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MATHEMATICS 2016 (A)
( NEW SYLLABUS )
πÂä∆Â
SCHEME OF VALUATION Subject Code : 35 (N/S)
Qn.
No.
Marks
Instructions :
1) For any alternate method, it should be valued and suitably
awarded.
2) All answers ( including extra, struck off and repeated ) should
be valued. Answer with maximum marks awarded must be
considered.
3) If the student had written wrong question number, write the
correct question number and be valued.
Code No. 35 (N/S) 2
Qn.
No.
Marks
PART - A
I. 1. Getting Answer =
!
"cot x " cosec!x + c 1
2. x = 6 or x = – 6 1
3. Writing
!
dy
dx= "
sin x
2 cos x 1
4. Writing
!
cos"
2= 0
1
5. Getting
!
OA = OB " AB = i#
" 3 j#
+ 3k#
1
6. Getting distance =
!
14
7
1
7.
!
x = 2 and y = 3 1
8. Getting
!
P(A" B) = 0 # 32 1
9. Not a binary operation 1
10. Definition 1
PART - B
II.
11. Getting
!
GE = tan"1
3
4" tan x
1 +3
4tan x
#
$
% % %
&
'
( ( (
=
!
tan"1 3
4" tan
"1tan x( ) = tan
"1 3
4" x
1
1
12. Writing
!
" = a b + c 1
b c + a 1
c a + b 1
Getting
!
" = 0
1
1
13. Getting
!
fog "1
2
#
$ %
&
' ( = 1
and
!
gof "1
2
#
$ %
&
' ( = 0
1
1
14. Putting x = cos θ and θ = cos
!
"1x
Proving the answer
1
1
15. Putting
!
x = cos " and getting y = 2
!
"
Getting
!
dy
dx=
"2
1" x2
1
1
3 Code No. 35 (N/S)
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Qn.
No.
Marks
16. Writing
!
y log x = x log a
Getting
!
y
x+ log x.
dy
dx= log a
and
!
dy
dx=
x log a " y
x log x
1
1
17. Getting
!
I =1
tan x+ tan x
" # $
% & '
( sec2
x dx
And
!
I = log|tan x|+tan2
x
2+ c
1
1
18. Writing
!
x = 27, "x = 2, x + "x = 25
!
f (x) = x1/3, f (x + "x) = (x + "x)
1/3
Getting
!
f /(x) =
1
3x2/3 and Answer = 2·926
1
1
19. Writing
!
"cos x dx
0
#
$
Getting Answer = 0
1
1
20. Writing
!
|a + b|2 = |a " b|2
And
!
|a|2 +|b|2 + 2(a . b) = |a|2 +|b|2 "2(a . b)
Getting
!
a . b = 0 and writing
!
a " to
!
b
1
1
21. Writing order = 4
Degree not defined
1
1
22. Getting
!
a . b = 1, |a|= 3, |b| = 3
Getting
!
cos" =1
3 and writing
!
" = cos#1 1
3
1
1
23. Getting
!
k =1
6
Getting
!
P(X " 2) = 1
1
1
24. Writing direction ratios 0, 1, 0
Equation of the line
!
x "1
0=
y "1
1=
z "1
0
1
1
Code No. 35 (N/S) 4
Qn.
No.
Marks
PART- C
III.
25. Writing LHS =
!
tan"1
1
2+
2
11
1"2
22
#
$
% % %
&
'
( ( (
+ tan"1 4
3
=
!
tan"1 3
4+ tan
"1 4
3
=
!
tan"1
3
4+ cot
"13
4=#
2= RHS
1
1
1
26. Writing A = IA
Getting any one non-diagonal element as zero
Getting the inverse =
!
1
5
3 1
"2 1
#
$ %
&
' (
1
1
1
27. Reflexive :
!
"a # A, (a $ a) = 0 which is
multiple of 4 :
!
(a, a) " R
Symmetric : Let
!
(a, b) " R #|a $ b| is multiple of 4
!
"|b # a| is multiple of
!
4 " (b, a) # R
Transitive : Let
!
(a, b) and ( b, c )
!
" R
!
" |a # b| and |b # c| are multiple of 4
!
" |a # b + b # c| = |a # c| is multiple of 4
!
" (a .c) # R
1
1
1
28. Writing
!
f (x) is continuous in [ 1, 3 ] and differentiable in ( 1, 3 )
!
f /(x) = 3x2
"10x " 3 or
!
f (3) = "27, f (1) = "7
and getting c = 7/3
Writing
!
c =7
3" (1, 3)
Note : If
!
c =7
3" (1, 3) is not written deduct one mark.
1
1
1
29. Getting
!
dx
d"= #3acos
2"sin "
!
dy
d"= 3asin
2"cos "
getting
!
dy
dx= "
sin #
cos #=
y
x3
1
1
1
5 Code No. 35 (N/S)
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Qn.
No.
Marks
30. Writing
!
P(E1) = P(E2) =1
2& P(A/ E1) = 1, P(A/ E2) =
1
2
Writing
!
P(E1/ A) =P(E1)P(A/ E1)
P(E1)P(A/ E1) + P(E2)P(A/ E2)
=
!
2
3
1
1
1
31. Writing
!
x
(x "1) (x " 2)=
A
x "1+
B
x " 2 and
!
x = A(x " 2) + B(x "1)
Getting A = – 1 and B = 2
Getting the answer =
!
" log(x "1) + 2 log(x " 2) + c
1
1
1
32. Getting
!
I =1
(t +1) (t + 2) dt"
=
!
1
t +1"
1
t + 2
# $ %
& ' (
) dt
=
!
log|t +1|" log|t + 2| + c
=
!
log|x2
+1|" log|x2
+ 2| + c
1
1
1
33. Writing
!
xy = 100 and
!
S = x + y is minimum
getting
!
ds
dx= 1"
100
x2
and
!
ds
dx= 0 " x = ± 10
getting
!
d2s
dx2
=200
x3
> 0 at x = 10
!
" x = 10 and y = 10
1
1
1
34. Finding the values x = 0 and x = 1
Writing area =
!
4x
0
1
" dx # 2x dx
0
1
"
Getting the area
!
=4
3"1 =
1
3 sq. unit
1
1
1
Code No. 35 (N/S) 6
Qn.
No.
Marks
35. Writing
!
(a " b) . (b " c) # (c " a){ }
Getting =
!
(a " b) . (b # c) " (b # a) + (c # a){ }
= 0
1
1
1
36. Getting
!
a2 " a1 = 2 i#
+ j#
" k#
and
!
b = 2 i"
+ 3 j"
+ 6k"
Writing distance
!
d = b " (a2 # a1)
|b|
=
!
9 i"
+14 j"
+ 4k"
49
!
d =293
7
1
1
1
37. Getting
!
a " b =
i#
j#
k#
1 2 2
3 2 6
= 8 i#
$ 4k#
Getting
!
a " b = 4 5, |a|= 3, |b|= 7
Getting
!
sin " = |a # b|
|a| |b| =
4 5
21
1
1
1
38. Writing
!
dy
dx=
x
y or
!
ydy = xdx
Getting general solution
!
ydy = xdx""
=
!
y2
2=
x2
2+ c
Putting x = 1, y = 1
!
" c = 0 and
Writing equation
!
x2= y2
1
1
1
7 Code No. 35 (N/S)
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Qn.
No.
Marks
PART - D IV.
39. Getting
!
AB =
"2 "6 12
4 12 "24
5 15 "30
#
$
% % % %
&
'
( ( ( (
Writing
!
A/ = "2 4 5 [ ]
!
B/
=1
3
"6
#
$
% %
&
'
( (
Getting
!
B/A
/=
"2 4 5
"6 12 15
12 "24 "30
#
$
% %
&
'
( (
Proving
!
(AB)/
= B/A
/
1 1 1 1 1
40. Writing
!
A =
2 "3 5
3 2 "4
1 1 "2
#
$
% % %
&
'
( ( (
, X =
x
y
z
#
$
% % %
&
'
( ( (
& B =
11
"5
"3
#
$
% % %
&
'
( ( (
Or
!
| A| = "1
Getting adj
!
A =
0 2 1
"1 "9 "5
2 23 13
#
$
% % %
&
'
( ( (
T
=
0 "1 2
2 "9 23
1 "5 13
#
$
% % %
&
'
( ( (
Writing
!
X = A"1B =1
| A| (adj A) B
Getting x = 1, y = 2 and z = 3
1 2 1 1
41. Getting
!
y = 4x2+12x +15 " x =
y # 6 # 3
2
Stating
!
g(y) =y " 6 " 3
2, y # S
or
!
g(x) = x " 6 " 3
2, x # S
Proving
!
gof (x) = 9(4x2+12x +15) = x and
writing
!
gof = IN
Proving
!
fog(y) = fy " 6 " 3
2
#
$ % %
&
' ( ( = y, y ) S
Or
!
fog(x) = fx " 6 " 3
2
#
$ % %
&
' ( ( = x, x ) S and writing
!
fog = IS
Writing
!
f "1(x) = x " 6 " 3
2or f "1 =
x " 6 " 3
2
1 1 1 1 1
Code No. 35 (N/S) 8
Qn.
No.
Marks
42. Figure
Writing
!
dx
dt= "3 cm/min and
!
dy
dt= 2 cm/min
Writing
!
P = 2x + 2y and getting
!
dP
dt= "2 cm/min
Writing
!
A = xy and getting
!
dA
dt= x
dy
dt+ y
dx
dt
Getting
!
dA
dt= 2 cm
!
2/min
1
1
1
1
1
43. Getting
!
dy
dx=
2sin"1 x
1" x2
Writing
!
1" x2 dy
dx= 2sin
"1 x
Getting
!
1" x2 d2y
dx2"
x
1" x2
dy
dx=
2
1" x2
Getting
!
(1" x2)d
2y
dx2" x
dy
dx= 2
1
1
1 + 1
1
44. Putting
!
x = a tan" # dx = asec2" d"
Getting
!
I =1
a1d"#
=
!
1
atan
"1 x
a
#
$ %
&
' ( + c
Writing
!
I =1
(x +1)2
+ 2
dx"
=
!
1
2
tan"1 x +1
2
#
$ %
&
' ( + c
1
1
1
1
1
9 Code No. 35 (N/S)
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Qn.
No.
Marks
45. Figure :
Getting equations of the sides
!
AB : y = 2(x "1), BC : y = 4 " x and AC : y =1
2(x "1)
Writing area =
!
2(x "1)dx + (4 " x)dx "1
2(x "1)dx
1
3
#2
3
#1
2
#
=
!
2 x
2
2" x
#
$ % %
&
' ( ( 1
2
+ 4x "x
2
2
#
$ % %
&
' ( ( 2
3
"1
2
x2
2" x
#
$ % %
&
' ( ( 1
3
Getting area =
!
3
2sq units
1 1
1 1 1
46. Figure :
Getting
!
AP " N and
!
AP . N = 0
Getting
!
( r " a ) . N = 0
Writing
!
a = x1 i"
+ y1 j"
+ z1 k"
, r = x i"
+ y j"
+ zk"
!
N = A i"
+ B j"
+ Ck"
and writing
!
[ (x " x1) i#
+ (y " y1) j#
+ (z " z1)k#
] [ A i#
+ B j#
+ Ck#
] = 0 Getting
!
A(x " x1) + B(y " y1) + C(z " z1) = 0
1
1
1
1
1
Code No. 35 (N/S) 10
Qn.
No.
Marks
47.
Writing
!
q =1
5, p =
4
5, n = 5
!
P(x = r) = 5Cr
4
5
"
# $ $
%
& ' '
r1
5
"
# $ $
%
& ' '
5(r
writing
!
P(x " 4) = P(x = 4) + P(x = 5)
getting
!
P(x " 4) =9.4
4
55
getting
!
P(x " 3) = 1# P(x $ 4) = 1#9.4
4
55
1 1 1 1 1
48. Writing
!
dx
dy+
x
y= ey
Getting
!
IF = e
1
ydy"
= elog
ey= y
Writing general solution
!
xy = eyydy"
Getting
!
xy = yey " eydy + c1#
!
xy = yey" ey
+ c
1
1
1 1
1
PART-E V.
49. a) Drawing graph
Getting corner points
!
A(60, 0), B(120, 0), C(60, 30) and
!
D(40, 20) Getting corresponding value of z at each corner point At A, Z = 300 At B, Z = 600, At C, Z = 600 and At D, Z = 400 Writing maximum value of Z is 600 at ( 120, 0 ) and ( 60, 30 ) Minimum value of Z is 300 at ( 60, 0 )
2 1 1 1 1
11 Code No. 35 (N/S)
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Qn.
No.
Marks
b) Writing
!
x"0
lim f (x) = f (o) = k
getting
!
x"0
lim
2sin2
x
2sin2 x
2
#
$ %
& %
'
( %
) %
= k
=
!
x"0
lim
sin2
x
x2
sin2
x
2
x2/4
# 4
$
%
& &
'
& &
(
)
& &
*
& &
= k
getting k = 4
1 1 1 1
50. a) Writing
!
I = f (x)dx = f (x)dx + f (x)dx
a
2a
"0
a
"0
2a
"
!
I = I1 + I2
getting
!
I2 = f (x)dx = f (2a " x)dx
0
a
#a
2a
#
writing
!
I = f (x)dx + f (2a " x)dx
0
a
#0
a
#
=
!
2 f (x) dx, f (2a " x) = f (x)
0
a
#
= 0,
!
f (2a " x) = " f (x)
writing
!
I = 2 cos5 xdx, f (2" # x) = f (x)
0
"
$
= 0,
!
f (" # x) = # f (x)
1 1
1 1 1 1
b) Getting LHS =
!
0 0 1
a " b b " c c
a3" b
3b
3" c
3c3
Getting LHS =
!
(a " b) (b " c)
0 0 1
1 1 c
a2
+ ab + b2
b2
+ bc + c2
c3
Getting LHS =
!
(a " b) (b " c) b2 + bc + c
2" a
2" ab " b
2[ ] Getting LHS =
!
(a " b) (b " c) (c " a) (a + b + c)
1 1 1 1