INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
=Question – 1 : The value of the expression 4sin20sin80−1sin10 is:
Answer – (A) : 2
Answer – (B) : 4
Answer – (C) : 1
Answer – (D) : 2
View Answer
Correct Answer : (D)
Explanation : As we know,2sinxsiny=cos(x−y)−cos(x+y)4sin20sin80−1=2×(2sin80sin20)−1=2×[cos(80−20)−cos(80+20)]−1=2×[cos60−cos100]−1=2cos60−2cos100−1=2×12−2cos(90+10)−1=1−2[−sin10]−1[∵cos(90+θ)=−sinθ]=2sin10Now,4sin20sin80−1sin10=2sin10sin10=2
Question – 2 : If x=2+223+213 , then what is the value of x3−6×2+6x ?
Answer – (A) : 2
Answer – (B) : 2
Answer – (C) : 1
Answer – (D) : 0
View Answer
Correct Answer : (B)
Explanation : Given,x=2+223+213⇒(x−2)=(223+213)Taking cube both the sides,x3−8−6×2+12x=4+2+3×223×213(223+213)⇒x3−14−6×2+12x=6(x−2)⇒x3−14−6×2+12x−6x+12=0⇒x3−6×2+6x=2
Question – 3 : If p+q+r=a+b+c=0, then the determinant |paqbrcqcrapbrbpcqa| equals:
Answer – (A) : 0
Answer – (B) : 1
Answer – (C) : pa+qb+rc
Answer – (D) : pa+qb+rc+a+b+c
View Answer
Correct Answer : (A)
Explanation : Let |paqbrcqcrapbrbpcqa|=R1Expanding R1=pa(ra×qa−pb×pc)−qb(qc×qa−pb×rb)+rc(qc×pc−ra×rb)=pa(a2qr−p2bc)−qb(q2ac−b2pr)+rc(c2pq−r2ab)=a3pqr−p3abc−q3abc+b3pqr+c3pqr−r3abc=pqr(a3+b3+c3)−abc(p3+q3+r3)Now we know that,a3+b3+c3=3abc, if a+b+c=0=pqr(3abc)−abc(3pqr)=0
Question – 4 : For a hyperbola x216−y29=1 then its equation of directrix is:
Answer – (A) : x=165
Answer – (B) : x=−45
Answer – (C) : x=165
Answer – (D) : x=175
View Answer
Correct Answer : (C)
Explanation : Given,x216−y29=1Compare with standard equation, we geta2=16 and b2=9Eccentricity e=1+b2a2=1+916=16+916=2516=54Now, equation of directrix,x=±ae=±4(54)=±165
Question – 5 : Find the intercepts cut off by the plane x+2y−4z=8 .
Answer – (A) : (8,4,−2)
Answer – (B) : (1,2,−4)
Answer – (C) : (8,4,2)
Answer – (D) : (8,4,−2)
View Answer
Correct Answer : (D)
Explanation : As we know, The intercept form of the plane is given by,xa+yb+zc=1Where a is the x -intercept and b is the y -intercept and c is the z -intercept. Given,x+2y−4z=8⇒x8+2y8−4z8=1⇒x8+y4+z−2=1As we know that, Intercept form of the plane is given by,xa+yb+zc=1So, the intercepts cut off by the given plane are (a,b,c)=(8,4,−2)
Question – 6 : Find the radius of circle which passes through the points (1,2) and (3,4) and the centre lies on the straight line y−3x+2=0 .
Answer – (A) : 3
Answer – (B) : 3
Answer – (C) : 5
Answer – (D) : 32
View Answer
Correct Answer : (A)
Explanation : As we know, Distance between two points (x1,y1) and (x2,y2) is given byd=(x2−x1)2+(y2−y1)2Centre lies on the line y−3x+2=0 . Let x=h⇒y=3h−2So the center is of the form (h,3h−2) . Distance of centre from (1,2) and (3,4) will be equal.⇒(h−1)2+(3h−2−2)2=(h−3)2+(3h−2−4)2⇒h2−2h+1+9h2−24h+16=h2−6h+9+9h2−36h+36⇒−26h+17=−42h+45⇒16h=28⇒h=74∴y=214−2=134So, the centre is (74,134) . Now, radius will be distance from any point (say (1,2) ) to the center of circle (74,134) .∴r2=(1−74)2+(2−134)2=(−34)2+(−54)2
Question – 7 : If a line makes 45∘,60∘ with the x -axis, y -axis respectively, then the angle made by that line with the z – axis is:
Answer – (A) : 60∘
Answer – (B) : 45∘
Answer – (C) : 60∘
Answer – (D) : 30∘
View Answer
Correct Answer : (C)
Explanation : Given, A line makes 45∘,60∘ with the x -axis, y -axis. Therefore, α=45∘ and β=60∘Let γ is the angles made by the line with the z -axis. As we know that,cos2α+cos2β+cos2γ=1⇒cos245∘+cos260∘+cos2γ=1⇒(12)2+(12)2+cos2γ=1⇒12+14+cos2γ=1⇒cos2γ=1−34=14⇒cosγ=±(12)∴γ=60∘ or 120∘ .
Question – 8 : If a,b,c are real numbers, then the value of the determinant |1−aa−b−cb+c1−bb−c−ac+a1−cc−a−ba+b| is:
Answer – (A) : 0
Answer – (B) : (a−b)(b−c)(c−a)
Answer – (C) : (a+b+c)2
Answer – (D) : (a+b+c)3
View Answer
Correct Answer : (A)
Explanation : Let Δ=|1−aa−b−cb+c1−bb−c−ac+a1−cc−a−ba+b|Applying C2→C2+C3, we get Δ=|1−aab+c1−bbc+a1−cca+b|Applying C1→C1+C2,C3→C3+C2 , we get Δ=|1aa+b+c1bb+c+a1cc+a+b|Taking common a+b+c from C3, we get Δ=(a+b+c)|1a11b11c1|We know that if two columns of a determinant are identical, the value of the determinant is zero.∴Δ=0
Question – 9 : If the vectors i^−xj^−yk^ and i^+xj^+yk^ are orthogonal to each other, then what is the locus of the point (x,y) ?
Answer – (A) : A circle
Answer – (B) : An ellipse
Answer – (C) : A circle
Answer – (D) : A straight line
View Answer
Correct Answer : (C)
Explanation : Given,i^−xj^−yk^ and i^+xj^+yk^ are orthogonal to each other.i^−xj^−yk^ and i^+xj^+yk^⇒(i^−xj^−yk^)⋅(i^+xj^+yk^)=0⇒1−x2−y2=0⇒x2+y2=1So, it is a circle.
Question – 10 : Let the average of three numbers be 16 . If two of the numbers are 8 and 10, what is the remaining number?
Answer – (A) : 30
Answer – (B) : 18
Answer – (C) : 12
Answer – (D) : 30
View Answer
Correct Answer : (D)
Explanation : Mean of ′n′ observations = Sum of observations nHere n=3Let the third number is x .∴16=x+8+103⇒x+18=48⇒x=30
Question – 11 : What is the value of (1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8) ?
Answer – (A) : 18
Answer – (B) : 12+122
Answer – (C) : 12−122
Answer – (D) : 18
View Answer
Correct Answer : (D)
Explanation : (1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8)=(1+cosπ8)(1+cos3π8)(1+cos(π−3π8))(1+cos(π−π8))=(1+cosπ8)(1+cos3π8)(1−cosπ8)(1−cos3π8)(∵cos(π−θ)=−cosθ)=(1−cos2(π8))(1−cos2(3π8))(∵(a+b)(a−b)=a2−b2)=sin2(π8)×sin2(3π8)(∵1−cos2θ=sin2θ)=(1−cosπ42)×(1−cos3π42)We know that,cosπ4=12 and cos3π4=−12=14(1−12)×(1+12)=18
Question – 12 : Let p,q and r be three distinct positive real numbers. If D=|pqrqrprpq| , then which one of the following is correct?
Answer – (A) : D≤0
Answer – (B) : D≤0
Answer – (C) : D>0
Answer – (D) : D≥0
View Answer
Correct Answer : (B)
Explanation : D=|pqrqrprpq|=p(qr−p2)−q(q2−pr)+r(pq−r2)=pqr−p3−q3+pqr+pqr−r3=3pqr−(p3+q3+r3)As we know, Arithmetic mean (AM) ≥ Geometric mean (GM) Now, AM of p3,q3 and r3=p3+q3+r33Now, AM of p3,q3 and r3=(p3×q3×r3)13=pqTherefore,p3+q3+r33≥pqr⇒p3+q3+r3≥3pqr⇒3pqr−(p3+q3+r3)≤0∴D≤0
Question – 13 : If 2×3−3y2=7 , what is dydx equal to (y≠0) :
Answer – (A) : x2y
Answer – (B) : x2y
Answer – (C) : x2y
Answer – (D) : None of the above
View Answer
Correct Answer : (C)
Explanation : Given,2×3−3y2=7Differentiating w.r.t. x, we get6×2−6ydydx=0⇒x2−ydydx=0⇒dydx=x2y
Question – 14 : If cosx+cosy+cosz=0 and sinx+siny+sinz=0 then find the value of cos(x−y) .
Answer – (A) : −12
Answer – (B) : 12
Answer – (C) : −12
Answer – (D) : 0
View Answer
Correct Answer : (C)
Explanation : As we know,sin2x+cos2y=1cos(x−y)=cosxcosy+sinxsinyGiven,cosx+cosy+cosz=0⇒cosx+cosy=−cosz…(1)sinx+siny+sinz=0⇒sinx+siny=−sinz…(2)Squaring and adding equation (1) and equation (2) , we get(cosx+cosy)2+(sinx+siny)2=(−cosz)2+(−sinz)2⇒cos2x+cos2y+2cosxcosy+sin2x+sin2y+2sinxsiny=cos2z+sin2z⇒(cos2x+sin2x)+(cos2y+sin2y)+2[cosxcosy+sinxsiny]=1⇒1+1+2cos(x−y)=1⇒2cos(x−y)=−1∴cos(x−y)=−12
Question – 15 : If A=[2−301] and B=[1230] then (B−1A−1)−1 is equal to:
Answer – (A) : [−7430]
Answer – (B) : [−7430]
Answer – (C) : [−7405]
Answer – (D) : [4−730]
View Answer
Correct Answer : (B)
Explanation : Given, A=[2−301] and B=[1230]⇒(B−1 A−1)−1=(A−1)−1( B−1)−1(∵(AB)−1=B−1 A−1)=AB(∵(A−1)−1=A) =[2−301]×[1230] =[(2×1)+(−3×3)(2×2)+(−3×0)(0×1)+(1×3)(0×2)+(1×0)] =[−7430]
Question – 16 : |a+bb+ccb+cc+aac+aa+bb|=k|abcbcacab| then k is equal to:
Answer – (A) : 1
Answer – (B) : 2
Answer – (C) : 4
Answer – (D) : 6
View Answer
Correct Answer : (A)
Explanation : Given, |a+bb+ccb+cc+aac+aa+bb|=k|abcbcacab|Let Δ=|a+bb+ccb+cc+aac+aa+bb|Apply C2→C2−C3, we get =|a+bbcb+ccac+aab|Apply C1→C1−C2, we get =|abcbcacab|Now, |a+bb+ccb+cc+aac+aa+bb|=k|abcbcacab| ⇒|abcbcacab|=k|abcbcacab|⇒k=1
Question – 17 : The angle between the planes 2x+y+z=7 and x−y+2z=9 is:
Answer – (A) : 60∘
Answer – (B) : 120∘
Answer – (C) : 90∘
Answer – (D) : 530∘
View Answer
Correct Answer : (A)
Explanation : The acute angle θ between two planes a1x+b1y+c1z=d1 and a2x+b2y+c2z=d2 is given by the formula,cosθ=a1a2+b1 b2+c1c2(a12+b12+c12)(a22+b22+c22) Given,2x+y+z=7 and x−y+2z=9That means,a1=2, b1=1,c1=1 and a2=1, b2=−1,c2=2Using the above formula for angle θ ,cosθ=2×1+1×(−1)+1×2(22+12+12)(12+(−1)2+22)=36×6=36=12⇒θ=60∘
Question – 18 : What is the value of ∫491xdx ?
Answer – (A) : 2
Answer – (B) : −2
Answer – (C) : 2
Answer – (D) : −1
View Answer
Correct Answer : (C)
Explanation : ∫1xdx=∫x−12dx=x−12+1−12+1+C=2x+C∴∫491xdx=[2x]49=2[9−4]=2(3−2)=2
Question – 19 : The length of latus rectum of the ellipse 3×2+y2−12x+2y+1=0 is:
Answer – (A) : 43
Answer – (B) : 12
Answer – (C) : 43
Answer – (D) : 32
View Answer
Correct Answer : (C)
Explanation : As we know,
Standard Equation of ellipse is
x2a2+y2b2=1
.
Length of latus rectum
=2 b2a,
when
a>b
and
2a2b,
when
a
Question – 20 : Find the value of a for which x+3x+a24 is a perfect square.
Answer – (A) : 3
Answer – (B) : 23
Answer – (C) : 33
Answer – (D) : 43
View Answer
Correct Answer : (A)
Explanation : Any quadratic equation ax2+bx+c=0 is a perfect square if the discriminate (D) is zero, where D=b2−4ac . Given, Equation, x+(3x)+a24=0Putting x=y2, we get,y2+y3+a24=0In order to be a perfect square,D=(3)2−(4×1×a24)=0⇒3−a2=0⇒a2=3∴a=3
Question – 21 : The two circles x2+y2=r2 and x2+y2−10x+16=0 intersect at two distinct points. Then which one of the following is correct?
Answer – (A) : 2
Answer – (B) : r=2 or r=8
Answer – (C) : r<2
Answer – (D) : r>2
View Answer
Correct Answer : (A)
Explanation : Given,
Circles are
x2+y2=r2
and
x2+y2−10x+16=0
.
Let center of two circles are
C1
and
C2
and radius are
r1
and
r2
.C1=(0,0)
and
r1=rC2=(5,0)
and
r2=52+02−16=3Now,C1C2=(5−0)2+(0−0)2=5When two circles will intersect at two points,r1−r2
Question – 22 : Focus of the parabola y2 − 8x + 6y + 1 = 0 is:
Answer – (A) : (1, -3)
Answer – (B) : (1, -3)
Answer – (C) : (8, 0)
Answer – (D) : None of the above
View Answer
Correct Answer : (B)
Explanation : Given,y2 − 8x + 6y + 1 = 0⇒ y2 + 6y + 9 – 9 – 8x + 1 = 0⇒ (y + 3)2 – 8x – 8 = 0⇒ (y + 3)2 = 8x + 8⇒ (y + 3)2 = 8 (x + 1)Let new coordinate axes be X and Y.Here X = x + 1 and Y = y + 3⇒ Y2 = 4aXNow comparing with above equation,∴ 4a = 8⇒ a = 2Focus: (a, 0)X = a and Y = 0⇒ x + 1 = 2 and y + 3 = 0⇒ x = 1 and y = -3∴ focus of parabola is (1, -3).Hence, the correct option is (C).
Question – 23 : If |a→|=2,|b→|=3 and |a→+b→|=6, then |a→−b→| equal to:
Answer – (A) : 2
Answer – (B) : 2
Answer – (C) : 3
Answer – (D) : 4
View Answer
Correct Answer : (B)
Explanation : Given,|a→|=2,|b→|=3 and |a→+b→|=6As we know,|a→+b→|2+|a→−b→|2=2(|a→|2+|b→|2)⇒(6)2+|a→−b→|2=2×[(2)2+(3)2]⇒|a→−b→|2=4∴|a→−b→|=2
Question – 24 : ∫2x+3dx is equal to:
Answer – (A) : (2x+3)323+c
Answer – (B) : (2x+3)322+c
Answer – (C) : (2x+3)323+c
Answer – (D) : None of the above
View Answer
Correct Answer : (C)
Explanation : ∫xndx=xn+1n+1+c I=∫2x+3dx Let 2x+3=t2Differenating with respect to x , we get2dx=2tdt⇒dx=tdtNow, I=∫t2×tdt=∫t2dt=t33+c=(2x+3)323+c
Question – 25 : If A and B are two independent events with P(A)=35 and P(B)=49, then P(A∩B) equals:
Answer – (A) : 415
Answer – (B) : 845
Answer – (C) : 13
Answer – (D) : 712
View Answer
Correct Answer : (A)
Explanation : Given,P(A)=35 and P(B)=49Since A and B are independent events: P(A∩B)=P(A)×P(B)⇒P(A∩B)=35×49=415
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
INDIAN NAVY AA SSR MATHEMATICS PRACTICE TEST
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