CLASS XII

MATHEMATICS
(PREPARED BY TEAM MATHS – DOE)

S.NO. OBJECTIVE TYPE QUESTIONS MARKS

(SECTION – I)
1.  1
4
2. 1 1
36
3. x=3 1
4. 0 1
5. 16 1
sq.units
3
OR
F(x) = |x – 1| + |x – 2| (Or any Correct Response) 1

6. 3 1
7. 1 unit 1
8. 1 1
x2
9. 0 1
OR
f ( x)  x.e x 1
10. 11 1
11. 1 1
15
12. 1 0  1
I  
0 1
OR 1
m+n=1
13. (1, 0, 0) 1
14. 2 1
OR
0 1
15. 1 1
OR
3 1
16. 1 1
7
SECTION – II
17. (i) (d) 1
17. (ii) (c) 1
17. (iii) (b) 1
17. (iv) (c) 1
17. (v) (a) 1
18. (i) (d) 1
18. (ii) (c) 1
18. (iii) (c) 1
18. (iv) (d) 1
18. (v) (c) 1
SECTION – III
19. Reflexive : Since, a+a=2a, which is even ½
∴ (a,a) ∈ 𝑅 ∀𝑎 ∈ Z Hence R is reflexive

Symmetric: If (a,b) ∈ R, then a+b = 2λ ⇒ b+a = 2 λ ⇒ (b,a) ∈R, ½

Hence R is symmetric

Transitive: If (a,b) ∈ R and (b,c,) ∈ R then

a+b = 2 λ---(1) and b+c =2 𝜇 ---- (2)
Adding (1) and (2) we get a+2b+c=2(λ + 𝜇) ⇒ a+c=2 (λ + 𝜇 − 𝑏) ⇒ a+c=2k ,where λ + μ − b = k 1
⇒ (a,c) ∈R, Hence R is transitive
20. NO. OF REFLEXIVE RELATIONS = 26 = 64 1

NO. OF SYMMETRIC RELATIONS = 23 = 8 1

21. x  4 y  7 z 1 1
Cartesian Equation :  
1 2 2
 1
Vector Equation : r  (4iˆ  7 ˆj  kˆ)   ( iˆ  2 ˆj  2kˆ)
OR
The given lines can be written as:
xq y0 z s x  q' y 0 z  s' 1½
  and  
p 1 r p' 1 r'
½
As lines are perpendicular then pp′ + rr′ + 1 = 0.
22.  
a  iˆ  ˆj  3kˆ and b  2iˆ  7 ˆj  kˆ
iˆ ˆj kˆ
 
a  b  1 1 3  20iˆ  5 ˆj  5kˆ 1

2 7 1
1   1 15
Area of parallelogram = | a  b | 400  25  25  2 sq.units 1
2 2 2
OR
As,Each one of them being perpendicular to the sum of other two
   ½
2( a.b  c .a  b .c )  0
and
        
| a  b  c |2  (a  b  c ).(a  b  c )
     
= |a |2  | b |2  | c |2 2( a.b  c .a  b .c )
1
= 9 + 16 + 25 + 0 =50
  
| a  b  c | 5 2 ½
23. f '( x)  4 x ( x  1)( x  2) 1
so, f ( x) is increasing for x  [0,1]  [2, )
1
24. dx 1 x  2 1 ½+1
I   log c
( x  2)  1 2
2 2
x  2 1
1 x 3 ½
I  log c
2 x 1

25. dy dx 1
 a sin  ,  a cos 
d d
dy a sin 
  tan  1
dx a cos 
26. As 10, 3   R,  3, 2   R but 10, 2   R 1½
so, R is not Transitive. ½
(OR Any Correct Response)
OR
2x  4 2 y  4 ½
Let , f ( x)  f ( y )   xy
3x  9 3 y  9
So, f ( x ) is one-one function.
2x  4 9y  4
Let , y  f ( x)  x
3x  9 3y  2
As, Range  Codomain 1

thus, f(x) is onto function.

So, f(x) is Bijective Function. ½
27.   ½
2 2
sin x cos x
I dx   dx
0
sin x  cos x 0
cos x  sin x
 
2
sin x  cos x 2

2I   dx   1.dx  1
0
sin x  cos x 0
2

I ½
4
28. 2
4 2 32 2 1½
A  2 2 2 x .dx  ( x )0
0
3
32
A sq.units
3 1/2

SECTION - IV
29.
Area of Ellipse = 4 (Area of BOC) ½ + ½ (FOR FIGURE)
6
1 2 1
A  4 6  x 2 .dx 
0
3
4 x 36 x
A  ( 6 2  x 2  sin 1 ( ))60
3 2 2 6
A  12 sq.units 1
30. Let the Events be
E1 : Choosing 1st Coin
E2 : Choosing 2nd Coin
E3 : Choosing 3rd Coin
A: Getting Heads 1
1
P ( E1 )  P( E2 )  P( E3 ) 
3
A 40 A 75 A 1
P( )  , P( )  , P( )  1
E1 100 E2 100 E3 2
1 1
.
E3 3 2 50 10
P( )   
A 1 40  75  50
( ) 165 33 1
3 100
OR
Let X represents the number of defective bulbs drawn.
½
∴ X can take values 0,1 ,2, 3 or 4
PROBABILITY DISTRIBUTION
6 5 4 3 360
P ( X  0)  ( . . . )  X P(X)
10 9 8 7 5040
360
6 5 4 4 1920 0
P ( X  1)  4( . . . )  5040
10 9 8 7 5040
1 1920
6 5 4 3 2160
P ( X  2)  6( . . . )  5040
10 9 8 7 5040 2 2160
6 4 3 2 576
P ( X  3)  4( . . . )  5040
10 9 8 7 5040 3 576
4 3 2 1 24
P ( X  4)  1( . . . )  5040
10 9 8 7 5040 4 24
5040
TOTAL 1 2½
31.

1
32. The function f(x)=2x−∣x∣ can be written as
f(x)={3x , x ≤ 0
x , x>0.
Now,
Lim x→0− f(x) = 0. 1½
And, Lim x→0+ f(x) = 0.
f(0) = 0
So we've, LHL = RHL = f(x=0), Thus the function f(x) is continuous at x = 0.

We observe LHD = 3, RHD = 1 thus LHD is not equal to RHD, thus f(x) is not 1½
differentiable at x = 0.
OR
x x 2
y  cot 1 (cot ) 
2 2
dy 1
 1
dx 2
33. dy 1 2
 y 2
dx x log x x
1
Integrating factor = log x

so, solution of given differential equation is,

2
y.log x   log x dx 1
x2
2
y.log x  (1  log x)  c
x 1
34. dy b
 cot 
dx a
b
so, slope of tangent = 1½
a
a
Slope of Normal =
b

Equation of Tangent: 2(bx  ay )  ab

Equation of Normal: 2 2(by  ax)  b 2  a 2 1½

35. 2 0 1 2 1
I   | x  x | dx   ( x  x )dx   ( x  x )dx   ( x 3  x)dx
3 3 3

1 1 0 1
4 2 4 2 4 2
x x x x x x
I [  ]01  [  ]10  [  ]12
4 2 4 2 4 2
11 2
I
4
OR
x 1 A Bx  C
Let ,   2
( x  3)( x  4) ( x  3) x  4
2

2 2 7
On solving we get, A = , B  ,C  1½
13 13 13
 2 1 7 x
I log( x  3)  log( x 2  4)  tan 1 ( )  C 1½
13 13 26 2
SECTION - V
36.  1 7 5 
1 
A   5 1 7 
1 2
18
 7 5 1 
 1 7 5 6 18 1
1 1     1    2½
X  A B   5 1 7  6   18  1
Then 18 18
 7 5 1  6 18 1
Thus, x  1, y  1, z  1 ½
OR
 7 8 3
1 
1
A  6 4 2  2
4  
 5 4 1 
 7 6 5   1   4  1
T 1 1 T 1     1    2½
X  (A ) B  (A ) B  8 4 4   4   4  1
4  4    
 3 2 1  7   4  1
Thus, x  1, y  1, z  1 ½
37. Let the required equation of plane passing through the intersection of planes
x + 3y + 6 =0 and 3x - y -4z = 0 be

2
 OR
2x = y = z
38.
Corner points are
A(20,180), B(40, 160) and
C(20, 80)

Z at A = 1,88,000
Z at B = 1,76,000
Z at C = 88,000

SO maximum Z = 1,88,000 C (20, 80)

At x = 20 and y = 180

OR
(i)

1½

𝑀𝑎𝑥 𝑍 = 12 𝑎𝑡 𝐸(4,0) 𝑀𝑖𝑛 𝑍 = −32 𝑎𝑡 𝐴(0,8) 1

(ii) Since maximum value of Z occurs at B(4,10) and C(6, 8)

∴ 4𝑝 + 10𝑞 = 6𝑝 + 8𝑞 2𝑞 = 2𝑝 𝑝 = 𝑞
2
Number of optimal solution are infinite
½