2016 (A) πÂä  - Karpue.kar.nic.in/PUE/PDF_files/exam/scv/2016M/35_N.pdf · [Turn over...
Transcript of 2016 (A) πÂä  - Karpue.kar.nic.in/PUE/PDF_files/exam/scv/2016M/35_N.pdf · [Turn over...
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[ Turn over
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.
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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
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3 Code No. 35 (N/S)
[ Turn over
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
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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
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5 Code No. 35 (N/S)
[ Turn over
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
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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
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7 Code No. 35 (N/S)
[ Turn over
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
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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
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9 Code No. 35 (N/S)
[ Turn over
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
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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
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11 Code No. 35 (N/S)
[ Turn over
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