(1031)dpp_11_matrices_determinant_and_trigonometry_b.pdf.tmp.pdf

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DATE : 11.05.2016 Part Test - 05 (PT-05)

S llabus : Permutation and Combination & Probabilit

TARGET : JEE (Main + Advanced) 2016

E

EST

INFORMATIO

Course : VIJETA(ADP) & VIJAY(ADR) Date : 11-05-2016

MMAATTHHEEMMAATTIICCSS

DPPDPPDPPDAILY PRACTICE PROBLEMS

NO. 11

TTEESSTT IINNFFOORRMMAATTIIOONN

Revision DPP of Matrices & Determinant, Trigonometry And Miscellaneous

Total Marks : 111 Max. Time : 125 min.Single choice Objective ('–1' negative marking) Q.1 to Q.15 (3 marks 3 min.) [45, 45]Multiple choice objective ('–1' negative marking) Q.16 to Q.31 (4 marks 3 min.) [34, 48]Subjective Questions ('–1' negative marking) Q.32 to Q.34 (3 marks 3 min.) [9, 9 ]Match the Following (no negative marking) (2 × 4) 35 (8 marks 8 min.) [8, 8]Comprehension ('–1' negative marking) Q.36 to Q.40 (3 marks 3 min.) [15, 15]

1_. Value ofcos61

1

cos1

.

cos621

cos2

.

cos631

cos3

………

cos1191

cos59

is.

cos611

cos1

.cos62

1cos2

.cos63

1cos3

………cos119

1cos59

dk eku gS

(A) –1 (B*) 1 (C) 2 (D) –2

Sol. cosA – cosB = –2 A B

sin2

. A B

sin2

cos(60 k)

1cosk

=

cosk cos(60 k)

cosk

=

sin(30 k)

cosk

ANSWERKEY OF DPP # 11 1. (B) 2. (A) 3. (C) 4. (C) 5. (C) 6. (C)

7. (A) 8. (A) 9. (D) 10. (D) 11. (A) 12. (C)

13. (B) 14. (D) 15. (A) 16_. (BC) 17. (ABD) 18. (AB)

19. (ABC) 20. (AB) 21. (ABCD) 22. (AC) 23. (ABC) 24. (BD)

25. (AD) 26. (ABD) 27. (ABC) 28. (BCD) 29. (BCD) 30. (BC)

31. (ACD) 32. 3 33. 81 34. 6

35. (AP,Q); (BS); (CP,R); (D R) 36. (C) 37. (D) 38. (B)

39. (C) 40. (A)


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cos61

1cos1

cos62

1cos2

………….cos119

1cos59

=59

k 1

sin(30 k)

cosk

=59

k 1

cos(60 k)

cosk

= 1

2_. Value of cot40o . cot20

o ( 4sin10

o –1) is

cot40o . cot20

o ( 4sin10

o –1) dk eku gS

(A*) –1 (B) 1 (C) 2 (D) 0

Sol.o o o

o o

cos40 .cos20 .(4sin10 1)

sin40 sin20

= –

o o o o o

o o

cos 40 .cos 20 4 sin10 cos 20 cos 40

sin40 sin20

= –o o o o o

o o

cos 40 .cos 20 2(sin30 sin10 ) cos 40

sin40 sin20

= –o o o o o

o o

cos 40 .cos 20 cos 40 2 sin10 .cos 40

sin40 sin20

= –o o o o o

o o

cos 40 .cos20 cos 40 sin50 sin30

sin40 sin20

= –o o o

o o

cos 40 .cos 20 sin 30

sin40 sin20

= –o o

o o

cos 40 .cos 20 cos 60

sin40 sin20

= –o o o o

o o

cos 40 .cos20 cos(40 20 )

sin40 sin20

= –1

3_. Value of sin + 2sin2 + 3sin3+ 4sin4 + …….+ nsin n issin + 2sin2 + 3sin3+ 4sin4 + …….+ nsin n dk eku gS

(A)2

(n 1)cosn ncos(n 1)

4sin2

(B)2

(n 1)cosn ncos(n 1)

4cos2

(C*)2

(n 1)sinn nsin(n 1)

4sin 2

(D)2

(n 1)sinn nsin(n 1)

4cos 2

Sol. Consider the series, ekuk fd Js .kh

cos + cos2 cos3 +…………..+ cosn =sin(n /2)

sin / 2

.n 1

cos2

Differentiate both sides w.r.t. '' nksuks rjQ '' ds lkis{k vodyu djus ij –sin – 2sin2 – 3sin3…. –n sinn

2

n n n 1 n 1 n 1 n n n 1 1sin cos .cos sin .sin sin .cos . cos

2 2 2 2 2 2 2 2 2 2 2

sin / 2

=2

2n 1 n n2nsin / 2.cos 2sin .cos2 2 2

4sin / 2

sin + 2 sin 2 + 3 sin 3 ……….+ n sin n 2

(n 1)sinn nsin(n 1)

4sin / 2


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4_. If f(x) = a + bx + cx2 a, b, cR and a, b, c are distinct, then value of

a b c

b c a

c a b

is (where & 2 are

complex cube roots of unity)

;fnf(x) = a + bx + cx2 a, b, cR rFkk a, b, c fofHkUu gS rca b c

b c a

c a b

dk eku gS ( tgka vkS j 2 bdkbZ ds

?kuewy gS)

(A) f(1) f() f( 2) (B) –f() f(2) (C*) –f(1) f() f(2) (D) f() f(2)

Sol.

a b c

b c a

c a b

= – (a3 + b

3 + c

3 –3abc)

= –(a + b + c)(a2 + b

2 + c

2 –ab – bc–ca) = –(a + b + c)(a + b + c2)(a + b2 + c) = –f(1) f() f(2)

5_. The general solution of sin10

x + cos10

x 29

16 cos

42x is

sin10

x + cos10

x 29

16 cos

42x dk O;kid gy gSA

(A)3

2n , 2n8 8

32n , 2n

8 8

; nI

(B) 2n , 2n8

; nI

(C*)3

n , n8 8

3

n , n8 8

; nI

(D)3

n , n4 4

3n , n

4 4

; nI

Sol. sin10

x + cos10

x 2916

cos42x

5

1 cos 2x

2

+

51 cos 2x

2

29

16 cos

42x

Put cos2x = t j[kus ij

5

1 t

2

+

51 t

2

29

16 t

4

24t4 –10t2 –1 2t2 –1) (12t2 + 1) (2cos22x –1)(12cos22x + 1)

(2cos2 2x –1) 12

cos2x 12

3

,4 4

3,

4 4

; = 2x

General solution is O;kid gy

2n 3 / 4,2n / 4 2n / 4,2n 3 / 4

x 3

n , n8 8

3n , n

8 8

; nI


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6_. If tan =5

2, tan =

7

9 where

3,

2

, ,

2

, then lies in the interval.

;fn tan =5

2, tan =

7

9 tgka

3,

2

, ,2

, rc vUrjky es a fLFkr gS

(A)3 7

,2 3

(B)13 5

,6 2

(C*)13 7

,6 3

(D) ,6 3

Sol. tan( ) =tan tan 31

1 tan tan 53

2n , 2n6 3

but3

,2

, ,2

3 5

,2 2

+ 13 7

,6 3

7. If a2 + b2 + c2 + ab + bc + ca 0 a, b, c R, then value of the determinant2 2 2

2 2 2

2 2 2

(a b 2) a b 1

1 (b c 2) b c

c a 1 (c a 2)

equals

;fn a2 + b2 + c2 + ab + bc + ca 0 a, b, c R rc lkjf.kd2 2 2

2 2 2

2 2 2

(a b 2) a b 1

1 (b c 2) b c

c a 1 (c a 2)

dk eku gksxkµ

(A*) 65 (B) a2 + b2 + c2 + 31(C) 4(a2 + b2 + c2 ) (D) 0

Sol. We have a2 + b2 + c2 + ab + bc + ca 0 (a + b)2 + (b + c)2 + (c + a)2 0

a + b = 0, b + c = 0, c + a = 0 a = b= c = 0 =4 0 11 4 0

0 1 4

= 65

8. A and B be 3 × 3 matrices such that AB + A + B = 0Statement-1 : AB = BAStatement-2 : PP –1 = I = P –1 P for every matrix P which is invertible.(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

A vkS j B, 3 × 3 Øe ds nks vkO;w g bl izdkj gS fd AB + A + B = 0.dFku 1 : AB = BA

dFku 2 : PP –1 = I = P –1P, izR;sd izfrykseh; P ds fy, (A*) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA(B) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA(C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA

Sol. Given fn;k x;k gS AB + A + B = O AB + A + B + I = I A(B + I) + (B + I) = I (A + I) (B + I) = I (A + I) and (B + I) are inverse of each other (A + I) (B + I) = (B + I) (A + I) AB = BA

(A + I) vkS j (B + I) ,d nqljs ds izfrykse gS (A + I) (B + I) = (B + I) (A + I) AB = BA


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9. The equation

2 2 2(1 x) (1 x) (2 x )

2x 1 3x 1 5x

x 1 2x 2 3x

+

2

2

(1 x) 2x 1 x 1

(1 x) 3x 2x

1 2x 3x 2 2x 3

= 0

(A) has no real root (B) has 4 real roots(C) has two real and two non-real roots (D*) has infinite number of roots

lehdj.k

2 2 2(1 x) (1 x) (2 x )

2x 1 3x 1 5x

x 1 2x 2 3x

+

2

2

(1 x) 2x 1 x 1

(1 x) 3x 2x

1 2x 3x 2 2x 3

= 0

(A) dksbZ okLrfod ewy ugha j[krk (B) ds 4 okLrfod ewy gS(C) ds nks okLrfod rFkk nks vokLrfod ewy gS (D*) ds vuUr ewy gS

Sol. 1st two columns of 1st determinant are same as 1st two rows of 2nd. Hence transpose the 2nd. Add the

two determinants and use C1 C1 + C3 = 0

10. In a square matrix A of order 3 each element aii is equal to the sum of the roots of the equation

x2 – (a + b)x + ab = 0, each ai, i+1 is equal to the product of the roots, each ai,i–1 is unity and the rest of

the elements are all zero. The value of the |A| is equal to

,d 3 × 3 Øe ds oxZ vkO;w g A ds lHkh ai i vo;o lehdj.k x2

– (a + b)x + ab = 0 ds ewyksa ds ;ksx ds cjkcj gS] lHkh ai , i + 1 blh lehdj.k ds ewyks a ds xq .kuQy ds cjkcj] lHkh ai , i – 1 vo;o bdkbZ gS a rFkk 'ks"k lHkh vo;o 'kwU; gS] rc |A| dk eku gksxkµ (A) 0 (B) (a + b)3 (C) a3 – b3 (D*) (a2 + b2)(a + b)

Sol. |A| =

a b ab 0

1 a b ab

0 1 a b

= (a2 + b2)(a + b)

11. If sin3x cos 3x + cos3x sin 3x =3

8 , then the value of sin 4x is

;fn sin3x cos 3x + cos3x sin 3x =3

8 gks] rc sin 4x dk eku gksxk

(A*) 1/2 (B) 1 (C) – 1/2 (D) 0Sol. We have 4 sin3x cos 3x + 4 cos3x sin 3x = 3/2

(3 sin x – sin 3x) cos 3x + (3 cos x + cos 3x) sin 3x =3

2

3 (sin x cos 3x + cos x sin 3x) =3

2 sin 4x =

1

2

12. The number of solutions of the equation cot (sinx + 3) = 1 in [0, 3 ] is lehdj.k cot (sinx + 3) = 1 ds gyks a dh la [;k gS] tc [0, 3] gks (A) 0 (B) 1 (C*) 4 (D) 2

Sol. sinx + 3 = n + /4 sinx = n + /4) – 3 sinx = (5/4) – 3 2 solutions in [0, ]

13. If A and B are two non-singular square matrices obeying commutative rule of multiplication then A3B3(B2 A4) –1 A =

;fn A rFkk B nks O;qRØe.kh; oxZ vkO;wg gS tks xq .kuQy esa Øe fofues ; fu;e dk ikyu djrs gS] rc A3B3(B2 A4) –1 A dk eku gSµ (A) A (B*) B (C) A2 (D) B2

Sol. A3B3 (A4) –1 (B2) –1 A = B3 A3(A4) –1(B2) –1 A = B3 A –1(B2) –1 A = B3(B2 A) –1 A = B3(AB2) –1 A = B3(B2) –1 A –1 A = B

14. The number of 2 × 2 matrices X satisfying the matrix equation X2 = I (I is 2 × 2 unit matrix) is

2 × 2 Øe ds oxZ vkO;w gksa X dh la[;k tks lehdj.k X2 = I dks la rq"V djrs gSµ( tgk¡ I, 2 × 2 Øe dk rRledvkO;w g gS)(A) 1 (B) 2 (C) 3 (D*) infinite vuUr


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Sol. X2 =a b

c d

a b

c d

=1 0

0 1

2

2

a bc ab bd

ac cd bc d

=

1 0

0 1

a2 + bc = 1 ....(1) b(a + d) = 0 ....(2)c(a + d) = 0 ....(3) bc + d2 = 1 ....(4)

case-I a + d 0 b = 0 and c = 0 a = ±1 and d = ±1 (a, d) = (1, 1), (–1, –1) X = I, –I

case-II a + d = 0 a2 + bc = 1 infinite matrices vuUr vkO;w g

15. If

x x 1 x 2r r 1 r 2

y y 1 y 2r r 1 r 2

z z 1 z 2r r 1 r 2

C C C

C C C

C C C

=

x x xr r 1 r 2

y y yr r 1 r 2

z z zr r 1 r 2

C C C

C C C

C C C

, then '' is equal to

;fn

x x 1 x 2r r 1 r 2

y y 1 y 2r r 1 r 2

z z 1 z 2r r 1 r 2

C C C

C C C

C C C

=

x x xr r 1 r 2

y y yr r 1 r 2

z z zr r 1 r 2

C C C

C C C

C C C

, rc '' cjkcj gS&

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

Sol. R.H.S. =

x x xr r 1 r 2

y y yr r 1 r 2

z z zr r 1 r 2

C C C

C C C

C C C

Apply C3 C3 + C2 x x x 1

r r 1 r 2

y y y 1r r 1 r 2

z z z 1r r 1 r 2

C C C

C C C

C C C

Apply C2 C2 + C1 x x 1 x 1

r r 1 r 2

y y 1 y 1r r 1 r 2

z z 1 z 1r r 1 r 2

C C C

C C C

C C C

Apply C3 C3 + C2 x x 1 x 2

r r 1 r 2

y y 1 y 2r r 1 r 2

z z 1 z 2r r 1 r 2

C C C

C C C

C C C

16_. If A is a non-singular square matrix of order 'n' such that 3ABA –1

+ A = 2A –1

BA, then

(A) A & B both are identity matrix (B*) |A + B| = 0(C*) |ABA

–1 – A

–1BA| = 0 (D) A + B is a non-singular matrix

;fn A 'n' Øe dk O;qRØe.kh; vkO;w g bl izdkj gS fd 3ABA –1 + A = 2A –1BA,(A) A vkS j B nksuksa loZ le vkO;w g gSA (B*) |A + B| = 0(C*) |ABA

–1 – A

–1BA| = 0 (D) A + B O;qØe.kh; vkO;w g gSA

Sol. 3ABA –1

+ 3A = 2A –1

BA + 2A 3A(BA –1 + I) = (A –1B + I)2A 3A(B + A)A –1 = A –1(B + A)2A |3A(A + B)A –1| = |2A –1(A + B)A| 3n |A| |A + B| |A –1| = 2n |A –1| |A + B| |A| (3n – 2n) |A + B| = 0 |A + B| = 0Now vc, Let M = ABA –1 – A –1BA AM = A2BA –1 – BA BA = A2BA –1 – AM


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3ABA –1

+ A = 2A –1

BA

3ABA –1 + A = 2A –1(A2BA –1 – AM) 3ABA –1 + A = 2ABA –1 – 2M –2M = ABA –1 + A –2M = A(B + A)A –1 (–2)n |M| = |A| |A + B| |A –1| = 0

17. The equation x3 –3

4 x = –

3

8 is satisfied by

lehdj.k x3 – 34

x = –3

8 ftu x ds ekuks a ls larq"V gks rh gS] og gSµ

(A*) x = cos5

18

(B*) x = cos7

18

(C) x = cos23

18

(D*) x = cos17

18

Hint: Let x = cos

4 cos3 – 3 cos = –3

2 cos 3 = cos

5

6

3 = 2n ±

5

6

=2n

3

±

5

18

18. The solution of the equation sin3x + sin x cos x + cos3x = 1 is (n N) lehdj.k sin3x + sin x cos x + cos3x = 1 dk gy gSµ ( tgk¡ n N)

(A*) 2n (B*) (4n + 1)2

(C) (2n + 1) (D) None of these buesa ls dksbZ ughSol. The given equation is sin3 x + cos3 x + sin x cos x = 1

(sin x + cos x) (sin2 x – sin x cos x + cos2 x ) + sin x cos x – 1 = 0 (1 – sin x cos x)(sin x + cos x – 1) = 0Either 1 – sin x cos x = 0 sin 2 x = 2 which is not possible

Or sin x + cos x – 1 = 0 cos (x – /4) =2

1 x – /4 = 2n ±

4

x = 2n or x = (4n + 1)/2

19. If A =1

3

1 2 2

2 1 2

x 2 y

and AAT = I then

;fn A =1

3

1 2 2

2 1 2

x 2 y

rFkk AAT = I gS] rc

(A*) x + 2y = 4 (B*) x – y = 1 (C*) 2x + y = 5 (D) 2x + y = –13

Sol. AAT = I

1 2 2 1 2 x 9 0 0

2 1 2 2 1 2 0 9 0x 2 y 2 2 y 0 0 9

x = 2, y = 1

20*. If there are three square matrix A, B, C of same order satisfying the equation A2 = A –1 and let B =n2 A

& C =(n 2)

2 A

then which of the following statements are true? (where n N)

;fn leku Øe ds rhu oxZ vkO;wg A, B rFkk C gS] A2 = A –1 gS rFkk ekuk B =n2 A & C =

(n 2)2 A

gS] rc fuEu esa ls lR; dFku gSµ( tgk¡ n N)(A*) |B – C| = 0 (B*) (B + C)(B – C) = 0

(C) |B – C| = 1 (D) None of these buesa ls dksbZ ugh


8/14




Sol. B =n2 A =

n 12 · 2 A

=n 12 2(A )

=n 11 2(A )

= n 1 12 A

=

n 2 12 · 2 A

= n 2 12 2(A )

= n 22

1 1(A )

=(n 2)

2 A

= C B – C = 0

21*. Let A =

–3 –7 –5

2 4 3

1 2 2

and B =

a

b

1

. If AB is a scalar multiple of B, then

ekuk A = –3 –7 –5

2 4 3

1 2 2

rFkk B =a

b

1

. ;fn AB vkO;w g B dk vfn'k xq .kt (scalar multiple) gS rc

(A*) 4a + 7b + 5 = 0 (B*) a+ b + 2 = 0 (C*) b – a = 4 (D*) a + 3b = 0

Sol. AB =

–3 –7 –5 a

2 4 3 b

1 2 2 1

AB =

–3a – 7b – 5 a

2a 4b 3 b

a 2b 2 1

3 a 7b 5 0

2a 4 – b 3 0a 2b 2 – 0

3 7 5

2 4 – 31 2 2 –

= 0

= 1 a = – 3 & b = 1

22*. Values of '' for which system of equations x + y + z = 1, x + 2y + 4z = and x + 4y + 10z = 2 isconsistent, are

'' ds eku ftlds fy, lehdj.k fudk; x + y + z = 1, x + 2y + 4z = rFkk x + 4y + 10z = 2 laxr gS] gksaxsa& (A*) 1 (B) 3 (C*) 2 (D) 0

Sol. =

1 1 1

1 2 4

1 4 10

= 0 1 =2

1 1 1

2 4

4 10

= 2(2 – 3 + 2) = 0

2 =2

1 1 1

1 4

1 10

= 3(2 – 3 + 2) = 0 3 =2

1 1 1

1 2

1 4

= 2 – 3 + 2 = 0

= 1, 2

23*. Consider a matrix M =

3 4 0

2 1 0

3 1 K

and the following statements:

Statement (S1) : Inverse of M exists.Statement (S2) : K 0,Which of the following in respect of the above matrix and statements is/are incorrect?(A*) S1 implies S2, but S2 does not imply S1.(B*) S2 implies S1, but S1 does not imply S2.(C*) Neither S1 implies S2 nor S2 implies S1.(D) S1 implies S2 as well as S2 implies S1.

vkO;w g M =3 4 0

2 1 0

3 1 K

ds fy,

dFku (S1) : M dk iz frykse fo|eku gS dFku (S2) : K 0,


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fuEu esa ls dkS ulk@dkSuls vkO;w g M rFkk fn, x, dFkuks a ds fy, vlR; gS?(A*) S1 S2, fdUrq S2 / S1. (B*) S2 S1, fdUrq S1 / S2.(C*) u rks S1 S2 u gh S2 S1. (D) S1 S2 rFkk S2 S1.

Sol. |M| =

3 4 0

2 1 0

3 1 K

= –5K

24*. The product of all the values of t, for which the system of equations (a – t)x + by + cz = 0,

bx + (c – t)y + az = 0, cx + ay + (b – t)z = 0 has non-trivial solution, ist ds mu lHkh ekuksa dk xq.kuQy ftuds fy, lehdj.k fudk; (a – t)x + by + cz = 0, bx + (c – t)y + az = 0,cx + ay + (b – t)z = 0 'kwU;rj gy j[krk gS] gksxkµ

(A)

a –c –b

–c b –a

–b –a c

(B*)

a b c

b c a

c a b

(C)

a c b

b a c

c b a

(D*)

a a b b c

b b c c a

c c a a b

Sol.

a – t b c

b c – t a

c a b – t

= – t3 + t2 + t + = 0

product of roots ewyksa dk xq .kuQy = =a b c

b c a

c a b

25*. Let A and B are square matrices of same order satisfying AB = A and BA = B, then (A2015

+ B2015

)2016

is

equal to

ekuk A rFkk B leku Øe dh oxZ vkO;w g bl izdkj gS fd AB = A rFkk BA = B rc (A2015 + B2015)2016 cjkcj gS& (A*) 2

2015 (A

3 + B

3) (B) 2

2016 (A

2 + B

2) (C) 2

2016 (A

3 + B

3) (D*) 2

2015 (A + B)

Sol. AB = A & vkS j BA = B AB.A = A2 & BA.B = B2 A.BA = A2 & B.AB = B2 AB = A2 & BA = B2

A = A2

& B = B2

An = A & Bn = BNow vc, (A2015 + B2015)2 = (A + B)2 = A2 + B2 + AB + BA = 2(A + B)(A+ B)

3 = 2(A + B)

2 = 4(A + B)

(A + B)4 = 4(A + B)

2 = 8(A + B) (A + B)n = 2n–1(A + B)

26*. If p, q, r are in A.P. then value of determinant

2 n 1 2 n 2 2

n n 1

2 n 2 n 1 2

a 2 2p b 2 3q c p

2 p 2 q 2q

a 2 p b 2 2q c – r

is

(A*) 0 (B*) Independent from a, b, c(C) a

2b

2c

2 – 2

n (D*) Independent from n

;fn p, q, r lekUrj Js


10/14




Sol. =

2 n 1 2 n 2 2

n n 1

2 n 2 n 1 2

a 2 2p b 2 3q c p

2 p 2 q 2q

a 2 p b 2 2q c – r

R1 R1 – (R2 + R3)

= n n 1

2 n 2 n 1 2

0 0 p r – 2q

2 p 2 q 2q

a 2 p b 2 2q c – r

= 0

27*. If

2

2

2

x – 5x 3 2x – 5 3

3x x 4 6x 1 9

7x – 6x 9 14x – 6 21

= ax3 + bx

2 + cx + d, then which of the following are correct ?

;fn

2

2

2

x – 5x 3 2x – 5 3

3x x 4 6x 1 9

7x – 6x 9 14x – 6 21

= ax3 + bx

2 + cx + d rc fuEu esa ls dkS ulk lR; gS?

(A*) a = 0 (B*) b = 0 (C*) c = 0 (D) d = 0

Sol. f (x) =

2

2

2

x – 5x 3 2 32x – 5 2x – 5 3

6x 1 6x 1 9 3x x 4 6 9

14x – 6 14x – 6 21 7x – 6x 9 14 21

= 0

f(x) is a constant polynomial & f(0) 0 d 0

Hindi. f (x) =

2

2

2

x – 5x 3 2 32x – 5 2x – 5 3

6x 1 6x 1 9 3x x 4 6 9

14x – 6 14x – 6 21 7x – 6x 9 14 21

= 0

f(x) vpj gS rFkk f(0) 0 d 0

28*. If the equations x + ay – z = 0, 2x – y + az = 0, ax + y + 2z = 0 have non-trivial solution, then a =

;fn lehdj.k fudk; x + ay – z = 0, 2x – y + az = 0, ax + y + 2z = 0 'kwU;rj gy j[krk gS rc a =

(A) 2 (B*) –2 (C*) 1 + 3 (D*) 1 – 3

Sol.

1 a –1

2 –1 a

a 1 2

= 0 (a + 2)(a2 – 2a – 2) = 0

29*. If the elements of a 2 × 2 matrix A are positive and distinct such that |A + AT|T = 0, then

;fn ,d 2 × 2 Øe dh A vkO;w g ds vo;o /kukRed rFkk fHkUu&fHkUu bl iz dkj gS fd |A + AT|T = 0 rc (A) |A| 0 (B*) |A| > 0 (C*) |A – AT| > 0 (D*) |AAT| > 0

Sol. Let ekuk A =a b

c d

|A + AT| =

2a b c

b c 2d

= 4ad – (b + c)

2 = 0

b c

2

= ad


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b c

2

> bc

ad bc ad > bc ad – bc > 0 |A| > 0

|A – AT| =

0 b – c

c – b 0 = (b – c)

2 > 0

30*. If M = {A : A is a 3 × 3 matrix whose entries are –1 and 1}, then(A) |A| lies from –1 to 1 (B*) |A| {–4, 0, 4}(C*) n(M) = 2

9 (D) n(M) = 3

9

;fn M = {A : A ,d 3 × 3 vkO;w g gS ftlds vo;o –1 rFkk 1 gS }, rc (A) |A| dk eku –1 ls 1 esa gSA (B*) |A| {–4, 0, 4}(C*) n(M) = 2

9 (D) n(M) = 3

9

Sol. Let A =

1 2 3

1 2 3

1 2 3

a a a

b b b

c c c

|A| = a1 b2 c3 + a2 b3 c1 + a3 b1 c2 – a1 b3 c2 – a2 b1 c3 – a3 b2 c1

det(A) = P1 + P2 + P3 – P4 – P5 – P6 where |Pi| = 1

|det (A)| |P1| + |P2| + |P3| + |P4| + |P5| + |P6| |det(A)| 6Hence option (A) is correct.

Now, applying C1 C1 + C2 & C2 C2 + C3, we getelements of 1

st and 2

nd column as even number

|A| = multiple of 4Hence option (B) is correct.

Hindi. ekuk A =1 2 3

1 2 3

1 2 3

a a a

b b b

c c c

|A| = a1 b2 c3 + a2 b3 c1 + a3 b1 c2 – a1 b3 c2 – a2 b1 c3 – a3 b2 c1

det(A) = P1 + P2 + P3 – P4 – P5 – P6 tgk¡ |Pi| = 1

|det (A)| |P1| + |P2| + |P3| + |P4| + |P5| + |P6| |det(A)| 6vr% fodYi (A) lR; gSA vc C1 C1 + C2 & C2 C2 + C3, yxkus ij ge ikrs gS 1

st o 2nd LrEHk ds vo;o le la[;k gSA

|A| = 4 dk xq .kt gSA vr% (B) lR; gSA

31*. A solution of the system of equations x – y =1

3 and cos

2x – sin2y =1

2 is given by

lehdj.k fudk; x – y =1

3 rFkk cos2x – sin2y =

1

2 dk ,d gy fuEu iz dkj fn;k tk ldrk gS&

(A*)7 5

,6 6

(B)8 1

,15 6

(C*) –5 –7

,6 6

(D*)1 1

,–6 6

Sol. cos2x – sin2(x – /3) =

1

2

cos2x –2

1 3 1sin x. – cos x.

2 2 2


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cos2x – 2 21 3 3 1

sin x. cos x. – sin2 x4 4 4 2

1

4(cos

2x – sin2x) +3

4sin2x =

1

2

1

2 cos2x +

3

2sin2x = 1

cos 2 x – 3 = 1

2x –3

= 2n

x = n +1

6 ; N I

32. If f(x) =

cos(x ) cos(x ) cos(x )

sin(x ) sin(x ) sin(x )

sin( – ) sin( – ) sin( – )

and f(0) =1

4, then

15

r 1

f(r)

is (where [.] is G.I.F.)

;fn f(x) =cos(x ) cos(x ) cos(x )

sin(x ) sin(x ) sin(x )

sin( – ) sin( – ) sin( – )

rFkk f(0) = 14

rc 15

r 1

f(r)

dk eku gS ( tgk¡ [.] egÙke

iw.kkZa d Qyu gS)Ans. 3

Sol. f'(x) = 0 f(x) is a constant function f(x) =1

4

Hindi. f'(x) = 0 f(x) ,d vpj Qyu gS f(x) =1

4

33. If the summation of the series

sin sin 3 sin9

cos 3 cos9 cos27

+ . . . . . . up to n terms is given by1

2(tan a – tan b), where

a

bis a two

digit natural number for some 'n', then the maximum value ofa

b is

;fn Js .khsin sin 3 sin9

cos 3 cos9 cos27

+ . . . . . . n inksa rd dk ;ksxQy

1

2(tan a – tan b) gS tgk¡

a

b 'n' ds

dqN ekuksa ds fy, ,d f} vadh; iz kd`Ùk la [;k gS rca

b dk vf/kdre eku Kkr dhft,A

Ans. 81

Hint. Sn =

r r n–1 n–1

r 1 r 1 r r 0 r 0

sin 3 sin 2.31

2cos 3 cos 3 cos 3

34. If sin14

sin

3

14

sin

5

14

. . . . . . . upto 7 factors =

n

1

2 , then find the value of ‘n’.

;fn sin14

sin

3

14

sin

5

14

. . . . . . . 7 xq .ku[k.M rd =

n

1

2 gks rc ‘n’ dk eku Kkr dhft,A

Ans. 6

Sol. sin sin3 sin5 sin7 sin9 sin11 sin13, where tgk¡ =14

= (sin sin 3 sin 5)2


13/14




= (cos 6 cos 4 cos 2)2 =2

2 3cos cos cos

7 7 7

Match the column Type Questions

LrEHk feyku izdkj iz'u

35. Consider a square matrix A of order 2 whose four distinct elements are 0,1,2 and 4. Let N denote thenumber of such matrices.

Column–I Column–II

(A) Possible non-negative value of |A| is (P) 2

(B) Sum of values of determinants corresponding to N matrices is (Q) 4

(C) If absolute value of |A| is least, then possible value of | adj(adj(adj A)) | is (R) –2

(D) If |A| is algebraically least, then possible value of |4A –1| is (S) 0

ekuk A Øe 2 dk ,d oxZ vkO;w g gS] ftlds pkj fHkUu vo;o 0,1,2 rFkk 4 gSA ekuk N ,sls gj laHko vkO;wgks a dh la [;k gSA

LrEHk –I LrEHk –II

(A) |A| dk laHkkfor v_.kkRed eku gSµ (P) 2

(B) lHkh N vkO;w gks a ds lkjf.kdks a ds ekuksa dk ;ksxQy gSµ (Q) 4

(C) ;fn |A| dk fujis{k eku U;wure gS] rc | adj(adj(adj A)) | gSµ (R) –2

(D) ;fn |A| dk eku U;wure gks] rc |4A –1| dk laHkkfor eku gksxkµ (S) 0

Ans. (A P, Q); (B S); (C P, R); (D R)Sol. Here 24 matrices are possible.

Values of determinants can be –8, – 4, – 2, 2, 4, 8(A) Possible non-negative values of |A| are 2, 4, 8

(B) Sum of these 24 determinants is 0(C) Mod. (det(A)) is least | A | = ± 2 | adj (adj (adj (A)) | =

3(n 1) A

= ± 2

(D) Least value of det.(A) is –8 Now | 4 A –1 | = 161

| A |=

16

8= –2

Comprehension # 1 (Q. No. 36 to 38)

Consider the system of equations

x – y + 2z = 10 ; 2x + y + z = 8; 3x + 2y + z =

vuqPnsn # 1 (iz0 la0 36 l s 38)

rhu lehdj.k fn;s x;s gSµ x – y + 2z = 10 ; 2x + y + z = 8; 3x + 2y + z =

36. If system has unique solution then the range of sin –12 1

+ cos –1

2 1

is

;fn lehdj.k fudk; dk vf}rh; gy gks] rks sin –12 1

+ cos –1

2 1

ds ekuks a dk ifjlj gksxkµ

(A)3

,2 2

(B)2

(C*)5

,6 6

(D) None of these buesa ls dksbZ ugh


14/14




37. If system has infinite solution then the range of 2 + 2 is ;fn lehdj.k fudk; ds vuUr gy gks] rks 2 + 2 ds ekuksa dk ifjlj gSµ (A) [0, ) (B) [101, ) (C) [0, 101] (D*) {101}

38. If system has no solution then the range of2 1

is

;fn lehdj.k fudk; dk dksbZ gy laHko ugh gS] rc2

1

ds ekuks a dk ifjlj gS

(A) (0, 1] –1

101

(B*) (0, 1]

(C) [0, 1) (D) None of these buesa ls dksbZ ugh Sol. (36)

System of equation can be written as AX = B

For unique solution | A | 0 1 and R

2 1

1 1,

2 2

sin –1

2 1

,6 6

2 1

1 1,

2 2

cos –1

2 1

2

,3 3

Range =5

,6 6

(37) = 0 and x = 0 = 1, = 10 2 + 2 = 101

(38) For no solution. = 1; 10 2 1

=

2

1

1 (0, 1] –

1

101

Comprehension (Q. No. 39 to 40)

vuqPNsn (iz'u la[;k 39 l s 40)

A square matrix A is said to be orthogonal if AT A = I = AA

T.

oxZ vkO;w g A yEcdks .kh; vkO;wg gksxh ;fn AT A = I = AAT.

39. Let A =29 –28

30 –29

and P is a orthogonal matrix of order 2. if Q = PT AP, then PQ

2016P

T =

ekuk A =29 –28

30 –29

rFkk P, 2 Øe dh yEcdks .kh; vkO;w g gS ;fn Q = PT AP rc PQ2016PT =

(A) 2015 A (B) A2016

(C*) I (D) A

40. P is an orthogonal matrix of order 3 and , , are direction angles of a straight line.

Let A =

2

2

2

sin sin sin sin sin

sin sin sin sin sin

sin sin sin sin sin

and Q = PT AP.

If PQ6P

T = 2

k A, then k =

(A*) 5 (B) 7 (C) 6 (D) 0

P, 3 Øe dh yEcdks .kh; vkO;w g gS rFkk , , ,d ljy js[kk ds fnd~dks .k g]S

ekuk A =

2

2

2

sin sin sin sin sin

sin sin sin sin sin

sin sin sin sin sin

rFkk Q = PT AP.

;fn PQ6PT = 2k A, rc k =(A*) 5 (B) 7 (C) 6 (D) 0

Sol. (39) Q2 = P'AP.P'AP = P'A

2P Q2016 = P'A2016P

PQ2016P' = PP'A2016PP' = A2016 = (A2)1008 = (I)1008 = (40) PQ

6P' = A

6

Now, A2 = 2A A3 = 2A.A = 4A A6 = 16A2 = 32A =25 A

(1031)dpp_11_matrices_determinant_and_trigonometry_b.pdf.tmp.pdf

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