part : i mathematics

®

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TEST TYPE : ADVANCED PATTERN CUMULATIVE TEST-1 (ACT-1)

TARGET : JEE (MAIN+ADVANCED) 2021

COURSE NAME : VIJAY (01JR, 01JRH, 08JD, 05JDH1 & 08JDH1)

SYLLABUS : FUNDAMENTALS OF MATHEMATICS, QUADRATIC

EQUATION, RELATION FUNCTION ITF

TEST DATE : 10-10-2021 TARGET DATE : 00-00-20

PART : I MATHEMATICS

SECTION 1 (Maximum Marks: 08)

This section contains TWO (02) questions.

Each question has TWO matching lists: LIST-I and LIST-II.

FOUR options are given representing matching of elements from LIST-I and LIST-II. ONLY ONE of

these four options corresponds to a correct matching.

For each question, choose the option corresponding to the correct matching.

For each question, marks will be awarded according to the following marking scheme:

Full Marks : +4 If ONLY the correct option is chosen.

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : –1 In all other cases.

[kaM 1 ¼vf/kdre vad% 08)

bl [kaM esa nks (02) ç'u gSA izR;sd ç'u esa nks lqesyu lwfp;k¡ (matching lists) gS % lwph&I vkSj lwph&II lwph–I vkSj lwph–II ds rRoksa ds lqesyuksa dks n'kkZrs gq, pkj fodYi fn, x, gSA bu pkj fodYiksa esa flQZ ,d

fodYi gh lgh lqesyu çnf'kZr djrk gSA

izR;sd iz'u ds fy, lgh lqesyu çnf'kZr djus okys fodYi dks pqusaA izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk%&

iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSaA 'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

PATTERN CODE

10-21

®




1. Match List I with List II and select the correct answer using the code given below the lists :

If f(x) = ax2 + bx + c is such that (f(0))2 + (f(1))2 + (f(–1))2 + 30 = 4f(0) + 10 f(1) + 2f(–1) then match

the items in columns I with those in column II (where [.] denotes greatest integer function)

List – I List – II

(P) [sin–1(f(x))] equal to (1) 0

(Q) 2 + [tan–1 (f(x)] can be equal to (2) 1

(R) [2–f(x)] can be equal to (3) 2

(S) 3 +

xf

1 can be equal to (4) 4

lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s % ;fn f(x) = ax2 + bx + c bl izdkj gS fd (f(0))2 + (f(1))2 + (f(–1))2 + 30 = 4f(0) + 10 f(1) + 2f(–1) rc

LrEHk-I dk LrEHk-II (tgk¡ [.] egÙke iw.kkZad Qyu gS)

lwph - I lwph- II

(P) [sin–1(f(x))] dk eku gks ldrk gSA (1) 0

(Q) 2 + [tan–1 (f(x)] dk eku gks ldrk gSA (2) 1

(R) [2–f(x)] dk eku gks ldrk gSA (3) 2

(S) 3 +

xf

1 dk eku gks ldrk gSA (4) 4

Code (P) (Q) (R) (S)

(A) 1 2 1 4

(B*) 2 3 1 4

(C) 2 1 3 4

(D) 1 1 2 3

Sol. (f(0))2 – 4f(0) + 4 + (f(1))2 – 10f(1) + 25 + (f(–1))2 – 2 f(–1) + 1 = 0

(f(0) – 2)2 + (f(1) – 5)2 + (f (–1) – 1)2 = 0

f(0) = 2, f(1) = 5, f(–1) = 1

f(x) = x2 + 2x + 2 = (x + 1)2 + 1

1 f(x) <

2. Match list-I with list-II and select the correct answer using the code given below the lists –

List– I List– II

(P) If x2 – 4x + log 1

2

(a – 1) = 0 does not have two distinct real roots, then (1) 17

maximum value of 16 a is (Q) Least value of expression x2 + 4y2 + 3z2 – 2x – 12y – 6z + 16 is k is then (2) 14 k + 13 is (R) If number of integral values of p for which the equation x3 – 3x + p = 0 has 3 real (3) 16 and distinct roots is k then k +11 is

(S) If 2x x 0 , x3 – 2x2 + 2x – 1 = 0 (, R) have two roots in common, (4) 15

then 15 is

lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s % lwph– I lwph – II

(P) ;fn x2 – 4x + log 1

2

(a – 1) = 0 ds nks fHkUu okLrfod ewy ugha gS] rc (1) 17

16a dk vf/kdre eku gS&

(Q) O;atd x2 + 4y2 + 3z2 – 2x – 12y – 6z + 16 dk U;wure eku k gS] rc (2) 14

k + 13 dk eku gS&

(R) ;fn lehdj.k x3 – 3x + p = 0 ds 3 okLrfod vkSj fHkUu fHkUu ewy gS] ds fy, (3) 16

®




p ds iw.kk±d ekuksa dh la[;k k gS] rc k +11 dk eku gS&

(S) ;fn 2x x 0 , x3 – 2x2 + 2x – 1 = 0 (, R) ds nks ewy mHk;fu"B gS] rc (4)

15

15 gS&

Code (P) (Q) (R) (S) (A) 4 3 2 1 (B*) 1 3 2 4 (C) 3 1 2 4 (D) 1 3 4 2

Sol. (P) 1

2

1716 4log (a 1) 0 1 a

16

(16a)max 17

(Q) (x – 1)2 + (2y – 3)2 + 3(z – 1)2 + 3 so blfy, k + 13 = 16

(R) f (x) = 0 x = 1

f(1). f (–1) < 0 –2 < p < 2 so blfy, k + 11 = 14

(S) 111

0 15 15

SECTION – 2 : (Maximum Marks : 16) This section contains FOUR (04) questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct

For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : +2 If three or more options are correct but ONLY two options are chosen

and both of which are correct. Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases.

For example, in a question, if (A),(B) and (D) are the ONLY three options corresponding to correct answers, then : Choosing ONLY (A),(B) and (D) will get +4 marks Choosing ONLY (A) and (B) will get +2 marks Choosing ONLY (A) and (D) will get +2 marks Choosing ONLY (B) and (D) will get +2 marks Choosing ONLY (A) will get +1 marks Choosing ONLY (B) will get +1 marks Choosing ONLY (D) will get +1 marks Choosing no option (i.e. the question is unanswered) will get 0 marks, and Choosing any other combination of options will get –1 mark

[kaM 2 : (vf/kdre vad : 16)

bl [kaM esa pkj (04) iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %

iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA vkaf'kd vad % +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSA vkaf'kd vad % +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gS ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj

nksauks pqus gq, fodYi lgh fodYi gSaA vkaf'kd vad % +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj

pquk gqvk fodYi lgh fodYi gSA 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

®




mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc % dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’ dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) pquus ij +1 vad feysaxs ’ dsoy fodYi (B) pquus ij +1 vad feysaxs ’ dsoy fodYi (D) pquus ij +1 vad feysaxs ’ dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj vU; fdlh fodYiksa ds la;kstu dks pquus ij –1 vad feysaxs ’

3. The value of x satisfying the equation 22x – 8·2x = – 12 is

lehdj.k 22x – 8·2x = – 12 dks larq"B djus okys x ds eku gS -

(A*) 1 + log3

log2 (B)

1

2log 6 (C) 1 + log

3

2 (D*) 1

Sol. 2x = t t2 – 8t + 12 = 0

(t – 2) (t – 6) = 0

t = 2 or t = 6

2x =2, 2x = 6

x = 1 x = log2 6 = log2 (2× 3)

= log2 2 + log2 3 = 1 + log3

log2

4. If p and q are the roots of the equation x2 + px + q = 0, then the possible values of p are

;fn p vkSj q lehdj.k x2 + px + q = 0 ds ewy gS] rc p ds laHkkfor eku gS&

(A*) 1 (B*) 0 (C) –1

2 (D) –2

Sol. p and q are the root of the equation x2 + px + q = 0.

p vkSj q lehdj.k x2 + px + q = 0 ds ewy gSA

p+q = –p and pq = q p = 0 or 1

5. The value(s) of satisfying the equation 1 1[sin(cot cos tan x)] = sin , ( x R & [.] show G.I.F.

is/are) is

lehdj.k 1 1[sin(cot cos tan x)] = sin , dks larq"B djus okys ds eku gS ( x R rFkk [.] egÙke iw.kkZad Qyu

n'kkZrk gS)

(A*) 0 (B) 2

(C*) (D)

3

2

Sol. –1tan x2 2

0 < cos(tan–1x) 1

–1 –1cot (cos(tan x))4 2

–1 –11sin(cot (cos(tan x))) 1

2

®




[sin(cot–1(cos(tan–1x)))] = 0

sin = n

= 0,

Alter :

1 1sin(cot cos tan x) = 1sin(cot cos ) , where 1tan x

= sin 1cot2

1

1 tan = sin 1cot

2

1

1 x

= 2

2

1 x

2 x

Now for all xR 2

2

1 x0

2 x

< 1 then

lHkh xR ds fy, 2

2

1 x0

2 x

< 1 rc -

1 1[sin(cos cos tan x)] = 0

sin= 0 n

6. Suppose A1, A2, ......, A30 are thirty sets each with five elements and B1, B2, ..... Bn are n sets each

with three elements. Let 30

1i

iA

= n

1j

jB

= S

Assume that each element of S belongs to exactly ten of the A's and exactly to nine of the B's.

Then correct option(s) is (are)

ekukfd A1, A2, ......, A30 rhl leqPp; gS tcfd izR;sd esa 5 vo;o gS rFkk B1, B2, ..... Bn n leqPp; gS ftlesa

izR;sd esa rhu vo;o gS -

ekuk 30

1i

iA

= n

1j

jB

= S

ekukfd S dk izR;sd vo;o esa A ds Bhd 10 leqPp;ksa ls rFkk B ds Bhd 9 leqPp;ksa ls lEcfU/kr gS rc fuEu

esa ls lgh fodYIk gS

(A*) Number of elements of set S is 15 (B) Number of elements of set S is 30

(C*) Number of elements of set B is 45 (D) Number of elements of set B is 60

(A) leqPp; S esa vo;oksa dh la[;k 15 gS (B) leqPp; S esa vo;oksa dh la[;k 30 gS

(C) leqPp; B esa vo;oksa dh la[;k 45 gS (D) leqPp; B esa vo;oksa dh la[;k 60 gS

Ans. (A,C)

Sol. Given A's are thirty sets with five elements each so

30

1i

i)A(n = 5 × 30 = 150 ....... (i)

If the m distinct element in S and each elements of S belong to exactly 10 of the Ai's so we have

30

1i

i)A(n = 10 m ...... (ii)

from (i) and (ii) we get 10 m = 150

m = 15 ....... (iii)

similarly

n

1j

j)B(n = 3n and

n

1j

j)B(n = 9 m

3n = 9m

®




n = 3

m9 = 3m = 3 × 15 = 45

Hindi fn;k x;k gS fd A ds rhl leqPp; gS ftless 5 vo;o gSA

30

1i

i)A(n = 5 × 30 = 150 ....... (i)

;fn m, S esa fHkUu fHkUu vo;o gS rFkk S ds izR;sd vo;o esa A ds Bhd 10 vo;o gSA

30

1i

i)A(n = 10 m ...... (ii)

(i) vkSj (ii) ls 10 m = 150

m = 15 ....... (iii)

blfy,

n

1j

j)B(n = 3n vkSj

n

1j

j)B(n = 9 m

3n = 9m

n = 3

m9 = 3m = 3 × 15 = 45

SECTION – 3 : (Maximum Marks : 36)

This section contains TWELVE (12) questions.

The answer to each question is SINGLE DIGIT INTEGER ranging from 0 to 9 with Positive /Negative.

If your answer is 5 then the correct answer chosen is 5.

If your answer is –5 mark correct answer chosen is –5.


Marking scheme :



Negative Marks : –1 In all other cases

[kaM 3 : (vf/kdre vad : 36) bl [kaM esa ckjg (12) iz'u gSA izR;sd iz'u dk mÙkj 0 ls 9 rd, nksuksa 'kkfey]ds chp dk ,d ,dy vadh; iw.kk±d gS tks /kukRed@_.kkRed gSA ;fn mÙkj 5 gS rc mÙkj pquus dk lgh rjhdk 5 gSA ;fn mÙkj –––5 gS rc mÙkj pquus dk lgh rjhdk –5 gSA ;fn mÙkj –0 gS rc mÙkj pquus dk lgh rjhdk 0 gSA vadu ;kstuk %&


7. The complete solution set of the inequality (|x –2| –2)(|x + 2| – 3) 0 is (–,a] [b, c] [d,) then a + b + c + d is

vlfedk (|x –2| –2)(|x + 2| – 3) 0 dk lEiw.kZ gy leqPp; (–,a] [b, c] [d,) rc a + b + c + d dk eku gS

Ans. 0

SOL. (|x –2| –2)(|x + 2| – 3) 0

fLFkfr -I x (–, –2)

– x (– x – 5) 0

®




x(x + 5) 0

x (–, –5] .......(i)

fLFkfr -II x [–2, 2]

– x (x – 1) 0

x(x – 1) 0

x [0, 1] .......(ii)

fLFkfr -III x (2, )

(x – 4)(x – 1) 0

x [4, ) .......(iii)

(i) (ii) (iii)

x (–, –5] [0, 1] [4, ) 8. If both the roots of equation k(6x2 + 3) + rx + 2x2 – 1 = 0 and 6k(2x2 + 1) + px + 4x2 – 2 = 0 are

common, then 2r – p =

;fn lehdj.k k(6x2 + 3) + rx + 2x2 – 1 = 0 vkSj 6k(2x2 + 1) + px + 4x2 – 2 = 0 ds nksuksa ewy mHk;fu"B gS]

rc 2r – p =

Ans. 0

Sol. Both the roots of equation k(6x2 + 3) + rx + 2x2 – 1 = 0

lehdj.k ds nksuksa ewy k(6x2 + 3) + rx + 2x2 – 1 = 0 gSA

i.e., (6k + 2)x2 + rx + 3k – 1 = 0

and (12k + 4)x2 + px + 6k – 2 = 0 are common.

vkSj (12k + 4)x2 + px + 6k – 2 = 0 mHk;fu"B gSA

6k 2

12k 4

= r

p =

3k – 1

6k – 2

so blfy, , r

p =

1

2

so blfy, , 2r – p = 0

9. If f(x) = 2(x – 1)

x2–x

9–x2–x

2

2 then number of integers in domain of |)x(|f–

;fn f(x) = 2(x – 1)

x2–x

9–x2–x

2

2 rc |)x(|f– ds izkUr esa iw.kk±dksa dh la[;k gS&

Ans. 4

Sol. 2(|x| – 1)

|x|2–x

9–|)x|2–x(2

22

0 )2–|x|)(|x(|

)3–|x)(|1–|x(| 0

0 < |x| 1 or 2 < |x|3

10. The number of negative integers lying in range of f(x) = sin–1x + x2 + 4x + 1 is

f(x) = sin–1x + x2 + 4x + 1 ds ifjlj esa mifLFkr _.kkRed iw.kkZdksa dh la[;k gS -

Ans. 3 Sol. Domain is [–1, 1] sin–1x + x2 + 4x + 1 is strictly increasing in [–1, 1]

range is – – 2, 62 2

®




Hindi izkUr [–1, 1] gSA sin–1x + x2 + 4x + 1, [–1, 1] esa fujUrj o/kZeku gS

ifjlj – – 2, 62 2

gSA

11. The number of irrational roots of the equation 2/335

5

3

422

xx

x

xx

xis

lehdj.k 2/335

5

3

422

xx

x

xx

xds vifjes; ewyksa dh la[;k gS - 01JR _ACT-1_10-10-21

Ans. 2

Sol. Here, x = 0 is not a root. Divide both the numerator and denominator by x and put x + 3/x = y to

obtain

3,52

3

5

5

1

4

y

yy

x + 3/x = – 5 has two irrational roots and x + 3/ x = 3 has imaginary roots

Hindi ;gk¡ x = 0 ewy ugh gS va'k o gj nksuksa dks x ls Hkkx nsus ij x + 3/x = y j[kus ij

3,52

3

5

5

1

4

y

yy

x + 3/x = – 5 nks vifjes; ewy gS rFkk x + 3/ x = 3 ds dkYifud ewy gSA

12. If point (a,a) lies in between the lines |x + y| = 6, and P = (sum of all possible values of [a] )2 then P is ([.] denotes G.I.F.)

;fn fcUnq (a, a) js[kkvksa |x + y| = 6 ds e/; fLFkr gS] rFkk P = ([a] ds lHkh laHkkfor ekuksa dk ;ksxQy)2 rc P

gS& ([.] egÙke iw.kk±d Qyu gS) Ans. 9 Sol. (a,a) lies above line x + y = –6 and below x + y = 6

(a,a) js[kk x + y = –6 ij fLFkr gS vkSj x + y = 6

a a 6 0 andvkSj a a 6 0 3 a 3

[a] = –3, –2, –1, 0, 1, 2

Sum;ksxQy = –3

P = 9

13. Let f(x) = 23 22sin x cos x and g(x) = 111 tan x

2

. If the number of real values of ‘x’ in interval

[ 10 , 20 ] satisfying the equation f(x) [g(x)] (where [.] denote greatest integer function) is p

then value of p

5is

ekuk f(x) = 23 22sin x cos x vkSj g(x) = 111 tan x

2

;fn vUrjky [ 10 , 20 ] esa lehdj.k f(x) [g(x)] dks

larq"B djus okys x ds okLrfod ekuksa dh la[;k p gS (tgk¡ [.] egÙke iw.kkZad Qyu gS) rc p

5dk eku gS -

Ans. 3

Sol. 0 tan–1|x| < 2

0 1

2tan–1|x| <

4

®




1 1+1

2tan–1|x| < 1 +

4

1 g(x) < 1 +4

[g(x)] = 1

f(x) = [g(x)] = 1

sin23x – cos22x = 1

sin23x = 1 + cos22x

sin23x = 1 and vkSj cos22x = 0

x = 2n + 2

number of solutions gyksa dh la[;k = 15

14. If + 1

and 2 – –

1

(, > 0) are the roots of the quadratic equation x2 – 2 (a + 1) x + a – 3 =

0 then find the sum of all the integral values of 'a'.

;fn + 1

vkSj 2 – –

1

(, > 0) f}?kkr lehdj.k x2 – 2 (a + 1) x + a – 3 = 0 ds ewy gS rc a ds lHkh

iw.kkZad ekuksa dk ;ksxQy gS

Ans. 5

Sol. + 1

2

– – 1

– 2 2 – –

1

0

Thus one root is greater than or equal to 2 and other roots is smaller than or equal to zero.

vr% ,d ewy 2 ls cM+k ;k cjkcj gS rFkk vU; ewy 0 ls NksVk ;k cjkcj gSA f(0) 0 and f(2) 0

a [–1, 3]

15. Given the relation R = {(1,2), (2,3)} on the set A = {1,2,3} the minimum number of ordered pairs which when added to R make it is an equivalence relation is

fn;k x;k gS fd lEcU/k R = {(1,2), (2,3)} leqPp; A = {1,2,3} ij ifjHkkf"kr gSA R dks rqY;rk lEcU/k cukus ds fy, vko';d vfrfjDr U;wure Øfer ;qXeksa dh la[;k gS&

Ans. 7 Sol. R is reflexive if it contains (1, 1), (2, 2), (3, 3) R LorqY; gksxk ;fn blesa (1, 1), (2, 2), (3, 3) fo|eku gksA

(1,2) R, (2,3) R.

R is symmetric if (2,1), (3,2) R.

R lefer gksxk ;fn (2,1), (3,2) R.

Now vc, R = {(1,1), (2,2), (3,3), (2,1), (3,2),(2,3),(1,2)}

R will be transitive if (3,1), (1,3) R. Thus R becomes an equivalence relation by adding (1,1), (2,2), (3,3) (2,1),(3,2), (1,3), (3,1).

R laØked gksxk ;fn (3,1), (1,3) R vr% R dks rqY;rk lEcU/k cukus ds fy, vfrfjDr vko';d Øfer ;qXe (1,1), (2,2), (3,3) (2,1),(3,2), (1,3), (3,1) gksxsaA

Hence the total no. of ordered pairs is 7.

vr% dqy Øfer ;qXeksa dh la[;k 7 gksxhA

®




16. Find the number of integral values in the range of the function

f(x) =

21

x2

+

21

x3

– 2

1 1x x

2 3

, where [.] denotes greatest integer function.

Qyu f(x) =

21

x2

+

21

x3

– 2

1 1x x

2 3

, tgk¡ [.] egÙke iw.kk±d Qyu gS] ds ifjlj esa iw.kk±d

ekuksa dh la[;k Kkr dhft,A Ans. 2

Sol. f(x) =

2

1 1x – x

2 3

(i) Let ekuk x = , then rc f(x) = 0

(ii) Let ekuk x = + f, 0 < f < 1

2, then rc f(x) = 0

(iii) Letekuk x = + f, 1

2 f <

2

3, then rc f(x) = 1

(iv) Letekuk x = + f, 2

3 < f 1, f(x) = 0 Range is ifjlj {0, 1}

17. Let P(x) be a cubic polynomial such that P(x) + x is divisible by (x – 2) and (x – 3). If P(1) = 9 and

P(–1) = – 11 then value of P(0) is 10 + k, then the value of k is

ekuk P(x) ,d ?kuh; cgqin bl izdkj gS fd P(x) + x, (x – 2) vkSj (x – 3) ls foHkkftr gS ;fn P(1) = 9 rFkk P(–1)

= – 11 rc P(0) dk eku 10 + k gS rc k dk eku gS -

Ans. (2)

Sol. Let P(x) + x = (x – 2)(x – 3) (ax + b)

P(1) = 9 10 = +2(a + b) .......... (1) a + b = 5 ........... (2)

and P(–1) = – 11 – 12 = 12(–a + b)

– a + b = – 1 ............ (3)

from equation (1) and (2) b = 2 and a = 3

P(x) + x = (x – 2)(x – 3)(3x + 2)

P(0) = (–2)(–3)(2) = 12

Hindi ekuk P(x) + x = (x – 2)(x – 3) (ax + b)

P(1) = 9 10 = +2(a + b) .......... (1) a + b = 5 ........... (2)

vkSj P(–1) = – 11 – 12 = 12(–a + b)

– a + b = – 1 ............ (3)

(1) vkSj (2) ls b = 2 vkSj a = 3

P(x) + x = (x – 2)(x – 3)(3x + 2)

P(0) = (–2)(–3)(2) = 12

®




18. If the product of the integral root of the equation (x + 2)(x + 3)(x + 8)(x + 12) = 4x2 is , then find

the value of 6

.

lehdj.k (x + 2)(x + 3)(x + 8)(x + 12) = 4x2 ds iw.kk±d ewyksa dk xq.kuQy gS] rc 6

dk eku gS&

Ans. 4

Sol. (x + 2)(x + 12)(x + 3)(x + 8) = 4x2

(x2 + 14x + 24)(x2 + 11x + 24) = 4x2

14

x

24x

11

x

24x = 4

Let ekuk x + x

24 = t

(t + 14)(t + 11) = 4

t2 + 25t + 150 = 0

(t + 15)(t + 10) = 0

15

x

24x

10

x

24x = 0

(x2 + 15x + 24) (x2 + 10x + 24) = 0

d is not perfect square integral root

d iw.kZ oxZ ugha gSA iw.kk±d ewy not an integral root

iw.kk±d ewy ugha Product xq.kuQy = 24

®


Website : www.resonance.ac.in | E-mail : [email protected] P201JRACT1101021C1-1


®

TEST TYPE : ADVANCED PATTERN CUMULATIVE TEST-1 (ACT-1)

TARGET : JEE (MAIN+ADVANCED) 2021

COURSE NAME : VIJAY (01JR, 01JRH, 08JD, 05JDH1 & 08JDH1)

TEST DATE : 10-10-2021 TARGET DATE : 00-00-20

Test Syllabus : Kinematics, Geometrical Optics, Newtons laws of motion

PAPER-2

SECTION 1 (Maximum Marks: 08) This section contains TWO (02) questions. Each question has TWO matching lists: LIST-I and LIST-II. FOUR options are given representing matching of elements from LIST-I and LIST-II. ONLY ONE of these

four options corresponds to a correct matching. For each question, choose the option corresponding to the correct matching. For each question, marks will be awarded according to the following marking scheme:

Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases.


bl [kaM esa nks (02) ç'u gSA izR;sd ç'u esa nks lqesyu lwfp;k¡ (matching lists) gS % lwph&I vkSj lwph&II

lwph–I vkSj lwph–II ds rRoksa ds lqesyuksa dks n'kkZrs gq, pkj fodYi fn, x, gSA bu pkj fodYiksa esa flQZ ,d fodYi gh

lgh lqesyu çnf'kZr djrk gSA

izR;sd iz'u ds fy, lgh lqesyu çnf'kZr djus okys fodYi dks pqusaA

izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk%&

iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSaA

'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A

_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

®




19. In list I, light ray is incident at an angle shown and corresponding angle of deviation in clockwise (CW) or Anti-clockwise (ACW) sense after first two events [reflection/refraction] are listed in list II; select the correct option.

fyLV-I esa fdj.kksa dk vkiru dks.k vkSj fyLV-2 mu fdj.kksa dk nks ?kVukvksa (ijkorZu@viorZu) dqy fopyu dks.k nf{k.kkorZ (CW) ;k ckeorZ (ACW) esa n’’’'kkZ;k gqvk gSSA lgh fodYi pqfu;ssaA

List – I List - II

(P)

60°30°

(1) 300º CW

(Q) (2) 330º ACW

(R)60°

60° 60°

µ= 245°

(3) 270º ACW

(S)

60°

30°

(4) 240º CW

Code : P Q R S (A) 1 2 3 4 (B*) 4 3 2 1 (C) 1 3 2 4 (D) 4 2 3 1

Sol.

60°30°30°

30°30°

= 120° + 120° CW

= 240° CW

60°

µ=1

µ=2 3

µ=2

®




60°

30°

TIR

30°

30°

= 30° + 240° ACW

= 270° ACW

60°

60° 60°

µ= 2

45° 45°

= 330° ACW

30°60°60°30°

30°

= 60° + 240° CW

= 300° CW

20. Refractive index of lens is 3/2. Then final image from ‘P’ is

ysal dk viorZukad 3/2 gSA rc ‘P’ ls vafre izfrfcEc dh nqjh gksxh

List-I List – II

(P)

30cm

R=20cm R=20cm

PO

(1) 6cm, left to P ds ck;s

(Q)

20cm

R=20cm R=20cm

PO

(2) 30

cm7

, right to P ds nk;s

(R)

30cm

R=20cm

PO

(3) 12cm, right to P ds nk;s

(S)

30cm

R=20cm

PO

(4) 60cm, left to P ds ck;s

Code : P Q R S

(A) 4 1 2 1

®




(B) 2 4 1 3 (C) 4 3 1 2 (D*) 2 1 4 3

Sol. Use following equations for each case. izR;sd fLFkfr esa fuEufyf[kr lehdj.kksa dk mi;ksx fdft,

eff

1 1 1

V ( 30) f

eff m L

1 1 2

f f f

L 1 2

1 3 1 11 .

f 2 R R

SECTION – 2 : (Maximum Marks : 16) This section contains FOUR (04) questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is(are) correct For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : +2 If three or more options are correct but ONLY two options are chosen

and both of which are correct. Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases.

For example, in a question, if (A),(B) and (D) are the ONLY three options corresponding to correct answers, then : Choosing ONLY (A),(B) and (D) will get +4 marks Choosing ONLY (A) and (B) will get +2 marks Choosing ONLY (A) and (D) will get +2 marks Choosing ONLY (B) and (D) will get +2 marks Choosing ONLY (A) will get +1 marks Choosing ONLY (B) will get +1 marks Choosing ONLY (D) will get +1 marks Choosing no option (i.e. the question is unanswered) will get 0 marks, and Choosing any other combination of options will get –1 mark






mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc % dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’ dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) pquus ij +1 vad feysaxs ’

®




dsoy fodYi (B) pquus ij +1 vad feysaxs ’ dsoy fodYi (D) pquus ij +1 vad feysaxs ’ dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj vU; fdlh fodYiksa ds la;kstu dks pquus ij –1 vad feysaxs ’

21. An object is moving in a circular path of radius 1m around (0, 10m) is xy plane with constant speed 1 m/s. Speed of image w.r.t. object is v.

,d fcUnq oLrq ,d ehVj f=kT;k ds o`Ùkkdkj iFk esa (0, 10m) dks dsUnz ysrs gq, xy ry esa fu;r pky 1 m/s ls xfr'khy gSA oLrq ds lkis{k izfrfcEc dh pky v gSA

= 37° x

y

(A*) vmax = 2 m/s (B) vmin = 1 m/s (C) v is maximum when object crosses y axis.

v vf/kdre gksxk tc oLrq y-v{k dks ikj djrh gSA (D*) v is minimum when object is closest or farthest from the mirror

v U;wure gksxk tc oLrq niZ.k ds fudVre~ ;k nwjLFk fLFkfr ij gSA Sol. When velocity of object is perpendicular to mirror, relative velocity is maximum and vmax = 2 m/s.

atc oLrq dk osx niZ.k ds yEcor~ gS] lkis{k osx vf/kdre ,oa vmax = 2 m/s gksxkA

When velocity of object is parallel to mirror, relative velocity is minimum and vmin = 0 m/s.

atc oLrq dk osx niZ.k ds lekUrj gS] lkis{k osx U;wure ,oa vmin = 0 m/s gksxkA

22. Rain is falling down on ground making certain angle ‘’ with vertical. A man starts walking on level ground. Mark the correct statement.

ckfj'k Å/okZ/kj ds lkFk ‘’ dks.k cukrs gq, /kjkry ij fxj jgh gSA ,d O;fDr /kjkry ij pyuk izkjEHk djrk gSA lR; dFkuksa dk p;u dhft;saA

(A*) If the man increases his speed of walking, then it is possible that the angle made by rain with vertical (as observed by man) goes on decreasing.

(B) If the man goes on increasing his speed, then with respect to him, it is sure that the angle made by rain with horizontal will go on decreasing.

(C*) Keeping his own velocity constant on horizontal, if the man observes rain to be falling vertically downwards but with variable magnitude then it is sure that velocity of rain with respect to ground is variable.

(D) None of these

(A*) ;fn O;fDr vius pyus dh pky dks c<+krk gS] rks ;g laHko gS fd Å/okZ/kj ds lkFk ckfj'k ds }kjk cuk;k x;k dks.k ¼O;fDr ds lkis{k½ ?kVrk tk;sxkA

(B) ;fn O;fDr viuh pky dks c<+krk tkrk gS rks mlds lkis{k ;g fuf'pr gS fd ckfj'k }kjk {kSfrt ds lkFk cuk;k x;k dks.k ?kVrk tk;sxkA

(C*) mlds Lo;a ds osx dks {kSfrt lrg ij fu;r j[krs gq, ;fn O;fDr o"kkZ dks Å/okZ/kj uhps dh vksj fxjrs gq, ns[krk gS] fdUrq ifjofrZr ifjek.k ds lkFk] rks ;g fuf'pr gS fd /kjkry ds lkis{k ckfj'k dk osx ifjorZu'khy gSA

(D) buesa ls dksbZ ugha

®




23. A particle is resting over a smooth horizontal floor. At t = 0, a horizontal force starts acting on it.

Magnitude of the force increases with time according to law F = t, where is a positive constant and t is time. For the figure shown which of the following statements is/are correct?

,d d.k fpdus {kSfrt Q'kZ ij j[kk gSA t = 0 ij] ,d {kSfrt cy bl ij dk;Z djuk izkjEHk djrk gSA cy dk ifjek.k le; ds lkFk F = t ds vuqlkj c<+rk gS] tgk¡ ,d /kukRed fu;rkad vkSj t le; gSA n'kkZ;s x;s fp=k ds fy, fuEu esa ls dkSulk@ls dFku lR; gS@gSa\

(A*) Curve 1 shows acceleration against time (B*) Curve 2 shows velocity against time (C*) Curve 2 shows velocity against acceleration (D) none of these

(A*) oØ 1 le; ds lkFk Roj.k dks n'kkZrk gSA (B*) oØ 2 le; ds lkFk osx dks n'kkZrk gSA (C*) oØ 2 Roj.k ds lkFk osx dks n'kkZrk gSA (D) buesa ls dksbZ ughaA

Sol. a = tm

dv

dt= t

m

v = 2t2m

v = 2m

a2

.

24. Considering a cubical slab as shown in figure.

fp=k ds vuqlkj ?kukdkj ifV~Vdk ij fopkj fdft,

(A*) Geometrical path length traversed by light in slab is 2m. (B*) Total deviation suffered by light ray is 30° (C) Angle of emergence from the slab is 45°. (D) None of these.

(A*) ifV~Vdk esa izdk’’'k dk T;kfefr; iFk 2m gSA (B*) izdk’'k dh fdj.k dk dqy fopyu 30° gSA (C) ifV~Vdk ls fuxZr dks.k 45° gSA (D) buesa ls dksbZ ugha Sol.

®




Total path length dqy iFk dh yEckbZ 1 + 1 = 2m

angle of deviation fopyu dks.k = 30°







Marking scheme :





bl [kaM esa ckjg (12) iz'u gSA

izR;sd iz'u dk mÙkj 0 ls 9 rd, nksuksa 'kkfey]ds chp dk ,d ,dy vadh; iw.kk±d gS tks /kukRed@_.kkRed gSA

;fn mÙkj 5 gS rc mÙkj pquus dk lgh rjhdk 5 gSA

;fn mÙkj –––5 gS rc mÙkj pquus dk lgh rjhdk –5 gSA

;fn mÙkj –0 gS rc mÙkj pquus dk lgh rjhdk 0 gSA

vadu ;kstuk %&


'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

®




25. In a river flowing with speed 4m/s, a swimmer crosses transversely a flowing stream of with ‘d’ to and fro in time t1 sec. The time taken to cover same distance up and down the stream is t2. If t3 is the time swimmer would take a distance ‘2d’ in still water. Find the value of 2t1–t2 – t3. Given speed of swimmer in still water is 5m/s; and width of river is 45m.

4m/s pky ls izokfgr ‘d’ pkSM+kbZ dh ,d unh esa ,d rSjkd lh/ks fdukjs ds yEcor tkus vkSj vkus esa t1 sec dk le; ysrk gSA mruh gh nwjh /kkjk ds vuqfn'k vkSj /kkjk ds foijhr vkus tkus esa t2 sec dk le; ysrk gSA ;fn 'kkUr ty esa ‘2d’ nwjh r; djus esa fy;k x;k le; t3 gks] rc 2t1–t2 – t3 dk eku Kkr djsA fn;k x;k gS] 'kkUr ty esa rSjkd dh pky 5m/s gS, vkSj unh dh pkSM+kbZ 45m gSA

Ans. –8

Sol. 12 2

2d 2 45t 30sec

9v u

2 2 2

d d 2vd 2 5 45t 50sec

v 4 v 4 9v u

3

2d 45t 18sec

v 5

1 2 32t t t 8

26. A small trolley of height 1.25 m and length 2m is moving along a straight road with constant velocity.

There is a bolt exactly at mid point of the Roof. The instant at which bolt comes out of the roof breaks are applied which produces constant retardation and bolt falls exactly on the edge of wall and floor. Then constant retardation is : (g = 10 m/s2)

,d NksVh xkM+h dh Å¡pkbZ 1.25 m rFkk yEckbZ 2m gSA ;g lh/kh lM+d ij fu;r osx ls xfr'khy gSA bldh Nr ds e/; fcUnq ij ,d cksYV fLFkr gSA tSls gh cksYV Nr ls uhps fxjrk gS] xkM+h ds czsd yxk, tkrs gSA tksfd fu;r eUnu çnku djrhs gS vkSj cksYV nhokj rFkk xkM+h ds ry ds fdukjs ij fxjrk gS rks fu;r eUnu Kkr djksA (g = 10

m/s2)

Ans. 8

Sol. w.r.t. trolley acceleration of bolt is xkM+h ds lkis{k cksYV dk Roj.k

tfalling

= 2 1.25 1

10 2

sec.

1.25

2m

and initial relative velocity is zero.

rFkk çkjfEHkd lkis{k osx 'kwU; gS

So vr% 1 =

21 1

a2 2

a = 8 m/s2

®




27. A lift with an open roof is moving upwards with some constant acceleration ‘α’ m/s2. A person in the lift

throws a ball from 2 meter height in upward direction with velocity 20 m/s relative to lift. It was observed that ball hits the floor of lift after 2 seconds of the throw with speed ꞵ m/s: relative to ground. If velocity of lift at the time of throw as 10 m/s upwards and g = 10 m/s2. Find ꞵ– α.

,d [kqyh gqbZ NÙk okyh fy¶V fu;r Roj.k ‘α’ m/s2 ls Åij dh vksj xfr dj jgh gSA fy¶V esa gh ,d O;fDr ,d xsan dks 2 ehVj Åps fcUnq ls 20 m/s dh pky ls Åij dh vksj fy¶V ds lkis{k esa Qsdrk gSA ;g ns[kk x;k fd xsan fy¶V ds ry dks 2 lsd.M ckn tehu ds lkis{k ꞵ m/s ls Vdjkrh gSaA ;fn Qsdus ds le; fy¶V dh pky Åij dh vksj 10 m/s Fkh] rFkk g = 10 m/s2. Find ꞵ– α Kkr djsaA

2m

20m/s

10m/s

a

Ans. –1

Sol. Relative to lift fy¶V ds lkis{k esa

S = ut + 1

2at2

−2 = 20 x 2 − 1

2(g+a) x 22

−42 = −2(g+a) a = 11 m/s2

In ground frame /kjkry ds lkis{k esa v = u + gt = 30 – 10 × 2 = 10

28. At t = 0, two particles A and B are present at points (0,0) and ( 5m,0 ) respectively. They start moving

with constant velocities Aˆ ˆv 5 i 5 j m / s

and B

ˆv 5 jm / s

. If ‘x’ is the minimum separation

between them and ‘y’ sec be the time when separation is minimum; then what is the value of 5

3x–y ?

t = 0 {k.k ij nks d.k A rFkk B Øe'k% fcUnq (0,0) rFkk ( 5m,0 ) ij fLFkr gSA ;g nksuks fu;r osx

Aˆ ˆv 5 i 5 j m / s

rFkk B

ˆv 5 jm / s

. ls xfr djuk izkjEHk djrs gSA ;fn ‘x’ muds chp dh U;qure nqjh gks

vkSj ;g ‘y’ sec ds ckn gksrk gks] rc 5

3x–y dk eku D;k gksxk ?

Ans. 1 Sol.

B5

5

5

A

B5m/s5m

5m

/s

A

2

minR

θ

Rmin = x = 5.sin

x= 2

5. m5

x = 2m

T = y = ref

5 cos 5 cos 1

v 55. 5

sec

3x 3 1

y 2 15 5 5

®




29. A thin equi-convex glass lens with radius of each face ‘R’ is placed between water and oil. Given

glass

3

2 , oil

5

2 and water

4

3 if incident ray is from oil, focal length of lens is ' ' and if incident

ray is from water side focal length is ' ' . What is the value of 5

R

?

,d iryk leksÙky ysal (glass) ftldh oØrk f=kT;k ‘R’, ikuh (water) vkSj rsy (oil) ds chp j[kk x;k gSA fn;k

x;k gS] glass

3

2 , oil

5

2 and water

4

3 . ;fn fdj.k dk vkiru rsy dh rjQ ls gksrk gS rc ysal dh Qksdl

nqjh ' ' gS vkSj ;fn vkiru ikuh dh rjQ ls gksrk rc Qksdl nqjh ' ' gSA 5

R

Kkr djs ?

Ans. –5 Sol.

32

52 3

4

4 533 5 4

3 3 22 2 2

R R

8R

5

32

52 3

4

534 4 5 3

3 2 32 2 2

R R

3R

5 8R 3R 5R

30. XZ plane acts as boundary between two transparent medium. A ray of light in medium I, given by vector

ˆ ˆn i 3 j

is incident on plane of separation. The reflected and refracted ray are perpendicular to each

other. Refractive index of medium I with y >0 is ‘’ where as refractive index of medium II with y <0 is 1.

What should be the value of 2 6 ?

XZ ry nks ijkn'khZ ek/;e ds chp dh lhek gSA ek/;e–I ls vkrk gqvk ˆ ˆn i 3 j

lfn'k ls vUrjkry ij vkifrr

gksrh gSA ijkofrZr vkSj viofrZr fdj.k ,d nqljs ls yEcor gSA ek/;e–I y >0 dk viorZukad ‘’ gS tcfd ek/;e–II

y <0 dk viorZukad 1 gSA 2 6 dk eku D;k gksxk ?

Ans. –3 Sol.

n

p !

r

>

>

>

!

>

r

1

µ

ˆ ˆ1.i 3 j

n2

®




ˆ ˆ ˆ ˆ ˆr n 2 n.p p

1 3ˆ ˆr i j2 2

Also rc gh; ˆ ˆn.p cos( i)

3

cosi ;i 302

2 6 3 6 3

sini 1 sin r

3

31. For an equilateral prism, a ray of light is incident at grazing incidence. It was observed that light also

emerges grazing the surface. The refractive index of prism is?

leckgq fizTe ij ,d fdj.k Li'kZ js[kh; vkifrr gksrh gSA ;g ik;k tkrk gS fd ;g fdj.k Li'kZ js[kh; fuxZr Hkh gksrh gSA fizTe dk viorZukad gksxk

Ans. 2 Sol.

60°

90° 90°c c

2C = 60° C 30

1.sin90 n.sin30

n 2 32. An object is kept on the principal axis of a diverging mirror of focal length 90 cm at a distance of 90 cm

from the pole. The object starts moving with velocity 2 cm/s towards the mirror at angle 30º with the

principal axis. If speed of the image (in cm/s) is x

y, find x – y. Here x and y are least possible

positive integers.

90 cm Qksdl nwjh ds ,d vilkjh niZ.k dh eq[; v{k ij niZ.k ds /kqzo ls 90 cm dh nwjh ij ,d oLrq j[kh gqbZ gSA oLrq niZ.k dh vksj eq[; v{k ls 30° dk dks.k cukrs gq, 2 cm/s ds osx ls xfr djuk izkjEHk djrh gSA ;fn

izfrfcEc dh pky cm/s esa x

y gS, rks x + y Kkr dhft,A ;gk¡ x o y U;wure lEHko /kukRed iw.kk±d gSA

Ans. –9

Sol. m = f 90 1

= = f – u 90 – (–90) 2

vix = m2v0x = – 3

4 cm/s

viy = mv0y = 1

2cm/s

vi = 7 x

v v = = 16 y

2 2

ix iy

33. A small object stuck on the surface of a glass sphere (n = 1.5) is viewed from the diametrically opposite

position. Find transverse magnification.

dk¡p ds xksys (n = 1.5) ds i"B ij fLFkr NksVh oLrq dks O;klr% foijhr fLFkfr ls ns[kk tkrk gSA vuqizLFk vko/kZu Kkr dhft,

Ans. 3

®




Sol. 2 1n n

v v = 2 1n n

R

1 1.5

v 2R

= 1 1.5

R

= –4R

m = 1

2

n v

n u =

1.5 ( 4R)

1 (2R)

m = 3. 34. In the figure shown a converging lens and a diverging lens of focal lengths 20 cm and –5 cm respectively

are separated by a distance such that a parallel beam of light of intensity 100 watt/m2 incident on the converging lens comes out as parallel beam from the diverging lens. The intensity of the outgoing beam

in kilowatt/m2 is 5

x, find x.

n'kkZ;s x;s fp=k esa Øe'k% 20 cm o –5 cm Qksdl nwjh ds ,d vfHklkjh ySUl o ,d vilkjh ySUl dqN nwjh ij bl çdkj j[ks gS fd 100 watt/m2 rhozrk dk lekUrj çdk'k iqat vfHklkjh ySUl ij vkifrr gksrk gS vkSj vilkjh ySUl ls lekUrj çdk'k iqUt ds :i esa ckgj fudyrk gSA fuxZr iqUt dh rhozrk watt/m2 esa Kkr djksA

Ans. 8

Sol.

( )o ut

=

2

0

2

0

100 ( r )

r

4

= 1600 W/m 2

35. A solid sphere of mass 10 kg is placed over two smooth inclined planes as shown in figure. The normal

reactions at 2 is 10x N. Find x (g = 10 m/s2)

10 kg nzO;eku dk ,d Bksl xksyk fp=kkuqlkj nks fpdus urryksa ij j[kk gSA 2 ij vfHkyEc çfrfØ;k 10x N gSA x

Kkr djksA (g = 10 m/s2)

2 160° 30°

Ans. 5 Sol.

®




2 160° 30°

30°60°

N2N1

N sin 60°2

N sin 30°1

N cos 60°2

N cos 30°1

N1 sin 30° = N2 sin 60°

N1 cos 30° + N2 cos 60° = mg

Solving above equation mijksDr lehdj.k dks gy djus ij

N2 = mg 10 10

502 2

36. A stone is thrown from the top of a tower at an angle of 30º up with the horizontal with a velocity of 16

m/s. After 4 sec. of flight it strikes the ground. Let the height of tower is H.meter. Then the value of H

12

is: (g = 10 m/s)

,d ehukj ds 'kh"kZ ls ,d iRFkj dks {kSfrt 30º dks.k cukrs gq, Åij dh vksj 16 m/s ds osx ls Qsadrs gSA 4 sec.

mM~M;udky ds i'pkr~ ;g lrg ls Vdjkrk gSA ekuk nhokj dh Å¡pkbZ H ehVj gSA rc H

12 dk eku gS : (g = 10

m/s) Ans. 4

Sol.

–H = 16 sin 30º × 4 – 1

2 (10) 42

H = 48 meter.

1

®

Course : 01JR (ACT-1) Test Date : 10-10-2021 Test Type : (JEE ADVANCED PATTERN) Target Date : 00.00.2021 Paper 1 Time Duration : Paper 2 Time Duration :

01JR, 01JRH, 08JD, 05JDH1 & 08JDH1 (PAPER LEVEL)

SYLLABUS : Mole Concept, Quantum Mechanical model of atom (QMM), Periodic

Table, Real Gases, Chemical Bonding (till taught)

SYLLABUS : IUPAC Nomenclature & Structural isomerism, Structural

Identification & POC-I, GOC-I, GOC-II (upto Basic strength of organic compounds)

S.No. Subject Nature of Questions No. of Questions Marks Partial Marks Negative Total

1 to 6 MCQ (Partial Marking) 6 4 +1 –2 24

7 to 14 Integer (Double digit) 8 3 - 0 24

15 to 18 Comprehension SCQ (2 Com. × 2 Q.) 4 3 - –1 12







54 180

S.No. Subject Nature of Questions No. of Questions Marks Partial Marks Negative Total

1 to 2 Match listing 2 4 - –1 8


7 to 18 Single digit integer with positive/negat 12 3 - –1 36







54 180

Chemistry

Total Total

Maths

Physics

Total Total

ACT-1 | PAPER-2 | PATTERN NO. 10-20

Physics

Chemistry

ACT-1 | PAPER-1 | PATTERN NO. 10-20

Maths

PAPER-2

SECTION 1 (Maximum Marks: 08)

This section contains TWO (02) questions.

Each question has TWO matching lists: LIST-I and LIST-II.

FOUR options are given representing matching of elements from LIST-I and LIST-II. ONLY ONE of these

four options corresponds to a correct matching.

For each question, choose the option corresponding to the correct matching.

For each question, marks will be awarded according to the following marking scheme:





bl [kaM esa nks (02) ç'u gSA izR;sd ç'u esa nks lqesyu lwfp;k¡ (matching lists) gS % lwph&I vkSj lwph&II

2

lwph–I vkSj lwph–II ds rRoksa ds lqesyuksa dks n'kkZrs gq, pkj fodYi fn, x, gSA bu pkj fodYiksa esa flQZ ,d fodYi gh

lgh lqesyu çnf'kZr djrk gSA

izR;sd iz'u ds fy, lgh lqesyu çnf'kZr djus okys fodYi dks pqusaA

izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk%&


'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A

_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

37. Match the following : List-I List-II

(P) For a Vander Waal's gas with b 0, a 0 (1) Z = 0.375 (Q) For a real gas at its Boyle's temperature in high pressure region (2) Z < 1 (R) For a Vander waal gas with very large molar volume (3) Z > 1 (S) For a Vander waal gas at the critical point (4) Z = 1

fuEu dks lqesfyr dhft, :

lwph–I lwph–I

(P) b0, a 0 ds lkFk ,d okUMj okWYl xSl ds fy, (1) Z = 0.375

(Q) mPp nkc {ks=k esa blds ckW;y rkieku ij ,d okLrfod xSl ds fy, (2) Z < 1

(R) vf/kd ogn~ eksyj vk;ru ds lkFk ,d okUMj okWYl xSl ds fy, (3) Z > 1

(S) ØkfUrd fcUnq ij ,d okUMj okWYl xSl ds fy, (4) Z = 1

Code :

dwV % P Q R S (A) 2 4 3 1 (B) 3 4 2 1 (C) 4 3 1 2 (D*) 2 3 4 1

38. Match the contents of List I with the contents of List II. LIST-I LIST-II

(P)

O

(1) Stablised by Hyperconjugation

(Q) (2) Destablised by + I effect

(R) (3) Stablised by + m effect

(S)

CH2H3C-

(4) Stablised by – m effect

lwph I dk lwph II ls feyku dhft;sA

lwph-I lwph-II

(P)

O (1) vfrla;qXeu izHkko }kjk LFkk;h

(Q) (2) + I izHkko }kjk vLFkk;h

(R) (3) + m izHkko }kjk LFkk;h

(S)

CH2H3C-

(4) – m izHkko }kjk LFkk;h

Code: (P) (Q) (R) (S) (A*) 3 4 1 2 (B) 1 2 4 3

3

(C) 2 3 1 4 (D) 4 3 2 1


This section contains FOUR (04) questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is(are) correct

For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are chosen

and both of which are correct.

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option.



For example, in a question, if (A),(B) and (D) are the ONLY three options corresponding to correct

answers, then :

Choosing ONLY (A),(B) and (D) will get +4 marks

Choosing ONLY (A) and (B) will get +2 marks

Choosing ONLY (A) and (D) will get +2 marks

Choosing ONLY (B) and (D) will get +2 marks

Choosing ONLY (A) will get +1 marks

Choosing ONLY (B) will get +1 marks

Choosing ONLY (D) will get +1 marks

Choosing no option (i.e. the question is unanswered) will get 0 marks, and

Choosing any other combination of options will get –1 mark

4






mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc % dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’ dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) pquus ij +1 vad feysaxs ’ dsoy fodYi (B) pquus ij +1 vad feysaxs ’ dsoy fodYi (D) pquus ij +1 vad feysaxs ’ dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj vU; fdlh fodYiksa ds la;kstu dks pquus ij –1 vad feysaxs ’

39. Which of the following orders are correct with respect to the properties mentioned alongwith : (A*) S2– < Cl– < K+ < Ca2+ (Ionisation energy) (B*) I– > Se2– > Br– > O2– > F– (Ionic radii) (C*) Cl > F > S > O (electron affinity) (D) Na < Mg < Al (Metallic character)

uhps of.kZr xq.k/keZ ds lanHkZ esa fuEu esa ls dkSuls Øe lgh gSa \ (A*) S2– < Cl– < K+ < Ca2+ (vk;uu ÅtkZ)

(B*) I– > Se2– > Br– > O2– > F– (vk;fud f=kT;k) (C*) Cl > F > S > O (bySDVªkWu ca/kqrk)

(D) Na < Mg < Al (/kkfRod vfHky{k.k) Sol. (A) S2– < Cl– < K+ < Ca2+ (Ionisation energy) as Z

eff increases.

(B) I– > Se2– > Br– > O2– > F– (Ionic radii) (C) Cl > F > S > O (electron affinity) (D) Na > Mg > Al (Metallic character)

(A) S2– < Cl– < K+ < Ca2+ (vk;uu ÅtkZ) lkFk gh Zeff

esa o`f) gksrh gSA (B) I– > Se2– > Br– > O2– > F– (vk;fud f=kT;k) (C) Cl > F > S > O (bySDVªkWu ca/kqrk) (D) Na > Mg > Al (/kkfRod vfHky{k.k) 40. Which of the following statement is/are correct regarding real gas ? (A*) In the vander Waals equation, ‘a’ signifies intermolecular attraction. (B*) At critical point attractive tendency dominate. (C) Below Boyle temperature gas shows negative deviation at all range of pressure. (D*) Boyle temperature of a gas depends upon its nature.

okLrfod xSl ds lanHkZ esa fuEu esa ls dkSulk@dkSuls dFku lgh gS@gSas \ (A*) ok.Mj okWy lehdj.k esa ‘a’ varjv.kqd vkd"kZ.k dks crkrk gSA

(B*) Økafrd fcUnq ij vkd"kZ.k izo`fr izHkkoh gksrh gSA (C) ckW;y rki ls de rki ij xSl nkc dh lHkh ijkl ij _.kkRed fopyu n'kkZrh gSA (D*) xSl dk ckW;y rki bldh izÑfr ij fuHkZj djrk gSA Sol. Below Boyle temperature gas shows negative deviation in low pressure region while in high pressure

region show positive deviation.

5

fuEu nkc {ks=k esa ckW;y rki ds uhps xSl _.kkRed fopyu n'kkZrh gSA tcfd mPp nkc {ks=k esa /kukRed fopyu n'kkZrh gSA

41. Which of the following is/are less basic (Kb) than (CH3)2NH.

fuEu esa ls dkSulk @dkSuls ;kSfxd (CH3)2NH dh rqyuk esa de {kkjh; (Kb) gksrs gS\

(A*) (B*) (C*) (D)

42. How many of the following carbocations can undergo rearrangement ?

fuEu esa ls fdrus dkcZ/kuk;uksa esa iquZfoU;kl lEHko gS \

(A*) (B) (C*) (D)

Sol. (A) (C)







Marking scheme :




[kaM 3 : (vf/kdre vad : 36) bl [kaM esa ckjg (12) iz'u gSA izR;sd iz'u dk mÙkj 0 ls 9 rd, nksuksa 'kkfey]ds chp dk ,d ,dy vadh; iw.kk±d gS tks /kukRed@_.kkRed gSA ;fn mÙkj 5 gS rc mÙkj pquus dk lgh rjhdk 5 gSA ;fn mÙkj –––5 gS rc mÙkj pquus dk lgh rjhdk –5 gSA ;fn mÙkj –0 gS rc mÙkj pquus dk lgh rjhdk 0 gSA vadu ;kstuk %&


43. Compressibility factor (Z) for N2 at – 23ºC and 25 atm pressure is 1.2. Calculate moles of N2 gas

required to fill a gas cylinder of 7 liter capacity under the given conditions : 1 11

Use R Latmmol K12

–23ºC rFkk 25 atm nkc ij] N2 ds fy,] lEihM~;rk xq.kkad (Z) 1.2 gSA rc nh xbZ ifjfLFkfr;ksa esa] ,d 7 yhVj {kerk okys xSl flys.Mj] dks Hkjus ds fy, vko';d N2 xSl ds eksyksa dh x.kuk dhft,A

1 11R Latmmol K

12

iz; qDr dhft,A

6

Ans. 7

Sol. Z =PV

nRT ;k or n =

PV

ZRT. =

25 77

11.2 250

12

44. For a gas 'X', the volume of molecules was negligible. Its Z vs m

1

TV graph was obtained as below. If

the value of its Vanderwaal's constant, a = x × 10–2 (L2 atm/mol2), find x (nearest integral value).

xSl 'X' ds fy, v.kqvksa dk vk;ru ux.; gSA blds fy, Z rFkk m

1

TV ds e/; xzkQ uhps fn;s vuqlkj izkIr gksrk

gSA ;fn blds okUMjokWy fu;rkad a dk eku x × 10–2 (L2 atm/mol2) gS rc x (fudVre iw.kk±d eku) dk eku Kkr dhft,A

Ans. 2

Sol. Z =m

a1

RTV

Slope (<ky) = a

R

a = 0.4

0.0821.64

= 0.02 = 2 × 10–2

45. An element has highest -ive electron gain enthalpy in periodic table. In its outer most shell total number

of electrons are x and it has total p-electrons y. What will be the value of y-x.

,d rRo tks vkoÙkZ lkj.kh esa –ve bysDVªkWu yfC/k ,sUFkSYih dk vf/kdre eku j[krk gS] blds lcls ckgjh dks'k esa] bysDVªkWuksa dh dqy la[;k dk eku x gS rFkk blesa mifLFkr dqy p-bysDVªkWu dk eku y gS] rks y-x dk eku Kkr dhft,saA

Ans. 4 Sol. The element is Cl, it has 7 electrons in its outermost shell (x). It has total p-electrons 11 because its

configuration is 1s2, 2s22p63s23p5. So, y – x = 4 11 – 7 = 4

gy % mijksDr rRo Cl gSA blds lcls ckgjh dks'k esa 7 bysDVªkWu mifLFkr gksrs gSA blesa dqy 11 p-bysDVªkWu mifLFkr gksrs gS] D;ksafd bldk bysDVªkWfud vfHkfoU;kl 1s2, 2s22p63s23p5 gSA

blfy,, y – x = 4

11 – 7 = 4

46. For an orbital the graph between (r ) and r (distance from nucleus) is :

,d d{kd ds fy, (r ) rFkk r (ukfHkd ls nwjh) ds e/; xzkQ gS&

7

Find the value of n + of the orbital.

d{kd ds n + dk eku Kkr dhft,A Ans. 1 Sol. It is 1s orbital.

;g 1s d{kd gSA 47. The number of completely filled orbitals in

29Cu which have atleast two nodes is/are :

29

Cu esa de ls de nks uksM j[kus okys iw.kZiwfjr d{kdksa dh la[;k gS@gSa % Ans. 9

Sol. 29

Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Nodes 0 1 1 2 2 2 3 (but incompletely filled) Orbitals 1 3 5

So, 9.

Sol. 29

Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1

uksM 0 1 1 2 2 2 3 (ysfdu viw.kZ Hkjs gS) d{kd 1 3 5

blfy,, 9.

48. In how many of the following species the central atoms have two lone pairs of electrons ?

fuEu esas ls fdruh Lih'kht esa] dsUnzh; ijek.kq ij nks ,dkdh bysDVªkWu ;qXe mifLFkr gSa \

XeF4 ClF

3 F

2SeO

2

XeF3

+ NH2

– ClOF3

ICl4– SCl

2 XeOF

2

Ans. 7

Sol. XeF4

F

ClF3

F2SeO

2 XeF

3+

Xe – F|

|

F

F

+

NH2

– ClOF3

8

ICl4–

Cl Cl

I

ClCl–

SCl2

SCl Cl

XeOF2

49. 2M of 100 ml Na2 SO4 is mixed with 3M of 100 ml NaCl solution and 1M of 200 ml CaCl2 solution. Then

the ratio of the concentration of cation and anion.

2M Na2 SO4 ds 100 ml dks] 3M NaCl ds 100 ml rFkk 1M CaCl2 ds 200 ml foy;u esa feykrs gS rks /kuk;u rFkk _.kk;u dh lkUnzrk dk vuqikr Kkr djksA

Ans. 1

Sol. Conc. of cation /kuk;u dh lkUnzrk = 400 300 200

400

conc. of anion _.kk;u dh lkUnzrk = 200 300 400

400

ratio of the conc. lkUnzrk dk vuqikr = 1

50. Write the number of carbocations which are more stable than 22CH CH CH

:

,sls dkcZ/kuk;uksa dh la[;k fyf[k, tks 22CH CH CH

ls vf/kd LFkk;h gS\

(i) (ii) (iii) (iv)

(v) (vi)

(vii) (viii) 3 2CH CH

(ix)

Ans. 4 Sol 4 (i, iii, v, vii)

51. How many electrons or p-electrons are present in conjugation in the following compound.

fdrus bysDVªkWu ;k p-bysDVªkWu fuEu ;kSfxd esa la;qXeu esa mifLFkr gSa \

N

NH2

Ans. 8 Sol. In the given compound lone pair present on double bonded nitrogen atom do not take part in conjugation.

Sol. fn;s x;s ;kSfxd esa ukbVªkstu ijek.kq ds f}ca/k ij mifLFkr ,dkdh ;qXe la;qXeu esa Hkkx ugha ysrk gSA 52. How many groups (each attached with benzene ring) show + m effect?

fuEu esa ls fdrus lewg (tks csUthu oy; ls tqM+s gSa) + m izHkko n'kkZrs gSa\

9

N(CH3)2

SH

C–OCH3

O

O–C–CH3

O

O–S–OH

O

CN

NHC–CH3

O

CN

CH3

CH3

Cl

Ans. 6

Sol.

N(CH3)2

SH

O–C–CH3

O

NHC–CH3

O

Cl

O–S–OH

O

have + m group.

;s lHkh + m lewg j[krs gSaA 53. How many structural alcohol of formula C5H12O can give instant turbidity with Lucas reagent

(HCl + ZnCl2)

lw=k C5H12O ds fdrus lajpukRed ,YdksgkWy yqdkl vfHkdeZd (HCl + ZnCl2) ds lkFk rhoz iafdyrk ¼turbidity½ ns ldrs gS \

Ans. 1 Sol. 2-methylbutan-2-ol. 2-esfFkyC;wVsu-2-vkWy 54. Observe the following molecule and write the total number of hydrogen atoms involved in

hyperconjugation?

fuEUkfyf[kr v.kq dk izs{k.k djds crkb;s fd blesa vfrla;qXeu esa dqy fdrus gkbMªkstu ijek.kq lfEefyr gS\

Ans. 9

Sol.

Total number of hydrogen involved in hyperconjugation = 9

vfrla;qXeu esa Hkkx ysus okys dqy gkbMªkstu ijek.kq dh la[;k = 9

part : i mathematics

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