IIT JEE Part Test 2

IIT JEE Part Test 2

Instructions:

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

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

Match the columns +1 for correct and -1 for wrong

Q1) Straight Objective Type

find the summation of the series to infinite terms
1/3 + 1/3 + 7/27 + 5/27 + ....

a) 1
b) 3/2
c) 2
d) none of these

Q2) The greatest value of x²y³z⁴ (x+2y+3z=1, x,y,z>0) is

a) 2¹⁰/3¹⁵
b) 2⁷/3¹⁹
c) 2⁶/3¹⁸
d) 2¹³/3¹²

Q3) For the expansion (1+x)¹⁰
C₀+C₃+C₆+C₉=

a) 340
b) 341
c) 342
d) none of these

Q4) for x,y in R (4xy?1)



find f'(1)

a) 1
b) 1/2
c) 1/5
d) 2/5

Q5)

a) 1
b) 1/e
c) e
d) none of these

Q6) find the area bounded by the curves
x²+y²=4
x²+y²-2|x|-2|y|+4=0

a) 4p
b) 8p-8
c) 4p+2
d) 8p-16

Q7) Let x,y be the real numbers satisfying the equation x²+2x+y²+2y-47=0
If the maximum and minimum values of x²+y²-4x-6y+13 are M and m respectively, then the numerical value of M-m is

a) 14
b) 10
c) 5
d) none of these

Q8) Let a,b,c be the unit vectors such that a+b+c=0. Then the area of the triangle formed by a,b and c is

a) 0
b) 1 units
c) can't be determined
d) none of these

Solution to IIT JEE Part Test 2

1 (b) it can be written as1/3 + 3/9 + 7/27 + 15/81....(2ⁿ-1)/3ⁿ=(2/3)ⁿ-1/3ⁿ(2/3)*1/(1-2/3) - (1/3)*1/(1-1/3)2-1/2=3/2

2 (b) AM??GM(x/2+x/2+2y/3+2y/3+2y/3+3z/4+3z/4+3z/4+3z/4)/9 ??[(x/2)²(2y/3)³(3z/4)⁴]^1/9

3 (b) putting x=1: C₀+C₁+C₂+.......+C₁₀=2¹⁰putting x=?? C₀+C₁??C₂??sup>2+.......+C₁₀??sup>10 =(1+??¹⁰=??sup>20=??sup>2putting x=??sup>2: C₀+C₁??sup>2+C₂??sup>4+.......+C₁₀??sup>20 =(1+??sup>²)¹⁰=??sup>10=??adding all three we get3(C₀+C₃+C₆+C₉)=2ⁿ+????sup>2= 2¹⁰-1Hence C₀+C₃+C₆+C₉=(2¹⁰-1)/3=1023/3=341

4 (d) putting x=y=0 we get f(0)=0putting y=-x we get f(x)+f(-x)=0 f'(x)=lim(h??) (f(x+h)-f(x))/h=(f(x+h)+f(-x))/h given f(x)=-f(-x)=f[(x+h-x)/(1+4(x+h)x)]=f[(h)/(1+4x²+4hx)]/h=As h tends to zerof'(x)=2/(1+4x²)f'(1)=2/5

5 (b) it is e^[tan(??/2)lntan(??/4)][tan(??/2)lntan(??/4)]=lntan(??/4)/cot(??/2)Using L'Hospital rule[sec²(??/4).??4]/[-tan(??/4).cosec²(??/2)]=-1

6 x²+y²-2|x|-2|y|+4??0is not a real circle this is being cancelled. For this circle solution isx²+y²-2|x|-2|y|-4??0brown area is to be found area of yellow + area of brown+area of yellow=2*2=4 units area of yellow + area of brown=??²/4=??rea of brown 2??4area will be in all 4 quadrants A=4(2??4)=8(??2)

7 (d) x²+y²-4x-6y+13=(x-2)²+(y-3)²P(2,3) lies inside the circlelet the distance between centre of circle and P be xlet M'=r+x and m'=r-xM=M'²=r²+x²+2rxm=m'²=r²+x²-2rxM-m=4rx=4*7*5=140

8 (d) It will be an equilateral triangle with the side length 1so area=??/4 units

9 (c) seating can be as 1,3,5; 1,3,6; 1,3,7; 1,3,8; 1,4,6; 1,4,7 ; 1,4,8; 1,5,7; 1,5,8; 1,6,8total no. of ways=103 seats can be chosen is ⁸C₃=56 ways probability is 10/56=5/28

10 (c) The line itself is a tangent to the circle

11 (a) [aX(bXc)]=(a.c)b-(a.b)c(bXc).[aX(bXc)]=(bXc).[(a.c)b-(a.b)c]=0

12 (c) |a+b+c|²??03+2(a.b+b.c+c.a)??0(a.b+b.c+c.a)??(-3/2)|a-b|²+|b-c|²+|c-a|²=2[(a.a)+(b.b)+(c.c)] -2(a.b+b.c+c.a)=6-2(a.b+b.c+c.a)

13 (a) the tangents are x??a=0

14 (b) A²-4A-5I=0

15 (a) A²-4A-5I=0multiplying by A^-1A-4I=5A^-1

16 (c) X=A^-1B

17 (d) sin^-1sin^-1x+cos^-1cos^-1xfirst one exists for[-0.84,0.84] and second one for [0.54,1]it exists for [0.54,0.84][??5,??4] lies in this interval

18 (d) it is true for x=1/??=0.707

19 (d) sin^-1sinx+cos^-1sinxsin^-1sinx+cos^-1cos(??2-x)x+??2-x=??2

20 (a) shifting the centre of circle to origin circle becomes x²+y²=25 and chord is x+y=5so coordinates are O(0,0), P(5,0), Q(0,5) and R(5,5)area OPQ=PQR=25/2centroid of OPQ (5/3,5/3) and that of PQR is (10/3,10/3)shifting origin back centroid of OPQ(14/3,-7/3) and of PQR (19/3,-2/3)

21 (a)

22 (d)

23 (c)

24 (b) first oneI=??sinxdx even function I=(0 to pi) 2??sinxdx=2??pi-x)sinxdx2I=2??i.sinxdxI=(0 to pi) pi??inxdxI=(0 to pi/2) 2pi??inxdx=2pisecond odd function so it is zerothird(-pi to 0){sinx}dx+(0 to pi){sinx}dx(-pi to 0)(1+sinxdx+(0 to pi)sinxdx(-pi to 0)1dx+(-pi to 0)sinxdx+(0 to pi)sinxdx(-pi to 0)1dx+(-pi to pi)sinxdxsecond is odd function so zero(-pi to 0)1 dx=pifourth one(-1 to 0) (-1)dx+(0 to 1)0.dx+(1 to 2)1dx=0

25 (d)

26 (a)

27 (d)

28 (b) A=30??, B=60?? and C=90??tan(B/2)tan(C/2)+tan(C/2)tan(A/2)+tan(A/2)tan(B/2)=1 for any angles4sin(A/2)sin(B/2)sin(C/2)=cosA+cosB+cosC-14cos(A/2)cos(B/2)cos(C/2)=sinA+sinB+sinCcot15??=2+??

29 (a)

30 (c)

31 (d)

32 (b) let x length of rod is inside watertaking moment about corner you can find the answerVdg.x/2=mg.l/2xAdg.x/2=lA??.l/2x²=l².(??d)=l²/2=1/2x=1/??height of water=1/2??cos^-1(1/??)=45??

33 (b) current at time t=0 is through inductor=6A-Ldi/dt=(R₁+R₂)i=-4idi/dt=-10idi/i=-10dtln(i₂/i₁)=-10tI=6e^-10t

34 (b) 1/u + 1/v =1/f-1/30 +1/v=1/20v=12 cm from polev'=42 V=42*2/3=28 cm

35 (a) 1.5/v1 -1.75/?? =-(1.5-1.75)/R1.5/v1=0.25/Rv1=6R1.75/v - 1.5/v1=(1.75-1.5)/R1.75/f -1.5/6R=0.25/R1.75/f=1/2Rf=3.5R1.75/f=(1/R)(1/4)

36 (c) PE=-Ke²/rKE=(1/2)mv²=Ke²/2rTE=PE+KE=-Ke²/2rso |KE|=|TE|TE in the first excited state=-13.6/4=-3.4 eVKE=3.4 eV