QUESTION PAPER CODE 65/3/B
EXPECTED ANSWERS/VALUE POINTS
SECTION - A
Marks
1.
dy
y
2


dx
x log x
x
(Standard form)
½m
I.F. = log x
2.
3.
4.
½m
x=2, y=9
(½ for correct x or y)
1
 x + y = 11
½m
order 3 , or degree 1
½m
 Degree + order = 4
½m
using sin θ 
 
ab


a  b
½m
 θ  0o
5.
6.
½m
 
ab0  x  –6
½m
y   40 or  2 10
½m
a 2 sin 2 α  a 2 sin 2 β  a 2 sin 2 γ
½m
= 2 a2
½m
SECTION - B
7.



AB  – 2î – 5k̂
AC  î – 2ĵ – k̂
1m
AB  AC  – 10î – 7ˆj  4k̂
AB  AC 
1m
165
½m
26
AB  AC
n̂ 
1m
AB  AC
=
8.
– 10î – 7ĵ  4k̂  or 10î  7ĵ – 4k̂
165
½m
165


1
1
a 1  – î , b1  î  ĵ – k̂
2
12


1
a 2  – 2ĵ  k̂ , b 2  î  ĵ  k̂
6





1m


a 2 – a 1  î – 2ˆj  k̂
½m
 
1
1
1
b1  b 2  î – ĵ  k̂
6
4
2
½m
 
7
b1  b 2 
12
1m

S.D. =

a 2 – a1   b1  b 2 
 
b1  b 2
 2
1m
OR
Foot of perpendicular are (0, b, c) & (a, 0, c)
Equ. of required plane
1m
x y z
0 b c  0
a 0 c

9.
2m
bcx + acy – abz = 0
1m
p (x = 2) = 9 . P(x = 3)
1m
 3C2 p2 q  9  3C3 p3  q0
1m
 3 p 2 (1 – p)  9 p3
1m
27

p 
1
4
1m
OR
Let H1 be the event that red ball is drawn
H2 be the event that black ball is drawn
E be the event that both balls are red
P(H1) =
3
5
, P(H2) =
8
8
P(E/H1) =
5C 2
1m
3
2
 , P(E/H ) = C 2  1
2
9
10 C2 15
10 C2
P(E) = P(H1) P(E/H1) + P(H2) . P(E/H2)
=
10.
3 2
5 1
1
  

8 9
8 15
8


x
1m
1m
y
 log x – log (a  b x) 
x
x
1m
½m
dy
–y
1
b
dx

–
2
x
x
abx
1m
dy
ax
– y 
dx
a  b x ................................... (i)
1m
Differentiating again,
x
d2 y
a2

dx 2
a  b x 2
2
1m
2
 ax 
d2 y
 dy

   x
x  2  
– y  (using (i))
dx
 dx

a  b x
3
28
½m
11.
 1 
du
  2 cos –1x 
u  sec –1  2
dx
 2x – 1 
1– x2 
v 
dv
dx

1
2
x
π
12.
Let I 
2

0
1½ m
1 – x2
–x
1m
1 – x2
2
 4
x
1½ m
5 sin x  3 cos x
dx .................. (i)
sin x  cos x
π

dv

dx
–2
I 
2

0
5 cos x  3 sin x
dx ........ (ii)
cos x  sin x
Adding (i) and (ii)
π
2I = 8




a

0
a
f(x) dx 

 f(a – x) dx 
1½ m
0
1+1 m
2
 1  dx
 4π
½m
0

I  2π
OR
put log x = t  x = et  dx = et dt

e
t

e
t
1m
1

 log t  2  dt
t 


1 1
1 
 log t – t    t  t 2  dt
 


1½ m
1

 e t  log t –   c
t

1m

1 
 x log log x  –
c
log x 

½m
29
13.
I 
x cos x
1m
 cos x  x sin x dx
put cos x + x sin x = t

=

x cos x dx = dt
1m
dt
t
1m
= log cos x  x sin x + c
14.
x 4 dx
 x – 1 x 2  1


1m

1
 x  1  x – 1 x

2

dx
 1 
1m

(using partial fractions)
1
15.
dx
=
x2
1
1
1
 x  log x – 1 – log x 2  1 – tan –1 x  c
2
2
4
2
 2
 100
[15000 15000] 
 x
 100



 = [1800]



300 + 150x = 1800

x = 10%
cot –1 (x  1)  sin –1
and tan –1 x  cos –1
x 1
dx
2
1
 x  1 dx  2  x – 1 – 2  x

1½ m
1½ m
2m
1m
yes : compassionate or any other relevant value
16.
1
=
1
1m
1½ m
1  (x  1) 2
1
1½ m
1  x2
30

1
 sin  sin –1

1  (x  1) 2



1
  cos  cos –1


1  x2


 1  x 2  2x  1  1  x 2  x  –
1
2




1m
OR
17.
2 sin –1
3
17
– tan –1
5
31
=
2 tan –1
3
17
– tan –1
4
31
1m
=
tan –1
24
17
– tan –1
7
31
1m
=
tan –1 1 
π
4
1+1 m
C1  C1  C 2  C 3 ,
1 b c
(a + b + c)
1 c a  0
1 a b
1m
R 2  R 2 – R1 , R 3  R 3 – R1

1
b
c
0 c–b a –c  0
( a  b  c  0)
2m
0 a–b b–c

– a2 – b2 – c2 + ab + bc + ca = 0

–

a=b=c
½m
1
[(a – b) 2 + (b – c) 2 + (c – a) 2] = 0
2
31
½m
18.
1 –1 0
1 0 0




 2 5 3  0 1 0  A
 0 2 1
0 0 1




1m
R 2  R 2 – 2 R1 ,
1 –1 0
 1 0 0




 0 7 3   – 2 1 0   A
0 2 1
 0 0 1




R2  R2 – 3 R3
1 –1 0
 1 0 0 




 0 1 0    – 2 1 – 3  A
0 2 1
 0 0 1 




R1  R 1  R 2 , R 3  R 3 – 2 R 2
1 0 0
 –1 1 – 3




 0 1 0    – 2 1 – 3  A
0 0 1
 4 – 2  7




 A
19.
–1
(2 marks for all operations)
 – 1 1 – 3


  – 2 1 – 3
 4 –2 7 


f (x) = x – x – x
2
1m
2x – x 2

 x – x (1 – x)   0
 x2

, –1  x  0
,
x 0
,
1m
0  x 1
f (x) being a polynomial is continuous on – 1, 0  0, 1
lt – f(x)  lt – (2x – x 2 )  0
½m
lt  f(x)  lt  x 2  0
½m
x 0
x 0
x 0
x0
Also, f(0) = 0
 lim– f(x)  f(0)  lim f(x)
x0
1m
 There is no point of discontinuity on [– 1, 1]
1m
x 0
32
SECTION - C
20.
1m
Required Area
2
=

2
2
4 – x dx –
0
 2 – x  dx
 x 4 – x2 4
 sin –1
= 
2
2

=
21.
π – 2
2m
0
2
2

x
x2 
 – 2 x – 
2
2 0
0 
sq. units
1+1 m
1m
dy
xy
 2
dx
x  y2
put y = v x 

dy
dv
 vx
dx
dx
1  v2
dx
 –
3
v
x
1m
1m
Integrating both sides
–
1
 log v  – log x  c
2 v2
1m
33
 –
x2
 log y  c
2y 2
when x = 1, y = 1  c = –
 log y 
1
2
1m
x 2 – y2
2y 2
when x = x0, y = e

x0 
½m
1½ m
3e
OR
I F = e  tan x dx  e log sec x  sec x

1m
d
(y  sec x)  3x 2 sec x  x 3 sec x tan x
dx
y sec x 

y = x3 + c cos x
when x 
 3x
2

sec x  dx  x 3 sec x –
 3x
1m
2
 sec x dx  c
π
– 2π 3
, y  0 ; we get c 
3
27
1m
2π 3
 y  x –
cos x
27
3
22.
2m
1m
Let us consider the man invested on x
electronic and y manually operated machines
Maximise P = 220 x + 180 y ............................... (i)
1m
subject to
x + y < 20
3600 x + 2400 y < 57600 
3x + 2y < 48
1½ m
x, y > 0
(1 mark for
34
plotting each
line) = 2 m
(½ to find the vertices
of feasible
region)
P
P
P
23.
 3520 Rs.
A (16, 0)
B (8, 12 )
 3920 Rs.
 3600 Rs.
C (0, 20 )
Maximum profit is Rs. 3920 at x = 8, y = 12
1m
x–3
y–4
z –1


2
–3
5
1m
Equation of line is
Equation of plane is
x – 2 y –1 z – 2
1
2

0
–3
–2
–1
 0
1m
2x + y + z – 7 = 0 ..................................... (i)
1m
general point on given line (2 λ  3, – 3 λ  4, 5 λ  1) lies on (i)

2 (2 λ  3)  (–3 λ  4)  (5 λ  1) – 7  0  λ  –
7
5
 Point of intersection  , 6, – 
3
3
24.
Let
1m
2
3
1m
1m
H1 : be the event 1, 2 appears
H2 : be the event 3, 4, 5, 6 appears
1m
E3 : be the event that head appears
35
P(H1) =
2 1
4 2
 , P(H2) = 
6 3
6 3
P(E/H1) =
P(H2/E) =
=
1m
3
1
P(E/H2) =
8
2
1m
P (H 2 )  P (E/H 2 )
P (H1 )  P (E/H1 )  P (H 2 ) P (E/H 2 )
1m
8
11
2m
OR
Let
H1 : be the event that 4 occurs
H2 : be the event that 4 does not occurs
1m
E : be the event that man reports 4 occurs
on a throw of dice
P(H1) =
1
,
6
P(E/H1) =
P(H1/E) =
=
25.
P(H2) =
3
5
5
6
1m
P(E/H2) = 1 –
3 2

5 5
1m
P (H1 )  P (E/H1 )
P (H1 )  P (E/H1 )  P (H 2 )  P (E/H 2 )
1m
3
13
2m
*
0
1
2
3
4
5
6
0
0
1
2
3
4
5
6
1
1
2
3
4
5
6
0
2
2
3
4
5
6
0
1
3
3
4
5
6
0
1
2
4
4
5
6
0
1
2
3
5
5
6
0
1
2
3
4
6
6
0
1
2
3
4
5
36
4m
 a  {0, 1, 2, 3, 4, 5, 6}
a * 0 = a = 0 * a  0 is identity
1m
 a  {1, 2, 3, 4, 5, 6}
a* b = 0=b*a
26.

a * (7 – a) = 0 = (7 – a) * a

(7 – a) is inverse of a
1m
A = y (x + 4)
x 2 y2

1
16 9
9
9
(16 – x 2 ) (x  4) 2  y 2 
(16 – x 2 ) .......... (i)
16
16
1m
9
(4 – x) (4  x) 3
16
1m
dz
9

(4  x) 2 (8 – 4x)
dx
16
1m
dz
 0  x  2
dx
1m
2
Let z = A 

d2z
9
9
 – (4  x) 2  (4  x) (8 – 4x)
2
dx
4
8
d2z
dx 2
 0
1m
x2
 Maximum value of A = 9 3 sq. units
37
1m