2 PRACTICE PAPER

[Time Allowed: 3 Hours] (Maximum Marks: 80]

General Instructions: As given in the Latest Sample Paper.

SECTION - A
(Multiple Choice Questions)
Each question carries I mark

1. Let f':R-(2} ’R- {1} be afunction defined by 4.

x-1
f(r) = , then f' is
r-2
543B + Cis equal to
(a) into function
(b) many one function -1 -1
(c) bijective function (a)20 20
(d) many one, into function.
(c)
Sol. (c)
2. The relation R in the set of real numbers defined as Sol. (d)
R= {(a, b) eR× R:1+ ab>0} is 5. IfAB = CthenA, B, C is equal to
(a) reflexive and transitive
(a) A,x , B, x» Czx3 (6) A,x2 Bx3, C; x3
(6) symmetric and transitive
(c) A;x3» B, x Cx3 (d) As x, Bz x3 Cyx2
(c) reflexive and symmetric
Sol. (b)
(d) equivalence relation
x+1
Sol. (c) 6. The function f, defined by fx) = is
1+/1+x
13 continuous at x = 0, when f(0) =
3. The value of cos cos 6
is
1
13r (a) 1 (b) -2 (c)
(a) (b) (c) (d) 2
6 6 4
Sol. (b), We know that the range of the principal value Sol. (c), The function f(x) is continuous at x = 0
branch of cos is (0, n. LHL = RHL = f(0)
13r
But lim f(r) = X-0
X-0
lim f(x) =f(0)
6
0-h+1 0+h+1
.. cos lim = lim
h-01+/1+0-h h-01+1+0+h
= cos - f0)
1 1
=6 e[0, ] - f0)
1+1 1+1
.. cos 1
 PracticePapers 475

[|-4| if x*4 12. The area enclosed bythecirclex+y'=

2is equal
Thefuunction
fr) =2(x - 4) is
to
0, if x=4
continuous at all x e R (a) 4r sq units (b) 2/2n sq units
(a)

discontinuous at x = 4 (c) 4n sq units (d) 2 sq units

lc)continuous only when x =4
Sol. (d), since area = 4//2
None of these
JO
(d) 2
(b). The
function iss defined when f(4) = 0(given) = 2r sq units.
.. fu)exists. /20
Consider LHL +
= 1is
13. The area enclosed by the ellipse a?
lim f(r) = lim |x-4| equalto
X-4
I-42(x- 4) (a) r´ab sq units (b) Tab sq units
= lim |4-h-4| (c) ra'bsq units (d) rab' sq units
h-02(4 -h-4)
= lim h -1 Sol. (b), since area = 4"2/a'-ta
Jo a
Consider RHL
h-02(-h) 2h
a' = Tab sq units
|x-4| V-+sin
2
lim f(r) = lim
r-4 x-42(r- 4) 14. Solution of the differential equation

= lim
|4+h -4 | |h | 1
x1-y'ar +y1-tay = 0is
= lim
h-02(4 +h4) h -0 2h 2

and f(4) = 0
(a) 1++/1-y' =C
lim f(o)= X-4
limf(r) *f(4) (6) V1-+/1-y² =C
Y4

»fx) is discontinuous at x=4 (c) 1--/1-y' =C

8. For the functiony = + 21, the value of x, when y (d) (V1-)(/1-y)=C
increases 75 times as fast as x, is Sol. (b)
(a) +3 (b) + 5/3 15. Integrating factor for the solution of differential
(d) None of these equation (r-y'dy +y dr = 0 is
(c) t5
1
Sol. (c) (a) (b) logy (c) y (d) y²
9. The function fr) = cos r-2px is decreasing for Sol. (c) Equation| is (-y) dy +y dr = 0
1 2
(a) p< 6) p> (c) p< 2 (d) p >
1
dr x-y dr 1
Sol. (6) dy y dy y

10.
+3)dr = Integrating factor = =eo =y
23
25 26 (a) 16. The general point on the line
a (6) (c)8 3
3
3 r=(2i+j-4k) +A(3i +j-k) is
(a) (2, 1, -4) (b) (3, 2, -1)
(c) (-1, 1, 3) (d) (2+ 32, 1 + 22, -4)
V1-sin 2x dr =
(a) V2 -1
(b) /2+1 Sol. (d), As given line is r = (2i +Ë-4k) +
(c) V2 (d) /3 (3i +2j k)
Sol
(a)
 476 gether wit Mathematics-12 ASSERTION-REASON BASED QUESTIONs
. position vector of a point through which line
assertion ta
passes. In the following questions, a statement of
Choose the core
is followed by a statement of reason (R).
r= (2+3A)i +(1+ 22)j +(-4- A)k answer out of the following choices.
.:. General point is (2 + 32, 1+ 2., -4 - 2) (a) BothAand Rare true and R is the correct explanatios
17. A die is tossed twice. The probability of getting 1, 2, of A.
3, 4 on first toss and 4,5, 6 on the second toss is (b) Both Aand Rare true but R iS not
the corre

(a) (b)
1
1 explanation of 4.
(c) (d)
2 (c) A is true but R is false.
Sol. (c), A
die is tossed twice.
(d) A is false but R is true.
.:.Total possible outcomes = 36
19. Assertion (4): Ifk is a scalar andAis a 3 x3square
Let E:1,2, 3, 4 appear on first die.
matrix. Then | k4 | is equal tok |A |.
4 2 Reason (R): If every element of a third order
P(E) = 6
:3
determinant of value A is multiplied by n, then the
Let F: 4, 5, 6 appear on second die. value of new determinant is n°A.
3 1
P(F) == Sol. (a), Both assertion (A) and reason (R) are true
6 2
and reason (R) is the correct explanation of
Here E and F are independent events.
assertion (R).
1 1
:.P(E OF) = PE) P(F) =x2 20. Assertion (4): The order of differential equation of
familyof curves + = l is 1.
18. A and Bare two independent events, PA) = 0.2 and
P(B) = 0.8, then P(An B) = Reason (R): The number of arbitraryconstant in the
(a) 0.03 (b) 0.04 (c) 0.4 (d) 0.3 general solution ofa differentialequation ofordern is
Sol. (b), Aand B are independent events. always n.
.:. P(An B) = P(A)" PB) Sol. (d), As we know that number of arbitrary constant in
Now PA n B) = P(A)· P(B) general solution is equal to order of differential
equation
= PA) [1-P(B)] So order = 2(as a and b are two arbitrary
= 0.2 (1-0.8) constants, given)
= 0.2 x 0.2 = 0.04 So, A is false, but R is true.
SECTION - B
(This section comprises of very short answer type-questions (VSA) of 2 marks each)
21. If cos x + cos y =2n, then find the value of 1
y100 +y200
+y0+ x0y200
l0 200 1
Sol. Consider cosx + cos y = 2n = I + T
-(-1)"+(-1)20+.
(-1)0(-1)20
=1+1+1=3
Also, maximum yalue of cos =I
22. Given matrix A= show that A+ A' is
cos x= T, COS y= 1
symmetric matrix and A -A' is a skew symmetric
X= COS T, V= C0S T matrix.
OR
x = -1, y = -1 IfA is a square matrix such that A = A, show that
(U+ A)=74+ I.
 Practice Papers 477
. GivenA
1
C= log| /2 -1|
Substituting in (i), we get
4+ 4 = logcosec x- cot x|
2y
|1+3 2+ 1-Jogl V2-1|
P whose one side
Now 24. Find the area of the parallelogram
coinitial vector
and a diagonal are represented by
Pis symmetric matrix
3i k and 2i +j - 4k.
+j -4k)|
Again
SOl. Area of parallelogram = |(3í -k) x(2i
4-4 3+1| =|3 0 -1||= |i+ 10 +3k|
*1-3 2- 21 4

-?---o
Qis skewsymmetric matrix.
= /1+100 +9 = /110 sq units
25. A and B throw apair of dice alternatively. A wins if
he throws 6 before B and B wins if
he throws 7 before
the odds in favour
OR A throws 6. IfAbegins, show that
ofA are 30 :31.
U+ A= (|+ A)(I + A)l + A)
OR
= (+ IA + AI + A)+ A)
random (without
=(|+A +A +A)(I + A) Two numbers are selected at
= (|+3A)(1 + A) replacemnent) from positive integers 2, 3, 4, 5, 6
numbers
and 7. Let X denotes the larger of the two
= I+ A+ 34I+ 34
obtained. Find the probability distribution of X.
= |+A+ 34 + 34 = I+ 74. 5
Sol. Let P(A) = 36
23. Solve the differential equation dy =y' cosecx, given (Probability of throwing a total of 6), P(A) = 36
31
dx

thaty)=2. [: P(X) + P() = 1]

6
Sol. Consider
dy =y cosec x P(B) = -=÷(Probability
36 6
ofthrowing a total of 7),
dx
5
P(B) = 6

Probability of A'swins =P(A)+ P(A)P(B)P(A) +..,

cotx + C 5 31
-2
= log|cosecx- 36 36 636
...) 5
C
x- cot x +
= logcosec
1 36 5 216 30
155 '.a+ ar + az
36 61 61
1
216
given ylä)= 2 Odds in favour of Aare P(A):
-cot +c P(A)
- loglcosec i.e., 30 30 31
61) 61 = 30:31
+C 61
=loglW2-1|
 478 agether wek Mathematics-12
distribution is as follow:
OR Table for probability
PX)
Given numbers are 2, 3, 4, 5, 6 and 7
1
Total possibilities of selecting 2numbers = °C,=15 3 15

X denote larger of the two numbers. 2

4 15
X can take values 3, 4, 5, 6, 7
3
For 3,onc possibility (2, 3) 5 15

For 4, two possibilities (2, 4), (3, 4) 4

6 15
For 5, three possibilities (2, 5),(3, 5), (4,5)
5
For 6, four possibilities (2, 6),(3, 6), (4, 6). (5, 6) 7 15
For 7,five possibilities (2, 7), (3, 7), (4,7), (5, 7),(6,7)
SECTION -C
(SA) of 3 marks each)
(This section comprises of short answer type questions
26. Let f: W’ # be defined as fn)=n-1,if nis odd 27. Ifr =asin 2t(1 + cos 2) and y =bcos 2(1- cos 2).
andf(n)=n+1,ifn is even. Show thatfis injective ay
show that
as well as surjective. Here, Wrepresents the set of d lat t =
whole numbers. Sol. Given x = a sin 21(1 + cos 2r);
n-1, n is odd Differentiating both sides w.r.t. x, we get
Sol. Given fn) =
n+1, n is even dx
dt
= asin 2(-2 sin 2r) + 2cos 2r(1 + cos 2))
For one-one:Letn,, n, e Wbe even.
= 2a(- sin'21 + cos 2t + cos2)
fn) = fn) =2a(cos 2: + cos 41) ..)
n, + 1= n, +1’ n, =n and y= bcos 21 (1 - cos 21)
Let n,,n, e Wbe odd. Differentiating both sides w.r.t. t, we get
fn,) =fn)» n,-1 =n,-1’n, =n, dy
dt
= b<cos 21(2 sin 21) -2sin 21(1-cos 21)
Let n, be even and n, be odd e W.
= 2b(cos 2t sin 2t - sin 2t + sin 2t cos 2)
n, n. Then n, + 1 is odd and n, -1 is even = 2b(2 sin 2/ cos 2t- sin 2)
fn,) f(n,) = 2b(sin 4t - sin 2) ..i)
From above, we notice f is one-one injective dy dy d 2b(sin 4t-sin21)
For onto:Let for m (even) eW(co-domain), there d dt dt 2a (cos 2+ cos 41)
exists n e W(domain) [from (i) and (iü)]
such that m = fn)
m = n-1 dy b(sin 1- sin ;) b(0-1) b
n= m +1e W
Jat t=* a(0-1) a
-+ coS
f(m + 1) =m + 1-1 =m ...()
For m (odd) e W(co-domain), there exists n e W
(domain) 28. Ify = log x, show that d'y 2 log x-3
such that m = fn)
m = n +1 n = m-1 OR

f(m - 1) =m-1 +1= m ...ü) Examine the differentiability of the function

.:. fisonto or surjective fr) =
xr], if 0<x < 2 atr = 2.
Asfis injective as well as surjective. la-1)x, if 2Sr < 3
 Practice Papers 479

=
log x log x 30. Evaluate (3 x-1 -e"dr.
sol. Giveny 1-log x
Again differentiating W.I.t. x, we get OR
Solve the differential equation:
(1- log x) (2r) dy = sin (x+ y) + cos (r + y).
du? x
dx
[a+1)-2|l
-X-2r +2x log x 2logx 3 Sol. Consider

1 2
OR
-2
x*] , if 0<x<2 1
Given function fx) = Let f(r) = (r+1)2 then f'()= (x+1)
(x-1)x, if 2<x<3
Using e{fr) +fu)}dt = e.fr) + Cwe get
LHD = Lf(2) = h-0
lim
f(2-h)- f(2)
-h X-1 1 +C.
(a+1)²
(2-h)[2-h]- 2
= lim
-h
= lim (2-h)(1)-2
h-0 h-0 -h (3 x-1
-h 16
= lim
h-0 -h lim(1(1) = 1.
= h-0
OR
and RHD = Rf(2) lim f(2+h)-f(2) dy
h-0
Consider equation de = sin ( + y) + cos (r+ y)..)
= lim
(2 + h-1)2 +A)-2
h-0
dy dt
Let x+y=t ’ 1+ dx dr
= lim (h + 1)(h + 2) -2
h-0 dt
From (i), -1 = sin t + cOS t
h'+ 2h +h+2 h+3h - lim (4+3)=3. de
= lim
h-0 im h
h-0 h-0 dt
= 1+ sin t + cos t
LHID RHD d
=2 X=2
dt
Since, LHD RHD =dx
1+ sint+ coS t
at
Hence, the given function is not differentiable
X=2. Integrating both sides, we get
r>-1is an dt
29. Show that y = log (1 + x) -
2+x 1+ sin t + cos t
domain.
increasing function ofx throughout its Let tan 2 =zi= 2 tan-' z
2x
Sol. Given y = log (1 +x) - 2+x dt =
2
dz
(2+x).2-2r.1] 1+22
dx 1+x 27 1-z?
sin t = COs t
4
(2+x)'-4(1+x) l+,2 1+z'
1+x (2+x)* (1 +x)(2 +x) dz 2
.. (1) .1-z² 1+2
4+ 4x +x'-4 4x _ 1+
=

(1+x)(2+x)? (1+x)(2+r) 1+2 1+²

positive, also 1+x > 0 dz
Nowx, (2 + ' are always
for x > -1. l+z+2+ls2dt
Hence, function is
.:. from (i), d >0forx>-1.
increasing for x> -1
 480 ogether wit Mathematics-12 OR

’ log |1 +z<=x+C
log|1+ tan =x+C, where Cis constant of 7

6
integration. B(4, 5)
5
log| 1+ tan
x+ y| +Cis requiredsolution. 4
2
C(6, 3)
area of the region in the
31. Using integration find the 3
the line y=x
first quadrant enclosed by the r-axis, 2
and the circle +y'= 32.
1
M
OR A(2, 0)
X'+ 4 5 6 7
area of the 3
Using the method of integration find the
2
are A(2, 0),
AABC, coordinates of whose vertices
and C(6, 3).
Given points are A(2, 0), B(4, 5)
B(4, 5) and C(6, 3).
triangle ABC.
Sol. Plotting the points, we get
x²+
y²=
32
4

ar(4CM)
A(4,4) ar(AABC) =ar(ABL) + ar(LBCM) -
Equation of AB:A(2, 0), B(4, 5)
X
4 J442 y-0=
S--)
4-2

y=-2) ..)

Eliminating y fromn+y= 32 and x=y, we get Equation of BC: B (4, 5), C (6, 3)
X= 4
3-5
y-5= (-4)
6-4
.. area = [32-x' dt y-5= -x +4
y= -x +9 ..()

- /32 4 Equation of CA:C (6, 3),A(2, 0)

=-2)
y-0

4
V32 - 16 + 16sin-1,
y= ..(ii)
Using(i), (iü) and (üi), we get
ar(a4BC) =.-2)d +| (-x+9)de
=8+ 16 x-2 x 4- 16 x
4 --2)dt
= 8n 4T = 4n sq units = 7 sq units

SECTION - D
(This section comprises of long answer-type questions (LA) of 5 marks each)
OR
2 3
32. IfA= |4 5, find A". How we can use A 4 4 4||1 -1 1
Determine the product -7 1 -2 -2
6 9 -20
2 3 3 2 1 3
10
to solve the system of equations: +-+ =2;
y and state how we can use it to solve the system of
4 6. 5 6 9 20 equations x -y +z =4;x-2y-2, = 9;
-+=5;+ -= -4?
2r +y + 3z = 1.
 2 3 10| Practice Papers 481
s Consider A= |4 -6 5 OR
6 9 - 20 -4 4
Consider -7 I 3
Wehave A ladj 4) 5 -3 -1|

|2 3 |1 -1
10| and B=1 -2 -2
|4|=4 -6
|2 1 3
6 9 -20
4 4 -1 1
= (120 -445) -3(- 80 30) +10(36 + 36) AB = -2 -2
5 -3 1 3
-150 + 330 + 720 = 1200 )
-4+4+8 4-8+4 -4-8+12|
Hence, A exists.
-7+1+6 7-2+3 -7-2+9|
Matrix formed by cofactors of each element in JA. 5-3-2 -5+6-1 5+6-3
75 110 72| 8 0 0|
150 -100 AB = 0 8 0 = 8/ ..)
75 30 -24| 0 0 8
75 110
Consider equations
72| X-y +z = 4
AdjA = 150 -100 0
X-2y - 2% = 9
75 30 -24|
2r +y + 3z = 1
75 150 75 Corresponding matrix equation is
110 -100 30
1 -1
|72 0 -24
|1 -2
75 150 75| 2 1
1
A= 110 -100 30 ..) We can note the matrix of coefficient
is A or B and
1200
72 0 -24| write accordingly
Consider equations BX = Cis matrix equation.
2 10 ...i)
Z24 = 2 Its solution is X=B'C
+ Now from (i), we can findB as
A6,5 =5-+
y
AB = 8I ’

6.9 20 B-! = 1 A
=4 8

Corresponding matrix equation is This value of B we can substitute in (ii) and after
writing matrix A, we can find X and hence x, y, z.

10 33. Show that lines r = (i+j-k) + a(3i -) and

1
|4 -6 r=(4i - ) + u(2i + 3k)) intersect. Also find their
|6 9 point of intersection.
OR
AX = B is matrix equation. Find the value of, so that the lines
Its solution is X = A'B ..ü) Sz- 10 and 7-7x_y-5_6-z
I-r_7y -14
32 5
We had already calculated A. So we can substitute 3 22 11

from (i) in (ii)and find X and hence x, y, z. are perpendicular to each other.
482 7ogether wt
Mathematics-12
Sol. Consider liner =(í+- ) + A(3i - ) Direction ratios of line are -105.
...i) 102, 77.
Here, a, =itj-k and b, =3i -j ...(i) Now consider, line 7-7x i)
32.
and the line r = (4Í -k) + u(2í + 3k) ...(iiüi) -7(x-1) y-5 -(z-6)
32
Here, a, =4i -k and b, = 2i + 3k ...(iv) -5 z-6
From (ü) and (iv) 32 -7 35
Direction ratios of line are 32, -7, 35
a, - 4, = 4i - k-i-j+k = 3i -j
If lines are perpendicular, then
|i k
(- 105) (32.) + (102) (-7) + 77 x 35 =0
b,x b, =|3 -1
-315. 70. + 2695 = 0
|2 3
385). = 2695 ’ =7.
=i-3)- j9)+k(2) 34. Solve the following linear
=-3i -9, +2k graphically. programming problem
Minimise Z= 3x+y+39500
(a, -a) -(6,xb) = (3i- '- (3i -g +2) subject to the constraints,
=-9 +9+0 = 0
x>0,y >0, x +ys7000, xs4500, y< 3000.
’ lines are coplanar and b, x b, 0. x+y 3500
So, lines are intersecting. Sol. Plotting the graph of above inequations, we notice
Position vector of general point on line (i) is shaded portion is optimum solution. Possible
points for minimum Z are A(3500, 0), B(4500, 0).
= (1+34)i +(1- 2)j-k C(4500, 2500), D(4000, 3000), E(500, 3000).
and position vector of general point on line (i) is
7000
r = (4+2u)i +(-1 +3u)k ...(vi)
6000

x+y=7O0
If (v) and (vi) represent same point for some , u,
X= 4500
then r 5000

1+ 32 =4 + 2u ...(vii) 4000
E(500, 3000) D(4000, 3000)
y= 3000
1-2=0 ’ =1 ...(vi) 3000
C(4500,2500)
-1=-1+ 3u ’ u=0 ...(ix) 2000 +
Substituting in (viül) for . and 4, we get
1 +3= 4+ 0, true
1000 x+y= 350 B(4500, 0)
Hence, for 2 = 1and u = 0, the lines intersect. 0 +000L t000z f0008 too09 toooz
Substituting for 2 in (v) or u in (vi), we get the position
vector of point of intersection as r = 4i - k. Points
Z= 3r + y + 39500 Values

Point of intersection is (4, 0, -1). 10500 + 0 + 39500 50,000

A(3500, 0)
OR + 39500 53000
B(4500, 0) 13500 + 0
7y- 14 5z -10 S5,500
Consider, line = 13500 + 2500 + 39500
22 11 C(4500, 2500)
54,500
-(x -1) 7(y- 2) 5(z - 2) D(4000, 3000) 12000 + 3000 +39500
44,000
3 22. 11 39500
E(500, 3000) 1500 + 3000 +
x-1 y- 2 z-2
3000), i.e.r = 500,y = 3000
Z is minimum for E(500.
=

- 105 102. 77
 35. Find the Practice Papers 483

equation tanparticular
x. dy = 2xsolution of the
differential
dx tan x + x-y: (tan x # 0) ’ sin x.y =sin x(2x tan x + x).cot xdx
given that y = 0 when x =
2 = | (2r sin x+r'cosx) d
Sol. Consider equation tan x. dxdy = 2x tan - (2r sinx dk + cosrd
dy x+x-y
= |2rsin xdu +x² sin x-(2r-sin xd
tanx.
dx +y = 2r tan +
dy sin x,y =r sinx +C ...(i)
dx + cot * y = (2r tan x +x cot x
Here P() = cot x, Q(r) Given y= 0, whenx= 2
el -sin+C
=(2x tan x + )
cot x
Integrating factor (L.E) = efcot xde ’C=
4
= e8sin x| = sin x Substituting in (i), we get
Solution is (I.E)y = (LE)Q0)d Sin xy=rsin y- is required solution.
SECTION E
(1his section comprises of 3case-study/passage-based auestions of 4 marks each with sub-parts. First two
case
Study questions have three sub-parts (i), (iü), (üi) of marks 1., 1. 2 respectively. The third case study questou
has two sub-parts of 2 marks each.)
36. Read the following passage and answer the (iü) Find the cost C for digging the tank in terms
given below.
questions
ofx and h.
(iüi) Find the cost C in terms of h only.
OR

(iüi) Find the value ofh for which cost Cis minimum.
Sol. Volume = xx.h= 250
xh =250
(ü) Cost (C) =50x+ 400 (h)?
= 50x + 400 h²
C= 50 250 + 400 h
(ii) [from ()]
12500
+ 400 h2
Material to prepare fertilizer h

OR
(iii) h = 2.5 m
37. Read the following passage and answer the
given below. questions
square base
A village panchayat wants to dug out a
tank. For preparing fertilizersand wantscapacity to
calculations it was found that
be 250 cu metres. On
meter and cost of
cost of the land is 50 per square
cost of the whole
digging increases with depth and
shown as
tank is 400 (depth). Tank is
base is x m and the
(i) If the side of the square Aperson has to reach a company for an interview. He
then establish
height of the tank is h m, has four options to reach the company i.e. by metro,
relation betweenx and h.
 484 Together witk Mathematics-12
by bus, by scooter or by other means of transport. 9+8+1 18
3 1 1 120 120 20
The probabilities of using these means are
10 5' 10 OR
2
and respectively. The probabilities that he will be
11 (ii) P(A/E) = 0.5
late he comes by metro. bus or scooter are 4'3
38. Read the following passage and answer tthe
and
1
12
respectively but if he comes by other means given below. questions
he will not be late. Using the above information A
answer the following questions.
() What is the conditional probability of
reaching late by other means of transport?
(ü) What is the probability that he travelled by
bus and was late? B

(üiü)) What is the probability of reaching late? A building is to be constructed in the form of a
OR triangular pyramid, ABCD as shown in the figure.
Let its angular points are A(0, 1, 2), B(3, 0, 1),
(iiü) What is the probability that he reached late to
C(4, 3, 6) and D(2, 3, 2) and G be the point of
the company and he used metro as means of
intersection of the medians of ABCD.
transport?
(i) Find the coordinates of point G.
Sol. Let A: comes by metro, B: by bus, C: by scooter,
D:by other means (iü) Find the length of vector AG.
E:he reaches late Sol. (i) Clearly, G be the centroid of ABCD, therefore
coordinates of G are
() P(EID) = 0.
(ü) Probability P(BlE) = P(B) P(E/B) 3+4+2 0+ 3+3 1+6+2 =(3,2,3)
3 3
1
(i) Since, A= (0, 1, 2) and G= (3,2, 3)
3 15
(iü) P(E) = P(A) P(ELA) + P (B) P(EB) :: AG = (3-0)i +(2-1)j +(3-2)k=3i +j +k
+ P (C) P(EJC) + P (D) P(E/D) |A| = 3+1+12
=
3 1 11,1+.0 = /9+1+1
104 53 1012 5 =/11