IIT JEE Part Test 1

IIT JEE Part Test 1

Instructions:

For straight objective type with only one correct answer. +3 for correct answer, -1 for wrong answer, 0 for not attempting

Objective type with one or more than one correct answer. +4 if all the correct answer are marked, there is no negative mark in this section

Reasoning type
If STATEMENT-I is true and STATEMENT-II is correct explanation of STATEMENT-I mark (a)
If STATEMENT-I is true and STATEMENT-II is true but not the correct explanation of STATEMENT-I mark (b)
If STATEMENT-I is true and STATEMENT-II is false mark (c)
If STATEMENT-I is false and STATEMENT-II is true mark (d)
+3 for correct, -1 for wrong, 0 for not attempting

Linked comprehension type. Answer the questions according to the information given. Only one option is correct. +4 for correct answer and -1 for wrong answer, 0 for not attempting

Q1) Straight Objective Type

=

a) π/2√2
b) π/√2
c) 1
d) π/4√2

Q2) Find lim(n→∞) 1/√(n²+n) + 1/√(n²+2n)+.....+ 1/√(n²+2n²)

a) 1
b) 2(√2 - 1)
c) 2(√3 - 1)
d) 2√2 - 1)

Q3) The number of different matrices that can be with the numbers 1,2,3,4 each having four elements is

a) 4⁴
b) 2.4⁴
c) 3.4⁴
d) none of these

Q4) The value of cot^-1(-3)+cot^-1(-2)=

a) π/4
b) 3π/4
c) 5π/4
d) none of these

Q5) a²+b²+c²=ca+ab√3
then the triangle is

a) equilateral
b) right angled
c) isosceles
d) none of these

Solution to IIT JEE Part Test 1

1 (d) putting x=tanθit simplifies tofirst terms simplifies to cos2θ 1+tan⁴θ=(sec⁴θ)(sin⁴θ+cos⁴θ)=(sec⁴θ)[(sin²θ+cos²θ)²-2sin²θcos²θ]=(sec⁴θ)[1-sin²(2θ)/2]so it becomesnow put sin2θ=√2sint and solve

2 (c) it can be written as

3 (c) four places can be filled up in 4*4*4*4=4⁴ places matrices possible are 2X2, 4X1 and 1X4so total no. of ways=3.4⁴

4 (d) cot^-1(-3)+cot^-1(-2)=π-cot^-13+π-cot^-12=2π-(cot^-13+cot^-12)=2π-π/4=7π/4

5 (b) a²+b²+c²=ca+ab√3(√3a/2-b)²+(a/2-c)²=0so √3a=2b=2√3c=k(assume)b²+c²=a²

6 (c) let A is (3,-4) and B be (5,-2)slope of AB=1, so the diagonals are parallel to the axes let O be the centre then O will be at distance 2 from A i.e. O is (3,-2)

7 (d) , (c) from second|Z-4|=|Z-8|implies Re(Z)=6from first|Z-16i|/|Z-8i|=5/3put Z=6+iy

8 (c) , (b) , (a) y(2x³-y²)dx+x(y²-x³)dy=0x³(2ydx-xdy)+y²(xdy-ydx)=0((2x/y)dx-(x²/y²)dy+((1/x)dy-(y/x)dx)=0d(x²/y)+d(y/x)=0d(x²/y)+(y/x))=0x²/y+y/x=C(1,1) lies of the curve so C=2(a), (b), (c) lies of this curve

9 (b) , (a) shifting origin to (2,-1) we getX=x-2 and Y=y+1circle as X²+Y²=5 and hyperbola as X²-5Y²=125the equation of common tangent be Y=mX?√(5(1+m²) and Y=m'X?√(125m²-25)now m'=m and 5(1+m²)=25(5m²-1)1+m²=25m²-524m²=6m=?1/2so the tangents are2Y=?X?5so the equation are2(y+1)=?(x-2)?5

10 (b) , (a) π/2-cos^-1x+π/2-cos^-1(1-x)=cos^-1x2cos^-1x=π-cos^-1(1-x)cos^-1(2x²-1)=cos^-1(x-1)2x²-1=x-1

11 (d) the solution will be 2π which is not in the domain RHS≥2LHS≤2so the equation will be 2sin²(x/4)cos²(x/2)=2sin²(x/4)=1 and cos²(x/2)=1

12 (d) it is not always isosceles it can be right angled toosin2A-sin2B=02cos(A+B)sin(A-B)=0so either A=B or A+B=π/2

13 (b) f'(x)=2xsin(1/x)-cos(1/x)f'(0)=lim(h→0)[f(0+h)-f(0)]/h=[h²sin(1/h)]/hlim(h→0)hsin(1/h)=0

14 (d) This point is not the closest as the line intersects the curve so minimum distance=0and point will be the intersection of line and curve

15 (c) The given equation is an ellipse with focus (?(2/3)√10,0) and a=√5ae=(2/3)√10e=(2/3)√2AlternatelyThe given equation is an ellipse so the eccentricity<1hence>

16 (c) it is square with coordinates z=?1/√2?i/√2area=2

17 (c) z=?1/√2?i/√2these are the roots of x⁴+1=0

18 (c) let t₁ and t₂ be P and Q then slope of OP=2/t₁ and that of OQ=2/t₂OQ is perpendicular to OPso t₁t₂=-4equation of PQ is (y-4t₁)/(x-2t₁²)=2/(t₁+t₂)putting y=0 it gives x=8so the point is (8,0)

19 (c) 2x=2(t₁²+t₂²) and 2y=4(t₁+t₂)x=t₁²+t₂²=(t₁+t₂)²-2t₁t₂x=(y/2)²+8

20 (c) A²=|(1/4)*(4t₁⁴+16t₁²)*(4t₂⁴+16t₂²)|A²=4(t₁t₂)²[16+4(t₁²+t₂²)+(t₁t₂)²]A²=64*4[t₁²+t₂²+8]A²=64*4[(t₁+t₂)²-2t₁t₂+8]A²=64*4[(t₁+t₂)²+16]for minimum area t₁+t₂=0A=8*2*4

21 (c) The plane passes through (-1,-2,-1) and normal of plane is perpendicular to the lines so the equation of plane isa(x+1)+b(y+2)+c(z+1)=0where3a+b+2c=0a+2b+3c=0

22 (d) The minimum distance is the distance between the line L₂ i.e. point(2,-2,3) and plane P₁|2-14-15+10|/√75

23 (c) The plane passes through (2,-2,3)and contain normal of the plane P₁ let the equation be a(x-2)+b(y+2)+c(z-3)=0DR of normal of lines is (1,7,-5)a+2b+3c=0a+7b-5c=0

24 (c) let x be the compression(1/2)kx²=mg(x+h)x=0.1 m

25 (b) Let the pressure at layer between 2ρ and 3ρ is P₁ and that between ρ and 2ρ is P₂P₂=P₀+ρg(h/2)=P₀+ρgh/2P₁=P₂+(2ρ)g(h/3)=P₂+2ρgh/3=P₀+ρgh/2+2ρgh/3P₀+(1/2)(3ρ)v²=P₁+(3ρ)g(h/4-h/6)=P₁+ρgh/4=P₀+ρgh/2+2ρgh/3+ρgh/4=P₀+(17/12)ρghv²=(17/18)gh

26 (c) heat required for ice to reach 0?C=1*0.5*20=10 cal heat required to melt ice=1*80=80 cal90 cal of heat is extracted from water, let the temperature of water be T5*1*(30-T)=90T=12?Ctwo liquids at temperature of 12?C and 0?C mix so the final temperature is5*1*(12-T)=1*1*TT=10?C

27 (c) current in the circuit is v/3Rapplying kirchoffs law in above loop iR+v-v-v'=0v'=iR=v/3

28 (c) v/u=v/f - 12=-x/f - 13=-(x+25)/f -1f=-25 cm and x=75 cm1/u₁=1/f - 1/v=-1/25 + 1/75=-2/751/u₂=-1/25 + 1/100=-3/100u₂-u₁=75/2 - 100/3 =(225-200)/6=25/6

29 (d) μsin30=sinθ=0.72θ=[ 1 ]+30[ 1 ]=θ-302[ 1 ]=2(θ-30)

30 (d) , (a) Let f1 be the friction between the plank and cylinder and f2 between cylinder and ground let the acceleration of CM of cylinder be af1+f2=m₁a=20a(f1-f2)R=Ia/R=m₁aR/2=10aRf1=15a and f2=5aF-f1=m₂*2a=10a50-15a=10aa=2 m/s²

31 (d) , (c) after collision velocity of A is vv(CM)=v/2for maximum compression, velocity of A=velocity of B=v/2KE of system=m(v/2)²=mv²/4mv²/4+(1/2)kx²=(1/2)mv²x=v√m/2k

32 (c) , (b) when entering, the area of the portion of loop in the field increases so current will be anticlockwise to oppose the increase in magnetic flux it is zero when loop is inside the field and clockwise when leaving the field