Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2...

8/4/2019 Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2 Sectiona

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Aakash IIT-JEE -Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 47623456 Fax : 47623472- 25 -

Chapter - 2 (Section A:Objetive Type Single Choice)

Q.No. Solutions

1. Answer (2)

cos1°.cos2°……cos179° = 0

(as there will be a term cos 90° in this series whose value is zero)

2. Answer (1)

13sec ,0

5 2

2

2 3cot

4 9 sec 1

=

52 3

1212

4 95

( 2 2sec 1 tan )

1213

5

=3/4 15

88/5 352

3. Answer (1)2(cosec .cosec cot .cot ) A B A B 2(cosec .cot cosec .cot ) A B B A

2 2 2 2 2 2cosec (cosec cot ) cot (cosec cot ) A B B A B B

2 2cosec cot 1 A A

4. Answer (2)

5 99sin , sin

13 101 A B

cos( ) cos .cos sin .sin A B A B A B

513

12

99

20

101

=12 20 5 99

. .13 101 13 101

=255

1313

A + B + C = A + B = – C

or cos ( A + B) = cos ( – C )

cos C = – cos ( A + B)

=255

1313




Q.No. Solutions

5. Answer (2)

tan cot a , 4 4tan cot ? 4 4tan cot = 4 2 2 2 2(tan cot ) 4 tan .cot .(tan cot ) 6 tan .cot

= 4 24{(tan cot ) 2 tan .cot } 6a

= 4 24 2a a

6. Answer (1)

cos sin 3a b sin cos 4a b

Squaring and adding, a2

+ b2

= 25

7. Answer (1)

cos , sina b c d

cos ,sinb d

a c

Squaring and adding,2 2

2 21

b d

a c

2 2 2 2 2 2b c a d a c

8. Answer (3)

2

sin10

x x , [ , ] x

Clearly the total number of points of intersection is 3.

9. Answer (3)2 2

max min(3 sec ) (4 tan ) 2 4 2 x y

10. Answer (1)

tan ,

p

q

sin cos

?sin cos

p q

p q

Divide numerator and denominator by cos

2 2

2 2

tan

tan

p q p q

p q p q

11. Answer (3)sin1° < sin 1 is correct

As a 1 radian =180

degree 13.14

sin is increasing in 1st

quadrant




Q.No. Solutions

sin1° < sin 1

12. Answer (2)

sin1 cos2 tan3 cot 4 sec 5 cosec 6

= negative

13. Answer (1, 4)Both the options are correct

1

2

tan 1 > tan 2 (from graph)

14. Answer (1)

3sin 4cos 5 4sin 3cos Squaring and adding,

29 16 25 0

15. Answer (2)

sec tana c d sec tana d c sec tanb d c sec tanb c d Squaring and adding,

2 2 2 2 2 2 2( )sec (1 tan ) (1 tan )a b d c

2 2 2 2a b c d

16. Answer (3)

1 sin sec tan( )

1 sin sec tan

f

2(sec tan ) |sec tan |

Now f () = g ()

If sec – tan 0

1 sin0

cos

cos > 0

,2 2

17. Answer (1)

12 2

2r l r where r is radius of the circle and l is the length of the arc i.e. l r

2r r r




Q.No. Solutions

( 2) radian

18. Answer (1)

1sin , 2

4 42

1cos ,

2 3 3

tan 3 , 23 3

Sum of all possible values of . , 2 22

8 –3 3

19. Answer (2)Let the angles be , ,d d

3 180 60 180 60

d

d

d = –30°Hence the angles are 30°, 60°, 90°.

20. Answer (1)2 3sin sin sin 1 0 x x x

2(sin 1)(sin 1) 0 x x

sin x = –1

cos x = 06 4 2cos 4cos cos 0 x x B x

21. Answer (1)

12tan 5 0 A , 5cos 3 0B

5

tan12

A ,3

cos5

B

ABCD is a cyclic quadrilateral

A + C = , B + D = C = – A, D = – B

cos –cosC A tan tan D B

=12

13 =

4

3

Equation whose roots are cos C and tan D is 2 12 4 12 4 013 3 13 3

x x

239 16 48 0 x x

22. Answer (2)

Let 4 4sin cos and 8 8sin cos

max a , max = b, a – b = ?

2 2 2 2 2(sin cos ) 2sin cos

211 sin 2

2




Q.No. Solutions

max = a = 1Similarly, max = b = 1

a – b = 0

23. Answer (2)

cos sec 2 , [0, 2 ]

cos sec 1

8 8sin cos = 0 + 1 = 1

24. Answer (1)

f () = 2 24cos 24sin .cos 6sin

=

1 cos2 1 cos2

4 12sin2 62 2

= 5cos2 12sin2 1 f max = 13 – 1 = 12f min = – 13 – 1 = –14

Range of f () is [–14, 12]

25. Answer (3)

4(sin4 x – cos

4 x ) + cos

2 x = k

1 cos2

4( cos2 )2

x x k

4(sin2 x – cos

2 x ) + cos

2 x = k

4 – 7cos2 x = k

4

0 17

k

Then

26. Answer (3)

3sin

3

x x

x , 0 x

1 LHS 1 RHS 2 or RHS 2

No solution

27. Answer (4)

Let a be the side of dodecagon i.e. 3 1a

And let b the side of hexagon and r be the radius of the circle in AOC, AOD = 45°

sin15 AD

AO

3 1 3 1

2 2 2r

2r In ,

sin302

b

r




Q.No. Solutions

12 2 2

2b

28. Answer (1)

cos(cos )y x

max1 , when

2y a x

mincos1 when 0y b x

b = cos a

29. Answer (2)

2sin

1 sin cos

Ak

A A

,1 sin cos

?

1 sin

A A

A

2 2

2sin .(1 sin cos )

(1 sin ) cos

A A Ak

A A

2 2

2sin .(1 sin cos )

(1 sin ) (1 sin )

A A Ak

A A

=2sin .(1 sin cot )

(1 sin ).{(1 sin ) (1 sin )}

A A A

A A A

=1 sin cos

1 sin

A A

A

30. Answer (1)2sin sin 1 x x

2sin cos x x 12 10 8 6cos 3cos 3cos cos 1 x x x x

= 6 6 4 2cos .(cos 3cos 3cos 1) 1 x x x x

= 2 2 3{cos (1 cos )} 1 x x

= 3{sin (1 sin )} 1 x x

= 2 3 3{sin sin } 1 (1) 1 0 x x

31. Answer (3)

, satisfy 2cos cos 0 x a x b and 2sin sin 0 x p x q cos cos a … (1)

cos .cos b … (2)

sin sin p … (3)

sin .sin q … (4)

Squaring and adding (1) & (2)2 22 2(cos cos sin sin ) a p

2 22 2( )b q a p 2 2 2 2( )a p b q

32. Answer (1)

sec A and cosec A are the roots of 2 0 x ax b




Q.No. Solutions

sec cosec A A a , sec .cosec A A bsin cos

sin .cos

A Aa

A A

,

1sin .cos A A

b

sin cosa

A Ab

Squaring,

2

21 2sin .cos

a A A

b

2

2

21

a

b b

2 22b b a

or 2

( 2)a b b

33. Answer (1)29

115

cos2

r

r S

29

215

sin2

r

r S

S1 =29

15

cos2

r

r

=15 16 29

cos cos .... cos2 2 2

= (sin cos sin cos ) ..... (Sum of consecutive terms is zero and these are 15 terms in all)

S1 = (sin cos sin ) = cos

S2 =29

15

sin2

r

r

= – cos sin cos = sin

1

2

S

S=

cos

sin

= cot

34. Answer (3)

f () = 23 2sin cos

= 23 2sin 1 sin

= 2(sin 2sin 1) 5

= 25 (sin 1)

f max = 5f min = 1

max

min

f

f =

55

1




Q.No. Solutions

35. Answer (1) 2 2cos cos(33 ) sin sin(45 )

= 2 2cos ( cos ) sin ( sin )

= 2 2cos ( cos ) sin (sin )

cos cos(cos ) 1 2 2cos 1 cos (cos ) 1

and 2 20 sin (sin ) sin 1

2 2 2 2cos 1 cos (cos ) sin (sin ) 1 sin 1

Maximum value = 21 sin 1 At2

36. Answer (3)4 4sin cos 1

2 3 5

x x

5(3tan4 x + 2) = 6 sec

4 x

15tan4 x + 10 = 6 + 6tan

4 x + 12 tan

2 x

9tan4 x – 12tan

2 x + 4 = 0

(3tan2 x – 2)

2= 0

25

3

2 2tan3

x

2 22 3sin & cos

5 5 x x

8 8sin sin

8 27

x x

=16 81

8 625 27 625

=5 1

625 125

37. Answer (1)

2

2sec tan , (2 1)

2sec tan x x k x x x x

=2

2

1 tan tan

1 tan tan

x x

x x Let tan x = t

t 2(k – 1) + t (k + 1) + (k – 1) = 0

D = (k + 1)2

– 4(k – 1) 0

If k = 1, we get2

2

11

1

t t

t t

3k

2– 10k + 3 0




Q.No. Solutions

max625

2t where sin2( –) = 1

43. Answer (4)

44. Answer (3)

1 2 1 3 1 4

1 1 1

A A A A A A

Let the polygon be inscribed in a circle of radius ‘r ’ A

4A

3

A2

D

A1

Or

Each side subtends angle2

n

at the centre

1 2 2 3 3 4

2 AOA A OA A OA

n

In OAD,

1 2 sin2

A A AD r

n

1 2 2 sin A A r n

Similarly 1 3 1 4

2 32 sin & 2 sin A A r A A r

n n

1 1 1

2 32 sin 2 sin 2 sinr r r

n n n

3 2sin sin

1

2 3sin sin sin

n n

n n n

3 2sin sin

1

3sin 2sin cos sin

n n

n n n n

4 2 3 2

sin sin sin sinn n n n

4 3

sin sinn n

4 3( 1)k k

n n

4 30,k

n n

4 31,k

n n



Aakash IIT-JEE -Regd Office : Aakash Tower Plot No 4 Sector 11 Dwarka New Delhi 75 Ph : 47623456 Fax : 47623472

Q.No. Solutions

77n

n

45. Answer (2)

asin2+ bcos

2= m …(1)

bsin2+ acos

2= n …(2)

atan = btan(ii) …(3)

Divide (1) by cos2 we, get

a tan2 + b = m sec

2

2tanm b

a m

…(4)

Divide (2) by cos2, we get

Btan2 + a = nsec

2

2tann a

b n

…(5)

From (3), (4), (5)

2 2m b n aa b

a m b n

a2(mb – mn – b

2+ bn) = b

2(an – a

2– mn – am)

abm(a – b) + abn(a – b) = mn(a2

– b2)

abm + abn = mn(a + b)Divide both sides by abmn, we get

1 1 1 1

n m a b

Iitjee Two Yrs Comprehensive Crs for Iit Jee 2013 (Csp,Cjp,Cjp+) Mathematics Target1 Chapter2...

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