DPP_RESO_12

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TARGET : JEE (Main + Advanced) 2017

 E ST INF ORM AT IO Course : VIKAAS(JA)

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DPP

DAILY PRACTICE PROBLEMS

NO. 50 TO 52

This DPP is to be discussed in the week (17-08-2015 to 22-08-2015)

DPP No. : 50 (JEE-Main)

Total Marks : 32 Max. Time : 34 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.8 (3 marks, 3 min.) [24, 24]

Subjective Questions ('–1' negative marking) Q.9 & Q.10 (4 marks, 5 min.) [8, 10]

1. An equilateral triangle has each of its sides of length 6 cm. If (x₁, y₁) ; (x₂, y₂) & (x₃, y₃) are its vertices, then the value of the determinant

2 1 1 2 2 3 3 x y 1 x y 1 x y 1 is equal to : (A) –972 (B) –486 (C) 486 (D*) 972

,d leckgq f=kHkqt dh izR;sd Hkqtk dh yEckbZ

6

lseh gSA ;fn

(x₁, y₁) ; (x₂, y₂)

,oa

(x₃, y₃)

bl f=kHkqt ds 'kh"kZ gks]

rks lkjf.kd

2 1 1 2 2 3 3 x y 1 x y 1 x y 1

dk eku gSµ

(A) –972 (B) –486 (C) 486 (D*) 972

Sol. Area of 

dk {ks=kQy

= 3

4 (side

Hkqtk

) 2₌ 3 4 × 36 = 9 3 ANSWER KEY DPP No. : 50 (JEE-Main) 1. (D) 2. (C) 3. (D) 4. (D) 5. (D) 6. (C) 7. (A) 8. (C) 9. (4n, 3  +4n)  5 3  nn2 10. (i) x y O (ii) x y 2 0 1 (iii) y x O (iv) y x O (v) y x O ₁ –1 1 DPP No. : 51 (JEE-Advanced)

1. (A) 2. (A)(D) 3. (A)(B)(C)(D) 4. (A)(D) 5. (B)(D) 6. (A)(C)(D)

7. (A)(B) 8. (A) (s) ; (B) (r) ; (C) (p) ; (D) (p)

DPP No. : 52 (JEE-Advanced)

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

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Now

vc

, | 1 1 2 2 3 3 x y 1 1 x y 1 2 x y 1 | = 9 3  2 1 1 2 2 3 3 x y 1 x y 1 x y 1 = 243 × 4 = 972

2. Values of x satisfying the equality |x2_{+ 8x + 7| = |x}2_{+ 4x + 4| + |4x + 3| for x  R are}

lHkh

x  R

ds fy, lehdj.k

|x2_{+ 8x + 7| = |x}2_{+ 4x + 4| + |4x + 3|}

_{dks larq"V djus okys}

_x

_{ds lHkh ekuksa dk}

leqPp; gS&

(A) (– 2, ) (B) 3, 4        {– 2} (C*) 3 , 4    _{ }     {– 2} (D) 4 , 3    _{ }    Sol. | x2_{+ 4x + 4 + 4x + 3 | = | x}2_{+ 4x + 4 | + | 4x + 3 |} | a + b | = |a| + |b|  a.b  0 (x2_{+ 4x + 4) (4x + 3)  0} x  3 4 

3. The set of solution of inequality [x]2_{= – [x], where [.] denotes greatest integer function is}

vlfedk

[x]2_{= – [x]}

_{ds gyksa dk leqPp;] tgk¡}

_[.]

_{egÙke iw.kkZ±d Qyu dks iznf'kZr djrk gS] gSµ}

(A) {–1, 0} (B) [–1, 0) (C) [0, 2) (D*) [–1, 1) Sol. [x]2_{= –[x]} x =  + f 2_{= – }  = 0 or  = – 1 Case-   = 0 x =  + f = f x  [0, 1) ...(i) Case-II   = – 1 – 1  x < 0, x  [–1, 0) ...(ii) by (i) and (ii)

x  [–1, 1) Hindi. [x]2_{= –[x]} x =  + f 2_{= – }  = 0

;k

 = – 1

fLFkfr

-   = 0 x =  + f = f x  [0, 1) ...(i)

fLFkfr

-II   = – 1 – 1  x < 0, x  [–1, 0) ...(ii) (i)

,oa

(ii)

ls

x  [–1, 1)

4. Total number of solutions of sinx . tan4x = cosx belonging to (0, ) are :

vUrjky

(0,)

esa

sinx . tan4x = cosx

ds gyksa dh dqy la[;k gS&

(A) 4 (B) 7 (C) 8 (D*) 5

Sol. tan 4x = cot x = tan x

2         4x = n + x 2         5x = n + 2   (2n + 1) 10  = x n = 0, x = 10 

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n = 1, x = 3 10  n = 3 x = 7 10  n = 2, x = 5 10  x = 4 x = 9 10 

5. The point (11 , 10) divides the line segment joining the points (5 ,  2) and (9 , 6) in the ratio : (A) 1 : 3 internally (B) 1 : 3 externally (C) 3 : 1 internally (D*) 3 : 1 externally

fcUnq

(11 , 10)

fcUnq

(5 ,  2)

vkSj

(9 , 6)

dks tksM+us okys js[kk[k.M dks djrk gSµ

(A) 1 : 3

esa vUr% foHkkftr

(B) 1 : 3

esa cká foHkkftr

(C) 3 : 1

esa vUr% foHkkftr

(D*) 3 : 1

esa cká foHkkftr

Sol.  1

(5, –2) (11, 10) (9, 6)

Let point (11, 10) divide (5, –2) and (9, 6) in the ratio  : 1 (5, –2)

rFkk

(9, 6)

dks fcUnq

(11, 10),  : 1

esa foHkkftr djrk gS

9 5 1     = 11 9 + 5 = 11 + 11 – 6 = 2   = –3 

  point (11, 10) divide externally in the ratio 3 : 1 

fcUnq

(11, 10), 3 : 1

esa ckg~; foHkktu djrk gSA

6. If A & B are the points (3, 4) and (2, 1), then the coordinates of the point C on AB produced such that AC = 2BC are :

;fn

A

,oa

B

ds funsZ'kkad Øe'k%

(3, 4)

o

(2, 1)

gS] rFkk fcUnq

C , AB

dks tksM+us okyh js[kk ij bl izdkj gS fd

AC = 2BC

gks] rks fcUnq

C

ds funsZ'kkad gS&

(A) (2, 4) (B) (3, 7) (C*) (7, 2) (D) 1 5, 2 2        Sol. A (–3, 4) B (2, 1) C (x1, y1) A (–3 , 4) B (2, 1)  AC = 2BC

 B is mid point of AC

dk e/; fcUnq

B

gS

1 3 x 2   = 2  x₁ = 7 , 4 y1 2  = 1  y₁ = – 2 co-ordinate c

ds funsZ'kkad

= (7, –2)

7. If in triangle ABC, A



(1, 10) , circumcentre _





1 2,



3 3

 and orthocentre _





11 4,



3 3 then the

co-ordinates of mid-point of side opposite to A is :

;fn f=kHkqt

ABC

esa

A



(1, 10) ,

ifjdsUnz 



1 2,



3 3



rFkk yEcdsUnz 



11 4,



3 3

gks] rks 'kh"kZ

A

ds lkeus okyh

Hkqtk ds e/; fcUnq ds funsZ'kkad gS&

(A*) (1,



11/3) (B) (1, 5) (C) (1,



3) (D) (1, 6)

Sol. A  (1, 10), circum centre

ifjdsUæ

1 2,

3 3

 



 

 

ortho centre

yEcdsUæ

 11 4,

3 3

 

 

 

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B _D(x₁_{, y}₁₎ C A(1, 10) 2 1 (11/3, 4/3) G (–1/3, 2/3) 1 2 Centroid

dsUæd

1,8 9       1 2y 10 8 3 9   , 2x1 1 3  = 1 y₁ = – 11 3 , x1 = 1 D  1 , 11 3       

8. Harmonic conjugate of the point (5, 13) with respect to (2, –5) and (3, 1) is

fcUnq

(5, 13)

dk fcUnq

(2, – 5)

,oa

(3, 1)

ds lkis{k gjkRed la;qXeh gS&

(A) 1,13 5       (B) 13 , 1 5       (C*) 13 7 , 5 5        (D) 7 13 , 5 5        Sol. (2, –5) (x, y) (3, 1) 2 3

Let point (5, 13) divides (2, –5) and (3, 1) in the ratio  : 1

3 2

1  

  = 5   = – 3/2

 point (5, 13) divide externally in the ratio 3 : 2

then harmonic conjugate divide in the ratio 3 : 2 internally. x = 9 4 5  = 13 5 y = 3 10 7 5 5     point is 13, 7 5 5       

Hindi. (2, – 5)

rFkk

(3, 1)

dks fcUnq

(5, 13)

vuqikr

 : 1

esa foHkkftr djrk gS

3 2

1  

  = 5   = – 3/2



fcUnq

(5, 13), 3 : 2

esa ckg~; foHkktu djrk gS

rc gjkRed la;qXeh

3 : 2

esa vUr%foHkktu djsxk

x = 9 4 5  = 13 5 y = 3 10 7 5 5    

fcUnq

13, 7 5 5       

gS

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9. Solve for x : log₂ sinx 2       < – 1

x

ds fy, gy dhft, %

log₂ sinx

2       < – 1 Ans. (4n, 3  +4n)  5 3  nn2 Sol. 1/2 0 /6 5/6  _x/2 log₂ sinx 2      < – 1 0 < sinx 2< 1 2 x 2 (2n , 2n   / 6) 5 2n , 2n 6             x (4n, 4n + 3  )  4n 5 , 4n 2 3            

10. Draw the graph of the followings :

fuEufyf[kr ds vkjs[k [khafp,&

(i) y =₂log x2 _(ii) _{y =}_xlog 2x _(iii) _{y =} 2 2 log x 2 (iv) y = ₂2log x2 _{(v) y =} 1 2 1 2 | log x | log x Ans. (i) _x y O (ii) x y 2 0 1 (iii) y x O (iv) y x O (v) y x O ₁ –1 1

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DPP No. : 51 (JEE-Advanced)

Comprehension Type ('–1' negative marking) Q.1,2 (5 marks 4 min.) [10, 8]

Multiple choice objective ('–1' negative marking) Q.3 to Q.7 (5 marks, 4 min.) [25, 20]

Match the Following (no negative marking) Q.8 (8 marks, 8 min.) [8, 8]

Comprehension (Q. No. 1 to 2)

vuqPNsn

(

iz'u la[;k

1

ls

2

rd

) Given that N = log49900

7 , A =₂log 42 _₃log 42 _₄log 22 _₄log 32 _{, D = (log}

5 49) (log7 125)

Then answer the following questions : (Using the values of N, A, D)

fn;k x;k gS fd

N = ₇log49900_{, A =}₂log 42 _₃log 42 _₄log 22 _₄log 32 _{D = (log}

5 49) (log7 125)

rks fuEufyf[kr iz'uksa ds mÙkj nhft,A

(N, A, D

ds ekuksa dk mi;ksx djrs gq,

)

1. If log_A D = a, then the value of log₆ 12 is (in terms of a)

;fn

log_A D = a

gks] rks

log₆ 12

dk eku

(a

ds inksa esa

)

gS &

(A*) 1 3a 3a  (B) 1 2a 3a  (C) 1 2a 2a  (D) 1 3a 2a  Sol. N = 30 A = 4 + 9 + 4 – 9 A = 8 D = 6 log₈6 = a log₆12 log₂6 = 3a .. (1) = 1 + log₆2 = 1 + 1 3a = 3a 1 3a  (A)

2. If the value obtained in previous question is1 ma na 

, then choose the correct option

;fn fiNys iz'u esa izkIr eku

1 ma

na 

gS] rks lgh fodYi pqfu, &

(A*) log_N m < log_m N (B) log_N m < log_n N < log_m N (C) log_m N < log_N m < log_n N (D*) log_m N = log_n N

Sol. m = 3, n = 3, N = 30

log₃30 > log₃₀3

 log_N m < log_m N = log_n N

3. If sec A = 17

8 and cosec B = 5

4, then sec(A + B) can have the value equal to

;fn

sec A = 17

8

rFkk

cosec B =

5

4

gks] rks

sec(A + B)

dk eku gks ldrk gS&

(A*) 85 36 (B*) – 85 36 (C*) – 85 84 (D*) 85 84 Sol. sec A = 17 8 , cosec B = 5 4 cos A = 8 17, sin B = 4 5  sin A = ± 15 17, cos B = ± 3 5 sec (A + B) = 1

cos A cosBsin A sinB =

1 8 3 4 15 17 5 5 17     = –85 36, 85 36, – 85 84, 85 84

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4. The vertices of a triangle are A(x₁, x₁ tan ), B(x₂, x₂ tan ) and C(x₃, x₃ tan ). If the circumcentre of triangle ABC coincides with the origin and H(a, b) be the orthocentre, then a

b =

fdlh f=kHkqt ds 'kh"kZ Øe'k%

A(x₁, x₁ tan ), B(x₂, x₂ tan )

rFkk

C(x₃, x₃ tan )

gSA

;fn bldk ifjdsUnz ewy fcUnq

ij gks rFkk

H(a, b)

f=kHkqt dk yEcdsUnz gks] rks

a

b

dk eku gS&

(A*) 1 2 3

1 2 3

x x x

x tan x tan x tan

 

     (B)

1 2 3

x cos x cos x cos

x sin x sin x sin

    

(C) tan tan tan

tan .tan .tan

    

   (D*)

cos cos cos

sin sin sin

          Sol. (0, 0) A (x1, x1)tan B(x2, x2)tan C(x3, x3)tan OA = OB = OC = R H (a, b) G (x, y) (0, 0) 2 1 x = a 3 , y = b 3 centroid

dsUnzd

= a b, 3 3      

also

lkFk gh

G  x1 x2 x3 , x tan1 x tan2 x tan3

3 3               a b = 1 2 3 1 2 3 x x x

x tan x tan x tan

 

    

5. The sides of a triangle are the straight lines x + y = 1 ; 7y = x and 3 y + x = 0 . Then which of the following is an interior point of the triangle ?

(A) circumcentre (B*) centroid (C*) incentre (D) orthocentre

fdlh f=kHkqt dh Hkqtkvksa ds lehdj.k

x + y = 1 ; 7y = x

,oa

3y + x = 0

gks] rks fuEu esa ls dkSulk fcUnq

f=kHkqt ds vUnj fLFkr gS&

(A)

ifjdsUnz

(B*)

dsUnzd

(C*)

vUr%dsUnz

(D)

yEcdsUnz

Sol. 3 xy1(m  1) 7y = x(m1 = 1/7) 2 3 yx0(m  1/ 3 ) B C _A tan A = 1 1 7 3 1 1 7 3   +ve tan B = 1 1 3 1 1 3    +ve tan C = 1 1 17 1 1 17    -ve

hence triangle is absolute angled so orthocentre and circumcentre lie outside.

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6. The two adjacent sides of a parallelogram are represented by the lines x – y + 1 = 0 and 4x – 3y – 2 = 0. If one of the diagonals of the parallelogram is along the line y =3x

2 , then the other diagonal has (A*) Slope =5 4 (B) y intercept = 1 4 (C*) x intercept =1

5 (D*) the poionts (2, 1) and (1, 9) on either side of it.

lekUrj prqHkqZt dh nks vklUu Hkqtk,¡] js[kkvksa

x – y + 1 = 0

vkSj

4x – 3y – 2 = 0

}kjk O;Dr gksrh gSA

;fn lekUrj

prqHkqZt dk ,d fod.kZ] js[kk

y =3x

2

ds vuqfn'k gks] rks nwljs fod.kZ ds fy,&

(A*)

izo.krk

= 5

4

gksxhA

(B) y

vUr%[k.M

=

1

4

gksxkA

(C*) x

vUr%[k.M

=1

5

gksxkA

(D*)

fcUnq

(2, 1)

rFkk

(1, 9)

,d gh vksj fLFkr gksaxsA

Sol. x – y + 1 = 0

and 4x 3y 2 = 0 intersect at P(5,6) and y = 3x

2 not posess through P and Q,S are (2,3) , (4, 6)  

 T 1 , 3 2         P Q R S T Required line PT = 5x  4y 1 = 0 7. If log a , a_x x/2_{, log}

bx are in GP. then x is equal to

;fn

log a , ax x/2, logbx

xq.kksÙkj Js<+h esa gS] rks

x

gS&

(A*) log_a (log_ba) (B*) log(loga) log(logb)

loga 

(C) log (log b) _b _a (D) None of these

buesa ls dksbZ ugha

Sol. Given log a , a_x x/2_{, log}

bx are in GP.

fn;k x;k gS fd

log a , a_x x/2_{, log} bx

xq.kksÙkj Js<+h esa gSA

So

vr%

ax₌ x log a . log_bx  ax_{= (log} ba)  x = loga (logba)  x = loga log logb loga      _  x = log(loga) log(logb) loga 

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8. Match the column

Column - I Column - II

(A) If tan  + tan 3          + tan – 3         = k tan 3 (p) 0 then k equals

(B) If cos 2x + 2 cos x = 1, then sin2_{x (2 – cos}2 _x) _(q) _{– 1}

is equal to

(C) The value of 2tan

10        + 3 sec 10        – 4 cos 10        ,is (r) 1

(D) If ,,  and be four angles of a cyclic quadrilateral, (s) 3

then the value of cos + cos + cos + cos , is

LrEHk feyku dhft,

LrEHk

- I

LrEHk

- II

(A)

;fn

tan  + tan 3         + tan –3        = k tan 3 (p) 0

gks



rks

k

dk eku gS&

(B)

;fn

cos 2x + 2 cos x = 1

gks

,

rks

sin2_{x (2 – cos}2 _x) _(q) _{– 1}

dk eku gS&

(C) 2tan 10        + 3 sec 10        – 4 cos 10       

dk eku gS&

(r) 1 (D)

;fn

,, 

rFkk



,d pØh; prqHkqZt ds pkj dks.k gS rks

(s) 3

cos + cos + cos + cos 

dk eku gS &

Ans. (A) (s) ; (B) (r) ; (C) (p) ; (D) (p)

Sol. (A) tan  + tan 3

1 3 tan  _{ }     _ _   + tan 3 1 3 tan  _{ }     _ _   = tan  + 2

(tan 3 )(1 3 tan ) (tan 3 )(1 3 tan )

1 3 tan  _{ } _ _{ } _{ } _ _          = tan  + 2 2 2

4 tan 3 tan 3 4 tan 3 tan 3

1 3 tan  _{ } _{ } _ _{ } _{ }          = 2 2

(1 3 tan ) tan 8 tan 1 3 tan        = 3 2 9 tan 3 tan 1 3 tan      = 3 3 2 3 tan tan 1 3 tan  _{ } _    _ _    = 3 tan 3 (B) cos 2x + 2 cos x = 1 2 cos2_{x – 1 + 2 cos x = 1} cos2_{x + cos x – 1 = 0} cos x = 1 5 2   cos x = 5 1 2  sin2_{x (2 – cos}2_{x) = 1} (C) 2 tan 10        + 3 sec 10        – 4 cos 10       

(D) , , ,  be the angles of cyclic quadrilateral  +  = ,  +  = 

 cos  + cos  + cos  + cos 

(D) , , , 

pØh; prqHkqZt ds dks.k gS

 +  = ,  +  = 

(10)

DPP No. : 52 (JEE-Advanced)

Single choice Objective ('–1' negative marking) Q.1,2 (3 marks, 3 min.) [6, 6]

Multiple choice objective ('–1' negative marking) Q.3 to Q.8 (5 marks, 4 min.) [30, 24]

1. The graphs of functions f₁(x) = 2

1, if x 0 x 1, if 0 x 2 5, if x 2         _  , f₂(x) = log_{1 2}(x3) , f₃(x) = 23 – x_{and f} 4(x) = e {x}

where { . } denote the fractional part function are given (not in order) as :

(A) 5 4 3 2 1 1 2 3 4 5 O x y (B) –2 –1 1 2 3 e (C) 8 (D) 3 4

The correct order of graphs of functions f₁(x), f₂(x), f₃(x) and f₄(x) is

(A) ABCD (B) ACBD (C) CBDA (D*) ADCB

Qyu

f₁(x) = 2 1, x 0 x 1, 0 x 2 5, x 2         _  ;fn ;fn ;fn , , f₂(x) = log_{1 2}(x3) , f₃(x) = 23 – x

_rFkk

_f 4(x) = e {x}

_tgk¡

_{.}

_fHkUukRed

HkkxQyu dks iznf'kZr djrk gS] ds vkjs[k fuEu ¼Øe esa ugha½ gSµ

(A) 5 4 3 2 1 1 2 3 4 5 O x y (B) –2 –1 1 2 3 e (C) 8 (D) 3 4

Qyu

f₁(x), f₂(x), f₃(x)

,oa

f₄(x)

ds vkjs[kksa dk lgh Øe gksxkµ

(A) ABCD (B) ACBD (C) CBDA (D*) ADCB

Sol. f₁(x) = 2 1 , x 0 x 1 , 0 x 2 5 , x 2         _  5 4 3 2 1 1 2 3 4 5 O x y f₂(x) =log_{1 2}(x3)

(11)

3 4 f₃(x) = 23 – x 8 f₄(x) = e{x} –2 –1 1 2 3 e 2. tan2 16  + tan2 2 16  + tan2 3 16  + tan25 16  + tan2 6 16  + tan2 7 16  is equal to

(A) 24 (B*) 34 (C) 44 (D) None of these

tan2 16  + tan2 2 16  + tan23 16  + tan25 16  + tan2 6 16  + tan2 7 16 

dk eku gS&

(A) 24 (B*) 34 (C) 44 (D)

buesa ls dksbZ ugha

Sol. tan2 16  + tan2 2 16  + tan2 3 16  + ... + tan2 7 16  2 sec 1 16         + sec22 1 16         + ... + sec27 1 16         2 2 sec cos ec 16 16          + 22 22 sec cos ec 16 16          + 23 23 sec cos ec 16 16          – 6 = 2 2 1 sin cos 16 16   + 2 2 1 2 2 sin cos 16 16   + 2 2 1 3 3 sin cos 16 16   – 6 = 2 4 sin 8  + 2 4 sin 4  + 2 4 3 sin 8  – 6 = 4 2 2 1 1 sin cos 8 8                + 4 × 2 – 6 = 4 2 2 1 sin cos 8 8               + 2 = 2 4 4 sin 4   + 2 = 2 + 32 = 34

3. For the straight lines 4x + 3y – 7 = 0 and 24x + 7y – 31 = 0, the equation of (A*) bisector of the obtuse angle between them is x – 2y + 1 = 0

(B*) bisector of the acute angle between them is 2x + y – 3 = 0 (C*) bisector of the angle containing origin is x – 2y + 1 = 0

(12)

ljy js[kkvksa

4x + 3y – 7 = 0

,oa

24x + 7y – 31 = 0

ds fy, lehdj.k

(A*) x – 2y + 1 = 0

vf/kd dks.k v)Zd gSA

(B*) 2x + y – 3 = 0

U;wudks.k v)Zd gSaA

(C*) x – 2y + 1 = 0

ewy fcUnq fufgr dks.k dk v)Zd gSA

(D*) x – 2y + 1 = 0

fcUnq

(1, –2)

fufgr dks.k v)Zd gSA

Sol. 4x + 3y – 7 = 0

24x + 7y – 31 = 0 a1a2 + b1b2 > 0 obtuse angle bisector

20x + 15y – 35 = 24x + 7y – 31 4x – 8y + 4 = 0  2x – 2y + 1 = 0 acute Angle bisector

4x 3y 7 24x 7y 31 5 25      20x + 15y – 35 + 24x + 74 – 31 = 0 49x + 22y – 66 = 0 2x + y – 3 = 0 for origin L₁ < 0 L₂ < 0

some sign so origin lies in x – 24 + 1 = 0 for point (1 , –2) L₁ < 0 L₂ < 0

so some sign so point (1, –2) lies in x – 2x₁ + 1 = 0 Hindi. 4x + 3y – 7 = 0 24x + 7y – 31 = 0 a1a2 + b1b2 > 0

vf/kd dks.k v)Zd

20x + 15y – 35 = 24x + 7y – 31 4x – 8y + 4 = 0  2x – 2y + 1 = 0

U;wu dks.k v)Zd

4x 3y 7 24x 7y 31 5 25      20x + 15y – 35 + 24x + 74 – 31 = 0 49x + 22y – 66 = 0 2x + y – 3 = 0

ewy fcUnq ds fy,

L₁ < 0 L₂ < 0

leku fpUg vr% ewy fcUnq

x – 24 + 1 = 0

esa fufgr gSA

fcUnq

(1 , –2)

ds fy,

L₁ < 0 L₂ < 0

leku fpUg

vr% fcUnq

(1, –2) , x – 2x₁ + 1 = 0

esa fufgr gSA

4. The equation (3 + cos x)2_{= 4 – 2 sin}8_{x has}

(A*) exactly one solution in x  (0, 3) (B) exactly three solutions (C*) exactly two solutions x  [0, 5) (D) infinite solutions

lehdj.k

(3 + cos x)2_{= 4 – 2 sin}8_x

_{ds fy,}

(A) x  (0, 3)

esa dsoy ,d gy gS

(B)

Bhd rhu gy gS

(C*) x  [0, 5)

esa nks gy gS

(D)

vuUr gy gS

Sol. 4  LHS  16 and

vkSj

2  RHS  4  LHS = RHS = 4  cosx = –1  x = (2n + 1) x = , 3



5. The coordinates of a point on the line x + y = 3 such that the point is at equal distances from the lines |x| = |y| are

js[kk

x + y = 3

ij ,d fcUnq bl izdkj ls gS fd ;g fcUnq js[kkvksa

|x| = |y|

ls cjkcj nwjh ij gSA bl fcUnq ds

funsZ'kkad gS &

(13)

Sol. x + y = 3 (3, 0) y = x (0,3) y = –x

Obviously (3,0) , (0,3) as they lies on angle bisectors of y = x and y = – x Hindi. x + y = 3 (3, 0) y = x (0,3) y = –x

Li"Vr% ;s fcUnq

(3,0), (0,3)

gSA D;ksafd ;s

y = x

rFkk

y = – x

ds e/; cus dks.kksa ds v/kZd ij fLFkr gSA

6. Consider the equation, sin4_{x – cos}2_{x sin x + 2 sin}2_{x + sin x = 0 in 0  x  3}

(A*) Number of solution of the equation is 4 (B*) Sum of the solution of the equation is 6 (C*) Product of the solution of the equation is 0 (D) 2 is not the solution

0  x  3

esa lehdj.k

sin4_{x – cos}2_{x sin x + 2 sin}2_{x + sin x = 0}

_{ij fopkj dhft,A}

(A*)

lehdj.k ds gyksa dh la[;k

4

gSA

(B*)

lehdj.k ds gyksa dk ;ksx

6

gSA

(C*)

lehdj.k ds gyksa dk xq.kuQy

0

gSA

(D) 2

gy ugha gSA

Sol. sinx (sin3_{x – cos}2_{x) + 2 sin x + 1) = 0}

sin x(sin3_{x + sin}2_{x + 2 sin x) = 0}

sin2_{x (sin}2_{x + sin x + 2) = 0}

sin x = 0 x = n, n in 0  x  3

esa

x = 0, , 2, 3

7. If the lines L₁ : 2x – 3y – 6 = 0, L₂ : x + y – 4 = 0 and L₃ : x + 2 = 0 taken pair wise in order constitute the angles A, B and C respectively of ABC, then

(A*) tan A = 15 2

(B*) tan A tan B = 14 (C*) tan A tan B tan C = 15

2

(14)

;fn js[kkvksa

L₁ : 2x – 3y – 6 = 0, L₂ : x + y – 4 = 0

,ao

L₃ : x + 2 = 0

dks ;qXeksa ds Øe esa ysus ij f=kHkqt

ABC

ds

dks.k Øe'k%

A, B

,oa

C

izkIr gksrs gSa] rks &

(A*) tan A = 15 2

(B*) tan A tan B = 14 (C*) tan A tan B tan C= 15

2

(D*) tan A, tan B , tan C

ewyksa okyh lehdj.k

2x3_{– 15x}2_{+ 28x – 15 = 0}

_gSA

Sol. m₁ = 2/3 = tan m₂ = –1 = tab m₃ = 

 – A +  =  A =  +  –  ...

tan A = tan tan

1 tan tan       = 2 ( 1) 3 1 2 / 3    = 5 1

tan (B + 90) = –1 cot B = 1 tan B = 1

tan (90 - C) = 2 3  cot C = 2 3  tan C = 3 2 x3_– _{5 1} 3 2         x2₊ _{5 1 1} 3 3 _{5 x} 2 2            – 5×1× 3 2 = 0 2x3_{– 15x}2_{+ 28x – 15 = 0}

8. Consider the circle x2_{+ y}2_{– 10x – 6y + 30 = 0. Let O be the centre of the circle and tangent at A(7, 3)}

and B(5, 1) meet at C. Let S = 0 represents family of circles passing through A and B, then (A*) area of quadrilateral OACB = 4

(B) the radical axis for the family of circles S = 0 is x + y = 10

(C*) the smallest possible circle of the family S = 0 is x2_{+ y}2_{– 12x – 4y + 38 = 0}

(D*) the coordinates of point C are (7, 1)

ekukfd o`Ùk

x2_{+ y}2_{– 10x – 6y + 30 = 0}

_{gSA ekuk o`Ùk dk dsUnz}

_O

_{gS rFkk}

_{A(7, 3)}

_vkSj

_{B(5, 1)}

_{ij [khaph xbZ Li'kZ}

js[kk,¡

C

ij feyrh gSA ;fn

S = 0, A

rFkk

B

ls xqtjus okys o`Ùkks ds fudk; dks iznf'kZr djrk gS] rks&

(A*)

prqHkqZt

OACB

dk {ks=kQy

= 4

(B)

o`Ùkksa

S = 0

ds ifjokj ds fy, ewyk{k

x + y = 10

gSA

(C*)

o`Ùkksa ds fudk;

S = 0

dk lcls NksVk o`Ùk

x2_{+ y}2_{– 12x – 4y + 38 = 0}

_gSA

(D*)

fcUnq

C

ds funsZ'kkad

(7, 1)

gSA

Sol. Coordinates of O are (5, 3) and radius = 2

Equation of tangent at A(7, 3) is 7x + 3y – 5(x + 7) – 3 (y + 3) + 30 = 0

i.e. 2x – 14 = 0 i.e. x = 7

Equation of tangent at B(5, 1) is 5x + y – 5(x + 5) – 3(y + 1) + 30 = 0

i.e. – 2y + 2 = 0 i.e. y = 1

 coordinate of C are (7, 1)  area of OACB = 4

Equation of AB is x – y = 4 (radical axis)

Equation of the smallest circle is (x – 7) (x – 5) + (y – 3) (y – 1) = 0 i.e. x2_{+ y}2_{– 12x – 4y + 38 = 0}

Hindi. O

ds funsZ'kkad

(5, 3)

gS rFkk f=kT;k

= 2

A(7, 3)

ij Li'kZ js[kk dk lehdj.k

7x + 3y – 5(x + 7) – 3 (y + 3) + 30 = 0

gSA

vFkkZr

2x – 14 = 0

vFkkZr

x = 7

B(5, 1)

ij Li'kZ js[kk dk lehdj.k

5x + y – 5(x + 5) – 3(y + 1) + 30 = 0

gSA

vFkkZr

– 2y + 2 = 0

vFkkZr

y = 1

 C

ds funsZ'kakd

(7, 1)

gSA

 OACB

dk {ks=kQy

= 4

AB

dk lehdj.k

x – y = 4 (

ewyk{k

)

gSA

o`Ùk fudk; ds lcls NksVs o`Ùk dk lehdj.k fuEu gS &

(x – 7) (x – 5) + (y – 3) (y – 1) = 0