INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
Question – 1 : The value of tan60∘cot30∘ is equal to:
Answer – (A) : 1
Answer – (B) : 1
Answer – (C) : 2
Answer – (D) : 3
View Answer
Correct Answer : (B)
Explanation : tan60∘=3 and cot30∘=3So, tan60∘cot30∘=33=1
Question – 2 : Find median and mode of the messages received on 9 consecutive days 15,11,9,5,18,4,15,13,17 .
Answer – (A) : 13,18
Answer – (B) : 13,18
Answer – (C) : 18,15
Answer – (D) : 15,16
View Answer
Correct Answer : (B)
Explanation : Arranging the terms in ascending order 4,5,9,11,13,14,15,18,18Median is (n+1)2 term as n=9 (odd)=(9+1)2=102=5th term which is 13, Median is 13Mode =18 which is repeated twice.
Question – 3 : A and B are two sets having 3 elements in common. If n(A)=5,n(B)=4, then n(A×B) is equal to:
Answer – (A) : 20
Answer – (B) : 9
Answer – (C) : 15
Answer – (D) : 20
View Answer
Correct Answer : (D)
Explanation : Given, n(A)=5,n(B)=4 and A and B have 3 elements in common.n(A×B)=n(A)n(B)=5×4n(A×B) =20
Question – 4 : If z=−2+5i then z2+4z+30 is:
Answer – (A) : 1
Answer – (B) : 0
Answer – (C) : 2
Answer – (D) : 4
View Answer
Correct Answer : (A)
Explanation : Given,z=−2+5i⇒z+2=5iSquaring both sides, we get,(z+2)2=(5i)2⇒z2+4z+4=−25⇒z2+4z+29=0Now, adding (1) both sides, we get,∴z2+4z+30=1
Question – 5 : For what value of x , the matrix A is singular? A=[3−x2224−x1−2−4−1−x]
Answer – (A) : x=0,3
Answer – (B) : x=1,2
Answer – (C) : x=2,3
Answer – (D) : x=0,3
View Answer
Correct Answer : (D)
Explanation : A=[3−x2224−x1−2−4−1−x]If the matrix is singular, its determinant has to be zero.⇒(3−x)[(4−x)(−1−x)+4]−2[2(−1−x)+2]+2[−8+2(4−x)]=0⇒(3−x)[−4−4x+x+x2+4]−2[−2−2x+2]+2[−8+8−2x]=0⇒(3−x)[x2−3x]+4x−4x=0⇒(3−x)x(x−3)=0⇒x=0,3
Question – 6 : The value of k which makesf(x)={sinx,x≠0k,x=0 continuous at x=0, is:
Answer – (A) : 0
Answer – (B) : 1
Answer – (C) : -1
Answer – (D) : 0
View Answer
Correct Answer : (D)
Explanation : f(x) is Continuous at x=0⇒limx→0+f(x)=limx→0−f(x)=f(0)⇒limx→0+sinx=limx→0−sinx=k⇒limh→0sin(0+h)=limh→0sin(0−h)=k⇒k=0Hence, the correct option is (D)
Question – 7 : The co-efficient of y in the expansion of (y2+c/y)5 is?
Answer – (A) : 10c3
Answer – (B) : 20c3
Answer – (C) : 10c
Answer – (D) : 20c
View Answer
Correct Answer : (A)
Explanation : (y2+c/y)5=5C0(cy)0(y2)5−0+5C1(cy)1(y2)5−1+…+5C5(cy)5(y2)5−5=∑r=055Cr(cy)r(y2)5−r …..(i) We need cofficient of y⇒2(5−r)−r=1⇒10−3r=1⇒r=3put r=3 in (i),=5C3(cy)3(y2)2=5C3c3ySo, cofficient of y=5C3⋅c3=10c3
Question – 8 : Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).
Answer – (A) : 12
Answer – (B) : 13
Answer – (C) : 14
Answer – (D) : 15
View Answer
Correct Answer : (A)
Explanation : Using the formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B)Then n(A ∩B) = n(A) + n(B) – n(A ∪B)= 20 + 28 – 36= 48 – 36n(A ∩ B) = 12
Question – 9 : The period of f(x)=sinπx2+2cosπx3−tanπx4 is:
Answer – (A) : 12
Answer – (B) : 3
Answer – (C) : 4
Answer – (D) : 12
View Answer
Correct Answer : (D)
Explanation : Period of sinπx2 is 2ππ/2=4=T1 Period of cosπx3 is 2ππ/3=6=T2 Period of tanπx4 is ππ/4=4=T3 Period of f is = L.C.M. of T1,T2,T3=12
Question – 10 : If f(x)=xcosx, then f′(0)= ?
Answer – (A) : 1
Answer – (B) : 0
Answer – (C) : 1
Answer – (D) : ∞
View Answer
Correct Answer : (C)
Explanation : f(x)=xcosxAs we know that D(uv)=uv′+vu′So, f′(x)=x(−sinx)+cosx(1)=−xsinx+cosxBy putting x=0, we havef′(0)=0+cos(0)(∵cos(0)=1)=1
Question – 11 : If nPr=3024 and nCr=126 then find n and r .
Answer – (A) : 9,4
Answer – (B) : 10,3
Answer – (C) : 12,4
Answer – (D) : 11,4
View Answer
Correct Answer : (A)
Explanation : nPrnCr=3024126nPr=n!(n−r)!nCr=n!(n−r)!×r!So, [n!(n−r)!]÷[n!(n−r)!×r!]=2424=r!So, r=4Now, nP4=3024n!(n−4)!=3024n(n−1)(n−2)(n−3)=9.8.7.6n=9
Question – 12 : The area of a parallelogram whose adjacent sides are represented by the vectors a=−i^−2j^−3k^ and b=−i^+2j^−3k^ is:
Answer – (A) : 410
Answer – (B) : 6
Answer – (C) : 4936
Answer – (D) : 410
View Answer
Correct Answer : (D)
Explanation : Area of parallelogram =|a×b| =|i^j^k^−1−2−3−12−3|=|i^(6+6)−j^(3−3)+k^(−2−2)|=|12i^−4k^|=122+42=410
Question – 13 : If O is the centre of the circle and △AOB is an equilateral triangle, then the measure of ∠ACB is:
Answer – (A) : 30∘
Answer – (B) : 30∘
Answer – (C) : 90∘
Answer – (D) : 75∘
View Answer
Correct Answer : (B)
Explanation : Given,O is the center of the circle and △AOB is an equilateral triangle. Thus, ∠AOB=60∘We know that the angle at the center of the circle is twice the angle at the circumference subtended by the same arc. Thus, ∠AOB=2∠ACB⇒60=2∠ACB⇒∠ACB=602⇒∠ACB=30∘
Question – 14 : If cosX=23 then tanX is equal to:
Answer – (A) : 54
Answer – (B) : 52
Answer – (C) : 54
Answer – (D) : 25
View Answer
Correct Answer : (C)
Explanation : By trigonometry identities, we know:1+tan2x=sec2xAnd secX=1cosX=1(23)=32So,1+tan2X=(32)2=94tan2X=94−1=54TanX=54
Question – 15 : Forces F1 and F2 act on a point mass in two mutual perpendicular directions. The resultant force on the point mass will be:
Answer – (A) : F12+F22
Answer – (B) : F1−F2
Answer – (C) : F12+F22
Answer – (D) : F12+F22
View Answer
Correct Answer : (C)
Explanation : We know that resultant,R=A2+B2+2ABcosθ∴R=F12+F22+2F1F2cos90=F12+F22[cos900=0]
Question – 16 : What is the degree of the differential equation (d3ydx3)32=(d2ydx2)2?
Answer – (A) : 3
Answer – (B) : 2
Answer – (C) : 3
Answer – (D) : 4
View Answer
Correct Answer : (C)
Explanation : We have, (daydx2)32=(d2ydx2)2Squaring both the sides, we get,(d3ydx3)3=(d2ydx2)4Here highest derivative is (d3ydx3)3 .∴ Degree = power of (d3ydx3)3=3
Question – 17 : Find the 4th term in the expansion of (x−2y)12 .
Answer – (A) : −1760x9y3
Answer – (B) : 1760x9y3
Answer – (C) : −1760x8y9
Answer – (D) : −1999x9y3
View Answer
Correct Answer : (A)
Explanation : It is known that (r+1) th term (Tr+1) in the binomial expansion of (a+b)n is given by,Tr+1=nCran−rbrThus the 4 th term in the expansion of (x+2y)12 ,T4=T3+1=12C3(x)12−3(−2y)3=(−1)312!3!9!⋅x9⋅(2)3⋅y3=−12.11.103.2⋅(2)3x9y3=−1760x9y3
Question – 18 : In the given figure O is the centre of the circle and ∠PQR=40∘ , find the measure of ∠POR .
Answer – (A) : 80∘
Answer – (B) : 60∘
Answer – (C) : 50∘
Answer – (D) : 40∘
View Answer
Correct Answer : (A)
Explanation : We know, the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle. So, in given circle,∠POR=2∠PQRTherefore ∠POR=2×40∘=80∘
Question – 19 : If the line y=x+k is a normal to the parabola y2=4x then find the value of k .
Answer – (A) : −3
Answer – (B) : 3
Answer – (C) : 2
Answer – (D) : −2
View Answer
Correct Answer : (A)
Explanation : Given equation:y=x+k …..(i) Equation of normal of parabola =y=mx−2am−am3 …..(ii) On comparing (i)and (ii),m=1,a=1put value in (ii),y=x−2−1y=x−3 …..(iii) Compare the equation (i)and (iii),k=−3
Question – 20 : Consider a dice with the property that that probability of a face with n dots showing up is proportional to n . The probability of face showing 4 dots is?
Answer – (A) : 421
Answer – (B) : 542
Answer – (C) : 121
Answer – (D) : 421
View Answer
Correct Answer : (D)
Explanation : P(n) is proportional to n where n= 1,2,3,…6 is random variable.P(n)=knP(1)+P(2)….P(6)=1K(1+2+3+4+5+6)=1K=121So, P(4)=4 K=421
Question – 21 : What is C(n,r)+2C(n,r−1)+C(n,r−2) equal to:
Answer – (A) : C(n+2,r)
Answer – (B) : C(n−1,r+1)
Answer – (C) : C(n,r+1)
Answer – (D) : C(n+2,r)
View Answer
Correct Answer : (D)
Explanation : Given,nCr+2nC(r−1)+nCr−2=nCr+nC(r−1)+nC(r−1)+nC(r−2)Using, nCr+nC(r−1)=(n+1)Cr=(n+1)Cr+(n+1)C(r−1)Again using nCr+nC(r−1)=(n+1)Cr=(n+2)Cr
Question – 22 : The fifth term of an AP of n terms, whose sum is n2−2n, is:
Answer – (A) : 7
Answer – (B) : 7
Answer – (C) : 8
Answer – (D) : 15
View Answer
Correct Answer : (B)
Explanation : Given, Sum of n terms of an AP=n2−2n∴ Sum of first 5 terms (S5)=52−2⋅(5)=25−10=15Similarly, Now, Sum of first 4 terms (S4)=42−2⋅(4)=16−8=8∴ The fifth term of an AP(T5)=S5−S4… (Using Tn=Sn−Sn−1)=15−8T5=7
Question – 23 : The number of commutative binary operations that can be defined on a set of 2 elements is:
Answer – (A) : 2
Answer – (B) : 6
Answer – (C) : 4
Answer – (D) : 2
View Answer
Correct Answer : (D)
Explanation : The number of commutative binary operations on a set of n elements is nn(n−1)2 . Therefore, Number of commutative binary operations on a set of 2 elements =22(2−1)2=21=2
Question – 24 : The area of a triangle with vertices (−3,0),(3,0) and (0,k) is 9 sq units, then the value of k will be:
Answer – (A) : 3
Answer – (B) : 3
Answer – (C) : -9
Answer – (D) : 6
View Answer
Correct Answer : (B)
Explanation : The area of a triangle with vertices (−3,0),(3,0) and (0,k) is 9 sq unitsArea of a triangle with vertices (x1,y1),(x2,y2) and (x3,y3) is given by, Δ=12|x1y11x2y21x3y31| ∴Δ=12|−3013010k1|9=12[−3(−k)−0+1(3k)]⇒18=3k+3k=6k∴k=186=3
Question – 25 : If f(x)=xsinx, then f′(0)= ?
Answer – (A) : 0
Answer – (B) : 0
Answer – (C) : 1
Answer – (D) : ∞
View Answer
Correct Answer : (B)
Explanation : f(x)=xsinxAs we know that,D(uv)=uv′+vu′So,f′(x)=x(cosx)+sinx(1)=xcosx+sinxBy putting x=0, we havef′(0)=0+sin(0)(∵sin(0)=1)=0
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST PREPARATION
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