CBSE Sample Paper Solutions for Class 11 Mathematics - Set E

SAMPLE QUESTION PAPER–10

Section-A

1. A =  = { }  P(A) = 20 = 1 element. 1 2. tan 13 12       = tan 12          = + tan ₁₂   tan 4 6          = tan tan 4 6 1 tan tan 4 6       = 1 1 3 1 1 3   = 3 1 3 1   . 1 3. 5 2 3 x  < 5 6 x   10 – 4x < x – 30  5x > 40  x > 8  x = (8, ). 1 4. (x + 3)8 T_r+1 = 8_C r x 8–r ₃r  Coefficient of x5 ₌8_C 3 × 3 3 ₌ 8 7 6 3 2    ×3×3×3 = 56 × 27 = 1512. 1 5. A (– 2, 6), B (4, 8) Slope of AB = 2 6 = 1 3 C (8, 12) and D (n, 24) Slope of CD = x 12 8   AB CD  x 12 1 83  = – 1 x – 8 = – 4   x4 1 6. If two integers a and b are such that (a – b) is always positive integer, then a > b. 1

Section-B

7. L.H.S. : Let x  (A B)  x  (A B)  x  A and x B ½

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 x  A and x  B ½

 x  A A B

Thus x (A  B)  x  (AB)

 (A B)  (AB) ....(i) 1 Now R.H.S. take y  (AB)  y  A and y B ½  y  A and y  B  y  (A B)  y  (A B) ½

Thus y (A B) y  (A B)

 (A B)  (A B) ....(ii) From equations (i) and (ii),

(A B) = A B Hence proved. 1

8. f(x) = 16x2, x R

Domain : f (x) is defined when

16 – x2 _{> 0 ½}  x2 _{– 16 < 0}  (x – 4) (x + 4) < 0 ½  – 4 < x < 4  Domain of f = [– 4, 4] 1 Range : f (x) > 0, x (– 4, 4) Let f (x) = y = 2 16x  y2 = 16 – x2  x2 = 16 – y2  x = _16y2 ₁ It is defined, if 16 – y2 > 0  y2 – 16 < 0  – 4 < y < 4 But y > 0  0 < y < 4 1  Range = [0, 4].

9. L.H.S. = cos 6x = 2 cos2 _{3x – 1, ( cos 2 x = 2 cos}2 _{x – 1) 1}

= 2 [ 4 cos3 x – 3 cos x]2 – 1 1

= 2 [ 16 cos6 x + 9 cos2 x – 24 cos4 x] – 1

= 32 cos6 _{x + 18 cos}2 _{x – 48 cos}4 _{x – 1 1}

= R.H.S. Hence proved. 1 OR Take x = 8   2x = 4  ½  tan 2x = tan 4 

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48 | Oswaal CBSE Mathematics, Class–XI  2 2 tan 1 tan x x  = 1 ½ Let tan x = y  2y = 1 – y2  y2 + 2y – 1 = 0  y = 2 4 4 2    1  y = 2 2 2 2    y =  1 2 1  tan x =  1 2  tan 8  =  1 2 But 8   0, 2        1  tan 8  > 0   tan 1 2 8     ½

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 ½

 According to question, x + x + 3 + 2x < 91  3 + 4x < 91 1  4x < 88 1  x < 22 cm. Also 2x > x + 3 + 5 1  x > 8. Hence, 8 < x < 22

 Smallest piece of board is more than or equal to 8 cm and less than or equal to 22 cm. ½

11. z = i i 1 cos sin 3 3     = i i 1 1 3 2 2         ½ = 2( 1) (1 3) (1 3) (1 3) i i i i      = i i i i 2 2 2 2[ 3 1 3 ] 1 ( 3)     1 = 2[( 3 1) i( 3 1)] 4   

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z = 3 1 3 1 2 i 2                Let 3 1 2  = r cos  and 3 1 2  = r sin  ½

 Squaring and adding,

r2 = 2 2 ( 3 1) ( 3 1) 4 4    = 1[3 1 3 1] 4    r2 _{= 2 1}  r = ₂ and tan  = 3 1 3 1   tan  = 1 3 1 3 1 3 1 3               = tan tan 4 6 1 tan . tan 4 6       tan  = tan 4 6    _       = 4 6    = 5 12  1  z = 2 cos5 isin5 12 2         

This is required polar form.

OR As m i i 1 1       _  = 1, (given)  1 1 1 1 m i i i i         _{ }  = 1 1  2 2 (1 ) 1 m i i         = 1  2 1 2 2 m i i  _{ }      = 1 1  1 1 2 2 m i         = 1 1

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50 | Oswaal CBSE Mathematics, Class–XI

 im _{= 1}

It is possible when least value of m is 4. 1

 i4 = 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 single word and remaining 7. 1

Letters are separate.

 Total such arrangements = 8 ! 5 !

2 ! 3 ! 4 !    = 8 7 6 5 4 ! 5 4 3 2 3 ! 4 !          ½ = 16800. ½

(ii) Word begin with I and end with P

i.e., I NDEENDENCE P ½

So, we have arrange only 10 letters. ½

 Total such arrangements = 10 !

4 !3 !2 ! ½ = 10 9 8 7 6 5 4 ! 4 ! 3 2 2          = 720 35 2  = 360 × 35 ½ = 12600. OR

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

It is 7 digit number and we also have 7 digits so all digits should be used once. 1

Total 7 digits numbers using these digits = 7 !

3 !2 != 7 6 5 4 3 ! 3 ! 2 1       = 420 1

But it includes numbers that also start with zero. So, number that start with zero = 6 !

2 !3 !=

6 5 4 3 !

2 3 !

  

 = 60 1

 Total 7 digits numbers more than 1000000

= 420 – 60 = 360. 1 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., (1 +x)k _{> 1 + kx ....(i) ½}

Now to prove P(k + 1) is true.

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i.e., (1 + x)k+1 > 1 + (k + 1)x ....(ii) ½ Taking L.H.S.

(1 + x) k+1 = (1 + x)k·(1 + x)

Using equation (i), 1

(1 +x)k+1 _{> (1 + kx) (1 + x)}

> kx2 _{+ kx + x + 1}

As x2 > 0 kx2 > 0 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.

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

2

ab

and G.M. = ab = G 1

Let a and b are roots of quadratic equation, then

x2 _{– (a + b) x + ab = 0 1}

But a + b = 2A, and ab = G2

 x2 _{– 2Ax + G}2 _{= 0}

Using quadratic formula, we get

 x = 2 2 2 4 4 2 A A  G 1 = 2 2 2 2 2 A A G  x = A + (AG) (AG) 1 i.e., a = A (AG) (AG)

and b = A (AG) (AG) Hence proved.

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

them. 1

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

i.e., in 22C₇ ways 1

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

i.e., in 22C₁₀ ways 1

So, total ways of selection of 10 students

=22C₇ + 22C₁₀. 1 16. Equation of line is

4x – y = 0 ....(i) It is passing through origin, we need to find distance of line (i). From Q (4, 1) along a line, makes angle 135° with + ve x-axis  Slope of PQ = tan 135° = tan (180 – 45°) = – tan 45°

= – 1 1  Equation of PQ , y – 1 = – 1 (x – 4) 1  x + y = 5 ....(ii) Y Y' X X' 4x – y =0 P (x, y) Q (4, 1) 135°

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52 | Oswaal CBSE Mathematics, Class–XI On solving (i) and (ii), we get

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

 PQ = _{(4 1)}2 _{(1 4)}2

   = 9 9

PQ = 3 2 units.

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

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.  2 4 8 3 a   = 0  2a + 4 = 0 a = – 2 1 2 3 14 3 b   = 0  3b + 16 = 0 b = – 16/3 1 and 6 10 2 3 c   = 0  – 4 + 2c = 0  c = 2 1 18. 5 0 ( 1) 1 lim x x x          ; 0 form 0       = x x x 5 5 0 ( 1) (1) lim ( 1) 1            1 Let x + 1 = y  y 1 1 = 5 5 1 1 lim 1 y y y          n n n x a x a na x a 1 lim =         _ _     1  5 (1)5–1 = 5. 1 OR sin ( ) lim ( ) x x x        _{  }    ; 0 fo r m 0       ½ Apply R.H.L., put x =  + h, h 0 ½ = 0 sin ( ) lim ( ) h h h         _{    }    1 = 0 sin ( ) lim ( ) h h h      _{ }    1 = 0 sin 1 lim h h h         1 = 1 ,  x x x 0 sin lim 1         Q _D R G (0, 0, 0) A (2a, 2, 6) (8, 14, 2c) (– 4, 3b, –10)

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19. z₁ = 2 – i and z₂ = 1 + i  z_z1 z_z2 _i 1 2 1      _{ }    = i i i i i 2 1 1 2 1          _{   }  1 = 4 1 i      _  = _₍₁4 (1_{i) (1}i_)_i)_ 1 = 4 (1 ₂) 1 i i       _  = 2 (1 + i) 1  1 2 1 2 1 z z z z i     = |2(1 + i)| = _{2 1}2_(1)2 = 2 2. 1

Section-C

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

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.

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

21. f(x) = _4x+ ₂ 1 1 x  , x R For domain : 4 – x > 0 x < 4 2 and x2 – 1 > 0 (x – 1) (x + 1) > 0  x < – 1 and x > 1 2 – _{–1 1 4}  x < – 1 _{x > 1 x 4} i.e., domain of f = (– , – 1)  (1, 4]. 2

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54 | Oswaal CBSE Mathematics, Class–XI 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 1 2 3 4 5 6 1 2 3 4 5 6 A R : A  A B 1½ Arrow diagram

22. There are 10 points in a plane, so straight lines can be formed using two at a time.

 Total straight lines = 10C₂ = 10 9 2 1



 = 45 1

But 4 points are collinear.

 Lines using these points =4C₂ = 4 3 2 1 

 = 6 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. So, Number of triangles =10_C

3 = 10 9 8 3 2 1     = 120. 1

Using 4 points taking 3 at a time.

No. of triangles = 4C₃ = 4 1

But collinear points can’t make only tirangle.

So, total no. of triangles = 120 – 4 = 116. 1 23. Expansion of 2 3 m x x        = m_C 0 x m _– m_C 1 x m–1 _× 2 2 2 2 2 3 _{m m} 3 C x x x     _{  }   + ...

 Coefficients of first three terms of this expansion are mC₀, – 3.mC₁ and 9.mC₂ ½

According to question, m_C 0 + (– 3). m_C 1 + 9. m_C 2= 559 ½  1 – 3m + 9 (m m 1) 2  = 559  9m2 – 9m + 2 – 6m = 2 × 559 1  9m2 – 15m + (2 – 2 × 559) = 0  9m2 _{– 15m – 2 × 558 = 0}  3m2 _{– 5m – 372 = 0}  3m2 _{– 36m + 31m – 372 = 0}  3m (m – 12) + 31 (m – 12) = 0 1

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 (m – 12)(3m + 31) = 0  m = 31 3  (Not possible) m N) and m = 12  Expansion is now 12 2 3 x x         T_r+1 = (–1)r·12C_r (n)12 – r 3₂ r x       1 = (–1)r·12C_r x12–r–2r. 3r For coefficients of x3, put

12 – 3r = 3  r = 3 1  Term containing, x3 _{= (– 1)}3 12_C 33 3_x3 = 12 11 10 3 2 1     × 3 × 3 × 3 × x 3 = – 132 × 45 × x3 = – 5940x3 1 OR a and b are distinct integers (given), then

an = (a – b + b)x 1

Using binomial themorem,

an _{= [(a – b)}n ₊ n_C 1 (a – b) n–1_{. b +} n_C 2 (a – b) n–2 _b2 + n_C 3 (a – b) n–3_{. b}3 _+....+ n_C n b n_{] 1} an _{– b}n _{= [(a – b)}n ₊ n_C 1 (a – b) n _{(a – b)}–1 _{b +} n_C 2 (a – b) n _{(a – b)}–2 _b2 _{+....] 1} an – bn = (a – b) [(a – b)n – 1 + nC₁ (a – b)n– 2 b + nC₂ (a – b)n–3 b2 +....] 1 Let  = (a – b)n–1 + nC₁ (a – b)n–2 b + nC₂ (a – b) n–3 b2 +....] 1  an – bn =  (a – b)  (a – b) is a factor of an – bn_{. 1}

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

 Equation of beam is x2 = 4ay 1

But B 6, 3 100       lies on it i.e., 36 = 4a × 3 100  4a = 1200 1  x2 = 1200 y ....(i) Let PQ = 1 cm deflection in beam at x cm distance from centre.

 Q x, 2 100       lies on (i). 1  x2 = 1200 × 2 100 1  x2 = 24  x = 24 12 m B X A (x, 2) Y x 3 cm P Q BE AM 3 6, 100      

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56 | Oswaal CBSE Mathematics, Class–XI

 x = 2 6 metres 1

 1 cm deflection is at 2 6 metres distance from centre.

OR Equation of ellipse, 9x2 + 4y2 = 36  2 2 4 9 x y  = 1 1  a2 = 4 a = + 2 1 and b2 _{= 9 b = + 3}

 y-axis is the major axis.

 c2 = b2 – a2 = 9 – 4 = 5 1

 c = + 5

Foci = (0, + 5) ½

Vertices = (0, + 3) ½

Length of major axis = 2b = 2 × 3 = 6 units 1

and Eccentricity i.e., e = c

b=

5

3 1

25. There are 52 playing cards.

One card is to be drawn.

(i) P (Diamond card) = 13 52=

1 .

4 1

(ii) A = Black card

 n(A) = 26

 P(A) = 26

52 = 1

2. 1

(iii) B = Not an ace

 n(B) = 52 – 4 = 48

 P(B) = 48

52 = 12

13 1

(iv) C = Not a diamond card

 n(C) = 52 – 13 = 39 cards

 P(C) = 39

52 =

3

4 1

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

 n(D) = 26

 P(D) = 26

52 = 1

2 1

(vi) E = A face card

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

 P(E) = 12 52 = 3 13 1

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

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 S.D. = _{variance = 100 = 10} 1

In firm B,

No. of wage earners = 648 Mean wages = 5253 and variance = 121  Total wages paid by firm

= 648 × 5253 1

= 34,03,944.

S.D. = variance = 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.

C.V. (Firm A) = S.D. × 100 Mean 1 = 10 100 5253 = 0·190 C.V. (Firm B) = 11 100 5253 = 0·2094 1

Hence, firm B shows greater variability between firms. It means firm A looks more consistent than firm B about their wages. 1



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