77. Let the population of rabbits surviving at a time tt be governed by the differential equation dp(t)dt=12p(t)−200.Ifp(0)=100,thenp(t)equalsdtdp(t)?=21?p(t)−200.Ifp(0)=100,thenp(t)equals
(1) 600−500et/2600−500et/2
(2) 400−300et/2400−300et/2
(3) 400−300et/2400−300et/2
(4) 300−200et/2300−200et/2
Answer (3)
Sol.dp(t)dt=12p(t)−200Sol.dtdp(t)?=21?p(t)−200
⋅{d(p(t))2p(t)−200}=∫ttdt⋅{2d(p(t))?p(t)−200}=∫tt?dt
\cdot \left.\frac{1}{2}\log \left(\frac{p(t)}{2} -200\right) = t + c
⋅p(t)2−200=t22k⋅2p(t)?−200=2t2?k
Using given condition p(t)=400−300et/2p(t)=400−300et/2
78. Let PSPS be the median of the triangle with vertices P(2,2)P(2,2) , Q(6,−1)Q(6,−1) and R(7,3)R(7,3) . The equation of the line passing through (1,−1)(1,−1) and parallel to PSPS is
(1) 4x+7y+3=04x+7y+3=0
(2) 2x−9y−11=02x−9y−11=0
(3) 4x−7y−11=04x−7y−11=0
(4) 2x+9y+7=02x+9y+7=0
Answer (4)
Sol.
S is mid-point of QRQR
So S=(7+62,3−12)S=(27+6?,23−1?)
= (132,1)(213?,1)
Slope of PS=2−12−132=−29PS=2−213?2−1?=−92?
Equation of line ⇒y−(−1)=−29(x−1)⇒y−(−1)=−92?(x−1)
9y+9=−2x+2⇒2x+9y+7=09y+9=−2x+2⇒2x+9y+7=0
79. Let a,b,ca,b,c and dd be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=04ax+2ay+c=0 and 5bx+2by+d=05bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes then
(1) 3bc−2ad=03bc−2ad=0
(2) 3bc+2ad=03bc+2ad=0
(3) 2bc−3ad=02bc−3ad=0
(4) 2bc+3ad=02bc+3ad=0
Answer (1)
Sol. Let (α,−α)(α,−α) be the point of intersection
∴4aα−2aα+c=0⇒α=−c2a∴4aα−2aα+c=0⇒α=−2ac?
and5bα−2bα+d=0⇒α=−d3band5bα−2bα+d=0⇒α=−3bd?
⇒3bc=2ad⇒3bc=2ad
⇒3bc−2ad=0⇒3bc−2ad=0
Alternative method:
The point of intersection will be
x2ad−2bc=−y4ad−5bc=18ab−10ab2ad−2bcx?=4ad−5bc−y?=8ab−10ab1?
⇒x=2(ad−bc)−2ab⇒x=−2ab2(ad−bc)?
⇒y=5bc−4ad−2ab⇒y=−2ab5bc−4ad?
Point of intersection is in fourth quadrant so xx is positive and yy is negative.
Also distance from axes is same
So x=−yx=−y ( ∴∴ distance from xx - axis is - yy as yy is negative)
2(ad−bc)−2ab=−(5bc−4ad)−2ab−2ab2(ad−bc)?=−2ab−(5bc−4ad)?
2ad−2bc=−5bc+4ad2ad−2bc=−5bc+4ad
⇒3bc−2ad=0⇒3bc−2ad=0 ...(i)
80. The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6x2+3y2=6 on any tangent to it is
(x2+y2)2=6x2+2y2(x2+y2)2=6x2+2y2(x2+y2)2=6x2−2y2(x2+y2)2=6x2−2y2(x2−y2)2=6x2+2y2(x2−y2)2=6x2+2y2(x2−y2)2−6x2−2y2(x2−y2)2−6x2−2y2
Answer (1)
Sol. Here ellipse is x2a2+y2b2=1a2x2?+b2y2?=1 , where a2=6a2=6 , b2=2b2=2
Now, equation of any variable tangent is
y=mx±a2m2+b2…(i)y=mx±a2m2+b2?…(i)
where mm is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
y=−xm…(ii)y=m−x?…(ii)
Eliminating mm , we get
(x2+y2)2=a2x2+b2y2(x2+y2)2=a2x2+b2y2⇒1(x2+y2)2=6x2+2y2⇒(x2+y2)21?=6x2+2y2
81. Let CC be the circle with centre at (1,1)(1,1) and radius =1=1 . If TT is the circle centred at (0,y)(0,y) passing through origin and touching the circle CC externally, then the radius of TT is equal to
(1)12(2)14(1)21?(2)41?(3)32(4)32(3)23??(4)23??
Answer (2)
Sol.
C≡(x−1)2+(y−1)2=1C≡(x−1)2+(y−1)2=1
Radius of T=?y?T=?y?
TT touches CC externally
(0−1)2+(y−1)2=(1+?y?)2(0−1)2+(y−1)2=(1+?y?)2
⇒1+y2+1−2y=1+y2+2?y?⇒1+y2+1−2y=1+y2+2?y?
82. The slope of the line touching both the parabolas y2=4xy2=4x and x2=−32yx2=−32y is
(1)18(2)23(1)81?(2)32?(3)12(4)32(3)21?(4)23?
Answer (3)
Sol. y2=4xy2=4x …(1)…(1)
x2=−32yx2=−32y …(2)…(2)
mm be slope of common tangent
Equation of tangent (1)
y=mx+1m…(i)y=mx+m1?…(i)
Equation of tangent (2)
y=mx+8m2…(ii)y=mx+8m2…(ii)
(i) and (ii) are identical
1m=8m2m1?=8m2⇒m3=18⇒m3=81?m=12m=21??
Alternative method:
Let tangent to y2=4xy2=4x be
y=mx+1my=mx+m1?
as this is also tangent to x2=−32yx2=−32y
Solvingx2+32mx+32m=0Solvingx2+32mx+m32?=0
Since roots are equal
∴D=0∴D=0⇒(32)2−4×32m=0⇒(32)2−4×m32?=0⇒m3=432⇒m3=324?⇒m=12⇒m=21?
83. The image of the line x−13=y−31=z−4−5in the plane2x−y+z+3=03x−1?=1y−3?=−5z−4?in the plane2x−y+z+3=0 is the line
x−33=y+51=z−2−53x−3?=1y+5?=−5z−2?x−3−3=y+5−1=z−25−3x−3?=−1y+5?=5z−2?x+33=y−51=z−2−53x+3?=1y−5?=−5z−2?x+3−3=y−5−1=z+25−3x+3?=−1y−5?=5z+2?
Answer (3)
Sol.
a−12=b−3−1=c−41=λ2a−1?=−1b−3?=1c−4?=λ⇒a=2λ+1⇒a=2λ+1b=3−λb=3−λc=4+λc=4+λP≡(λ+1,3−λ2,4+λ2)P≡(λ+1,3−2λ?,4+2λ?)2(λ+1)−(3−λ2)+(4+λ2)+3=02(λ+1)−(3−2λ?)+(4+2λ?)+3=02λ+2−3+λ2+4+λ2+3=02λ+2−3+2λ?+4+2λ?+3=03λ+6=0⇒λ=−23λ+6=0⇒λ=−2a=−3,b=5,c=2a=−3,b=5,c=2
So the equation of the required line is
x+33=y−51=z−2−53x+3?=1y−5?=−5z−2?
84. The angle between the lines whose direction consists satisfy the equations l+m+n=0l+m+n=0 and l2=m2+n2l2=m2+n2 is
π6(2)π26π?(2)2π?π3(4)π43π?(4)4π?
Answer (3)
Sol. l+m+n=0l+m+n=0
l2=m2+n2l2=m2+n2
Now,(−m−n)2=m2+n2Now,(−m−n)2=m2+n2
⇒mn=0⇒mn=0
m=0orn=0m=0orn=0
If m=0m=0
then l=−nl=−n
l2+m2+n2=1l2+m2+n2=1
Gives
⇒n=±12⇒n=±2?1?
i.e.(l1,m1,n1)i.e.(l1?,m1?,n1?)
=(−12,0,12)=(−2?1?,0,2?1?)
∴cos?θ=12∴cosθ=21?
θ=π3θ=3π?
85. If [a?×b?,b?×c?,c?×a?]=λ[a?,b?,c?]2[a?×b?,b?×c?,c?×a?]=λ[a?,b?,c?]2 then λλ is equal to
(1) 0
(2) 1
(3) 2
(4) 3
Answer (2)
Sol. L.H.S.
=(a?×b?)⋅[(b?×c?)×(c?×a?)]=(a?×b?)⋅[(b?×c?)×(c?×a?)]
=(a?×b?)⋅[(b?×c?⋅a?)c?−(b?×c?⋅c?)a?]=(a?×b?)⋅[(b?×c?⋅a?)c?−(b?×c?⋅c?)a?]
=(a?×b?)⋅[[b?c?a?c?][⋅b?×c?⋅c?=0]=(a?×b?)⋅[[b?c?a?c?][⋅b?×c?⋅c?=0]
=[a?b?c?]⋅(a?×b?⋅c?)=[a?b?c?]2=[a?b?c?]⋅(a?×b?⋅c?)=[a?b?c?]2
[a?×b?b?×c?c?×a?]=[a?b?c?]2[a?×b?b?×c?c?×a?]=[a?b?c?]2
Soλ=1Soλ=1
86. Let AA and BB be two events such that P(A∪B‾)=16,P(A∩B)=14P(A∪B)=61?,P(A∩B)=41? and P(A‾)=14P(A)=41?, where A‾A stands for the complement of the event A. Then the events A and B are
(1) Independent but not equally likely
(2) Independent and equally likely
(3) Mutually exclusive and independent
(4) Equally likely but not independent
Answer (1)
Sol. P(A∪B‾)=16⇒P(A∪B)=1−16=56P(A∪B)=61?⇒P(A∪B)=1−61?=65?
P(A‾)=14⇒P(A)=1−14=34P(A)=41?⇒P(A)=1−41?=43?
∴ P(A∪B)=P(A)+P(B)−P(A∩B)P(A∪B)=P(A)+P(B)−P(A∩B)
56=34+P(B)−1465?=43?+P(B)−41?
P(B)=13P(B)=31?
∴ P(A)≠P(B)P(A)?=P(B) so they are not equally likely.
Also P(A)×P(B)=34×13=14P(A)×P(B)=43?×31?=41?
=P(A∩B)=P(A∩B)
∴ P(A∩B)=P(A)⋅P(B)P(A∩B)=P(A)⋅P(B) so A & B are independent.
87. The variance of first 50 even natural numbers is
(1) 437
(2) 43744437?
(3) 83344833?
(4) 833
Answer (4)
Sol. Variance = ∑xi2N−(x?)2N∑xi2??−(x?)2
⇒σ2=22+42+…+100250−(2+4+…+10050)2⇒σ2=5022+42+…+1002?−(502+4+…+100?)2
= 4(12+22+32+…+502)50−(51)2504(12+22+32+…+502)?−(51)2
= 4(50×51×10150×6)−(51)24(50×650×51×101?)−(51)2
= 3434−26013434−2601
⇒σ2=833⇒σ2=833
88. Let fk(x)=1k(sin?kx+cos?kx)fk?(x)=k1?(sinkx+coskx) where x∈Rx∈R and k≥1k≥1. Then f4(x)−f6(x)f4?(x)−f6?(x) equals
(1) 1441?
(2) 112121?
(3) 1661?
(4) 1331?
Answer (2)
Sol. fk(x)=1k(sin?kx+cos?kx)fk?(x)=k1?(sinkx+coskx)
f4(x)−f6(x)=14(sin?4x+cos?4x)−16(sin?6x+cos?6x)f4?(x)−f6?(x)=41?(sin4x+cos4x)−61?(sin6x+cos6x)
= 14[1−2sin?2xcos?2x]−16[1−3sin?2xcos?2x]41?[1−2sin2xcos2x]−61?[1−3sin2xcos2x]
= 14−16=11241?−61?=121?
89. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is
(1) 202202?
(2) 20(3−1)20(3?−1)
(3) 40(2−1)40(2?−1)
(4) 40(3−2)40(3?−2?)
Answer (2)
Sol.
From figure tan?45?=20xtan45?=x20?
and tan?30?=20x+ytan30?=x+y20?
so, y=20(3−1)y=20(3?−1)
i.e., speed = 20(3−1)20(3?−1) m/s.
90. The statement ~(p ↔ ~q) is
(1) A tautology
(2) A fallacy
(3) Equivalent to p ↔ q
(4) Equivalent to ~ p ↔ q
Answer (3)
Sol. ~(p ↔ ~q)
| p |
q |
~q |
p ↔ ~q |
~(p ↔ ~q) |
| F |
F |
T |
F |
T |
| F |
T |
F |
T |
F |
| T |
F |
T |
T |
F |
| T |
T |
F |
F |
T |
Clearly equivalent to p ↔ q
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The test is of 3 hours duration.
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The Test Booklet consists of 90 questions. The maximum marks are 360.
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There are three parts in the question paper A, B, C consisting of Physics, Mathematics and Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.
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Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each question. 1441? (one-fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.
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There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 4 above.
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Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited.
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No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination room/hall.
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The CODE for this Booklet is D. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet.
refer page number 187 to 210 from given PDF
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