Put c=1 in (i), So, the equation of the curve will be x2−y2=1
Question – 15 : What is 1log2N+1log3N+1log4N+…+1log100N equal to (N≠1) ?
Answer – (A) : 1log100!N
Answer – (B) : 1log99!N
Answer – (C) : 99log100N
Answer – (D) : 99log99!N
View Answer
Correct Answer : (A)
Explanation : 1log2 N+1log3 N+…+1log100 N=logN2+logN3+…+logN100 =logN(2×3×⋯×100)=logN(100!) =1log100!N
Question – 16 : If α,β are the different complex numbers with |β|=1, then find |β−α1−α¯β|?
Answer – (A) : 1
Answer – (B) : 2
Answer – (C) : 1
Answer – (D) : 0
View Answer
Correct Answer : (C)
Explanation : |β−α1−αβ|2=(β−α1−α¯β)(β−α1−αβ)⇒|β−α1−αβ|2=ββ−βα−αβ+αα(1−αβ)(1−αβ)⇒|β−α1−α¯β|2=|β|2−βα−αβ+|α|21−αβ−αβ+|α|2|β|2⇒|β−α1−αβ|2=|α|2−βα¯−αβ¯+11−αβ−αβ+|α|2 , [∵|β|=1]⇒|β−α1−αβ|=1
Question – 17 : How many numbers between 100 and 1000 can be formed with the digits 5,6,7,8,9, if the repetition of digits is not allowed?
Answer – (A) : 60
Answer – (B) : 53
Answer – (C) : 120
Answer – (D) : 60
View Answer
Correct Answer : (D)
Explanation : Number lying between 100 and 1000 formed by 3 digits. And every digit have 5 option (1,2,3,4,5) to select a number. But repetition is not allowed. So, number =5P3=5!(5−3)!=5!2!5P3 =60
Question – 18 : If n∈N, then 121n−25n+1900n−(−4)n is divisible by which one of the following?
Answer – (A) : 2000
Answer – (B) : 2000
Answer – (C) : 2002
Answer – (D) : 2006
View Answer
Correct Answer : (B)
Explanation : (121)n−25n+1900n−(−4)nFor n=1 =121−25+1900−(−4)=121−25+1900+4=2025−25=2000
Question – 19 : If cosT=35 and if sinR=817, where T is in the fourth quadrant and R is in second quadrant, then cos(T−R) is equal to:
Answer – (A) : −7785
Answer – (B) : 1385
Answer – (C) : −1385
Answer – (D) : −7785
View Answer
Correct Answer : (D)
Explanation : We have,cosT=35sinT=1−cos2TsinT=1−(35)2=1−925=1625=−45 (since T is in IV quadrant)sinR=817cosR=1−sin2RcosR=1−(817)2cosR=1−64289=225289=−1517 (since R is in II quadrant) Now,∵cos(T−R)=cosTcosR+sinTsinR=35×−1517+−45×817=[−45−3285]=−7785
Question – 20 : If |x218x|=|623×6| , then x is equal to:
Answer – (A) : 6
Answer – (B) : ±6
Answer – (C) : -6
Answer – (D) : 0
View Answer
Correct Answer : (A)
Explanation : Given: |x218x|=|623×6|We have to find the value of x . ⇒x2−36=36−6x([abcd]=ad−bc)⇒x2+6x−72=0Using Shridharachaarye rule,⇒x=−6+62+4×722=−6±36(1+8)2x=−3±9=6 or −12
Question – 21 : Find the co-ordinates of in-centre of the triangle whose vertices are (0,6),(8,12) and (8,0):
Answer – (A) : (5,6)
Answer – (B) : (8,11)
Answer – (C) : (−4,3)
Answer – (D) : (5,6)
View Answer
Correct Answer : (D)
Explanation : Let, A(x1,y1)=(0,6),B(x2,y2)=(8,12) and C(x3,y3)=(8,0) be the vertices of triangle ABC.Distance =(X1−X2)2+(Y1−Y2)2Then, c=AB for (0,6)(8,12)=(0−8)2+(6−12)2AB=10b=CA for (0,6)(8,0)=(0−8)2+(6−0)2CA=10a=BC for (8,12)(8,0)=(8−8)2+(12−0)2BC=12The co-ordinates of the in-centre are (ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)⇒(12×0+10×8+10×812+10+10,12×6+10×12+10×012+10+10)⇒(16032,19232)⇒(5,6)
Question – 22 : In a class, 54 students are good in Hindi only, 63 students are good in Mathematics only and 41 students are good in English only. There are 18 students who are good in both Hindi and Mathematics. 10 students are good in all three subjects. What is the number of students who are good in Hindi and Mathematics but not in English?
Answer – (A) : 8
Answer – (B) : 12
Answer – (C) : 10
Answer – (D) : 8
View Answer
Correct Answer : (D)
Explanation : Let H,M,E denote the set of students studying Hindi, Mathematics and English.n(H∩M)−n(H∩M∩E)=18−10 =8
Question – 23 : In a class, 54 students are good in Hindi only, 63 students are good in Mathematics only and 41 students are good in English only. There are 18 students who are good in both Hindi and Mathematics. 10 students are good in all three subjects.What is the number of students who are good in either Hindi or mathematics but not in English?
Answer – (A) : 125
Answer – (B) : 107
Answer – (C) : 125
Answer – (D) : 130
View Answer
Correct Answer : (C)
Explanation : Let, H,M,E denote the set of students studying Hindi, Mathematics and English.n(H)+n(M)+n(H∩M)−n(H∩M∩E)[ Excluding English]=54+63+18−10=125
Question – 24 : The number of non-zero integral solutions of the equation |1−2i|x=5x is:
Answer – (A) : 0
Answer – (B) : 1
Answer – (C) : 2
Answer – (D) : 3
View Answer
Correct Answer : (A)
Explanation : Giving, |1−2i|x=5xWe need to find the value of x for which should be an integer but not zero.Let first find value of |1−2i|⇒ |1−2i|=(1)2+(2)2=5[∵z=x+iy⇒|z|=x2+y2]So,|1−2i|x=5x⇒(5)x=5x⇒ (5)x=(5)2xComparing powers,⇒x=2x⇒x−2x=0⇒−x=0⇒x=0
Question – 25 : If n=(2017)!, then what is 1log2n+1log3n+1log4n+…+1log2017n equal to:
Answer – (A) : 1
Answer – (B) : 1
Answer – (C) : n2
Answer – (D) : n
View Answer
Correct Answer : (B)
Explanation : If n=(2017)!1log2π+1log3n+…+1log2017n………… (i) Now we know that 1lognb=logba∴ We can rewrite (i) as, 1log2n+1log3n+…+1log2017n=logn2+logn3+logn4+…+logn2017=logn(2.3.4…….2017) as [logab+logac=loga(b×c)]∵n=(2017)!∴logn(2.3.4….2017)=log2017!(2017)!=1So,1log2n+1log3n+…+1log2017n=1
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
INDIAN NAVY AA SSR MATHEMATICS MOCK TEST
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