Solution

(1)

Corporate Office (New Campus): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005

DATE : 06-07-2014

CUMULATIVE TEST-2 (CT-2)

(JEE ADVANCE PATTERN)

TARGET : JEE (MAIN+ADVANCED) 2015

COURSE : VIJETA (JP)

HINTS & SOLUTIONS

_{¼lad sr ,oa gy ½}

Paper-1

Part-I Mathematics

1. Which of the following ……….

fuEufy f[kr Q y uksa d s ;qXeksa

……….

Sol. In (B) option, Domain of f and g is same

and f(x) = g(x)

fod Yi

(B)

esa

, f

rFkk

g

d k izkUr l eku gSA

vkSj

f(x) = g(x)

2. The value of tan 100º + 4 sin ………. tan 100º + 4 sin 100º

d k

………. Sol. tan 100º + 4 sin 100º =

sin100

2 sin 200

cos100

 



=

sin100

sin 200

cos100

 



=

2 sin150 cos 50

sin 20

cos100



 



=

cos 50

cos 70

sin10

 







=

2 sin 60 .sin10

sin10







= –

3

3. Th e s um o f t he s er ie s 1 +

3

5

+ 2

5

+ ……….

J s.kh

1 +

3

5

+ 2

5

+ 3

7

5

+ . .. .... .. ... .... .. . . S ol . Sn = 1 +

3

5

+ 2

5

+ 3

7

5

+ .. .. .. + n–1

(2n

– 1)

5

1

5

Sn =

1

5

+ 2

3

5

+ 3

5

+ .. .. . + n–1

(2n

– 3)

5

+ n

(2n

– 1)

5

S ub tra c ti ng

?kVkus ij

4

5

Sn = 1 +

2

5

+ 2

2

5

+… . . + n–1

2

5

– n

(2n

– 1)

5

=

3

2

– n

(4n

3)

2.5 

4. Let f(x) = xx ; x (1, ) and g(x) be inverse ……….

ekuk

f(x) = xx ; x (1, )

rFkk

……….... Sol. we have

fn;k x;k gS

f(g(x)) = g(x) g(x) = x also

iqu%

g (f (x) ) = x g (f (x)) . f(x) = 1 g (f (x)) =

1 f (x)



g (f (x)) = x

1 x .(1

 

n x)

  g  (f(g(x))) =





g(x)

1 g(x)

.(1

 

n(g(x)))

g (x) =

1 x (1

 

n g(x))

5. If matrix 11 12 13 21 22 23 31 32 33

a





















is the inverse ……….

;fn vkO;wg

11 12 13 21 22 23 31 32 33

a





















………. Sol. |A| = 10 a23 = th

cofactor of (3, 2)

entry of given matrix

| A |

(3, 2)

| A |

ok avo; o d k lg[k.M

    a23 =

1

1 –

0

1

10 

10 a23 = –1

Alternate solution

fod Yi gy %

1 2

1 1 0 0

0

2

1

0 1 0

0 0

5

0

1 





















~

1 2

1

0

0 1 1/ 2 0 1/ 2

0 0 0

1

0

0 1/ 5























~

1 2 0

1

0 1/ 5

0 1 0 0 1/ 2

1/10

0 0

1 0

0 1/ 5























 10a23 = – 1

(2)

6. If (tan–1_x)2_{+ (cot}–1_x)2 ……….

;fn

(tan–1 x)2 + (cot –1 x)2 = 2

5

8 

,

rc

………. Sol. (tan–1 x + cot –1 x)2 – 2tan–1x

– tan

–1

x

2 













= 2

5

8 

2(tan–1 x)2 – 2

2 













tan–1 (x) – 2

3

8 

= 0 tan–1 x = –

4 

x = – 1

7. If the function f : [1, )  [1, ) is defined ……….

;fn d ksbZ Q y u

f : [1, )  [1, )

bl izd kj

………. Sol. y = f(x) = 2x(x – 1)  x(x – 1) = log2y  x2– x – log2 y = 0  x = 2

1 1 4log y

2 



x  1 x =

1 1 4log y

2

2 



 f–1(x) =

1

2 (1



1 4 log x )



2

8. The value of k for which roots ………. k

d s eku ft ud s fy , l ehd j.k

……….

Sol. Let

ekuk

f(x) = 4x2– 4kx + k2 + 3k – 66 (i) f(2) > 0  k2– 5k – 50 > 0 (k + 5) (k – 10) > 0 k  (–, –5)  (10, ) (ii)

–

(

–4k)

2(4)

> 2  k > 4 (iii) D 0  16k2– 4.4 (k2 + 3k – 66)  0 – 16.3 (k – 22)  0  k  22 Hence

vr%

k  (10, 22] 9. Number of values of x ……….

lehd j.k

|2x – 3| + ……….

Sol. Let

ekuk

f(x) = |2x – 3| + |2x + 3|

g(x) = – |x| + 6

Only one solution at x = 0

d soy ,d gy

x = 0

ij

10. If the product (sin1º) (sin3º) (sin5º) (sin7º)...(sin89º) ……….

;fn xq.kuQ y

(sin1º) (sin3º) (sin5º) ……….

Sol. (sin1º) (sin3º) (sin5º) (sin7º)...(sin89º) = _n

1

2

[sin = cos(90 –)]

(cos1º) (cos3º) (cos5º) ...(cos89º) =

n

1

2



cos1

º cos 2º cos 3º...cos 88º cos 89º

cos 2

º cos 4º cos 6º...cos 88º



44

(2cos1

º sin1º )(2 cos 2º sin 2º )...(2 cos 4

4 º sin 44º )

2 cos 2

º cos 4º cos 6º...cos 88º

1 sin 45

º

2 













 ₄₄

sin 2

º sin 4º ...sin 88º

2 cos 2

º cos 4º...cos 88º

×

1

2

 44

1

2

× 1/ 2

1

2

 89 / 2

1

2

 n =

89

2

11. The curve y  exy + x = 0 has a vertical ……….

fuEu esa ls ft l fcUnq ij oØ

y  exy + x = 0

d s

……….

Sol. y  exy + x = 0

Differentiating w.r.t. to y y

d s lkis{k vod y u d jus ij

1 – exy

dx

. y

x

0 dy

dy

















dx

dy

= 0 1 – xexy = 0 xexy = 1 x = 1 , y = 0 Point is

fcUnq

(1, 0) 12. If x

lim

 3 2 3 2 3 2 4 4 4 2 a(2x – x ) b(x 5x – 1) – c(3x x ) a(5x – x) – bx c (4x 1) 2x 5x        ….

;fn

x

lim

 ………..

Sol. For limit to exist and equal to 1

coefficient of x4 in denominator = 0 (as degree of x in Dr > Nr) Now degree of x in Dr is 2 and degree of x in Nr is 3.  coefficient of x3 in Nr = 0 otherwise L  1 and will be infinity and 2 r 2 r

coefficient of x in N

coefficient of x in D

= 1 Now coefficient of x4 in Dr = 5a – b + 4c = 0 ...(1)  coefficient of x3 in Nr = 2a + b – 3c = 0 ...(2)   

–a

5b – c

2 

= 1 ;  a - 5b + c + 2 = 0 ...(3) On solving (1), (2) and (3)

(3)

Corporate Office (New Campus): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 a =

–2

109

, b =

46

109

and c =

14

109

 a + b + c =

58

109

=

p

q

 p + q = 167

Hindi.

_{lhek d s fo|eku gksus rFkk bld k eku}

1

_{gksus d s fy ,}

gj esa

x4

d k xq.kkad

= 0 (

paw

fd va'k esa

x

d h ?kkr

> Nr

)

vc gj esa

x

d h ?kkr

2

gS rFkk va'k esa

x

d h ?kkr

3

gSA

. 

va'k esa

x3

d k xq.kkad

= 0

vU;Fkk

L  1

rFkk vuUr gksxkA

rFkk

2 2

x

va'k e sa d k xq.kkad

gj e sa

d k xq.kkad

= 1

vc gj es

a

x4

d k xq.kkad

= 5a – b + 4c = 0 ...(1) 

va'k esa

x3

d k xq.kkad

= 2a + b – 3c = 0 ...(2)   

–a

5b – c

2 

= 1 ;  a - 5b + c + 2 = 0 ...(3) (1), (2)

rFkk

(3)

gy d jus ij

a =

–2

109

, b =

46

109 rFkk

c =

14

109

 a + b + c =

58

109

=

p

q

 p + q = 167

13. The coordinates of the feet of ……….

fcUnq

(14, 7)

ls

y2– 16x – 8y = 0 ……….

Sol. Given curve is y2– 16x – 8y = 0 ...(1) Let P  (14, 7) Equation (1) is (y – 4)2 = 16(x + 1) Equation of normal at (–1 + 4t2, 4 + 8t) is tx + y – 4t3– 7t – 4 = 0 passes through, (14, 7)  t = –1,

3

2

,

1

2 

foot are (0, 0) (3, – 4) (8, 16) Hindi.

fn;k x;k oØ

y2 – 16x – 8y = 0 ...(1)

ekuk

P  (14, 7)

lehd j.k

(1)

_gS

(y – 4)2 = 16(x + 1) (–1 + 4t2, 4 + 8t)

ij vfHky Ec d h lehd j.k

tx + y – 4t3– 7t – 4 = 0

gS t ks

(14, 7)

ls xqt jrh gSA

   t = –1,

3

2

,

1

2 

ikn gksxs

(0, 0) (3, – 4) (8, 16) 14. If f(n,) = n 2 r r 1

1 – tan

2



















then the value ……….

;fn

f(n,) = n 2 r r 1

1 – tan

2



















………. Sol. n 2 r r 1

1 – tan

2 

















=

2 tan



. 2

2 tan

2 



. 3 2

2 tan

2 



... n n–1

2 tan

2 



f(n,) = n n

2 tan



=

tan



as n  0

lim



tan















= 0 as

D;ksafd

tan >  as  0

15. The locus of point of intersection ………. y2

= 4x

d h mu Li'kZ js

[kkvksa

d s izfrPNsn

……….

Sol.

2 m

m

1 2



y = mx +

m

1

k = mh +

m

1

km = m2 h + 1 m2h – km + 1 = 0

m

1

m

2 m1 + m2 =

h

k

m1m2 =

h

1

m1 + 2m1 =

h

k

 m1 =

h

3 k

h

1 m

2

12



h

1 h

9 k

2

2 2





h = 0 or

;k

2k2 = 9h x = 0 or

;k

2y2 = 9x 16. If pth, qth, rth terms of an A.P. ……….

;fn fd lh l ekUrj Js<h

………..………. Sol. Given

_{fn;k x;k gS}

, k =

a

(q 1)d

a

(p 1)d









=

a

(r

1)d

a

(q 1)d









=

a

(q 1) d a

(r

1)d

a

(p 1)d a

(q 1)d









=

(q r)d

(p

q)d



=

(q r)

p

q



Since one root is 1, therefore, other root =

p

q

q r



=

1 k

pawfd ,d ewy

1

gS vr% nw

l jk ew

y

=

p

q

q r



=

1 k

(4)

17. The values of  satisfying sin7 = sin4–……….

lehd j.k

sin7 = sin4– sin

d ks l arq"V d jus

……….

Sol. sin 7 + sin = sin 4 2 sin 4 cos 3 = sin 4 2 sin 4cos 3– sin 4 = 0 sin 4 [2 cos 3– 1] = 0 sin 4 = 0 or

;k

1 cos 3

2  

4 



 

and

rFkk

9 

 

,

9

4 



 

18. If

g(x)

2h(x) | h(x) |







where ……….

;fn

g(x)

2h(x) | h(x) |







t gk¡

………. Sol. n n n n

2(sin x

sin x) | sin x

sin x |

g(x)

2(sin x

sin x) | sin x

sin x |











for 0 < n < 1

d s fy ,

, sinx < sinn x, g(x) =

1

3

and for n > 1 sinx > sinn

x

rFkk

n > 1

d s fy ,

sinx > sinn x g(x) = 3   for n > 1f(x) = 3, x (0, )  n > 1

d s fy ,

f(x) = 3, x (0, )  f(x) is continuous and differentiable at x =

2 



x

2 



ij

f(x)

lrr~ ,oa vod y uh; gksxkA

and for0 < n < 1

rFkk

0 < n < 1

d s fy ,

   

1 ;

x

0,

,

3

2

2 f(x)

3 ;

x

2 





































 









 





_





   

0 ;

x

0,

,

2

2 f(x)

3 ;

x

2 































_

_

_

_

 











    f(x) is not continuous at x =

2 

. Hence f(x) is not differentiable at x =

2 

 x =

2 

ij

f(x)

lrr~~ ugha gS vr%

f(x), x =

2 

ij vod y uh;

ugha gSA

19. I f f : R+  R+ a fu nc ti o n su c h ……….

;fn

f : R+  R+

Q y u bl

………. S ol. P u t x = 1 , y = 1 (f (1 ) )2 = f ( 1 ) + 6  f (1) = 3 , – 2 f (1) = 3 [ S in c e f(x) > 0 ] P u t y = 1 i n g iv en r el a ti o n f (x) f( 1 ) = f( x) + 2 (x + 2 ) 2 f(x ) = 2x + 4 f (x) = x + 2 f (2) = 4 Hi ndi . x = 1 , y = 1

j[kus ij

(f (1 ) )2 = f ( 1 ) + 6  f (1) = 3 , – 2 f (1) = 3 [

p w¡fd

f( x) > 0]

fn;s x;s l EcU/k esa

y = 1

j[kus ij

f (x) f( 1 ) = f( x) + 2 (x + 2 ) 2 f(x ) = 2x + 4 f (x) = x + 2 f (2) = 4 20. If f(x) = tan–1

3 sin x

4 5 cos x















, then ……….

;fn

f(x) = tan–1

3 sin x

4 5 cos x















,

rc

………. Sol. f (x) = 2 2

1 3cos x(4 5 cos x)– 3 sin x(–5 sin x) (4 5 cos x) 3 sin x 1 4 5 cos x             f (x) =





2 2 2 2 2 2

(4 5 cos x) 12cos x 15 cos x 15 sin x (4 5 cos x) (4 5 cos x) 3 sin x         f (x) = ₂ ₂

3(4 cos x

5)

16 40 cos x

25 cos x

9(1

– cos x)



 f (x) = ₂

3(4 cos x

5)

(4 cos x

5)



=

3 4 cos x



5

 f 

3 













=

3

7

21. If  = x 2012

[x]

im

{x}





and m = x 2012

{x}

im

[x]





……….

;fn

 = x 2012

[x]

im

{x}





rFkk¡

………. Sol. For  h 0 h 0 h 0 h 0

[2012 h]

2012

RHL

lim

not finite

{2012 h}

h

[2012

– h]

2011

LHL

lim

2011

{2012

– h}

1 – h

   







_

















  does not exist

for m  h 0 h 0

h

RHL

lim

0 2012

1 – h

1 LHL

lim

2011

 





_













(5)

Corporate Office (New Campus): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Hindi 

d s fy ,

h 0 h 0 h 0 h 0

[2012 h]

2012

RHL

lim

{2012 h}

h

[2012

– h]

2011

LHL

lim

2011

{2012

– h}

1 – h

   







_















ifjfer ugha



 

fo| eku ugha gSA

m

d s fy ,

h 0 h 0

h

RHL

lim

0 2012

1 – h

1 LHL

lim

2011

 





_













 m

fo| eku

ugha gSA

22. The locus of a point that divides a chord ……….

,d fcUnq d k fcUnqiFk t ksfd ijoy ;

……….

S ol . Le t th e eq u a ti on o f t he l i ne b e

ekuk js[kk d k l ehd j.k gSA

y – 2t = 2 ( x – t2) y = 2x + 2 t – 2t2 , t he n

rc

y2 = 2 ( y – 2 t + 2t2) y2 – 2 y + 4t – 4 t2 = 0 (y – 2 t) (y + 2t – 2 ) = 0  y = 2 – 2 t i s t he o th e r en d.  y = 2 – 2 t

nwl jk fl jk gksxkA

 Th e en ds of t he c ho r d a r e (t2, 2 t) a nd ( ( 1 – t)2 , 2(1 – t) ) t he po i nt whi c h d i v i de s th i s c h ord i n th e rat i o 1 : 2 i nt er na ll y is 2 2

2t

(1 t)

4t

2 2t

,

3

3 



















 

v r% t hok d s fl js

( t2, 2 t)

rFkk

( (1 – t )2 , 2 ( 1 – t ) )

gS bl t hok d ks

1 : 2

d s v uqikr esa v Ur% foHkkft r

d jus oky k fcUnq

2 2

2t

(1 t)

4t

2 2t

,

3

3 



















gSA

it s l oc us is giv en b y el i mi n at in g t be t we en

t

d ks foHkkft r d jus ij v Hkh"V fcUnqiFk izkIr gksxkA

x = 2

3t

2t

1

3 



, y =

2t

2

3 

2

8 y

9 













=

4

9

2 x

9 













a =

2

9

b =

8

9

  81

(a



b)

= 90 23. If roots of x2 – ax + b = 0 differ by 2 then ……….

;fn

x2 – ax + b = 0

d s ewy ksa

……….

Sol. Let roots be ,  + 2  +  + 2 = a

ekuk ewy

,  + 2  +  + 2 = a 

a

– 2

2  

and

rFkk

(

2)

b

a

– 2

a – 2

2 b

2

2 



  





_



_







,  a2–4 = 4b  a2 = 4(b + 1) locus of (a,b) is x2

= 4(y + 1) which is a parabola having vertex at (0, –1).

(a,b)

d k fcUnqiFk

x2 = 4(y + 1)

gS t ks

(0, –1)

'kh"kZ oky k

ijoy ; gSA

24. If cos(x – y), cos x, cos(x + y) ……….

;fn

cos(x – y), cos x, ……….

Sol. cos2x = cos(x + y) cos(x – y) cos2x = cos2x – sin2y

 sin y = 0  y = n,n  r =

cos x

cos(x

– y)

= secy = 1 or – 1 sum

;ksx

= 1 – 1 = 0

25. P is any point on the parabola ……….

P

ijoy ;

y2 = 4ax

ij fLFkr

………. Sol. M  (– a, 2at) D 

2a

a,

t















Slope of SM

d h izo.krk

m1 =

2at

a



= – t ; Slope of SD

d h izo.krk

m2 =

2a / t

a



=

1 t

Here

;gk¡

m1m2 = – 1 MSD =

2 

(6)

Part-II

Physics

26. In given arrangement all resistors are of 1...

iznf'kZr ifjiFk esa l Hkh izfrjks/kksa

d k eku

1...

Sol. A B    X _Y

By symmetry we can say that current through an will be equal to current through XB so current through XY will be zero.

lefefr ls ge d g ld rs gS fd

AX

ls t kus oky h /kkjk

XB

l s

t kus oky h /kkjk d s cjkcj gSA vr%

XY

ls t kus oky h /kkjk 'kwU;

gksxhA

A

B



Now lets say equivalent resistance between A and B is P.

vc ge d g ld rs gS fd

A

o

B

d s

e/; rqY; izfrjks/k

R

gSA

A B R 1 1 1 1 R =

2 (2

R)

2 (2 R)





4R + R2 = 4 + 2R R2 + 2R – 4 = 0 R = (

5

– 1) 

27. Figure shows an inclined plane of inclination 37°...

fp=k esa

37°

urd ks.k d k ur ry iznf'kZr gSA ft ld k

...

Sol: Consider the motion of particle at y

y

ij d .k d h xfr d ks y sus ij

x =

1

2

g(sin –cos)t 2 3y2 =

1

2

10(

3

5

–

4

5

)4 3y2 = (3–4)4 3 –

3

4

y 2 = 4 





2

3 4

y

16 

28. Figure shows two half coplanar rings of radii...

fp=k esa

a

rFkk

b

f=kT;k d h nks lery h; v) Z oy ;

... Sol. Method- : dQ =

Q



d dp =

Q



(b – a) d p = 0

Q

(b

a)sin d





 





=

2Q(b



a)



 bd Method- :

Considering charge to be concentrated at centre of mass

vkos'k d ks nzO;eku d s

fUnzr ekuus ij

2a  + – 2b  p p =

2Q(b



a)



29. A tube of length 30 cm has its inner lateral...

30 cm

y Ech uy h d h vkUrfjd ik'oZ

(lateral) ...

Sol.  /4  4 1 2



=

1

16

(7)

30. Refractive index of a glass cube is

2

. A ray of...

?kukd kj d k¡p d k viorZukad

2 gSA iz

d k'k d h

...

Sol. Glass cube 45°  60°  45° d k¡p d k ?ku Applying snell's law at first surface

izFke l rg ij Lusy d k fu;e y xkus ij

1 sin 45° =

2

× sin  = 30°

Since angle of incidence on horizontal face is more than 45° so the ray will suffer TIR on horizontal face as shown in the figure.

pwafd {kSfrt Q y d ij vkiru d ks.k

45°

ls vf/kd gS vr% fp=k esa

n'kkZ;suqlkj {kSfrt Q y d ij fd j.k d k iw.kZ vkUrfjd ijkorZu

gksxkA

So emergent ray will be parallel to incident ray.

vr% fuxZr fd j.k vkifrr fd j.k d s lekUrj gksxhA

31. In the figure shown, the coefficient of friction...

fp=k esa n'kkZ;s x, Cy kWd

B

o

C

rFkk

B

o

A... Sol. N0 N N mg  N N N F N FBD of C FBD of B F = mg + (cos + sin) N’ ....(1) N’ cos N’sin+ N N’ 

mg

cos

  

sin



....(2) (1) & (2)

tan

F

mg

775N

1 tan



  





_

 

_



 







32. Consider a triangular surface whose vertices...

,d f=kHkqt kd kj lrg d h d Yiuk d hft ,A ft ld s

... Sol. C A B y x z O net = 0

ABC = –AOB + BOC + COA]

=

2 a

E

0 2 + 2Ea2 + 3.E0a2 =

E

₀

a

2

11

33. When an ideal ammeter is inserted in series...

;fn ,d vkn'kZ vehVj ifjiFk esa Js.khØ e esa

y xk;k

...

Sol. Resistance

izfrjks/k

X = 1

12V 1 r  V =

12 1 r



× 1 = 10 1 + r = 1.2 r = 0.2

34. In the given circuit an ideal voltmeter reads 3V...

fn;s x;s ifjiFk esa vkn'kZ oksYVehVj d k ikB~;kad

3V...

Sol.

3 R

1

3

1

5 .

4

1

0

3

6

5 .

4 E



























9 R

9

3

9

2

3

9

18

9 E

2

2E –27R + 3 = 0

For calculation of current equivalent circuit

/kkjk d h x.kuk d s fy , rqY; ifjiFk

E 4.5 3 R R 6  R 3 R 3   = 2A E –



















R

3 R

3

5 .

4

2

3 R

R

6

solving there two we set E = 12V

(8)

35. Two blocks of 4 kg and 6 kg are attached... 4 kg

rFkk

6 kg

nzO;eku d s nks Cy kWd ,d

... Sol. 10g 6g 4g 6g 6g 6g a =

6g

4

= 15 m/s 2

36. A particle of mass 1 kg & charge

1

3

C is projected... 1 kg

nzO;eku rFkk

1

3

C

vkos'k d k ,d d .k] t M+or~

... Sol.

u

v

R

q

4

1 mv

2

1 )

u

(

1

2

1

2 0 2 2









m × u × 0.5 = mv × 1 v = u/2 u2 –

4 u

2 = 9 × 109× 2 ×

3

1

×

3

1

× ₃ 6

10

1

10

 



2

4 u

3

2



u =

3

2

37. A uniform solid sphere of mass ‘m’ rests... ‘m’

nzO;eku d k ,d le: i Bksl xksy k

...

Sol.  – mg T  N





T cos









mgsin







mgsin

T

cos









38. Consider a spherical symmetric distribution...

,d l fEer xksy h; vkos'k forj.k d h d Yiuk

...

Sol.  = 0 r 4 Q = 00 r 4

dQ

dV

= 2

dQ

4 r dr



= 3 0 0 2

4r

4 r

 



 = 0 0

r

 



39. A block of mass m slides on a frictionless... m

nzO;eku d k Cy kWd ?k"kZ.kghu esat ij fQ ly

...

Sol.

dV

dt

= – 2

V





V 2 V₀

dV

V





= t 0

dt







0

1

1 V



V

=

t





1 V

= 0

1 V

0

V t

1 

















V = 0 0

V

V t

1 

















40. An object is moving in front of two mirrors...

,d oLrq nks niZ.kksa d s lkeus xfr'khy

...

Sol. M2 V0 V0sin  V0cos  V0sin  V0 V0cos 2V0sin  V0   V0cos  V0cos 

41. The diagram shows three infinitely long...

iznf'kZr fp=k esa rhu vuUr js[kh; le: i

...

Sol.

y

z

x

2 

3 



W3 =

n

2

3

0







W2 = 0 W =

n

2

0







WT = 0

2 n





(9)

42. In given arrangement all contacts are smooth...

iznf'kZr O;oLFkk esa l Hkh l Eid Z fpd us gS

...

Sol. 2kg 2kg A B xA = 2m vA = 4m/s a = 4m/s2 T = 8N 4N xB = 4m vB = 8m/s b = 2a = 8m/s2 20 – T = 2(2a) 2T = 2a

Solving above equations a = 4 m/s2

mijksDr l ehd j.kksa d ks gy d jus ij

a = 4 m/s2 Wg = 20 × 4 = +80 J

WTB = 4(4)(–1) = –16J WTA = 8 × 2 × 1 = +16J WTA + WTB = 0

Wall = K = 80

43. Speed of a particle moving on circular...

o`Ùkh; iFk ij xfr'khy d .k d h pky

,

le;

...

Sol. at =

dv

2 dt



ar = 2 2 2

v

t

R



1 

Fnet = m 2 2 t r

a



a

= 1

2 

t

2 P =

ma .v

t



= 1 × 2× 2t = 4t W =

Pdt



= 1 1 ₂ 0 0

4t

4tdt

2J

2 





_

_









Work done in one cycle is

m

(

2 as

)

2

1

= ma 2r

,d pØ esa fd ;k x;k d k;Z

m

(

2 as

)

2

1

= ma 2r = 1 × 2 × 2 × 3.14 = 12.56 J

44. A block is moved very slowly from A to D...

,d t M+or~ oØ kd kj iFk ij ,d Cy kW

d

A

ls

D...

Sol. Work done by friction

?k"kZ.k }kjk fd ;k x;k d k;Z

=









mg

cos

d



=

mg

dx







= – 3 mg



45. Temperature coefficient of specific resistance...

csy ukd kj rkj d s inkFkZ d k fof'k"V izfrjks/k

...

Sol. R = 

2 

R



=





+





–

A



R



= (2 + 1– 22) t = ( 1–2) t

46. Consider a conducting medium having...

,d pky d ek/;e d h d Yiuk d hft , ft lesa

...

Sol.  =

J. da







=

_



E . da





 =

1 

E. da







 =  = 10 V.m

47. A straight Nichrome wire is initially...

,d lh/kk ukbØ kse d k rkj çkjEHk esa d {kh;

...

Sol. For steady state

LFkkbZ voLFkk d s fy ,

in

dt

dQ













= out

dt

dQ













(V) (iss) = 45(T – 20) (500) (4.5) = 45(T – 20) Tss = 70ºC.

48. A uniform rope of length L and mass M...

[kqjnjh {kSfrt lrg ij fLFkr ,d

L

y EckbZ

... Sol. f = Mg

Mg/2

F

F –

Mg

2

= Ma a =

g

2 Mg/4

T

M/2

T –

Mg

4

=

M

2 g

2

T =

Mg

2

49. Two Stars of mass 4m and m respectively... 4m

rFkk

m

nzO;eku d s nks rkjs ,d &nwljs d s

... Sol. m1v1 = m2v2 1 2

K

= 2 1

m

=

1

4

50. A satellite of mass m orbits the earth...

,d

m

nzO;eku d k mixzg i`Foh d s pkjksa

...

(10)

Part-III

Chemistry

51. A mixture of CO and CO2………

20 mL CO

rFkk

CO2

d s feJ.k

………

Sol. let

ekuk

CO = a mL & CO2 = b mL

So

vr%

(a + b) = 20 mL CO +

1

2

O2





CO2 Initially a x b

izkjEHk esa

–

a

x

2 













(b+a) 

a

x

2 













+ a + b = 16 + x b +

a

2

= 16 

a

2

= 4 or a = 8 mL b = 12 mL

40 mL of given mixture contain CO =

40

20

x 8 = 16 mL

fn;s x;s

40 mL

feJ.k esa mifLFkr

CO =

40

20

x 8 = 16 mL

remaining after absorption of CO2 by NaOH

NaOH

}kjk

CO2

d s vo'kks"k.k d s i'pkr~

16 mL

vk;ru 'ks"k jgrk

gSA

54. Which of the following galvanic cells ………

fuEu fy f[kr xSYosfud lSy ksa esa ls d kSulk

………

Sol. Since cathode and anode are present in the same solution

hence liquid – liquid junction potential is elimenated in option (A).

D;ksafd d SFkksM rFkk ,uksM nksuksa leku foy ;u esa mifLFkr gSaA

blfy ;s fod Yi

(A)

esa nzo nzo t aD'ku foHko ugha ik;k t krk gSA

55.

G

0_A = molar Gibbs energy ………

0 A

G

=

?kVd

A

d h eksy j fxCl

……… Sol. 2A

_{}

G 200 B

_{}

G ? 2C G 50 kJ/ mol    



D 0 A

G



100kJ / mol

x y 0 D

G



50 kJ / mol

For first step For third step x – 2 x 100 = 200 50 – 2y = –50

x = 400 kJ/mol y = 50 kJ/mol

Therefore for second step G° = 50 x 2 – 400 = – 300 kJ/mol 2A

_{}

G 200 B

_{}

  G ? 2C G 50 kJ / mol    



D 0 A

G



100kJ / mol

x y

G

0_D



50 kJ / mol

izFke in d s fy ;s

r`rh; in d s fy ;s

x – 2 x 100 = 200 50 – 2y = –50 x = 400 kJ/mol y = 50 kJ/mol

vr% f}rh; in d s fy ;s

G° = 50 x 2 – 400 = – 300 kJ/mol

56. How does the pressure to density ratio ……… 273 K

rki ij nkc d s lkFk

He

d s

nkc

……… Sol. P(Vm– b) = RT P(V –nb) = nRT P(V –

W

M

b) = nRT P[1–

W

V

.

b

M

] =

W

M

x

RT

V

P[1–

bd

M

] =

dRT

M













P –

bpd

M

=

RTd

M

d

RT

bp

M















= P

P

d













=

RT

M

+

b

M













x P

P

d













=

b

M













x P +

RT

M

y = mx + C

58. In the reaction [CoCl2(NH3)4] + ………

vfHkfØ ;k

[CoCl2(NH3)4] + + Cl– ……… Sol. Cl  symmetrical

lefer

only single product

d soy ,d y mRikn

Cl





two isomers product

nks l eko;oh mRikn

replacable positions

izfrLFkkiu fLFkfr

(11)

59. An aqueous solution (2 litre volume) ………

,d t y h;

(2

y hVj vk;ru

)

foy ;u

……… Sol. Ka = 2 2 2

[H ] [A

]

[H A]

  = 2 2

[4C] [A

]

C

 So

blfy ,

A2– =

k

a

16C

60. The number of OH units directly linked ……… Na2B4O7 . 10H2O

esa cksjkWu ijek.kq

………

Sol. The correct formula of borax is Na2[B4O5(OH)4].8H2O.

cksjsDl d k lgh l w=k

Na2[B4O5(OH)4].8H2O

gSA

61. Diborane upon hydrolysis gives ………

MkbZ

cksjsu t y vi?kVu ij nsrk gS

………

Sol. B2H6 + 6H2O





2H3BO3 + 6H2

Orthoboric acid

vkFkksZcksfjd vEy

64. The most reactive carbonyl compound towards ………

KCN/H

d s lkFk lk;uksgkbMªhu fuekZ.k gsrq

……… Sol. The

_–

_C

_–

O

group in

CH

3

–

C

O

H

is most electrophilic.

CH

3

–

C

O

H

esa

–

C

–

O

lewg vf/kd by sDVªkWuLusgh gSA

65. Which of the following compound does not produce ………

fuEu esa ls d kSulk ;kSfxd t y h; vEy h; ek/;e

………

Sol. On hydrolysis “D” gives a ketone.

t y vi?kVu ij

“D”

d hVksu nsrk gSA

66. Which of the following is not correct ………

fuEu esa ls d kSulk Ø e czsd sV esa nh xbZ vfHkfØ ;k

………

Sol. In SN2Ar, the reactivity order is F > Cl > Br > I.

SN2Ar

esa

,

fØ ;k'khy rk d k Ø e gS %

F > Cl > Br > I.

67. Which of the following ester produces single alcohol ………

fuEu esa ls d kSulk ,LVj

MeMgBr

d s vkf/kD; d s lkFk

………

Sol. (D) gives only one product t-butylalcohol, while all other gives

a mixture of product.

(D)

;kSfxd d soy ,d gh mRikn

t-

C;wfVy ,Yd ksgkWy nsrk gS] t cfd

lHkh vU; mRiknksa d k feJ.k nsrs gS

A

68. Which of the following statement is incorrect ………

fuEu esa ls d kSulk d Fku lgh ugha gS

………

Sol. Addition of HBr to alkene is electrophilic addition reaction

occurring through a open carbocation, hence it is not stereospecific.

,Yd hu ij

HBr

d k ;ksx [kqy s d kcZ/kuk;u }kjk gks

us oky h

by sDVªkWuLusgh ;ksx vfHkfØ ;k gS] vr% ;g ,d f=kfoefof'k"B

vfHkfØ ;k ugha gSA

69. In the following reaction sequence which is/are ………

fuEufy f[kr vfHkfØ ;k izØ e esa

d kSul h vfHkfØ ;k

………

Sol. II is not possible since Cl¯ is a much weaker base than

2

NH



. IV is possible as acid base reaction but not as SN2Th

since carboxylate ion formed is not attacked by a nucleophilie.

vfHkfØ ;k

II

l EHko ugha gS D;ksafd

Cl¯

vk;u]

NH

₂

vk;u d h

rqy uk esa cgqr vf/kd nqcZy gksrk gSA vfHkfØ ;k

IV

lEHko ugha gS t ks

vEy &{kkj vfHkfØ ;k gS y sfd u

SN2Th

ugha gS D;ksafd vfHkfØ ;k esa

fufeZr d kcksZfDly SV vk;u ukfHkd Lusgh ij vkØ e.k ugha d jrk gSA

70. Amongst the given series of compounds which lies ………

fn;s x;s ;kSfxd d h lfØ ;rk d s ?kVrs Ø e d h Js.kh

……… Sol. Order

Ø e

A > B > D > C 72. Ph—CC—Ph





x





y ……… Sol. Ph – C  C – Ph _Na₍_NH ₍ ₎₎ X 3





4 2 CCl Br    

73. Consider the following compounds ………

fuEu ;kSfxd ksa ij fopkj d hft ,

………

Sol. Reactivity of SN1 is proportional to the stability of carbocation.

SN1

vfHkfØ ;k d h fØ ;k'khy rk d kcZ/kuk;u d s LFkkf;Ro d s

lekuqikrh gksrh gSA

74. Which of the following is not biomolecular ………

fuEu es

a ls d kSulh vfHkfØ ;k f}v.kqd ukfHkd Lusgh

………

Sol. Reaction (E) is acid-base reaction in Ist

step.

O

H

OC

2

H

5 – 2 5 C H O





O

OC

2

H

5 to further reaction





vfHkfØ ;k

(E) Ist

in esa vEy {kkj vfHkfØ ;k gSA

O

H

OC

2

H

5 – 2 5 C H O





O

OC

2

H

5



vf/kd vfHkfØ ;k



75. Which of the following statement is true for the reaction

………

uhps nh xbZ vfHkfØ ;k d s l UnHkZ esa fuEu es ls

………

Sol. cis-2-Butene undergoes hydroxylation by syn addition

forming meso product.

lei{k

-2-

C;wVhu d s ;ksx }kjk gkbMªksfDly hd j.k ij felks&mRikn

curk gSA

(12)

Paper-2

Part-I Mathematics

1. If A and B are symmetric ………

;fn

A

rFkk

B

l eku Ø e

………

Sol. Given

fn;k x;k gS

BT = B and

rFkk

AT = A and

rFkk

AB = BA  ABA–1 = BAA–1  B = ABA–1  A–1B = A–1ABA–1 __A–1B = BA–1 Similarly

blh izd kj

AB–1 = B–1A Now

vc

(A) (A–1B)T = BT(A–1)T = BT(AT)–1 = BA–1 = A–1B (B) (AB–1)T = (B–1)TAT = (BT)–1 AT = B–1A = AB–1 (C) (A–1B–1)T = (B–1)T (A–1)T= (BT)–1 (AT)–1 = B–

1A–1 = A–1B–1

2. Square of the slope of common ……… y2 = 4x

rFkk

x2 + y2 + 8x + 7 = 0

d h

……… Sol. y = mx +

m

1

x3 + y2 + 8x + 7 = 0 centre

d s

Unz

= (–4, 0) r = 3

3 m

1 m

4

2







16m2 + 2

m

1

– 8 = 9 (1 + m2) 7m2 + 2

m

1

– 17 = 0 7m4– 17m2 + 1 = 0 m2 =

14

28

289

17 



 m2 =

14

261

17 

3. If sin(cos ) = cos( sin ) ………

;fn

sin(cos ) = cos( sin ) ………

Sol. sin

( cos )

sin

2 







 

_

 



_





cos

sin

2 



 

 



(cos

sin )

2 







2

1 (cos

sin )

4 





1 2 sin .cos

1

4 





3

3 sin 2

sin 2

4

4 









 

4. If a, b, c are in H.P., then which of the ………

;fn

a, b, c

gjkRed Js<h esa

gS] rks

……… Sol. (A)

a

1 2a



,

b

1 2b



,

c

1 2c



are in H.P.

gjkRed Js<h esa

gSA

   

1 2a

a



,

1 2b

b



,

1 2c

c



are in A.P.

lekUrj

Js<h esa

gSA



1 a

,

1 b

,

1 c

are in A.P.

lekUrj Js<h esa gSA



   a, b, c are in H.P.

gjkRed Js<h esa gS

A

(B)a –

b

2

,

b

2

, c –

b

2

are in G.P.    

xq.kksÙkj Js<h esa gSA

   n

b

a

2 













, n

b

2

, n

b

c

2 













are in A.P.

lekUrj Js<h esa gSA

 (C) c –

b

2

,

b

2

, a –

b

2

are in G.P.

xq.kksÙkj Js<h esa gSA

(D)

1 a

,

1 b

,

1 c

are in A.P.

lekUrj Js<h esa gSA

   e1/a, e1/b, e1/c are in G.P.

xq.kksÙkj Js<h esa gSA

5. Let a and b be the number of ………

ekuk

a

rFkk

b

lehd j.k

………

Sol. Consider

ekuk

2x

x 1



– | x | = 2

x

| x 1|



i.e. 2 – |x – 1| = |x| , where

t gk¡

x 1 or

;k

|x| = 0 case

fLFfkr

- x < 0, then

rc

2 + x – 1 = – x i.e. x = –

1

2

case

fLFkfr

- 0 < x < 1

2 + x – 1 = x (not possible)

laHko ugha

case

fLFkfr

-III x  1

2 – x + 1 = x i.e. x =

3

2

but x = 0 is also solution

ijUrq

x = 0

Hkh gy gSA

Hence

vr%

a = 3, b = 1

(13)

6. Let f(x) = max {1 + sinx, 1, 1 – cosx} ………

ekuk

f(x) = max {1 + sinx, 1, 1 – cosx} ………

Sol. f(x) =

1 sin x ,

0 x

3 / 4

1 – cos x ,

3 / 4

x

3 / 2

1 ,

3 / 2

x

2 



















_

_

_



g(x) =

1 – x ,

x

0

1 ,

0 x

2 x

– 1 ,

x

2 











_



f(0) = 1  g(f(0)) = 1 f(1) = 1 + sin1  0 < 1 <

3

4 

 g(f(1)) = 1  1 < 1 + sin1 < 2 again

iqu%

g(1) = 1 f(g(1)) = 1 + sin1  g(0) = 1  f(g(0)) = 1 + sin1

7. Which of the following function(s) ………

fuEufy f[kr es

a ls d kSu ls Q y u]

………

Sol. Option (A), (C) obvious.

fod Yi

(A), (C)

Li"Vr%

Option (D) limiting value = sin 1

fod Yi

(D)

lhekUr eku

= sin 1

But functional value = 1, so it is discontinuous.

y sfd u Q y u d k eku

= 1,

vr% ;g vlrr~ gSA

8. If p is a constant, f(x) ………

;fn

p

d ksbZ vpj gS]

f(x) ……… Sol. f(x) = 2 2 3 2 6x 12x 2 3 6 p p p f(x) = 0   12(2p2– 3p) x2– 6x (2p3– 6p) + 6p2(p – 2) = 0   12p(2p – 3) x2– 12p (p2– 3) x + 6p2(p – 2) = 0 compare with ax2 + bx + c = 0 ax2 + bx + c = 0

l s rqy uk d jus ij

at p = 0, it is an identity (a = b = c = 0) p = 0

ij ;g ,d loZlfed k gS

(a = b = c = 0) at p =

3

,

b

a

= 0. So roots are opposite in sign and equal

in magnitude

p =

3 ij

b

a

= 0.

vr% ewy ifjek.k esa leku ,oa foijhr

fpUg d s gksxsaA

at p = 2,

c

a

= 0, so product of roots is zero

p = 2

ij

c

a

= 0,

vr% ewy ksa

d k xq.kuQ y 'kwU; gS

A

at p = –

3

,

c

a

= 2

6p (p

2)

12p(2p

3)



= – ve, p = –

3 ij

c

a

= 2

6p (p

2)

12p(2p

3)



=

_ .kkRed

so product of roots is negative

vr% ewy ksa d k xq.kuQ y

_ .kkRed gksxkA

9. Matrix ………

vkO;wg

C

gS

………

10. The value of |B| ……… |B|

d k eku fuEu d s

………

Sol. (9 to 10) Let

_ekuk

A =

11 12 13 21 22 23 31 32 33

a





















Now

vc

AC = B  C = A–1B =

1 | A |

.adjA.B  C =

1 | A |

11 21 31 12 22 32 13 23 33 C C C C C C C C C           12 13 11 13 11 12 22 23 21 23 21 22 32 33 31 33 31 32

a



























=

1 | A |

.

0 | A | | A |

| A |

0 | A |

| A | | A |

0 



















 C =

0

1

1 1 0

1 1 1 0





















which is symmetric

t ks l efer gS

A

 |C| = 2  |A–1B| = 2  |B| = 2|A| 11. The domain of ………

Q y u

 

f x

g x

d k

……… 12. The domain of ………

Q y u

_

_

1 g( x )

f(x)

d k

………

Sol. Let

ekuk

g'(1) = a & g"(2) = b then

rc

f(x) = x2

+ ax + b

and

rFkk

g(x) = (1 + a + b)x2 + x(2x + a) + 2 = (a + b + 3)x2 + ax + 2

 g'(x) = 2(a + b + 3)x + a &

rFkk

g"(x) = 2(a + b + 3)  g'(1) = 2(a + b + 3) + a = a &

rFkk

g"(2) = 2(a + b + 3) = b

 a + b + 3 = 0 & 2a + b + 6 = 0  f(x) = x2– 3x and

rFkk

g(x) = –3x + 2

(14)

13. The ordered pair ………

Ø fer ;qXe

……… 14. Range of f(x), where ……… f(x)

d k ifjlj t gk¡

x  [–2, –1] ………

Sol. 13. –a + b = –a – 1 + 2b using continuity

–a + b = –a – 1 + 2b

l rr~rk~ d s mi;ksx ls

 b = 1 f (x) = ₂

a

,

x

1 3ax

1 , x

1  









 





 a = 3a + 1  a = –

1

2

14. y = f(x) = 3

x

1 , x

1

2 x

x

2 , x

1

2 





 







_

_

_{ }





For x  [–2, –1], f(x) is decreasing x  [–2, –1], f(x)

á leku gSA

f(–2) = 2 and

rFkk

f(–1) =

3

2

 f(x) 

3 , 2

2 











... (i) for x  (1, 2), f(x) is decreasing x  (1, 2), f(x)

á l eku gSA

f(1) =

5

2

and

rFkk

f(2) = 0  f(x) 

5 0,

2 











... (ii)

from (i) and (ii), we get f(x) 

5 0,

2 











(i)

rFkk

(ii)

ls

f(x) 

5 0,

2 











Hence the range is

0,

5

2 











vr% ifjlj

0,

5

2 











gSA

15. Let A = 2

3x

1 6x





















, B = [a b c] and C ………

ekuk

A = 2

3x

1 6x





















, B = [a b c]

rFkk

……… Sol. tr(AB) = tr(C) 3ax2 + b + 6cx = 6x2 + 6x + 4  x R a = 2, b = 4 and

rFkk

c = 1 16. Two tangents y = m1x + c1………

oØ

5x2– y2 = 5

ij fcUnq

(2, 8)

ls

……… Sol. y – 8 = m(x – 2)  y = mx + 8 – 2m

Condition of tangency

Li'kZ

gsrw izfrcU/k

 (8 – 2m)2 = m2– 5  m = 3,

23

3

 c = 2, c =

22

3 

respectively  m = 3

ij

c = 2, m =

23

3 ij

c =

22

3 

m1c1 + m2c2 = 6 –

23 22

9 

_{= 6} –

506

9

=

54

506

9 

=

452

9

 1 1 2 2

9 (m c

m c )

113 

= 4 17. Two curves C1 : y = x 2 – 3 and C2 : ………

nks oØ

C1 : y = x 2 – 3

rFkk

………

Sol. Point A lies on C1 and C2

fcUnq

A, C1

rFkk

C2

ij fLFkr gSA

y1 = a2– 3 y1 = ka2  a2– 3 = ka2 ...(1) Now

vc

y = kx2 

dy

dx

= 2 kx  1 (a, y )

dy

dx













= 2ka =

y

2

y

1

1 a



But y2 = 1 –3 = –2  2 ka = 2

1 a



= 1+ a ...(2) Substituting

k = 2 2

a

3 a



j[kus ij

 2 2

2a(a

3)

a



= 1 + a  a = +3 and

rFkk

a = – 2 Rejected

vekU;

(15)

Corporate Office (New Campus): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 18. If function f :

0,

5 _

a, b

2 

















……… f :

0,

5 _

a, b

2 

















,d Q y u

……… Sol.(3) f(x) =

1

0

1

2

1

2

3

5

2 – 1

2 







_



















x

3

2

1 ½

1

2

5 2  range is

ifjl j

1 , 3

2 











 (a, b] =

1 , 3

2 





_





 a =

1

2

, b = 3 [a] + [b] = 3

19. If CF is perpendicular from the ………

nh?kZo`Ùk

2

x

49

+ 2

y

25

= 1

ij fLFkr

………

Sol. Let a point P = (7 cos, 5 sin) on the ellipse

ekuk ,d fcUnq

P = (7 cos, 5 sin)

nh?kZo`Ùk ij fLFkr gSA

Equation of tangent at P is

x

7

cos  +

y

5

sin – 1 = 0 P

ij Li'kZ js[kk d k lehd j.k

x

7

cos  +

y

5

sin – 1 = 0  CF = 2 2

1 cos

sin

49

25 





= 2 2

35 25 cos

 

49 sin



Equation of normal at P is P

ij vfHky Ec d k l ehd j.k

7x sec – 5y cosec  = 24 coordinates of G 

24 0,

sin

5 















G

_{d s funsZ'kkad}



24 0,

sin

5 















 PG2 = (7 cos )2 + 2

24 5 sin

sin

5 



 











= 49 cos2  + 2 2

49 sin

25 

PG2 =

49

_

25 cos

2

49 sin

2

_

25  





CF. PG

=

7

35

5 

= 7

20. If a conic passing through origin has ………

;fn 'kkad o ewy fcUnq ls xqt jrk

………

Sol. Let S = (3, 3), S(– 4, 4), P(0, 0), then centre C

1 7

–

,

2 2













ekuk

S = (3, 3), S(– 4, 4), P(0, 0)

rc d sUnz

C

1 7

–

,

2 2













SP =

3 2

, SP =

4 2

If conic is an ellipse SP + SP = 2a

;fn nh?kZo`Ùk 'kkad o gS ;fn

SP + SP = 2a

7 2

= 2a 

b



2 3

director circle is

fu;ked o`Ùk

2 2

1

7

73 x

y

–

2

2 



























(2x + 1)2 + (2y – 7)2 = 146

If conic is hyperbola

;fn 'kkad o vfrijoy ; gS

|SP – SP| = 2a 

2

= 2a 

b



2 3

Director circle does not exists (as b > a).

(16)

21. If three equations (a + 1)3

x + (a + 2)3

y = ………

;fn rhu l ehd j.ksa

(a + 1)3 x + (a + 2)3 y = ………

Sol. Since the equations are consistent D = 0

pawfd lehd j.k fud k; laxr gS

D = 0

3 3 3

(a 1)

(a

2)

–(a

3)

(a 1)

(a

2)

–(a

3)

0

1

1 –1





Put u = (a + 1)

j[kus ij

, v = a + 2, w = a + 3 u – v = – 1, v – w = – 1, w – u = 2 u + v + w = 3a + 6  3 3 3

u

v

w

u

v

w

1

= 0 (u – v) (v – w) (w – u) (u + v + w) = 0 (–1) (–1) (2) (3a + 6) = 0 i.e. |a| = 2

22. A tangent is drawn to the parabola ………

ijoy ;

y2 = 4x

d s fcUnq

……… Sol. 2y .

dy

dx

= 4 y1 =

dy

dx

=

2 y

y – y1 = 1

2 y

(x – x1) Area of triangle =

1

2

(base)(height)

f=kHkqt d k {ks=kQ y

=

1

2

(

vk/kkj

)(

Å apkbZ

) =

1

2

(2x1)y1 = x1y1 = x1 . 2x1 1/2 = 2x1 3/2 = (2x1)y1 = x1y1 = x1 . 2x1 1/2 = 2x1 3/2

 minimum area of triangle happens when x1 = 1

x1 = 1

ij f=kHkqt d k U;wure {ks=kQ y gksxkA

 Required area

vHkh"V {ks=kQ y

= 2(1)3/2 = 2

Part-II

Physics

23. Block of mass 5kg is moving with...

5kg

nzO;eku d k Cy kWd

–5

k

ˆ

(m/s)

osx

... Sol. a = r r

1

25

50 f

f

´

₂

₄

m





=

5

2 

2

= 5 m/s 2 S = 2

5

10

=

5 m

2

t =

5

= 1 sec.

24. In the figure the light is incident...

fp=k esa

izd k'k



d ks.k

(

Ø kfUrd d ks.k ls FkksM+

k&lk

...

Sol.

In first case

çFke çd j.k esa

sinC = 1

2

n

In second case if

f}rh; çd j.k esa

n3 < n1 (A) sin C = 3 2

n

< sin C C < C

TIR will done at surface AB

i`"B

AB

ij

TIR

?kfVr gksxkA

(B) If

;fn

n3 > n1

sin C > sin C

C> C

ray will not undergo TIR at AB

(17)

Corporate Office (New Campus): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 (C) If

;fn

n3 > n1 n2 sinC = n3sinr sinr = 2 3

n

sinC = 2 1 3 2

n



n

sinr = 1 3

n

= sinC

ray will be undergo TIR at CD. CD

ij fd j.k d k

TIR

gksxkA

25. Small blocks A and B are simultaneously...

NksVs Cy kWd

A

rFkk

B

,d lkFk fpd us ost d s

...

Sol.

gresultant of B with respect to A is =

2 2

g

g 3

2

2 







 







_

_





_

_

=

g

2

1

3

4

4 













= g By energy conservation, For B mBgh =

1

2

mvB 2 For A mAgh =

1

2

mA.VA 2 . t = 2

2h

g sin



, so time taken is different.

Sol. B

d k

A

d s

lkis{k

g_ifj.kkeh = 2 2

g

g 3

2

2 







 







_

_





_

_

=

g

2

1

3

4

4 













= g

Å t kZ laj{k.k }kjk

– B

d s fy ,

mBgh =

1

2

mvB 2 A

d s fy ,

mAgh =

1

2

mA.VA 2 . t = 2

2h

g sin



,

blfy , le; vy x&vy x gSA

26. In the given circuit the point A...

fn;s x;s ifjiFk esa fcUnq

A

_ij

...

Sol.

I

24 15

6 R 1 2 1











A B

V



V



6 Ir



33

9

6 R

4 





R

7 





 

BC

V

15 3 2

9V









BD

V

30V





27. A cavity of radius R is taken out...



vkos'k ?kuRo ls ,d leku vkosf'kr

...

Sol. 0 r E 3   

[for inside points] &

3 2 0 R for which 3 r  _       0 r E 3    [

vkUrfjd fcUnqvksa d s fy ,

]

rFkk

3 2 0 R 3 r  _     _   

ckgjh fcUnqvksad sfy ,

1 2 3 0 R E E E 3     

 





3 4 2 0 ₀ R R E 3 ₃ _2R      _ 0 0 0 R R R 3 3 4 4          