New Scientist - 23 April 2016

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TARGET : JEE (Main + Advanced) 2017

EST INFORMATIO

Course : VIKAAS(JA)

MMAATTHHEEMMAATTIICCSS

DPP DPP DPPDAILY PRACTICE PROBLEMS

NO. 85 TO 87

This DPP is to be discussed in the week (14-12-2015 to 19-12-2015)

DPP No. : 85 (JEE-Main)

Total Marks : 31 Max. Time : 32 min.Single choice Objective ('–1' negative marking) Q.1 to 9 (3 marks, 3 min.) [27, 27]Subjective Questions ('–1' negative marking) Q.10 (4 marks, 5 min.) [4, 5]

1. A five digits number of the form x y

z

y

x is choosen, probability that x <

y is

x y

z

y

x ds izdkj dh 5 vadksa dh la[;k pquh tkrh gS izkf;drk Kkr dhft, fd x <

y gksµ

(A)35

90 (B*)

6

15 (C)

19

45 (D)

13

30

Sol. Probability izkf;drk =360

9 10 10

2. If P(A) = 0.59, P(B) = 0.30, P(A B) = 0.21, then P(A B) is equal to ;fn P(A) = 0.59, P(B) = 0.30, P(A B) = 0.21, rc P(A B) dk eku gSµ(A) 0.79 (B) 0.11 (C*) 0.32 (D) 0.38

Sol. (A B) = 1 – P(A B) = 1 – (P(A) + P(B) – P(A B))= 1 – (0.59 + 0.30 – 0.21) = 1 – (0.59 + 0.09) = 1 – 0.68 = 0.32

3_. The algebraic sum of deviations of 10 observations measured from 15 is 7. The mean is

10 izs{k.kks a ds 15 ds lkis{k fopyuks a dk chth; ;ksx 7 gSA rks ek/; gksxk& (A) 105 (B) 70 (C*) 15.7 (D) none of these buesa ls dksbZ ugha

Sol. 10

i

i 1

x – 15 = 7 xi – 150 = 7 i

x

10

= 15.7

ANSWER KEY

DPP No. : 85 (JEE-Main)

1. (B) 2. (C) 3. (C) 4. (A) 5. (B) 6. (B)7. (A) 8. (B) 9. (A) 10. (1) 243 (2) 1, 10, 40, 80, 80, 32

DPP No. : 86 (JEE-Advanced)

1. (C) 2. (C) 3. (B,C) 4. (A,B,C)

5. (A,C) 6. (A,B) 7. (A,B,C,D) 8. 2


1. (A,B,C) 2. (A,B,C) 3. (A,B) 4. (A,B,C)5. (A,B,D) 6. 9 7. p = 2


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4_. If mean of squares of deviations of a set of n observations about –2 and 2 are 18 and 10 respectivelythen standard deviation of this set of observations is(A*) 3 (B) 2 (C) 1 (D) None of these

n izs{k.kksa ds ,d leqPp; ds – 2 ,oa 2 ds lkis{k fopyuks a ds oxksZa dk ek/; Øe'k% 18 ,oa 10 gS rc iz s{k.kks a ds bl leqPp; dk ekud fopyu gksxk & (A*) 3 (B) 2 (C) 1 (D) bues a ls dksbZ ugha

Sol.

2i(x 2)

n

= 18 and

2i(x – 2)

n

= 10 (x

i2 + 4 + 4x

i) =18n and (x

12 + 4 – 4x

i) = 10n

2( xi2)+ 8n = 28n and 8x

i = 8n

2ix

n

= 10 and

ix

n

= 1

S.D. = = 10–1 = 32 2i ix x –

n n

5_. The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallestone. Then the sides of the triangle are

,d f=kHkqt dh Hkqtk,sa rhu Øekxr izkd`frd la [;k,sa gS rFkk lcls cMk dks .k lcls Nks Vs dk nksxquk gS] rks f=kHkqt dhHkqtk,s a gSµ (A) 3, 4, 5 (B*) 4, 5, 6 (C) 2, 5, 3 (D) 4, 3, 7

Sol. Let the sides be n, n + 1, n + 2 cms. i.e. AC = n, AB = n + 1, BC = n + 2Smallest angle is B and the largest one is A

B n + 2

n + 1

A

C

n

A = 2B Also A + B + C = 180°

3B + C = 180° C = 180° – 3B

We havesinA sinB sinC

n 2 n n 1

sin 180 3Bsin2B sinBn 2 n n 1

sin2B sinB sin3B

n 2 n n 1

cosB =

n 2

2n

and

sin2B

n 2=

33sinB 4sin B

n 1

22cosB 4cos B 1

n 2 n 1

1

n=

2

2

n 21

nn 1

Hindi. ekuk fd Hkqtk,sa n, n + 1, n + 2 cms. vFkok AC = n, AB = n + 1, BC = n + 2 lcls NksVk dks .k B gS rFkk lcls cM+k A gSA

B n + 2

n + 1

A

C

n

A = 2B A + B + C = 180°

3B + C = 180° C = 180° – 3B

sinA sinB sinC

n 2 n n 1

sin 180 3Bsin2B sinB

n 2 n n 1


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sin2B sinB sin3B

n 2 n n 1

cosB =

n 2

2n

and

sin2B

n 2=

33sinB 4sin B

n 1

22cosB 4cos B 1

n 2 n 1

1

n=

2

2

n 21

nn 1

n = 4 vr% Hkqtk,sa 4, 5, 6 gSaA

6. Number of diagonals in sixteen sided regular polygon are(A) 16C2 (B*)

16C2 – 16 (C) 16 (D) none of these

lksyg Hkqtk okys lecgqHkqt esa fod.kks ± dh la [;k gS& (A) 16C2 (B*)

16C2 – 16 (C) 16 (D) bues a ls dksbZ ughaSol. Total number of sides = 16C2

number of diagonals = 16C2 – 16

7_. Which of the following is false?

(A*) (p q) (~q ~p) is a contradiction (B) p (~p) is a tautology(C) ~(~p) p is a tautology (D) p (~p) is a contradiction

fuEufyf[kr es a ls dkSulk vlR; gSµ

(A*) (p q) (~q ~p) ,d fojks/kkHkkl gSA (B) p (~p) ,d iqu:fDr gSA (C) ~(~p) p ,d iqu:fDr gSA (D) p (~p) ,d fojks/kkHkkl gSA Sol. p q is logically equivalent to ~q ~p, therefore, (p q) (~q ~p) is a tautology but not a

ontradictionSo (A) is false

Hindi. p q rkfdZd :i ls ~q ~p ds rqY; gS vr% (p q) (~q ~p) ,d iqu:fDr gS ys fdu fojks/kkHkkl ugh gSA vr% (A) vlR; gSA

8_. If p : Ashok works hard ; q : Ashok gets good grade

The verbal form for (~p q) is ;fn p : v'kksd dfBu ifjJe djrk gSA ; q : v'kksd vPNh xzs M izkIr djrk gSA

(~p q) dk ekSf[kd :i gSµ (A) If Ashok works hard then gets good grade(B*) If Ashok does not work hard then he gets good grade(C) If Ashok does not work hard then he does not get good grade(D) Ashok works hard if and only if he gets grade

(A) ;fn v'kksd dfBu ifjJe djrk gS rks og vPNh xsz M izkIr djrk gSA (B*) ;fn v'kksd dfBu ifjJe ugha djrk gS rks og vPNh xzsM izkIr dj ldrk gSA (C) ;fn v'kksd dfBu ifjJe ugha djrk gS rks og vPNh xzsM izkIr ugha djrk gSA (D) v'kksd dfBu ifjJe djrk gS ;fn vkS j ds oy ;fn og vPNh xszM izkIr djrk gSA

Sol. ~p : Ashok does not work hard

Use '' symbol for then(~p q) mean = If Ashok does not work hard then he gets good grade

Hindi. ~p :v'kksd dfBu ifjJe ugha djrk gSA

'' fpUg dk mi;ksx djus ij(~p q) dk vFkZ = ;fn v'kksd dfBu ifjJe ugha djrk gS rks og vPNh xzsM izkIr djrk gSA

9_. If p : Ram is tall

q : Ram is intelligent

then the symbolic statement ~p q means ;fn p : jke yEck gSA

q : jke cq f)eku gSA rks dFku ~p q dk vFkZ gSµ (A*) Ram is not tall or he is intelligent (B) Ram is tall or he is intelligent(C) Ram is not tall and he is intelligent (D) Ram is not tall then he is intelligent


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(A*) jke yEck ugha gS vkS j og cq f)eku gSA (B) jke yEck gS ;k og cqf)eku gSA (C) jke yEck ugha gS vkS j og cq f)eku gSA (D) jke yEck ugha gS rc og cqf)eku gSA

Sol. ~p : Ram is not tall Use '' symbol for orSo ~p q means = Ram is not tall or he is intelligent

Hindi. ~p : jke yEck ugh gS '' dk iz ;ksx ' ;k' ds fy;s djus ijvr% ~p q dk vFkZ gS % jke yEck ugh gS ;k og cqf)eku gSA

10. A forecast is to be made of the results of five cricket matches, each of which can be a win, a draw or aloss for Indian team. Find(1) the number of different possible forecasts.(2) the number of forecasts contianing 0, 1, 2, 3, 4, & 5 errors respectively.

ik¡p fØds V eSpksa ds fy, iwokZuqeku yxk;k tkrk gS ftles a Hkkjrh; Vhe ds fy, izR;sd eSp thrk gqvk] gkjk gqvk ;k u thrk u gkjk gqvk ¼Mª kW½ gks ldrk gSA Kkr dhft,&(1) dqy laHko fofHkUu iwokZuqekuksa dh la [;kA(2) iwokZuqekuks a dh la [;k ftuesa Øe'k% 0, 1, 2, 3, 4 rFkk 5 xyfr;k¡ gksA

Ans. (1) 243 (2) 1, 10, 40, 80, 80, 32Sol. (1) 35

(2) The number of forecasts containing 0, 1, 2, 3, 4 and 5 errors

respectively 1, 5C1 (2)1, 5C2 (2)2 , 5C3 23 , 5C4 24 , 25


Total Marks : 35 Max. Time : 31 min.Single choice Objective ('–1' negative marking) Q.1, 2 (3 marks, 3 min.) [6, 6]Multiple choice objective ('–1' negative marking) Q.3 to 7 (5 marks, 4 min.) [25, 20]Subjective Questions ('–1' negative marking) Q. 8 (4 marks, 5 min.) [4, 5]

1. The number of non negative integral solution of the equation, x + y + 3z = 33 is:

lehdj. k x + y + 3z = 33 ds v_.kkRed iw .kk± d gyks a dh la [;k gS&

(A) 120 B) 135 (C*) 210 (D) 520Sol. x + y + 3z = 33

x + y = 33 – 3z

11

34 3

1

0

z

x

C

34 C1 + 31 + 28 ......... 1

6[1 + 34] = 210

2. Two non-negative integers are chosen at random, then the probability that the sum of their squares isdivisible by 5 is

nks v_.kkRed iw .kk±d ;kn`fPNd :i ls pqus tkrs gSA rks muds oxks aZ ds ;ksxQy ds 5 ls foHkkftr gksus dh izkf;drk

gS &

(A)7

25 (B)

8

25 (C*)

9

25 (D)

5

25

Sol. Let two non-negative number x and y

x = 5a + , y = 5b + where 0 4 and 0 4Now x2 + y2 = (5a + )2 + (5b + )2 = 25(a2 + b2) + 10 (a + b) + 2 + 2 x2 + y2 is divisible by 5 if and only if2 + 2 is divisible by 5total ways of choosing and = 5 × 5 = 25and 2 + 2 will be divisible by 5 if(, ) {(0, 0), (1, 2), (1, 3), (2, 1), (2, 4), (3, 1), (3, 4), (4, 2), (4, 3)}


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Favarable cases = 9

Probability =9

25

Sol. ekuk nks v_.kkRed la [;k,s a x rFkk yx = 5a + , y = 5b +

tgk¡ 0 4 rFkk 0 4vc x2 + y2 = (5a + )2 + (5b + )2 = 25(a2 + b2) + 10 (a + b) + 2 + 2

x2

+ y2

, 5 ls foHkkftr gksxk ;fn vkSj ds oy ;fn 2 + 2 , 5 ls foHkkftr gS rFkk dks pquus ds dqy rjhds = 5 × 5 = 25vkS j 2 + 2 , 5 ls foftr gksxk ;fn (, ) {(0, 0), (1, 2), (1, 3), (2, 1), (2, 4), (3, 1), (3, 4), (4, 2), (4, 3)}vuqdqy fLFkfr;k¡ = 9

izkf;drk =9

25

3. If in a triangle ABC,cos A 2cosC

cosA 2cosB

=

sinB

sinC then the triangle ABC can be

,d f=kHkqt ABC es a cos A 2cosCcosA 2cosB

= sinBsinC

gks] rks f=kHkqt ABC gks ldrk gSA

(A) right angle (B*) isosceles (C*) equilateral (D) all above

(A) ledks .k (B*) lef}ckgq (C*) leckgq (D) mijksDr lHkh Sol. cos A(sin B – sin C) + (sin 2B – sin 2C) = 0

cos A.(sin B – sin C) + 2 cos(B + C) sin (B – C) = 0 B + C = – A

cos A.(sin B – sin C) – 2 cos A.sin (B – C) = 0 cos A[(sin B – sin C) – 2(sin B cos C– cos Bsin C)] = 0 either ;k rks cos A = 0 A = 90° right angled ledks .k f=kHkqt gksxk or ;k (sin B – sin C) – 2(sin B cos C – cos B sin C) = 0

(b – c) – 2

2 2 2 2 2 2a b c a c bb. c.2ab 2ac

= 0

a(b – c) – 2(b2 – c2) = 0(b – c) [a – 2(b + c)] = 0

b – c = 0 b = c isosceles vr% f=kHkq t lef}ckgq f=kHkq t gksxkA

4. Number of ways in which 5 colours be selected out of 8 different colours including red, blue and green

(A*) if blue and green are always to be included is 20(B*) if red is always excluded is 21(C*) if red & blue are always included but green excluded is 10(D) None of these

8 fofH kUu j axk s a ] ftues a yky] uhyk rFk k gjk 'kkf ey gS ] e s a l s 5 j axk s a dks pquu s ds rjhdks a dh la[; kµ

(A*) 20 gS ;fn uhy k rFk k gj k ges 'kk 'kk fey fd;k tk; sA (B*) 21 gS ;fn yky ges 'kk 'kkf ey ug h a fd;k tk; sA(C*) 10 g S ;fn yky rFk k uhyk ges 'kk 'kk fey fd;k tk; s fdUrq gj k 'kk fey ug h a fd;k tk; sA (D) bues a ls dks bZ ug ha

Sol. (1) Total ways dqy rjhd s = 6C3 = 20(2) Total ways dqy rjhd s = 7C5 = 21(3) Total ways dqy rjhd s = 5C3 = 10

5. There are two groups of subjects one of which consists of 5 science subjects and 3 engineeringsubjects and the other consists of 3 science and 5 engineering subjects. An unbiased die is cast. Ifnumber 3 or number 5 turns up, a subject is selected at random from the first group, otherwise the


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8. There are N + 1 identical boxes each containing N wall clocks. r th box contains (r – 1) defective and

(Nr+1) effective clocks for 1 r N + 1. A box is selected at random and from this box a clock is

chosen at random and is found to be effective. the probability that it is from kth box is2

2 N 2 k

N N

,

then find .N nhokj ?kfM+ ;ksa okys N + 1 ckWDl gSA 1 r N + 1 ds fy, r osa ckWDl es a (r – 1) nks"k;qDr ,oa (Nr+1) nks"kjfgr

?kfM+ ;k¡ gSA ;kn`PN;k ,d ckWDl dk p;u djds blesa ls ;kn`fPNd :Ik ls ,d ?kM+h dk p;u fd;k tkrk gS vkS j ik;k

tkrk gS fd ;g nks"kjfgr gSA bl nhokj ?kM+h ds k osa ckWDl esa ls gks us dh çkf;drk2

2 N 2 kN N

gS] rks dk

eku Kkr dhft,A Ans. 2

Sol. Prob. of selection of any box is ,d cDls ds p;u dh izkf;drk =1

N 1

Let E be the event that the wall clock selected is effective then

ekukfd p;fur ?kM+ h ds nks"kjfgr gksus dh ?kVuk E gS] rks P(E) = P(B

1) P(E/B

1) + P(B

2) P(E/B

2) + .......... + P(B

N + 1) P(E/B

N + 1)

=1 N 1 N 2 1

1 ......... 0(N 1) N N N

=1 2 .......... N 1

N (N 1) 2

P (Bk/ E) = k k

1 (N K 1)P(B ).P(E/B ) N 1 N

P(E) 1/ 2

=

2

2N 2K 2

N N

.


Total Marks : 37 Max. Time : 35 min.Multiple choice objective ('–1' negative marking) Q.1 to Q.5 (5 marks, 4 min.) [25, 20]Subjective Questions ('–1' negative marking) Q.6 to Q.8 (4 marks, 5 min.) [12, 15]

1. Different words are formed by arranging the letters of the word ''SUCCESS'', then(A*) The number of words in which C are together but S's are seperated, is 24(B*) The number of words in which no two C's and no two S's are together is 96(C*) The number of words in which the consonants appear in alphabetic order is 42(D) None of these

'kCn ' 'SUCCESS'' ds v{ kj ks a dk s O;ofL Fkr djds fH kUu&fHk Uu 'kCn cuk; s tkr s g S ] rk s Kkr dhft,&(A*) 'kCnk s a dh la [; k ftues a C lkF k&lkF k ys fdu S vyx&vyx g S ] 24 gk sxh&(B*) 'kCnk s a dh la [; k ftues a dk sbZ Hk h nks C rFk k dks bZ Hkh nk s S lkFk ug ha g S a ] 96 gk sxh& (C*) 'kCn ks a dh la[;k ftues a O; atu ] o. kZ ekyk ds Øe es a g S ] 42 gk sxh& (D) bues a l s dks bZ ugha A

Sol. (A) 3! 4C3 = 24

(B) Required permutation = All S’s are seperated – All S’s are seperated but C’stogether

= 5C3 × 4!

2! – 4C

3 × 3! = 120 – 24 = 96

(B) vHkh"V Øep; = lHkh S i` Fkd – lHkh S i` Fkd fdUr q C lkFk- lkFk

= 5C3 ×

4!

2! – 4C

3 × 3! = 120 – 24 = 96

(C) Required words vHkh"V 'kCn = 7C5 × 2! = 21 × 2 = 42

2. If two events A and B are such that P(Ac) = 0.3. P(B) = 0.4 and P(A Bc) = 0.5, then nks ?kVuk,a A vkS j B bl izdkj gS fd P(Ac) = 0.3. P(B) = 0.4 rFkk P(A Bc) = 0.5 rc&(A*) P(A BC) = 0.8 (B*) P[B (A BC)] = 0.2(C*) P(B/A Bc) = 0.25 (D) P(B/A Bc) = 0.3


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Sol. P(B/A Bc) =c

c

P[B (A B )]

P(A B )

P[B (A BC)] = P(A B) = P(A) – P(A Bc) = 0.7 – 0.5 = 0.2 Again iqu% P(A Bc) = P(A) + P(Bc) – P(A Bc) = 0.7 + (1 – 0.4) – 0.5 = 0.8

P(B/A Bc) =0.2

0.8 = 0.25

3. The equation of the line passing through the point (2, 3) & making intercept of length2 units between the lines y + 2x = 3 & y + 2x = 5 is

ml ljy js [kk dk lehdj.k Kkr dhft, tks fcUnq (2, 3) ls xqtjrh gS rFkk js [kkvksa y + 2x = 3 ,oa y + 2x = 5 ds e/; 2 bZdkbZ yEckbZ dk vUr[k.M% dkVrh gS] gksxk&

(A*) 3x + 4y = 18 (B*) x = 2 (C) 3x – 4y = 18 (D) x = – 2

Sol. Distance of between lines =2

5

2

5

A

B

2

y + 2x – 3 = 0

y + 2x – 5 = 0

Equation of line AB y = mx + 3 – 2m

point A2m 6 m

,m 2 m 2

& B

22m 2 2m

m 2 m 2

2

2 2

4m 4

(m 2) (m 2)

= 4

4(m2+1) = 4(m+2)2 m2 + 1 = m2 +4 + 4m

4m = –3 m = –3

4

& x = 2

y = –3

4x + 3 – 2

3

4

y =3

4 –x + 3 +

3

2

3x + 4y = 18

4. If each root of x2 – ax – b = 0, a, b R has absolute value < 1, then(A*) |b| < 1 (B*) a + b < 1 (C*) b – a < 1 (D) a – b < 1

;fn lehdj.k x2 – ax – b = 0, a, b R ds izR;sd ewy dk fujis{k eku < 1 gks] rks&(A*) |b| < 1 (B*) a + b < 1 (C*) b – a < 1 (D) a – b < 1

Sol. = – b || = |–b| < 1 |b| < 1f(x) = x2 – ax – b

f(1) > 0 a + b < 1f(–1) > 0 b – a < 1

5. N fair coins are flipped once. The probability that at most 2 of the coins show up as heads is1

2 then

the value of N is not

N fu"i{kikrh flDdks a dks ,d ckj Qsa dk tkrk gSA vf/kd ls vf/kd 2 flDdks a ij 'kh"kZ vkus dh izkf;drk1

2 gSA rc

N dk eku ugha gS&


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(A*) 1 (B*) 3 (C) 5 (D*) 7

Sol.

no

y

32

5

y = 2n

y = n +n+22

P(0 heads) =

N1

2

P(1 head) =

N 1N1 1

·12 2

= N

N1

2

P(2 head) =

2 N 2N1 1

·22 2

=

NN N 1 12 2

P (at most 2 heads) = P(0 heads) + P(1 head) + P(2 heads)

1

2=

N N NN N 11 1 1N

2 2 2 2

N2 + N + 2 = 2N N = 5

Hindi.

no

y

32

5

y = 2n

y = n +n+22

P(0 'kh"kZ) =N

1

2

P(1 'kh"kZ) =N 1

N1 1·

12 2

= N

N1

2

P(2 'kh"kZ) =2 N 2

N1 1·

22 2

=

NN N 1 12 2

P (vf/kd ls vf/kd 2 'kh"kZ) = P(0 'kh"kZ) + P(1 'kh"kZ) + P(2 'kh"kZ)

1

2=

N N NN N 11 1 1N

2 2 2 2

N2 + N + 2 = 2N N = 5

6. A triangle ABC satisfies the relation 2sec4C + sin22A + sinB = 0 and a point P is taken on the longest

side of the triangle such that it divides the side in the ratio 1 : 3. Let Q and R be the circumcenter and

orthocenter of ABC. If PQ : QR : RP = 1 : : , then find the value of 2 + 2.Ans. 9

,d f=kHkqt ABC lEcU/k 2sec4C + sin22A + sinB = 0 dks larq"V djrk gS rFkk f=kHkqt dh lcls cM+h Hkqtk ij ,d fcUnq P bl izdkj fy;k tkrk gS fd ;g Hkqtk dks 1 : 3 ds vuqikr esa foHkkftr djrk gSA ekukfd Q rFkk R f=kHkqt ABC

ds ifjdsUnz rFkk yEcdsUnz gSA ;fn PQ : QR : RP = 1 : :

gks] rks 2 + 2

dk eku Kkr dhft,A


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Ans. 9

Sol.

2 sec 4C + sin22A + sinB = 0

A = 45º, B = 90º and C = 45º

Let AQ = a, then PQ =a

2 and QR = a

PR =2

2 aa4

=5a

2

1 : : =a

2 : a :

5a

2= 1 : 2 : 5

= 2 and = 5 2 + 2 = 9

Hindi

2 sec 4C + sin22A + sinB= 0

A = 45º, B = 90º rFkk C = 45º

ekukfd AQ = a gks] rks PQ =a

2 rFkk QR = a

PR =2

2 aa4

= 5a2

1 : : =a

2: a :

5a

2= 1 : 2 : 5

= 2 rFkk = 5 2 + 2 = 9

7. Find the smallest positive integer 'p' for which the equation cos (p sinx) = sin (p cosx) has a solution

in [0, 2].Ans. p = 2

U;wure /kukRed iw .kk±d p dk eku Kkr dhft, ftlds fy, lehdj.k cos (p sinx) = sin (p cosx) dk ,d gy

vUrjky [0, 2] esa gksA Ans. p = 2

Sol. Here cos (p sin x) = cos pcosx2

p sin x = 2n ± pcosx2

p(sin x ± cos x) = 2n 2

=

(4n 1)

2

or sin x ± cos x =(4n 1)

2p


11/12




2 sin x4

=

(4n 1)

2p

or sin x4

=

(4n 1)

2 2 p

............(i)

As |sin | 1 (i) will have solution if4n 1

p

2 2

1

or |p| |4n ± 1| .2 2

where n Z

the smallest positive value of p is2 2

and the smallest positive integral values of p is 2.

Hindi ;gk¡ cos (p sin x) = cos pcosx2

p sin x = 2n ± pcos x2

p(sin x ± cos x) = 2n 2

=

(4n 1)

2

;k sin x ± cos x =(4n 1)

2p

2 sin x4

=

(4n 1)

2p

;k sin x4

=

(4n 1)

2 2 p

............(i)

pw¡fd |sin | 1 lehdj.k (i) dk gy gksxk ;fn4n 1

p

2 2

1

;k |p| |4n ± 1| .

2 2

tgk¡ n Z

p dk U;wure /kukRed eku2 2

gSA

rFkk p dk U;w ure /kukRed iw .kk±d eku 2 gSA

8. In any ABC, prove that f=kHkqt ABC esa fl) dhft, fd &

(i) (r 3+ r

1) (r

3+ r

2) sin C = 2 r

3 2 3 3 1 1 2r r r r r r

(ii) A B2 2 2

Ctan tan tan 1

(a b)(a c) (b a) (b c) (c a) (c b)

(iii) (r + r 1) tan B C2 + (r + r 2) tan C A

2 + (r + r 3) tan A B

2 = 0

(iv) 2 2 2 21 2 3r r r r = 16R2 – a2 –b2 – c2.

Sol. (i) L.H.S. ck;ka i{k = (r 3+ r

1) (r

3+ r

2) sin C

=b

(s–a)(s–c)

a

(s–c)(s–b)

sin C =

2ab

(s– a)(s– b)(s – c)(s – c)

sin C

=ab s(s– a)(s– b)(s –c )

(s – a)(s –b)(s– c)(s – c) sin C =

abssinC

s – c =

2 .s

(s–c )

= 2 sr

3

R.H.S. nk;ka i{k = 2r 3 2 3 3 1 1 2r r r r r r = 2r 3

2s = 2sr

3

L.H.S. ck;ka i{k = R.H.S. nk;ka i{k


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(ii) L.H.S. ck;ka i{k = –1

(s – b)(s – c) (s – a)(s – c) (s – a)(s – b)

(a – b)(c – a) (a – b)(b – c) (c – a)(b – c)

= –1

(s – b)(s – c )(b – c) (s – a)(s – c )(c – a) (s – a)(s – b)(a – b)

(a – b)(b – c)(c – a)

=1

= R.H.S. nk;ka i{k

(iii) First term ( çFke in) = (r + r 1) tan

B – C

2

=b – c

s s – a b c

cot

A

2 =(2s – a)

s(s– a)

.

b – c

b c

.

s(s– a)

(s–b)(s–c) = b – c

similarly second term ( blh çdkj f}rh; in)= c – a & third term (vkS j r`rh; in) = a – b L.H.S. ck;ka i{k = b – c + c – a + a – b = 0 = R.H.S. nk;ka i{k

(iv) r 1 + r

2 + r

3 – r = 4R

(r 1 + r

2 + r

3 – r)2 = r

12 + r

22 + r

32 + r 2 – 2r (r

1 + r

2 + r

3) + 2(r

1r

2 + r

2r 3 + r

3 r

1) ........(i)

r(r 1 + r

2 + r

3) = ab + bc + ca – s2 and (vkS j) r

1r

2 + r

2r

3 + r

3r

1 = s2

from equation (i) lehdj.k (i) s ls 16R2 = r 2 + r

12 + r

22 + r

32 – 2 (ab + bc + ca – s2) + 2s2

r 2 + r 12 + r

22 + r

32 = 16 R2 – 4 s2 + 2 (ab + bc + ca)

= 16R2 – (a + b + c)2 + 2 (ab + bc + ca) = 16R2 – a2 – b2 – c2

New Scientist - 23 April 2016

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Transcript of New Scientist - 23 April 2016