Mock Test-1 (Solution)

 MOCK TEST-1

SOLUTION

ENTHUSIAST COURSE
MATHEMATICS SOLUTION
SECTION – A
1. (D) [1]
We have (BA)–1 = C ⇒ A–1 B–1 = C ⇒ A–1 = CB
 −1 0 1   2 6 4   −3 −5 −5
∴  –1
A = 1 1 3   1 0 1 = 0 9 2

 2 0 2   −1 1 −1  2 14 6 
     
2. (A) [1]

®
Let the order of A is 2 × 2
Given that; aij = i2 – j2
a11 = 12 – 12 = 0 , a12 = 12 – 22 = – 3, a21 = 22 – 12 = 3 , a22 = 22 – 22 = 0
0 −3
∴ A=
 3 0 
 0 3
AT = 
 −3 0 
0 −3
AT = −  =–A
 3 0 
So, AT = – A
|A| = 0(0) – (–3)(3) = 0 + 9 = 9 which is a perfect square number.
3. (D) [1]
In ∆BCD A B
  
BC + CD = BD
   
Given that=
BC b,= CD c F C
 
And BD is parallel to AE
  
⇒ AE= b + c E D
4. (C) [1]
Let the events be
A = Failure of the first component,
B = Failure of the second component
C = Failure of the third component
So, P(A) = 0.2, P(A) = 0.3, P(C) = 0.5
As the events are independent. The machine will operate only when all the components work,
i.e., P(A' ∩ B' ∩ C') = (1 – 0.2) (1 – 0.3) (1 – 0.5)
In rest of the cases, it won’t work,
So P(A ∪ B ∪ C) = 1 – P(A' ∩ B' ∩ C') = 1 – (0.8).(0.7).(0.5) = 1 – 0.28 = 0.72

E 1
Mathematics

 
5. (A) Given | a |= 1 , | b |= 1 [1]
                
| a − b |2 =(a − b)·(a − b) = a.a − a.b − b.a + b.b= | a |2 −2a ⋅ b + | b | 2 ( | a |2 = a ·a)
         
⇒ | a − b |2 = 1 − 2 | a || b| cos θ + 1 ⇒ | a − b |2 =2 − 2cos θ (= | a | 1,= = b.a)
| b | 1,a.b
    θ   θ
⇒ | a − b |2 = 2(1 − cos θ) ⇒ | a − b |2 =2 × 2sin 2 ⇒ | a − b |= 2sin
2 2
6. (A) [1]
dy 1  dx 
(1 + x 2 )(1 + y 2 ) ⇒ ∫
= = ∫ (1 + x )dx Since ∫ 1= tan −1 x + c 
2
Given that; dy
dx 1 + y2  +x 2

x3
⇒ tan–1y = x + +C
3
λ(x 2 + 2), if x ≤ 0

®
7. (C) Given, f(x) =  is continuous at x = 0 [1]
 4x + 6, if x > 0
Since f(x) is continuous at x = 0. Therefore, we have,
LHL = RHL = f(0) = 2λ .....(1)
=
Here, RHL = lim+
f(x) lim+ (4x + 6)
x →0 x →0

= lim[4(0 + h) + 6] = lim(4h + 6) = 4 × 0 + 6 = 6
h →0 h →0

From Eq.(1), RHL = f(0)

⇒ 2λ = 6 ⇒ λ = 3
8. (B) [1]
  
Given a = ˆi − ˆj + kˆ , c =− ˆi − ˆj and a.b = 1

Let b = b1 ˆi + b 2 ˆj + b3 kˆ

 (iˆ − ˆj + k).(b
ˆ ˆ 1⇒ b −b +b =
1 i + b 2 j + b 3 k) =
ˆ ˆ
1 2 3 1 .....(1)
ˆi ˆj kˆ
 
and a×=
b 1 −1 1 = − ˆi(b3 + b 2 ) + ˆj(b1 − b3 ) + k(b
ˆ
2 + b1 )

b1 b 2 b3
  
 a×b =c
⇒ − ˆi(b3 + b 2 ) + ˆj(b1 − b3 ) + k(b
ˆ
2 + b1 ) =− i − j
ˆ ˆ

Comparing the coefficients of ˆi, ˆj and k̂ respectively,

we get b2 + b3 =
1 …..(2)
b1 − b 3 =
−1 …..(3)
b 2 + b1 =
0 …..(4)
By solving the equations (1), (2), (3) and (4), we get b1 = 0, b 2 = 0 and b 3 = 1

2 E
 Mock Test-1 (Solution)

 x + 1 –3 4 
9. (D) Given, A =  –5 x + 2 2 
 [1]
 4 1 x – 6 
A is not singular if |A| ≠ 0
x +1 –3 4
∴ –5 x+2 2 ≠0
4 1 x–6
(x + 1) [(x + 2) (x – 6) – 2] + 3 (–5x + 30 – 8) + 4 (–5 –4x –8) ≠ 0
⇒ (x + 1) (x2 – 4x – 14) + 3 (–5x + 22) + 4 (–4x – 13) ≠ 0
⇒ x3 – 4x2 – 14x + x2 – 4x – 14 – 15x + 66 – 16x – 52 ≠ 0
⇒ x3 – 3x2 – 49x ≠ 0

®
⇒ x(x2 – 3x – 49) ≠ 0
⇒ x ≠ 0 or x2 – 3x – 49 ≠ 0
1
⇒ x ≠ 0 or x ≠ (3 ± 205)
2
10. (B) [1]
x b b
d
Here, ∆1 = a x b = x 3 − 3abx ⇒ ( ∆1=) 3(x 2 − ab) .....(1)
a a x dx

x b
and ∆ 2 = = x 2 − ab .....(2)
a x
d
from (1) & (2) ( ∆1 ) =3∆ 2
dx
2 1 5 3  1 0 
11. (B)=
Let A =  , B =  ,C  [1]
 5 3 3 2  0 1 
Then the given equation becomes AXB = I ⇒ X = A–1B–1
now |A| = 6 – 5 = 1 ≠ 0 and |B| = 10 – 9 = 1 ≠ 0 (A–1 and B–1are exists)
adj(A) 1  3 −1 adj(B)  2 −3
A −1
∴ = =  =
and B−1
= 
|A| 1  −5 2  | B|  −3 5 
 3 −1  2 −3  6 + 3 −9 − 5   9 −14 
⇒ X = A–1B–1 =     =  =
 −5 2   −3 5   −10 − 6 15 + 10   −16 25 
 9 −14 
Hence, X = 
 −16 25 
12. (D) [1]
3 −1 1 3
Let I = ∫
−2
|1 − x 2 | dx= ∫
−2
(x 2 − 1)dx + ∫ (1 − x 2 )dx + ∫ (x 2 − 1)dx
−1 1
−1 1 3
 x3   x3   x3  2 2 2  1  10 28
=  − x  +  x −  +  − x  = + + 2   + (9 − 3) −  − 1  = +6 = .
3  −2  3  −1  3 1 3 3 3 3  3 3

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Mathematics

13. (C) Maximum Z = 22x + 18y [1]

subject to constraints

x + y < 20

360 x + 240 y < 5,760

Now plot the straight lines on the graph and find the corner points of feasible region.

Corner points of feasible region are Y

A(16,0), P(8,12), C(0,20) and O (0,0) 24 D(0,24)

20

®
Corner Points Z = 22x + 18y
16
O(0,0) 0
12 P(8,12)
A(16,0) 352
8
P(8,12) 392 A(16,0)
4
C(0,20) 360 C(20,0)
X' O X
4 8 12 16 20 24 28
Hence, Max. Z = Rs. 392 at P(8,12)
Y' 360+240y = 5760 x+y = 20

14. (C) (b, b), (b, a), (c, a), (b, c), (c, b) and (d, d) must be included to make R an equivalence relation.
∴ Minimum number of elements of A × A which must be adjoined to R to make it an
equivalence relation is 6. [1]
15. (C) [1]
2
dy  dy 
Given, y = x + a 1 +  
dx  dx 
2
dy  
2 2 
  dy  
 y − x dx  =a 1 +  dx  (On squaring both sides)
     
 

 dy 
2
  dy 2 
 y − x dx  = a 1 +   
2

    dx  
2 2
 dy  dy  dy 
y 2 + x 2   − 2.xy =
a2 + a2  
 dx  dx  dx 
2
 dy  dy
(x – a )   − 2xy + y 2 − a 2 =
2 2
0
 dx  dx
order = 1; degree = 2
Now, sum is 2+1 = 3

4 E
 Mock Test-1 (Solution)

16. (A) [1]
 3
AD= ˆi + 5 kˆ ⇒ AD = 34
2 2 2
17. (D) We have to maximize Z = x + y [1]
subject to x + 2y ≤ 70,
2x + y ≤ 95
x, y ≥ 0
Corner–points Z=x+y Y
O (0, 0) Z=0

 95 
A  2 , 0 ZA = 47.5

®
B (40, 15) ZB = 55 (0,35)C
B(40,15)
C (0, 35) ZC = 35
X' X
Hence, Z is maximum at (40, 15) (0,0)O A
x+2y=70
2x+y=95
Y'

18. (D)
p = 8, q = 4, r = (–3) [1]
Given line m is parallel to line n.
∴ Direction vector of both lines are same
−x − 2 y + 3 2z − 6
Line m is = =
4 −2 3
x +2 y +3 z−3
⇒ = =
−4 −2 3
2
x +1 y + 3 z − 3
⇒ = =
8 4 −3
p = 8, q = 4, r = –3
19. (A) [1]
We have, f(x) = x2 – 4x + 6 or f '(x) = 2x – 4 = 2(x – 2)

–∞ 2 +∞
Therefore, f '(x) = 0 gives x = 2 .
Now, the point x = 2 divides the real line into two disjoint intervals name: (–∞, 2) and (2, ∞)
In the interval (–∞, 2), f '(x) = 2x – 4 < 0.
Therefore, f(x) strictly decreasing in this interval.
Also, in the interval (2, ∞), f '(x) > 0 and so the function f(x) is strictly increasing in this interval.
Hence, both the statements are true but Reason is correct explanation of Assertion.
E 5
Mathematics

20. (C) [1]
 2 −2 
Assertion: Let A = 
4 3 

2 −2
We have, | A | = = 6 − (−8) = 14
4 3

Cofactors of |A| are A11 = 3, A12 = – 4, A21 = 2 and A22 = 2

3 − 4  '  3 2
∴ =
adj(A) =   
2 2   −4 2 

1 1  3 2
Now, A −1
= =
14  −4 2 
(adj A)
|A|

®
⇒ So, Assertion is true.
a b 
Reason: Let A =  1 1 
a 2 b2 
We have, |A| = a1b2 – a2b1
Now, cofactors of |A| are A11 = b2, A12 = –a2, A21 = – b1 and A22 = a1
b − b1 
∴ adj (A) =  2
 −a 2 a1 

1 1  b2 − b1 
=
Now, A −1 =  a1 
(adj A)
|A| a1b 2 − a 2 b1  −a 2

So, Reason is false.

Hence A is true but R is false.

SECTION – B
 1  −1   π 
21. tan −1  − −1
 + tan (− 3) + tan  sin  − 2  
 3   
  π 
⇒
 1 
tan −1  −
 3
−1
( −1
)
 + tan − 3 + tan ( −1)  sin  − 2  =

−1


 1  π
 tan −1  − =−
 3 6

π π
− and tan −1 (−1) =
tan −1 (− 3) = − [1]
3 4
 1  π π π 3π
⇒ tan −1  − −1 −1
 + tan (− 3) + tan (−1) =− − − =− [1]
 3 6 3 4 4

6 E
 Mock Test-1 (Solution)

OR
 x 
 tan −1 
2 
1+ 1− x 
Put x = sin θ ⇒ θ = sin–1 x [½]
 θ θ
 2 sin cos 
 sin θ  −1  sin θ  −1 2 2
⇒ tan −1   = tan  1 + cos θ  = tan   [1]
 1 + 1 − sin θ 
2
 2 cos 2 θ 
 2 
 θ θ 1
= tan −1  tan  = = sin −1 x  θ =sin −1 x  [½]
 2 2 2
22. We have,

®
f(x) = x3 – 9kx2 + 27x + 30
⇒ f'(x) = 3x2 – 18kx + 27 = 3(x2 – 6kx + 9) [½]
If f(x) is increasing on R, then
f'(x) > 0 for all x
⇒ 3(x2 – 6kx + 9) > 0 for all x
⇒ x2 – 6kx + 9 > 0 for all x [½]
⇒ 2
36k – 36 < 0
⇒ k2 – 1 < 0
⇒ –1<k<1 [ ax2 + bx + c > 0 for all x ⇒ a > 0 and b2 – 4ac < 0] [1]
23. In isosceles ∆ABC,
let AB = AC = a and BC = b (given) A
da
= −3 cm/sec
dt
a a
b2
=
AD a − 2
4
1 b/2 b/2
A = Area of ∆ABC = (BC) AD) B C
2 D
1 b2
=   (b) a2 − [1]
2 4
−1/2
dA  b  1   2 b2   da  b(2a)(−3) −3ab
=
⇒ a − 4  ·  2a  = =
dt  2 
 2     dt  b2 b2
4 a2 − 2 a2 −
4 4
 dA  −3b2 −3b2
⇒  dt  = = = − 3b
 at a = b b2 2 3b
2 b2 −
4 2
∴ Area is decreasing at the rate of 3b cm/sec. [1]

E 7
Mathematics

OR
Let the side of a cube be x unit.
Volume of cube = V = x3
dV 2 dx
= 3x = k (constant)
dt dt

dx k
= 2 [1]
dt 3x
Surface area = S = 6x2
dS dx
= 12x
dt dt
dS k k
= 12x
=

®
4 
x
2
dt 3s
Hence, the surface area of the cube varies inversely as length of side. [1]
 sin 4x − 4   2sin 2x cos 2x − 4 
24. Let I = ∫ e x   dx = ∫ e x   dx
 1 − cos 4x   2sin 2 2x 
 2sin 2x cos 2x 4 
= ∫ ex  − dx
2sin 2x 
2 2
[1]
 2sin 2x

= ∫ e (cot 2x – 2 cosec 2x) dx

x 2

= ∫ ex (f(x) + f'(x)) dx [where f(x) = cot 2x ⇒ f'(x) = –2 cosec2 2x]

x x
= e f(x) + C = e cot 2x + C [1]

25. Let α, β, γ are the direction angle made by the line OP with positive direction of x, y & z axis
respectively. then,
l = cos 60° m = cos 45° n = cos 60°
1 1 1
⇒ l= m= n= [1]
2 2 2

Again let a, b, c are the direction ratio's of line OP
a  1
 l =  ⇒ a = l × OP ⇒ a =× 2 2 =2
OP 2

a  1
m =  ⇒ b = m × OP ⇒ b = × 2 2 =2
OP 2

c  1
n =  ⇒ c = n × OP ⇒ c =× 2 2 =2 [1]
OP 2

8 E
 Mock Test-1 (Solution)

SECTION – C

x2
26. I=∫ dx
( x 2 + 1)( 3x 2 + 4 )
Let x2 = y
x2 y A B
Let = = +
( x 2 + 1)( 3x 2 + 4 ) ( y + 1)( 3y + 4 ) ( y + 1) ( 3y + 4 )
⇒ y A ( 3y + 4 ) + B(y + 1)
= …..(1)

from equation (1)

Put y = –1 ⇒ A = –1

®
Put y = –4/3 ⇒ B = 4 [1]
y −1 4
∴ = +
( y + 1)( 3y + 4 ) y + 1 3y + 4
x2 −1 4
i.e. = 2 + 2 [1]
(x + 1)(3x + 4) x + 1 3x + 4
2 2

 −1 4  1 4 1
∴
= I ∫  x
2
+ 2
+ 1 3x + 4 
−∫ 2
 dx =
x +1
dx + ∫ 2
3 x +4/3
dx

4 3 x 3 2 x 3
− tan −1 x +
= · tan −1  =  + C tan −1  −1
 − tan x + C [1]
3 2  2  3  2 
 2 + sin x  dy 1 cos x
27. We have,   = − cos x ⇒ dy = − dx (Variable separation )
 1 + y  dx 1+ y 2 + sin x
On integrating both sides, we get,
1 cos x
∫ 1 + y dx = −∫ 2 + sin x dx
on right hand side using 
⇒ log|1 + y| = – log|2 + sinx| + logC  f '(x) 
 = 
 ∫ f(x)
dx log(f(x)) + c


⇒ log|1 + y| + log|2 + sinx| = log C

⇒ log (1 + y )( 2 + sin x ) =
log C [log m + log n = log mn]

⇒ (1 + y) (2 + sin x) = C .....(1) [2]

Also, given that at x = 0, y = 1.
On putting x = 0 and y = 1 in Eq. (1), we get,
(1 + 1) (2 + sin 0) = C ⇒ C = 4

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Mathematics

On putting C = 4 in Eq. (1), we get
4
(1 + y) (2 + sin x) = 4 ⇒ 1 + y =
2 + sin x
4 4 − 2 − sin x 2 − sin x
=
⇒ y −1 ⇒ y= ⇒ y=
2 + sin x 2 + sin x 2 + sin x
π
2 − sin
π π 2 π 1  π 
=
Now, at x =,y  ∴ y  =  sin = 1 [1]
2 2 π 2 3  2 
2 + sin  
2
OR
dy
+ 2y = x 2 (x ≠ 0)

®
We have, x
dx
dy  2 
⇒ + ·y =
dx  x 
x ......(1)

dy
This is linear differential equation of the form + Py =
Q,
dx
2
Here P = and Q = x.
x

∴ ∫ Pdx e ∫ (2/x)dx
I.F. = e= = e= 2 log x
e=
log x 2
x2 [1]
The general solution is given by

y . I.F. = ∫ (I.F. × Q) dx + C

⇒ y·x 2 = ∫ ( x 2 × x ) dx + C

⇒ =
y.x 2 ∫ x dx + C
3

x4
∴ y.x 2 = +C .....(2)
4
On putting x = 2, y = 1 in Eq. (2), we get
24
1·22 = +C⇒4=4+C⇒C=0
4

x4
∴ y·x 2 = [from Eq. (2)]
4

x2
⇒ y=
4
which is the required particular solution. [2]
10 E
 Mock Test-1 (Solution)

28. Consider the following events:
A = Drawing 3 white balls in first draw, B = Drawing 3 black balls in the second draw.
Total number of balls = 5 white + 8 black = 13 balls
 B
Required probability P (A ∩ B) = P(A)P   .....(1) [1]
A
 B
5 8
C3 5 C3 7
=
P(A) = and P =  = [1]
A
13 10
C3 143 C3 15
Substituting these values in (1), we obtain
 B 5 7 7
Required probability P(A ∩ B) = P(A) P   = × = [1]
 A  143 15 429

®
29. Let I = ∫ cos(log e x) dx .....(1) (using by parts)

put loge x = t
⇒ x = et
⇒ dx = et dt
I = ∫ e t cos t dt
II I

⇒ I cos t e t − ∫ e t (− sin t) dt
= [1]

=
⇒ I cos t e t + ∫ e t (sin t) dt (Again using by parts)
II I

⇒ I =cos t e t + e t (sin t) − ∫ e t cos t dt

⇒ I =cos t e t + e t (sin t) − I [from (1)] [1]

⇒ 2I =cos t e t + e t (sin t) + C

et
⇒ =
I
2
[cos t + sin t ] + C
x
=
⇒ I [cos (log e x) + sin (log e x)] + C [1]
2
OR
π
4xsinx
I=∫ dx …..(1)
0
1 + cos 2 x
π a a
4(π − x)sin(π − x)
=I ∫0 1 + cos2 (π − x) dx [Applying ∫f (x)dx
0
= ∫f (a − x)dx]
0

π π
4πsinx 4xsinx
=I ∫ dx − ∫ dx
0
1 + cos x
2
0
1 + cos 2 x

E 11
Mathematics

π
4πsinx
=I ∫ 1 + cos x dx − I
0
2
[From equation (1)] [1]

π
sinx
2I = 4π ∫ dx
0
1 + cos 2 x

π/2 2a a
sinx
=
2I 4π.2 × ∫
0
1 + cos 2 x
dx =
{Applying ∫f ( x ) dx 2 ∫f ( x ) dx=
0 0
if f(2a − x) f(x)}

put cosx = t ⇒ –sinx dx = dt

when x = 0 and when x = π/2

then t = 1, then t = 0

®
0
−dt dt
1
 b a

∴ I = 4π ∫ = 4π ∫  ∫
 f (x)dx = – ∫ f (x)dx  [1]
1
1+ t 2
0
1+ t2  a b 
1
I = 4π  tan −1 t 
0

I = 4π[tan–11 – tan–10]
π
I = 4π × = π2 [1]
4

30. We have e y + xy =
e.

dy dy
ey +y+x =0 (Differentiating w.r.t. x) .....(1) [1]
dx dx
2
d2y y  dy  dy d2y
ey
+e   +2 +x 2 = 0 (Again differentiating w.r.t. x) .....(2) [1]
dx 2  dx  dx dx

Putting x = 0 in e y + xy =
e, we get y = 1

Putting=x 0,=y 1 in Eq. (1), we get

dy
e +1 =0
dx
dy 1
⇒ = −
dx e
dy 1
Putting x = 0, y = 1, = − in Eq.(2),
dx e

d2y 1 1 d2y d2y 1

we get, e 2 + e 2 − 2 + 0. 2 =0 ⇒ 2 = 2 [1]
dx e e dx dx e

12 E
 Mock Test-1 (Solution)

31. Converting the inequations into equations, we obtain the following equations:
x + y = 4, 3x + 8y = 24, 10x + 7y = 35, x, y ≥ 0.
The feasible region of the LPP is shaded in Figure.

5 (0, 5)
4 (0, 4)
B(0, 3) 3 7 5
P , 
2 3 3
1

®
(4, 0) (8, 0) [1]
X' X
O 1 2 3 4 5 6 7 8 3x + 8y = 24
x+y=4
A(3.5, 0)
Y' 10x + 7y = 35

The coordinates of the corner points of the feasible region OAPQB are O (0, 0), A(3.5, 0),
7 5  8 12 
P  ,  , Q  ,  and B(0,3).
3 3 5 5 
Corner points Z = 5x + 7y
O (0, 0) ZO = 0
A (3.5, 0) ZA = 17.5
 8 12 
7 5 70 The maximum value of Z is 24.8 at  , 
P ,  ZP = = 23.23 5 5  [2]
3 3 3

 8 12  124
Q ,  ZQ = = 24.8
5 5  5
B(0, 3) ZB = 21
OR
Minimize Z = 5 x + 7y

Subject to the constraints

2x + y ≥ 8,

x + 2y ≥ 10,

x, y ≥ 0

E 13
Mathematics

The feasible region determined by the system is shown in the figure.

Corner points of feasible region are

A(0,8), B(2,4) and C(10,0) Y
Corner Points Z = 5 x + 7y 12
A(0, 8) 56
10
B(2, 4) 38 (min.)
C(10, 0) 50 8 A(0,8)
6
Since the feasible region is (0,5) B(2,4)
unbounded. So we draw the graph of [2]
2
inequality 5x + 7y < 38 and obtain (4, 0) C(10,0)
X' X
open half plane. Here open half O 2 4 6 8 10 12

®
plane has no common point with
feasible region. (2x+y =8) 5x+7y = 38
Y'
Thus, the minimum value of Z is 38
attained at the point (2, 4).
[1]

SECTION – D
32. Let A(2, 5), B(4, 7) and C(6, 2) be the vertices of a ∆ABC.

Now, Equation of line AB is y = x + 3

−5
Equation of line BC is y = x + 17
2
−3 13
Equation of line CA =
is y x+ [1]
4 2

(2, 0) (4, 0) (6, 0)

[Correct fig. 1 mark]

14 E
 Mock Test-1 (Solution)

Area of ∆ABC = Area of trapezium ABEF + Area of trapezium BCDE – Area of trapezium ACDF

 −5   −3 13 
4 6 6
=∫ (x + 3) dx + ∫  x + 17  dx − ∫  x +  dx [1]
2 4
 2  2
 4 2
4
 x2   5
6 6
  −3 13 
=  + 3x  +  − x 2 + 17x  −  x 2 + x  [1]
2 2  4 4  8 2 2

 51 23 
= (20 – 8) + (57 – 48) –  − 
 2 2 
= 12 + 9 – 14 = 7 square units. [1]
33. Given that, R = {(1, 39), (2, 37), (3, 35).....(19, 3), (20, 1)}

®
Domain = {1, 2, 3, ........., 20} [1]
Range = {39, 37, 35, .........., 9, 7, 5, 3, 1} [1]
 R is not reflexive as (2, 2) ∉ R
2 × 2 + 2 ≠ 41 [1]
 R is not symmetric
as (1, 39) ∈ R but (39, 1) ∉ R [1]
 R is not transitive
as (11, 19) ∈ R and (19, 3) ∈ R
But (11, 3) ∉ R
Hence, R is neither reflexive, nor symmetric nor transitive. [1]

OR
We observe the following properties of relation R.
Reflexive : Let (a, b) be an arbitrary element of N × N. Then, (a, b) ∈ N × N
⇒ a, b ∈ N
⇒ a + b = b + a [by commutativity of addition on N]
⇒ (a, b) R (a, b)
Thus,(a, b) R (a, b) for all (a, b) ∈ N × N. So, R is reflexive on N × N. [1]
Symmetric : Let (a, b), (c, d) ∈ N × N. be such that (a, b) R (c, d) Then, (a, b) R (c, d)
⇒ a+d=b+c
⇒ c + b = d + a [By commutativity of addition on N]
⇒ (c, d) R (a, b) [By definition of R]
Thus (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N.
So, R is symmetric on N × N. [1]

E 15
Mathematics

Transitive : Let (a, b), (c, d), (e, f) ∈ N × N such that (a, b) R (c, d) and (c, d) R (e, f). Then,
(a, b) R (c, d) ⇒ a + d = b + c .....(1)
(c, d) R (e, f) ⇒ c + f = d + e .....(2)
On adding eq. (1) & (2)
(a + d) + (c + f) = (b + c) + (d + e) ⇒ a + f = b + e ⇒ (a, b) R (e, f)
Thus, (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d) ∈ (e, f) ∈ N × N.
So, R is transitive on N × N. [1]
Hence, R being reflexive, symmetric and transitive, is an equivalence relation on N × N.
[(2, 3)] = {(x, y) ∈ N × N : (x, y) R (2, 3)}
⇒ [(2, 3)] = {(x, y) ∈ N × N : x + 3 = y + 2} = {(x, y) ∈ N × N : y = x + 1}

®
= {(x, x + 1) : x ∈ N}
= {(1, 2),(2, 3),(3, 4),(4, 5),..} [1]
[(7, 3)] = {(x, y) ∈ N × N : (x, y) R(7, 3)}
= {{x, y) ∈ N × N : x + 3 = y + 7]
= {(x,y) ∈ N × N : y = x – 4}
= {(x, x – 4 ) ∈ N × N : x ∈ N]
= {(5, 1), (6, 2), (7, 3), (8, 4), (9, 5),..} [1]

34. We have, r = ( 8 + 3λ ) ˆi + ( −9 − 16λ ) ˆj + (10 + 7λ ) kˆ

r= 8iˆ − 9ˆj + 10kˆ + λ ( 3iˆ − 16 ˆj + 7kˆ ) .....(1)

Also, =r 15iˆ + 29ˆj + 5kˆ + µ ( 3iˆ + 8ˆj − 5kˆ ) .....(2)
     
on comparing (1) & (2) with r= a1 + λb1 and r= a 2 + µb 2
 
⇒ a1 = 8iˆ − 9ˆj + 10kˆ and b1 =3iˆ − 16ˆj + 7kˆ
 
⇒ a 2 = 15iˆ + 29ˆj + 5kˆ and b2 = 3iˆ + 8ˆj − 5kˆ
 
Now, shortest distance between two lines is given by
( b × b ) . (a
1 2 2

− a1 )
[1]
b1 × b 2

ˆi ˆj kˆ
 
∴ b1 × b 2 = 3 −16 7 = ˆi ( 80 − 56 ) − ˆj ( −15 − 21) + kˆ ( 24 + 48 ) = 24iˆ + 36 ˆj + 72kˆ [1]
3 8 −5
 
Now, b1 × b=
2
( 24 )2 + ( 36 )2 + ( 72=
)2 12 2 2 + 32 + 6=
2
84
 
And ( a 2 − a1 ) = (15 – 8) î + (29 + 9) ĵ + (5 – 10) k̂ = 7 î + 38 ĵ – 5 k̂ [1]

16 E
 Mock Test-1 (Solution)


∴ Shortest distance =
( 24iˆ + 36ˆj + 72kˆ ) ⋅ ( 7iˆ + 38ˆj − 5kˆ )
84

168 + 1368 − 360 1176

= = = 14 units [2]
84 84
OR
Equation of line in vector form

Line-I: r= ( ˆi − ˆj + 0kˆ ) + λ ( 2iˆ + 0ˆj + kˆ ) .....(1)

Line-II: =
r ( 2iˆ − ˆj + 0kˆ ) + µ ( ˆi + ˆj − kˆ ) .....(2)
     
an comparing (1) & (2) with r= a1 + λb1 & r= a 2 + µb 2

®
 
a1 = ˆi − ˆj + 0kˆ and a 2 = 2iˆ − ˆj + 0kˆ
 
b1 = 2iˆ + 0 ˆj + kˆ and b 2 = ˆi + ˆj − kˆ

We know that the shortest distance between lines is:

   
d=
( a 2 − a1 ) ·( b1 × b 2 )
  [1]
b1 × b 2
 
(a 2 − a1 ) = ( 2iˆ − ˆj ) − ( ˆi − ˆj + 0kˆ )
 
(a 2 − a1 ) =i + 0 j + 0k
ˆ ˆ ˆ

ˆl ˆj kˆ
 
b1 × b 2 = 2 0 1
1 1 −1
 
b1 × b 2 =( 0 − 1) ˆi − ( −2 − 1) ˆj + ( 2 − 0 ) kˆ
 
⇒ b1 × b 2 = – ˆi + 3ˆj + 2kˆ [1]
 
b1 × b 2 = ( −1) + 32 + 2 2
2

 
⇒ b1 × b 2 =14

∴ d=
( ˆi + 0ˆj + 0kˆ ) ·( −ˆi + 3ˆj + 2kˆ ) = 1
units [2]
14 14

Shortest distance d between the lines is not 0. Hence the given lines are not intersecting. [1]

E 17
Mathematics

1 1 1 
35. Given; A = 1 2 −3
2 −1 3 
 

To prove : A 3 – 6A 2 + 5A + 11I =
O

1 1 1  1 1 1 1 + 1 + 2 1 + 2 – 1 1 – 3 + 3 
A2 = A.A = 1 2 
–3 1 .  2 –3= 1 + 2 – 6 1 + 4 + 3 1 – 6 – 9 
2   3  2 – 1 + 6 2 – 2 – 3 2 + 3 + 9 
 –1 3  2 – 1  

 4 2 1
⇒ A2 =  –3 8 – 14  [1]
 7 –3 14 


 4 2 1 1 1 1  4 + 2 + 2 4 + 4 –1 4–6+3 

®
3
2
and A = A .A = –3 8 
– 14 . 1  2  
– 3 = –3 + 8 – 28 – 3 + 16 + 14 – 3 – 24 – 42 
 7 –3
 14  2 – 1 3   7 – 3 + 28 7 – 6 – 14 7 + 9 + 42 

 8 7 1 
⇒ A = –23 27 –69 
3
 [1]
 32 –13 58 

L.H.S. = A3 – 6A2 + 5A + 11I

 8 7 1  4 2 1  1 1 1  1 0 0 
= –23 27 –69 – 6 –3 8 –14 +5 1 2 –3 + 11 0 1 0 
     
 32 –13 58      0 0 1 
   7 –3 14  2 –1 3   

 8 – 24 + 5 + 11 7 – 12 + 5 + 0 1 – 6 + 5 + 0  0 0 0 
=  –23 + 18 + 5 + 0 27 – 48 + 10 + 11 –69 + 84 – 15 + 0  = 0 0 0  = O (Zero Matrix) [1]
32 – 42 + 10 + 0 –13 + 18 – 5 + 0 58 – 84 + 15 + 11  0 0 0 
   

Now, A3–6A2 + 5A + 11I = O

⇒ A3A–1 –6A2A–1+5AA–1+11IA–1 = OA–1 (Post multiplying both sides by A–1)
⇒ A2 (AA–1) –6A (AA–1) +5 (AA–1) +11 IA–1 = OA–1
⇒ A2 – 6A + 5 I + 11 A–1 = O [ AA–1 = I and OA–1 = O]

 4 2 1  1 1 1  1 0 0  
–1 –1   
=
⇒ A = 8 –14 – 6 1 2 –3 + 5 0 1 0  
  
–1
⋅ (A 2 − 6A + 5I)  –3 [1]
11 11   –3 14   2 –1 3  0 0 1  
 7    

–1  4 – 6 + 5 2 – 6 + 0 1– 6 +0   –3 / 11 4 / 11 5 / 11 
⇒ A –1 =  9 / 11 –1 / 11 –4 / 11
 –3 – 6 + 0 8 – 12 + 5 –14 + 18 + 0  ⇒ A –1 = [1]
11  7 – 12 + 0 –3 + 6 + 0 14 – 18 + 5   5 / 11 –3 / 11 –1 / 11
 

18 E
 Mock Test-1 (Solution)

SECTION – E
36. (i) Since 'a' ft be the length of tank and 'b' ft be the height of the tank. The width of the tank is
equal to 10ft. (given)
∴ The area of two sides will be 10b square foot and the area of other two sides will be ab
square foot. So, the total area of the sides is equal to (20b + 2ab) square foot cost of the
sides is Rs. 20 per square foot.
So, the cost to build the sides is (20b + 2ab) × 20 = Rs. (400b + 40ab)
Also cost of base = (10a) × 20 = Rs. 200a
∴ Total cost of the tank in Rs. is given by
C = 400b + 40ab + 200a

®
 Volume of tank = 100 ft3

10
∴ 10ab = 100ft3 ⇒ a =
b
 10   10  2000
⇒ C(b) = 400b + 40   (b) + 200   = 400b + 400 + [1]
 b  b b
2000
(ii) Since C(b) = 400b + 400 + ,b>0 ( b is height)
b
differentiate w.r.to b
+ – – +
2000  b2 − 5 
C'(b) = 400 − 2 = 400  2  0
b  b 
when b∈ [ 5, ∞) , C'(b) ≥ 0
⇒ C(b) is increasing function
when b ∈ (0, 5] ,C'(b) ≤ 0
⇒ C(b) is decreasing function [1]

2000
(iii) C'(b) = 400 −
b2
4000
again differentiate w.r.t to b, C"(b) =
b3
To minimise the cost, put C'(b) = 0
2000
⇒ 400 – =0
b2
⇒ 400b2 = 2000 ⇒ b2 = 5 ⇒ b = 5 ( b > 0)
10
⇒ b = 5 and a = [ height can't be negative and C"(b) > 0 at b = 5]
5
The value of b is 5 for which C(b) is minimum

E 19
Mathematics

Now total surface area of the tank
= 20b + 2ab + 10a
2 × 10 10.10 100 200 + 20 5
= 20 5 + · 5+ = 20 5 + 20 + = sq. units [2]
5 5 5 5
OR
2000
(iii) C'(b) = 400 −
b2
4000
again differentiate w.r.t to b, C"(b) =
b3
To minimise the cost, put C'(b) = 0
2000
⇒ 400 – = 0 ⇒ 400b2 = 2000 ⇒ b2 = 5 ⇒ b = 5 ( b > 0)

®
b2
⇒ b= 5 [ height can't be negative and C"(b) > 0 at b = 5]
So, the cost of least expensive tank will be :
2000
C( 5) = 400 5 + 400 + = 400 5 + 400 + 400 =
5 400 + 800 5
5
⇒ = ( )
C 5 Rs. 400(1 + 2 5) [2]

37. (i) AB =(2 − 1)iˆ + (1 − 1)ˆj + (3 − 1)kˆ =ˆi + 2kˆ

∴ AB = 12 + 22 = 5 units

AC = (3 − 1)iˆ + (2 − 1)ˆj + (2 − 1)kˆ

= 2iˆ + ˆj + kˆ

∴ AC = 22 + 12 + 12

= 4 + 1 + 1= 6 units [1]

 
(ii)  BD =ˆi + 2 ˆj + kˆ and BC = 2iˆ + ˆj − kˆ
 
∴ Projection of BD on BC is
 
BD · BC (iˆ + 2 ˆj + k)
ˆ ·(2iˆ + ˆj − k)
ˆ 2 + 2 −1 3
 = = = units [1]
BC 4 +1+1 6 6

(iii) BC = (3 − 2)iˆ + (2 − 1)ˆj + (2 − 3)kˆ = 2iˆ + ˆj − kˆ

and CD =(3 − 3)iˆ + (3 − 2)ˆj + (4 − 2)kˆ =ˆj + 2kˆ
ˆ ˆ ˆ
  i j k
∴ BC × CD= 1 1 −1 = ˆi(2 + 1) − ˆj(2 − 0) + k(1
ˆ − 0) = 3iˆ − 2 ˆj + kˆ
0 1 2

20 E
 Mock Test-1 (Solution)

1  
∴ Area of ∆BCD = BC × CD
2
1 2 1 14
= 3 + 2 2 + 12 = × 14 = sq. units [2]
2 2 2
OR
 
AB= ˆi + 2kˆ and AC = 2iˆ + ˆj + kˆ
ˆ ˆ ˆ
  i j k
∴ AB × AC =1 0 2 =−2iˆ + 3ˆj + kˆ
2 1 1

1   1 1 14
∴ Area of ∆ABC = AB × AC= 4 + 9 + 1 = × 14 = sq. units

®
2 2 2 2
Here, Area of ∆BCD is not greater than area of ∆ABC. [2]
38. (i) E denote the event that the student has failed Economics and M denote the event that the
student has failed in Mathematics.
50 1 35 7 25 1
∴ P(E) = = , P(M) = = and P(E ∩ M) = =
100 2 100 20 100 4
The probability that the selected student has failed in Economics if it is known that he has
failed in Mathematics.
1
 E  P(E ∩ M) 4 1 20 5
Required probability = P   = = = × = [2]
M P(M) 7 4 7 7
20
(ii) Let E denote the even that student has failed in Economics and M denote the event that
student has failed in Mathematics.
50 1 35 7 25 1
∴ P(E) = = , P(M) = = and P(E ∩ M) = =
100 2 100 20 100 4
The probability that the selected student has failed in Mathematics if it is known that he has
failed in Economics.
1
 M  P(M ∩ E) 4 1
Required probability = P   = = = [2]
E P(E) 1 2
2

E 21