CBSE Sample Paper Solutions for Class 11 Mathematics - Set E

AglaSem Schools

SAMPLE QUESTION PAPER–10 Section-A 1.

A=={ } P(A) = 20 = 1 element.



1

   13  tan  = tan     = + tan     12  12 12

2.

  tan  tan     4 6 tan    =  4 6   1  tan tan 4 6



1 3 = = 1 1 3 5  2x x < 5 3 6 1

3. 

10 – 4x < x – 30



5x > 40

3 1 3 1

.

m o c

m e s

.

a l g

a . s

x > 8  x = (8, ).

 4. (x + 3)8

1

1

l o o

Tr+1 = 8Cr x8–r 3r

876 ×3×3×3 32 = 56 × 27 = 1512.

Coefficient of x5 = 8C3 × 33 =



h c s

.

5. A (– 2, 6), B (4, 8)

w

w

Slope of AB =

2 1 = 6 3

Slope of CD =

12 x8

1

C (8, 12) and D (n, 24)

w



AB CD 

12 1  =–1 x8 3

x–8=–4 

 x  4

6. If two integers a and b are such that (a – b) is always positive integer, then a > b.

1 1

Section-B 7. L.H.S. : Let x  (A B) 

x  (A B)



x  A and x B

½



x  A and x  B



x  A A B

Thus  Now R.H.S. take

AglaSem Schools ½

x (A  B)  x  (AB) (A B)  (AB)

y  y  y  y  y Thus y (A B) y   (A B)  From equations (i) and (ii), (A B) = 8.

....(i) 1

(AB) A and y B A and y  B (A B) (A B) (A B) (A B)

½

½ ....(ii)

A B

.

Domain : f (x) is defined when     Domain of f = [– 4, 4] Range :

16 – x2 x2 – 16 (x – 4) (x + 4) –4 0 y2 – 16 < 0 –40 8

tan

 But  

AglaSem Schools

=1

tan



c s .

w

w

w

m e s

a . s

½

< 91 91 88 22 cm. x+3+5 8. x < 22

1 1 1

 Smallest piece of board is more than or equal to 8 cm and less than or equal to 22 cm. 11.

z=

=

=

½

a l g

ol

o h

1

.

  1  2 8

10. A man has a board of length 91 cm. Let length of smallest piece = x cm  Length of second piece = (x + 3) and length third piece = 2x  According to question, x + x + 3 + 2x  3 + 4x <  4x <  x< Also 2x >  x> Hence, 8<

m o c

½

i1   cos  i sin 3 3

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



½

(1  i 3) (1  i 3)

=

2[ i  i2 3  1  i 3] 12  ( i 3)2

=

2[( 3  1)  i ( 3  1)] 4

1

AglaSem Schools

 3  1  3  1 z=   i  2   2 

3 1 = r cos  2 3 1 = r sin  2

Let and

½

 Squaring and adding,

( 3  1)2 ( 3  1)2  4 4

r2 =

1 [3  1  3  1] 4 r2 = 2

=

r=

 and

1

2

m o c

3 1

tan  =

3 1

1   3 1    3 1   3 1    3

tan  =

m e s

.

a l g

   tan 4 6 =   1  tan . tan 4 6 tan

a . s

l o o

 tan  = tan   4



h c s

.

w



=

  6

  5  = 4 6 12

z=

w

1

5 5   2  cos  i sin   12 2

This is required polar form.

w

OR  1  i   1  i

m

As

 1  i 1  i    1  i 1  i

m



= 1,

(given)

=1

1

m



 (1  i)2    =1  1  i2 



 1  i2  2i    =1  2 



 1  1  2i    2

m

1

m

=1

1

AglaSem Schools

im = 1



1

It is possible when least value of m is 4. 4



i = 1.

12. We have a word INDEPENDENCE consists 12 letters with N 3, D 2, E 4 and others are different. (i)

There are 5 vowels in it with E 4 and I, they should occur together i.e., EEEEI as a 1

single word and remaining 7. Letters are separate. 

8!  5! 2 ! 3 !  4 !

Total such arrangements =

87654 ! 543 23! 4 ! = 16800.

½

=

½

m o c

(ii) Word begin with I and end with P i.e., I NDEENDENCE P

.

So, we have arrange only 10 letters. 

m e s

10 ! 4!  3! 2!

Total such arrangements =

½ ½ ½

a l g

10  9  8  7  6  5  4 ! 4 !  3 2 2

=

a . s

720  35 = 360 × 35 2 = 12600.

ol

=

o h

½

OR

c s .

We need to use digits 0, 1, 2, 2, 2, 4, 4, to obtain a number more than 1000000. 1

It is 7 digit number and we also have 7 digits so all digits should be used once.

w

w

Total 7 digits numbers using these digits =

w

7! 76543! = 3!  2! 3!  2 1

1

= 420

But it includes numbers that also start with zero. So,

number that start with zero =

6! 6 543! = = 60 2! 3! 2  3!

1

 Total 7 digits numbers more than 1000000 1

= 420 – 60 = 360. 13.

P (n) : (1 + x)

n

> 1 + nx, x N and x > – 1.

Put n = 1 P (1) : (1 + x) > 1 + x (True) So, P(1) in true. Let P(k) is true i.e., Now to prove P(k + 1) is true.

(1 +x)k > 1 + kx

....(i)

½

AglaSem Schools ....(ii) ½

(1 + x)k+1 > 1 + (k + 1)x

i.e., Taking L.H.S.

(1 + x) k+1 = (1 + x)k·(1 + x) 1

Using equation (i), (1 +x)

k+1

> (1 + kx) (1 + x) > kx2 + kx + x + 1

x2 > 0 kx2 > 0

As

1

(1 + x)k+1 > (kx + x + 1)



(1 + x)k+1 > [1 + (k + 1) x]

1

i.e., P (k + 1) is true. Hence P(n) is true for all n N. ab and G.M. = 2 Let a and b are roots of quadratic equation, then

14. Let a and b are two numbers then A.M. = A =

x2 – (a + b) x + ab = 0 But

a + b = 2A, and ab = G x – 2Ax + G2 = 0



Using quadratic formula, we get

a . s

=

l o o

i.e.,

h c s

and

4 A 2  4G 2 2

1

2 A  2 A2  G 2 2

x=A+



1

a l g

2A

x=



m o c

.

2

m e s

2

1

ab = G

1

( A  G) ( A  G)

a = A  ( A  G) ( A  G) Hence proved.

b = A  ( A  G) ( A  G)

.

15. There are 25 students, 10 are to be chosen, 3 students decided to join all of them or none of them. 1

w

w

So, when all these 3 students are joining them we need to select 7 students out of 22. i.e., in

22

w

1

C10 ways

1

C7 ways

and when these 3 dicided not to go, then select 10 students out of 22. i.e., in

22

So, total ways of selection of 10 students =

22

C7 +

22

1

C10. Y

16. Equation of line is 4x – y = 0

....(i)

4x – y =0

It is passing through origin, we need to find distance of line (i).

P (x, y)

From Q (4, 1) along a line, makes angle 135° with + ve x-axis 

Slope of PQ = tan 135° = tan (180 – 45°) = – tan 45° Q (4, 1) 135°

= –1  Equation of PQ ,

y – 1 = – 1 (x – 4)



x+y= 5

X'

....(ii)

1 X

Y'

1

AglaSem Schools

On solving (i) and (ii), we get

1

n = 1, y = 4 i.e., P (1, 4) (4  1)2  (1  4)2 =

PQ =



99

PQ = 3 2 units. 1

i.e., Distance of line (i) from (4, 1) = 3 2 units.

17. The vertices of triangle PQR are P (2a, 2, 6), Q (– 4, 3b, – 10) and R (8, 14, 2c) and centroid is at orign. A (2a, 2, 6)

2a  4  8 =0 3



1

2a + 4 = 0 a = – 2



G (0, 0, 0)

2  3b  14 =0 3



6  10  2c =0 3 – 4 + 2c = 0



c=2

and

18.

om

Q (– 4, 3b, –10)

3b + 16 = 0 b = – 16/3



D

R (8, 14, 2c)

.c

 ( x  1) 5  1  0  ;  form  lim   0  x 0  x

m e s

a l g

a . s

 ( x  1)5  (1)5  = lim   x 0  ( x  1)  1 

ol

o h

Let x + 1 = y  y 1

.

sc

w



1

1 1

 y5  15  = lim   y 1  y  1 

   xn  an  n 1  lim    = na xa  x a 

5 (1)5–1 = 5.

w

1

1 1

OR

w

 sin (   x )   0  lim  fo r m  ;  x    (  x)    0

½ ½

Apply R.H.L., put x =  + h, h 0 sin (    h )  = lim  h  0   (    h )  

1

 sin (  h )  = lim  h  0   (  h)  

1

=

1  sin h  lim  h  0  h 

=

1, 

1 sin x    1  lim  x 0 x

19.

AglaSem Schools

z1 = 2 – i and z2 = 1 + i  z1  z2  1   2  i  1  i  1  z  z  i  =   2  i  1  i  i  1 2



1

4  =   1  i 

 4 (1  i)  =  (1  i)(1  i) 

1

 4 (1  i)  =   1  i2 

1

= 2 (1 + i) z1  z2  1 = |2(1 + i)| z1  z2  i



m o c

= 2 12  (1) 2 = 2 2.

.

m e s

Section-C 20. Let, A = Number of students in Chemistry class.

1

a l g

i.e., n(A) = 40 and B = No. of students in Physics class. 1 i.e., n(B) = 60 (i) When two classes meet at the same hour than there is no common student in these classes i.e., n(A  B) = 0 1  n(A  B) = n(A) + n(B) – n(A  B) = 40 + 60 – 0 n(A B) = 100 1 (ii) When classes meet at different hour, then there are 20 common students. i.e., n(A  B) = 20 1  n(A B) = 40 + 60 – 20 = 100 – 20 n(A  B) = 20. 1 Parents encourage their wards to these streams because of (i) Jobs oriented streams.

a . s

l o o

h c s

.

w

w

w

(ii) Develope a scientific skills. (iii) Technical education. 1

21.

f(x) =

4x +

For domain :

4 – x > 0 x < 4

and

2

2

x 1

, x R 2

x – 1 > 0 (x – 1) (x + 1) > 0 2

x < – 1 and x > 1

 x1

x4 1

 4 2

AglaSem Schools

OR A = {1, 2, 3, 4, 5, 6} R = {(x, y) : y = x + 1; x, y A}. i.e.,

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

2

Domain = {1, 2, 3, 4, 5}

1

Co-domain = {1, 2, 3, 4, 5, 6}

1

 and

Range = {2, 3, 4, 5, 6}

½ B

A

R:A A

1 2 3 4 5 6

1 2 3 4 5 6

1½

m o c

Arrow diagram

.

22. There are 10 points in a plane, so straight lines can be formed using two at a time. 10  9  Total straight lines = 10C2 = = 45 21 But 4 points are collinear. 43  Lines using these points = 4C2 = =6 2 1 But these 4 points can make a single line. So, total different lines = 45 – 6 + 1 = 40 Now triangle can be formed using 3 points at a time. 10  9  8 So, Number of triangles = 10C3 = 3 21 = 120. Using 4 points taking 3 at a time. No. of triangles = 4C3 = 4 But collinear points can’t make only tirangle. So, total no. of triangles = 120 – 4 = 116.

em

s a l

g a .

1

1

s l o

o h

c s .

w

w

3  23. Expansion of  x  2   x 

m

w

3

m

    

C0 + (– 3).mC1 + 9.mC2 = 559 9m(m  1) 1 – 3m + = 559 2 9m2 – 9m + 2 – 6m = 2 × 559 9m2 – 15m + (2 – 2 × 559) = 0 9m2 – 15m – 2 × 558 = 0 3m2 – 5m – 372 = 0 2 3m – 36m + 31m – 372 = 0



3m (m – 12) + 31 (m – 12) = 0



1 1

2

 3  mC2 xm 2   2  + ..... 2 x  x  Coefficients of first three terms of this expansion are mC0, – 3.mC1 and 9.mC2 According to question, = mC0 xm – mC1 xm–1 ×

1

½ ½

1

1



(m – 12)(3m + 31) = 0



m= 

31 3 m = 12

and

3   Expansion is now  x  2   x 

AglaSem Schools    m N)

(Not possible)

12

r

 3 Tr+1 = (–1)r·12Cr (n)12 – r  2  x 



1

= (–1)r·12Cr x12–r–2r. 3r For coefficients of x3, put 1

12 – 3r = 3  r = 3 3

 Term containing,

x = (– 1) =

3 12

3 3

C3 3 x

12  11  10 × 3 × 3 × 3 × x3 3 21

= – 132 × 45 × x3 = – 5940x3

.

m e s

OR a and b are distinct integers (given), then

m o c

1

la

an = (a – b + b)x

g a .

Using binomial themorem,

1

an = [(a – b)n + nC1 (a – b)n–1. b + nC2 (a – b)n–2 b2

ls

+ nC3 (a – b)n–3. b3 +....+ nCn bn] 1

an – bn = [(a – b)n + nC1 (a – b)n (a – b)–1 b + nC2 (a – b)n (a – b)–2 b2 +....] 1

o o h

an – bn = (a – b) [(a – b)n – 1 + nC1 (a – b)n– 2 b + nC2 (a – b)n–3 b2 +....] Let 

 = (a – b) n

n–1

c s .

n

n

+ C1 (a – b)

n–2

n

b + C2 (a – b)

n–3

2

b

1 1

+....]

a – b =  (a – b)

 (a – b) is a factor of an – bn.

1

w

24. AOB is a beam supported at the ends A and B and deflection is 3 cm at centre and beam is in parabolic shape. 1 Y

w

w

 Equation of beam is x2 = 4ay

1

3  But B  6, lies on it  100 

i.e.,

36 = 4a ×



4a = 1200



3 100

BE AM

2

x = 1200 y

Let PQ = 1 cm deflection in beam at x cm distance from centre. 2    Q  x, lies on (i).  100  2  x2 = 1200 × 100  x2 = 24 



12 m

P

A

x=

24

3 

B  6, 100 

3 cm x

Q (x, 2) X

1 ....(i)

1 1

AglaSem Schools 1

x = 2 6 metres



 1 cm deflection is at 2 6 metres distance from centre. OR Equation of ellipse, 9x2 + 4y2 = 36

x 2 y2  =1 4 9 a2 = 4 a = + 2

 

1 1

b2 = 9 b = + 3

and  y-axis is the major axis.

c2 = b2 – a2 = 9 – 4 = 5



1

c = + 5



Foci = (0, + 5)

m o c

Vertices = (0, + 3) Length of major axis = 2b = 2 × 3 = 6 units and

c 5 = b 3

Eccentricity i.e., e =

(i)

g a .

13 1 = . 52 4 A = Black card

P (Diamond card) =

½ 1 1

s a l

25. There are 52 playing cards. One card is to be drawn.

em

.

½

1

s l o

(ii) 

n(A) = 26



P(A) =

o h

c s .

(iii) 

w

w



w

26 1 = . 52 2 B = Not an ace

1

n(B) = 52 – 4 = 48

P(B) =

48 12 = 52 13

1

(iv) C = Not a diamond card   (v)

n(C) = 52 – 13 = 39 cards P(C) =

3 39 = 4 52

D = Not a black card i.e., red cards



n(D) = 26



P(D) =

(vi)

1

1 26 = 52 2

1

E = A face card



n(E) = 12 cards (4; king, queen and jack each)



P(E) =

12 3 = 52 13

1

26. There are two firms belonging to the same industry say A and B.

AglaSem Schools

In firm A, No. of earners = 586 Mean wages of earners = 5253 and variance = 100  Wages (total) paid by firm = 586 × 5253 = 30,78,258 and In firm B,

S.D. =

No. of wage earners = Mean wages = and variance =  Total wages paid by firm = =

variance =

1

100 = 10

648 5253 121 1

648 × 5253 34,03,944. variance =

S.D. =

m o c

121 = 11

1

 Firm B pays larger amount of wages to their workers. Now, to compare variability between two firms we need to find coefficients of variation. S.D. ×100 C.V. (Firm A) = 1 Mean 10  100 = 0·190 = 5253 11 C.V. (Firm B) = 1  100 = 0·2094 5253 Hence, firm B shows greater variability between firms. It means firm A looks more consistent than firm B about their wages. 1

.

m e s

a l g

a . s

l o o

.

h c s

w

w

w

