11 Mathematics Solved 02 New Sol Jse

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Sample Paper-02 (solved)

Mathematics

Class – XI

ANSWERS

Section A

1. Soltion

They are parallel since

0

2 2

a b

a b

−

=−

2. Soltion

Area of a triangle

2 2 0 61

5 2 3 62

− −

− − =

0 61

3 32

−

− = 9

!. Soltion

2 225 x y+ =

". Soltion #

( ) x f x a=

( ) y f y a=

( ). ( ) . ( ). ( ) x y x y f x f y a a a f x f y+

= = =

$. Soltion #

When 0, 1 x y= = in both cases. Hence

( ) {0,1} A B∩ =

%. Soltion # 2 pq

Section &

'. Soltion #

Let a satisfy the relation ( ) 3 f a =

( ( )).(1 ( )) ( ) f f a f a f a+ = −

(3).(4) 3 f = −

3(3)

4 f = −

. Soltion#





tan tantan( )

1 tan tan

1 13 2

1 11 .

3 2

1

45

2( ) 90

sin 90 1

A B A B

A B

A B

A B

++ =

−

+

=

−

=

+ =

+ =

=

. Soltion:

For a !"a#ratic e!"ation s" of $hose roots are %& an# pro#"ct of the roots is '(

2(30) 81 0 x x− + =

23 27 81 0 x x x− − + =

( 3) 27( 3) x x x− − −

( 3)( 27) 0 x x− − =

Hence the n"bers are % an# )*

10. Soltion#

Let : f R R→ be a f"nction gi+en by2

( ) 2 f x x= + fin#1

(27) f −

2

2

2

1

( ) 2

2 27

25

5

(27) { 5 , 5}

f x x

x

x x

f −

= +

+ =

=

= ±

= −

11. Soltion#

The f"nction is #efine# for all +al"es of , $here the #enoinator is not e!"al to -ero

1 0a x+ − ≠

Hence #oain =

{( 1)} R a− +

Range of f





Let ( ) y f x=

1

( 1)

( 1) ( 1)

( 1) 2

1

x a y

a x

a y xy x a

x y a y a

a y x

y

−=

+ −

+ − = −

+ = + +

+ +=

+

Range of f { 1} R= − −

12. Soltion

Rationali-e the n"erator

0

0

0

lim

( )( )lim

( )

lim( )

1

2

x

x

x

a x a

x

a x a a x a

x a x a

x

x a x a

a

→

→

→

+ −

+ − + +=

+ +

=+ +

=

1!. Soltion#

0 0

0 0

0 0 0 0

0 0

0

sin 75 cos 7 5

1 12 sin 75 cos 75

2 2

2(cos 45 sin 75 sin 45 cos75 )

2 sin(75 45 )

2 sin120

(

+

= +

= +

= +

=

Hence sign is positi+e an# +al"e is 2. 3

2 = 6

2

1". Soltion#

2cos 3 cos

3

23 2

3

2 2,

3 9

x

x n

n x n Z

π

π π

π π

=

= ±

= ± ∈





1$. Soltion#

Let ( )P n be the stateent gi+en by( 1)

1 2 32

n nn

++ + + ………+ =

2

1(1 1)(1)

2

1 ,

( 1)1 2 3

2

( 1)1 2 3 ( 1) ( 1)

2

( 1)

( 1) ( 1)2

3 2( 1)

2

( 1)( 2)( 1)

2

( ) ( 1)

P

True

Let it betrue for n m

m mm

m mm m m

m m

P m m

m mP m

m mP m

Thus P m is true P m is True

+=

=

=

++ + + ……+ =

++ + + ……+ + + = + +

++ = + +

+ ++ =

+ ++ =

⇒ +

ɺ

ɺ

1%. Soltion#

Let z = 8i−

( ) ( ), ( ) 0

2 2

8 0 8 08 , ( ) 0

2 2

(2 2 )

{ } { }

{ }

z Re z z Re z z i Im z

i i Im z

i

− −= ± − <

+ −− = ± − <

= ± −

1'. Soltion

2 53 0

2

2 5 3 60

2

110

2

110

2

( 11)( 2) 0

(2,11]

x

x

x x

x

x

x

x

x

x x

x

+− ≥

−

+ − += ≥

−

− += ≥

−

−= ≤

−

= − − ≤

∈





1. Soltion

4 12

2 8

4

x x

x

x

+ + =

=

=

1. Soltion

Let p be the probability of $inning ar / ( )P C

( )

( ) 2

( ) 6

( ) ( ) ( ) 1

2 6 1

9 1

1

9

1( )

9

2( )

9

6( )

9

P C p

P B p

P A p

P A P B P C

p p p

p

p

P C

P B

P A

=

=

=

+ + =

+ + =

=

=

=

=

=

Section C

20. Soltion

Let the ratios be

:a b

2

1

1

1

2

2

1

2 2 2

2 2 2

1

0

( )

( )

( )

( )

x p x q

a b p

a b p

a b qa b q

a b p

a b p

a b q

a b q

a b p

a b p

α α

β β

α α β β

α

β

α

β

α

β

+ + =

+ = −

+ = −

× =× =

+ = −

+ = −

=

=

+=

+





2 2

2 2

1

2

21

2

2

1 1

p

p

q

q

p q

p q

α

β

α

β

=

=

=

2 2

1 1 p q p q=

21. Soltion :

11 1

282 4

1 1 1 1 11 2

1 12 4 81

2 2

. . . .a a a a a

+ + + +……∞ = = =

−

……∞ =

22. Soltion

0t is gi+en that

( ) 700 , ( ) 200 , ( ) 295 , ( ) 115n U n A n B n A B= = = ∩ =

We nee# to fin# o"t

( )n A B′ ′∩

( ) ( )

( ) ( )

( ) { ( ) ( ) ( )}

700 {200 295 115}

320

n A B n A B

n U n A B

n U n A n B n A B

′ ′ ′

∩ = ∪

= − ∪

= − + − ∩

= − + −

=

2!. Soltion#

There are 1 gro"ps an# fo"r gro"ps can be arrange# in 4!$ays. lass () can be arrange# in 3!$ays /

lass (( can be arrange# in 4!lass (& can be arrange# in 4! . lass 9 can be arrange# in 2!$ays

Hence Total n"ber of $ays that they can be arrange# in a ro$ 4! 3! 4! 4! 2! 165888× × × × =

0n a circ"lar seating arrangeent the fo"r gro"ps can be arrange# only in 3! $ays only. Hence the

total n"ber of $ays that they can be seate# at a ro"n# table = 3! 3! 4! 4! 2! 41472× × × × =

2". Soltion

The ne$ coor#inates of the centre in the ne$ position are

( 4 , )a r bπ +

2 2 2{ ( 4 )} ( ) x a r y b r π − + + − =





2$. Soltion

2 2

2 2

2 2

2 2

2 2

4 4 16 16 0

4 4 4 16 16 4

( 2) 4( 2) 4

( 2) ( 2)1

2 1

x y x y

x x y y

x y

x y

+ + + + =

+ + + + + =

+ + + =

+ ++ =

This e!"ation represents an ellipse.

2%. Soltion

,i f i f i ,i 2,i 3(42 f i 2,i 3(42

) () )1 (% (45

(4 5 9& & &(* () )&1 ) )1

)% 9 )&* ' *)

)* 4 (%4 () 5&

44i

N f = Σ = 660i i

f xΣ = 15 312i i f xΣ − =

1 660( ) 15

44i i Mean X f x

N = = Σ = =

1 312

. ( 15 ) 7.090944i i MeanDeviation M D f x N = = Σ − = =

11 Mathematics Solved 02 New Sol Jse

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