Major Test 1 Mains

September 24, 2017 | Author: Nidhi Sisodia | Category: Chlorine, Physics, Physics & Mathematics, Physical Sciences, Science
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TARGET IIT - JEE MAJOR TEST (MAIN)

PAPER – I

BOOKLET CODE : A CLASS – XI [SPEED]

Date :– 06– 02 - 2013

Duration :3 Hours

Max. Marks : 270

_________________________________________________________________________________________ INSTRUCTIONS Do not break the seal of the question paper booklet before instructed to do so by the invigilator In each part of the paper, Section-A contains 30 questions. Total number of pages are 24. Please ensure that the Questions paper you have received contains ALL THE QUESTIONS in each section and PAGES. SECTION - A 1.

Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which only one is correct & carry 3 marks each. 1 mark will be deducted for each wrong answer.

NOTE : GENERAL INSTRUCTION FOR FILLING THE OMR ARE GIVEN BELOW. 1.

Use only HB pencil or blue/black pen (avoid gel pen) for darkening the bubble.

2.

Indicate the correct answer for each question by filling appropriate bubble in your OMR answer sheet.

3.

The Answer sheet will be checked through computer hence, the answer of the question must be marked by shading the circles against the question by dark HB pencil or blue/black pen.

4.

While filling the bubbles please be careful about SECTIONS [i.e. Section-A (include single correct)]

Booklet Test Code

D

1 0 01

A

0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

B C D E F G H I J

Name

Batch

Roll Number

10+1

28 3 2 3

10+1 10+2 10+3 Crash

Paper 5 5 5 5 6 6 6 6 1 7 7 7 7 Paper 1 8 8 8 8 9 9 9 9 Paper 2

0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9

For example if only 'A' choice is For example If Correct match correct then, the correct method for (A) is P; for (B) is R, S; for for filling the bubble is (C) is Q; for (D) is Q, S, T then A B C D E the correct method for filling For example if only 'A & C' the bubble is choices are correct then, the P Q R S T correct method for filling the bubble is A B C D E

the wrong method for filling the F I R S T N A M E M I D D bubble are

L E N A ME Test Date

L A S T N A ME

D D MMY Y

The answer of the questions in wrong or any other manner will be treated as wrong.

Ensure that all columns are filled. Answers, having blank column will be treated as incorrect. Insert leading zero(s) if required : '6' should be filled as 0006

'86' should be filled as 0086

0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4

5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

0 0 0 1 1 1 2 2 2 3 3 3

0 1 2 3

4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 (IVR No. 0744-2439051, 52, 53) www. motioniitjee.com , [email protected]

Page # 2

MAJOR TEST (MAIN)

PART - I [MATHEMATICS]

SECTION - A [STRAIGHT OBJECTIVE TYPE] Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct 1.

If the points (0, 0), (2, 2 3 ), and (p, q) are the vertices of an equilateral triangle, then (p, q) is

;fn fcUnq(0, 0), (2, 2 3 ), rFkk (p, q) leckgqf=kHkqt d s'kh"kZgS] rc (p, q) gS (A) (0, –4) 2.

(B) (4, 4)

(C) (4, 0)

(D) (5, 0)

If x, and x2 are the roots of the equation e2. xlnx = x3 with x1 > x2, then

;fn x, rFkk x2 lehd j.k e2. xlnx = x3 d sewy gSt gk¡x1 > x2] rc (B) x1 = x22

(A) x1 = 2x2 3.

(C) 2x1 = x22

(D) x12 = x23

The least value of 6 tan2 + 54 cot2 + 18 is I: 54 when A.M.  G.M. is applicable for 6 tan2, 54 cot2, 18 II: 54 when A.M.  G.M. is applicable for 6 tan2, 54 cot2 and 18 added further. III: 78 when tan2  = cot2 . (A) I is correct

(B) I and II are correct

(C) III is correct

(D) none of the above is correct

6

tan2

I: 54

+ 54

cot2

+ 18

d k U;wure eku gS

t c A.M.  G.M., 6 tan2, 54 cot2, 18 d sfy , iz;qDr gksrk gS

II: 54

t c A.M.  G.M., 6 tan2, 54 cot2 rFkk 18 ;ksx d jusd sckn] iz;qDr gksrk gS

III: 78

t c tan2  = cot2 .

lgh gS (C) III lghgS

rFkkII lghgS (D) mijks Dr esalsd ksbZHkh lgh ugh gS

(A) I

4.

Value of

(B) I

3  cot 80 cot 20 is equal to cot 80  cot 20

3  cot 80 cot 20 cot 80  cot 20 (A) cot 20°

d k eku cjkcj gS (B) tan 50°

(C) cot 50°

(D) cot

20

(SPACE FOR ROUGH WORK)

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

MAJOR TEST (MAIN)

Page # 3

5.

 r1  If in a triangle, 1  r  2 

 r  1  1  = 2, then the triangle is r3  

(A) right angled

(B) isosceles

 



(C) equilateral

(D) None of these

(C) leckgq

(D)



r1 r1 ;fn f=kHkqt esa, 1  r  1  r  = 2, rc f=kHkqt gS 

(A) led ks .kh; 6.

2

 

3



(B) lef}ckgq

buesalsd ksbZugha

Which of the following statement is incorrect ? (A) Graph of |f (x)| is symmetric about x-axis. (B) Graph of even function f (x) is same as that of f (|x|). (C) If a function is symmetric about the line x = a, then f (a – x) = f(a + x). (D) Graph of odd degree polynomial will always cut the x-axis.

fuEu esalsd kSulk d Fku vlR; gS\ (A) |f (x)| d k vkjs [k x-v{k d slkis{k lfEer gSA (B) le Q y u f (x) d k vkjs [k] f (|x|) d sleku gSA (C) ;fn d ks bZQ y u js[kk x = a d slkis{k lfEer gS]rc f (a – x) = f(a + x). (D) fo"ke ?kkr oky scgq O;at d d k vkjs[k lnSo x-v{k d ksd kVrk gSA 7.

If ‘O’ is the circumcentre of ABC and R1, R2 and R3 are the radii of the circumcircles of triangles

a b c OBC, OCA and OAB, respectively, then R + R + R has the value equal to 1 2 3 a

b

c

;fn ‘O’ f=kHkqt ABC d kifjd sUnzgSrFkkR1, R2 o R3 Ø e'k%f=kHkqt OBC, OCA rFkkOAB d hf=kT;k,¡gS]rc R + R + R 1 2 3 d k eku cjkcj gS abc (A)

2R

3

(B)

R3 abc



4 (C)

R

2

(SPACE FOR ROUGH WORK)

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

(D)

4R2

Page # 4

MAJOR TEST (MAIN)

8.

1  If the expression  mx  1   is non-negative for all positive real x, then the minimum value of m x 

must be 

1





;fn O;at d  mx  1  x  lHkh /kukRed okLrfod x d sfy , v_ .kkRed gS]rc m d k U;wure eku gksuk pkfg, (A)

9.

1 2

(B) 0

(C)

1 4

(D)

1 2

The product of r consecutive positive integers, divided by r! is (A) A proper fraction (B) equal to r (C) A positive integer

(D) none of these

Ø ekxr /kukRed iw.kkZad ksad k xq.kuQ y ]r! lsHkkx nsusij gS& (A) mfPPr fHkUu (B) r d scjkcj (C) ,d /kukRed iw .kkZad

(D)

r

10.

buesalsd ksbZugha

A rectangular billiard table has vertices at P (0,0), Q(0, 7), R(10, 7) and S(10, 0). A small billiard ball starts at M(3,4) and moves in a straight line to the top of the table, bounces to the right side of the table, then comes to rest at N(7, 1). The y-coordinate of the point where it hits the right side, is

,d vk;rh; fcfy ;MZVscy d s'kh"kZP (0,0), Q(0, 7), R(10, 7) rFkkS(10, 0) gSA ,d NksVhfcfy ;MZxsan M(3,4) ls'kq: gksrhgSrFkkVscy d sÅ ij ,d ljy js[kkesaxfr d jrhgqbZ]Vscy d snka;hrjQ mNy rhgSA vkSj mld sckn N(7, 1), ij fojke y srh gSA fcUnqd k y-funsZ'kkad t gk¡bl nk;savksj ekjk x;k gks& (A) 3.7 11.

(B) 3.8

(D) 4

The number of real roots of the equation 2sin2 x  2cos2 x  1 is (A) 2 (B) 1 (C) inifnite

lehd j.k 2sin

2

(A) 2 12.

(C) 3.9

x

2

 2cos

x

(D) none of these

y ksad h la[;k gS  1 d sokLrfod ew (B) 1

(C)

vuar

(D)

buesalsd ksbZugha

A flight of stairs has 10 steps. A person can go up the steps one at a time, two at a time, or any combination of 1’s and 2’s. Find the total number of ways in which the person can go up the stairs.

,d gokbZt gkt esa10 lhf 4

(D) ( 3 , )

(D) None of these (D)

buesalsd ksbZugha

(SPACE FOR ROUGH WORK)

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

MAJOR TEST (MAIN)

Page # 7

7  5 cot  23.

If tan , 2 tan  + 2,3 tan  + 3 are in G.P., then the value of

;fn tan , 2 tan  + 2,3 tan  + 3 xq.kksRrj Js.kh esagS] rc (A) 12/5

24.

If sin x + cos x =

(B) –33/28

9  4 sec2   1

7  5 cot  9  4 sec2   1

(C) 33/100

is

d k eku gS&

(D) 12/13

 1  y   , x  [0, ], then y  

1





;fn sin x + cos x =  y  y  , x  [0, ], rc 3  ,y=1 (B) y = 0 (C) y = 2 (D) x = 4 4 Let a, b, c be real numbers and a  0. If  is a root of a2x2 + bx + c = 0,  is a root of a2x2 – bx – c = 0 and 0 <  < , then the equation a2x2 + 2bx + 2c = 0 has a root  that always satisfies (A) x = 25.

ekuk a, b, c okLrfod la[;k,¡gSrFkk a  0A ;fn a2x2 + bx + c = 0 d k ewy  gS, a2x2 – bx – c = 0 d k ewy  gS rFkk 0 <  < , rc lehd j.k a2x2 + 2bx + 2c = 0 d k ,d ewy  gSt ksfd lnSo larq"V d jrk gS& (A)  = 26.

1 ( + ) 2

(B)  =  +

 2

(C)  = 

(D)  <  < 

In a triangle ABC, 2a2 + 4b2 + c2 = 4ab + 2ac, then the numerical value of cos B is equal to

fd lh f=kHkqt ABC esa, 2a2 + 4b2 + c2 = 4ab + 2ac, rc cos B d k la[;kRed eku gS& (A) 0

27.

(B)

3 8

(C)

5 8

Number of positive integers n for which n2 + 96 is a perfect square, is (A) 4 (B) 8 (C) 12

/kukRed iw.kkZad n d h la[;k ft ld sfy , (A) 4

(B) 8

n2

+ 96

(D)

7 8

(D) Infinite

iw.kZoxZgS& (C) 12

(SPACE FOR ROUGH WORK)

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

(D) vua r

Page # 8

MAJOR TEST (MAIN) 28.

Let p(x) = 0 be a polynomial equation of least possible degree, with rational coefficients, having 3

7 + 3 49 as one of its roots. Then, the products of all the roots of p(x) = 0 is

ekuk p(x) = 0 U;wure y Ecor~?kkr oky k cgqO;at d lehd j.k gS]ft ld k ,d ewy

3

7 + 3 49

gSA rc p(x) = 0 d slHkh

ewy ksad k xq.kuQ y gS (A) 7 29.

(B) 49

(C) 56

(D) 63

If S = {x  R: (log0.6 0.216) log5 (5 – 2x)  0}, then S is equal to

;fn S = {x  R: (log0.6 0.216) log5 (5 – 2x)  0}, rc S cjkcj gS& (A) [2.5, ) 30.

(B) [2, 2.5)

(C) (2, 2.5)

(D) (0, 2.5)

Let x1, x2, x3 be the roots of the equation x3 – x2 + x +  = 0, which are in AP, then x + y = 1 passes through the fixed point

ekuk x1, x2, x3 lehd j.k x3 – x2 + x +  = 0 d sewy gS]t ksfd lekUrj Js.kh esagS]rc x + y = 1 fuf'pr fcUnqls gksd j xqt jrkgSA  9 27   (A)  , 2 2 

9 7 (B)  ,  2 2

 5 27   (C)  , 2 2 

9 5 (D)  ,  2 2

(SPACE FOR ROUGH WORK)

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]

MAJOR TEST (MAIN)

Page # 9

PART - II [PHYSICS]

SECTION - A [STRAIGHT OBJECTIVE TYPE] Q.31 to Q.60 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct 31.

A rocket is fired vertically from the ground with a resultant vertical acceleration of 10m/s2. The fuel finishes in 1 minute and it continues to move up. After how much time from then (when fuel finished) will the maximum height be reached (g = 9.8 m/s2)

,d jkWd sV d ks/kjkry ls10m/s2 d sifj.kkehÅ /okZ/kj Roj.klsÅ /okZ/kjr%iz{ksfir fd ;kt krkgSA bld kbZa/ku 1 feuV esalekIr gkst krkgSrFkk;g Å ij d hvksj xfr d jrkjgrkgSA rc ls¼bZ/ku lekIr gksusd sckn½ fd rusle; i'pkr ;g vf/kd re Å ¡p kbZ ij igqap t k,xk\ (g = 9.8 m/s2) (A) 61.2 s

32.

(B) 64.2 s

(C) 68.2 s

(D) 70 s

A ball is thrown upward at an angle 37° to the horizontal and

v0

lands on the top edge of a building that is 20 m away and 10 m high. How fast was the ball thrown ? (g=10 m/s2)

10m

,d xsan d ks/kjkry ls37° d sd ks.kij Å ij d hvksj iz{ksfir fd ;kt krkgSrFkk ;g 20 m nwj ,oa10 m Å ¡p hbZekjr d hpksVhij Vd jkrhgSA xsan d siz{ksi.kosx

37° O 20m

d hx.kukd hft ,A (g=10 m/s2) (A) 10 m/s

33.

(B) 25 m/s

(C) 40 m/s

(D) 80 m/s

A ball rolls off the top of a stairway with a horizontal velocity of magnitude 1.8 m/s. The steps are 0.20 m high and 0.20 m wide. Which step will the ball hit first ?

,d xsan ,d lh
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