CBSE Sample Paper Solutions for Class 11 Mathematics - Set A

AglaSem Schools

SAMPLE QUESTION PAPER–6 Section-A 1. We have R = {(x, y) : (x – y) is odd natural number x A, y  B}.

1

R = {5, 4}

 2.

1

f (x) =

.

5 x

It is defined when 5 – x > 0 x < 5. Domain of f = {x : x  R , x < 5}



Range = (0, ).

1

m o c

  tan 8 x     8 x   8 x  8 tan 8 x = lim  lim = . 3 x  0   sin 3 x  x  0 sin 3 x    3 x  3x  

.

3.

m e s

4.

1

cosec (– 1410°) = cosec (– 360° × 4 + 30)

a l g

1

= cosec 30° = 2. x2 = 2 x= y3 = 2 y–3=

5. 

a . s

2

4+2=6

ol

and

o h



sc

A (–2, –3)

–5

O (2, –5)

– 10

y = – 10 + 3 = – 7 1

 Co-ordinates of B are (6, – 7).

.

6. If two lines do not intersect in plane, then they are parallel.

w

w

w

7.

B (x, y)

1

Section - B L.H.S. =

cos A cos B cos C   a b c 2 2 2 2 b c a a  c2  b2 a2  b2  c2   = 2abc 2abc 2abc

1 2

b2  c2  a2  a2  c2  b2  a2  b2  c2 1 2abc a2  b2  c2 = = R.H.S. Hence proved. 1 2abc =

OR 1 5 and sec  =  2 3

Given that,

cot =



tan = 2 and tan =

1 2

 5     1 3

AglaSem Schools 25  9 9

tan = tan  = + tan ( +) =



4 3

 [ B   ,    2 2 

tan   tan  1  tan  tan 

1

2 = = 3 4 11 12 3 3 2

tan ( + ) =



4 3

1

2 11

1 2 1 c os 7 2 = 1 s in 7 2 1 1 2 cos 7 .cos 7 2 2 = 1 1 2 sin 7 . cos 7 2 2 2 1 2 cos 7 2 = sin 15 

8.

L.H.S. = c ot 7

m o c

m e s

.

1

a l g

a . s

l o o

.

h c s

w

=

1  cos 15  sin 1 5 

=

1  cos (45   30  ) sin (45   30) 

=

w

w

=

1  cos 45  cos 30   sin 45  sin 30  sin 45. cos 30  cos 45  sin 30   1 3 1 1 1       2 2 2 2  1 3 1 1      2 2 2 2

 ( 3  1)   1   2 2  =  3  1   2 2 

 2 2  3  1 =    3 1  = =

(2 2  3  1)( 3  1) ( 3  1)( 3  1) 2 6  3  3  2 2  3 1 ( 3)2  12

1

1

AglaSem Schools 2 6 2 32 2 4 = 2 = 6 3 22 =

6 3 2 4 Hence proved. 1

= R.H.S. 9. We have

2kx – 5k = (x – 3).x

2

(k – 3).x2 – 2kx + 5k = 0



= (–2 k)2 – 4 (k – 3). 5k



1

= 4k2 – 20k (k – 3) = 4k (k – 5k + 15) = 4k (– 4k + 15)

om

= 4k (4k – 15) For real and unequal root

.c

>0 

– 4k (4k – 15) > 0



4k (4k – 15) < 0

a l g

15 4

00

For domain, as 

x2 + 1 > 0.



f(x) is defined for all x  R.



Domain of f = R.

l o o

ch

.s



  14.

m e s

, x R

½

a l g

½

f(x) = y





1

.c

a . s

For range, let

1

w

y=

x2 =

w

w

x=

x2 2

x 1

y 1 y

y 1 y

1

½

1

1 – y > 0  y < 1 and y > 0. Range : 0 < y < 1.

½

f (x) = sin (5x – 8) 

f (x) = lim

h 0

f ( x  h)  f ( x) h

= lim

sin {5 ( x  h)  8}  sin (5 x  8) h

= lim

2 cos h

h 0

h 0

= 2 lim

h 0

1 h

1

  5 x  5h  8  5 x  8   5 x  5h  8  5 x  8    ·sin      2 2  

 5h    5h   cos  5x  8  2  ·sin  2    

1

AglaSem Schools

 5h  sin    2 5h   = 5 lim cos  (5 x  8     5 h 0 2  h   2

1

 5h  sin    2  = 5.cos (5x – 8). lim h0  5h    2 sin x    1 = 5 cos (5x – 8),  lim x  0 x  

1

1 ,x  R x  [ x] 0 < (x – [x]) < 1, x  R

15.

f(x) = As,

1

m o c

But when x  Z x – [x] = 0 

0 < (x – [x]) < 1,  x  R – Z



Domain of f = R – Z.

m e s

For range, as

0 < x – [x] < 1, x  R – Z



0<



1<



1 < f(x) < 

1

o h

and x, b, c, y are in G.P.

w

w

Let r be common ratio.

w

.

sc

a=

x y 2

1

....(i) 1

b = xr, c = xr2, y = xr3 1/3



 y r=    x



 y b = x   x

and

1

0, y > 0 ....(ii) Take

3x

3x + 3y = 60

At

x = 0, y = 20

and

y = 0, x = 20

+

1

3y

30 =

60

2 20

m o c

At (0, 0); 0 + 0 < 60, (True) Now take

C(0, 10)

1

i.e., (0, 0) is included.

10

x + 3y = 30

 At

x = 0, y = 10

and

y = 0, x = 30.

X'

At (0, 0); 0 + 0 < 30, (True)

25.

L.H.S. =

10

20 A(20, 5)

30

40

.

w

w

w

X 3y = 30

a . s

1·22  2·32  ....  n(n  1)2 12 ·2  22 ·3  ....  n2 (n  1)

l o o

h c s

x+

a l g

2

So, shaded region OABC is the solution region.

.

m e s

0 Y'

So, origin is also included.

B(15, 5)

=

 [ n(n  1)2 ]  [ n2 (n  1)]

=

[ n3  2n2  n]  [ n3  n2 ]

=

n3  2 n2  n n3  n2

1

1

 n2 (n  1)2 n(n  1)(2 n  1) n(n  1)  2    4 6 2  =   n2 (n  1)2 n(n  1)(2 n  1)     4 6  

 3n2 (n  1)2  4n(n  1)(2n  1)  6n(n  1)    12 =  3n2 (n  1)2  2n(n  1)(2n  1)    12

1

=

n(n  1)(3n2  3n  8n  4  6) n( n  1)(3n2  3n  4n  2)

1

=

3n2  11n  10 3n2  7n  6

1

AglaSem Schools

(3n  5)(n  2) = (3n  1)(n  2) =

3n  5 = R.H.S. 3n  1

Hence proved. 1

26. Given that, 

(a + ib) (c + id) (e + if) (g + ih) = A + iB

1

(a – ib) (c – id) (e – if) (g – (ih) = A – iB

2

(A + iB) (A – iB) = (a + ib) (a – ib) (c + id)



(c – id) (e + if) (e – if) (g + ih) (g – ih) 2 2

2

2

2

2

A + B = (a + b ) (c + d2) (e2 + f2) (g2 + h2).

1

OR We have

z2i = 5, where z is only complex number. z  5  4i z 2i z  5  4i

2

x  iy  2  i x  iy  5  4i

2

( x  2)  i( y  1) ( x  5)  i( y  4)

2







1

1

l o o

 ( x  2)  i( y  1)   ( x  2)  i( y  1)   ( x  5)  i ( y  4)   ( x  5)  i( y  4)  = 25

h c s

 ( x  2)  i( y  1)][( x  2)  i( y  1) [( x  5)  i( y  4)] [( x  5)  i( y  4)]

.

w

= 25

1

(x + 2)2 + (y – 1)2 = 25 [(x + 5)2 + (y + 4)2]

x2 + 4 + 4x + y2 + 1 – 2y = 25 (x2 + 25 + 10x + y2 + 16 + 8y]

 

a l g

1

= 25

a . s

 

.

m e s

= 25

m o c

|z|2 = zz

As, 

= 25

w

25x2 + 25y2 + 250x + 200y + 1025 – x2 – y2 – 4x + 2y – 5 = 0

w

2

1

2

24x + 24y + 246x + 202y + 1020 = 0 12x2 + 12y2 + 123x + 101y + 510 = 0.

1

