II VHSE : ANSWER KEY SECOND TERMINAL EXAMINATION

II VHSE : ANSWER KEY
SECOND TERMINAL EXAMINATION- 2014
Qn.No.
1 (i)
(ii)
2
(i)
(ii)
3 (i)
(ii)
(iii)
4
(i)
Scoring Indicators
(hog)(y)=h(3y+4)=Sin(3y+4)
(gof)(x)=g(2x )= 6x+4
(hog)of = (hog)(2x)=Sin(6x+4)
ho(gof) = h(6x+4) = Sin(6x+4)
(hog)of = ho(gof)
π/3
(iii)
5(i)
(ii)
tan-1x + tan-1y = tan-1(
-1 1
6 (i)
(ii)
-1
tan ( )+ tan (
2
5 0
� Y=
1 4
25 0
X2 =�
�
9 16
X =�
2
11
x+y
1−𝑥𝑥𝑥𝑥
)
1
A =½ � 1
−1
1
2
1
1
0
�
1
4
Y2 = �
3
�
0
�
15
4
0
�
1
−1
1
1
1
−1�
1
X=A-1B ; x=1, y= -1 z=2
Graph
Not continuous at x=1
lim𝑥𝑥→1−𝑓𝑓(𝑥𝑥) ≠ lim𝑥𝑥→1+𝑓𝑓(𝑥𝑥)
dx
4
)/ = tan ( )
1 0 𝑥𝑥
0
𝑦𝑦
=
�
�
�
�
1 1
1�
𝑧𝑧
0 1
3
1
�0
1
dx
1
1
1
1
1
-1 3
21
Yes ; (X+Y).(X-Y) =�
6
21 0
2
2
�
X -Y = �
6 15
dy
Total score
1
-1
(ii)
Split
score
=
𝑦𝑦
𝑥𝑥
.
(y−xlogy )
(x−y logx )
=a𝜃𝜃 cos𝜃𝜃 ;
d𝜃𝜃
dy
dx
=tan𝜃𝜃
dy
d𝜃𝜃
2
1
1
5
1
2
2
5
2
1+2
3
1
=a𝜃𝜃 sin𝜃𝜃
3
2
5
(iii)
7(i)
(ii)
1
Let x = tan𝜃𝜃
dy
dx
=
2
2
9
1+x2
Formula
C=3/2
dy
dx
1
1
=A Cosx –B Sinx
1
proof
8(i)
(ii) (a)
(b)
(i)
(ii)
1
increasing in (-∞,-3/2) &decreasing in (-3/2,∞)
1+1
slope= -10
1
10 x+y-5=0
OR
1/3
26 =2.9629
V(x) =x (3-2x)(8-2x)
2
Largest volume =200/27 m
9(i)
(ii)
(iii)
(iv)
10 (i)
(ii)
(iii)
11 (i)
(ii)
3
1
2
- logl cosx-sinx l +C
2
log l x+1 +√𝑥𝑥 2 + 2𝑥𝑥 + 2 l +c
x2
x 1
tan-1x. - + tan-1x +C
3
formula
1
2
2
2
1
3
3
2
∫0 x dx = 1/3
2
π/4
3
3
8
y 1 dx + ∫4 y 2 dx
(8+3 π )
9
1
(0,0) &(4,4)
4
12
3
29
∫0
5
3
1
2
4
12 (i)
5
2
(log x) 1-m +C
1-m
x
4
Order = 2; degree =1
1
2
1
4
(ii)(a)
(b)
13 (i)
(ii)
(iii)
14 (i)
(ii)
(iii)
1+x2
2
-1
y(1+x2) = tan x +C
1
when x= 1 y=0, soln.is y(1+x2) = tan-1x- π/4
1
�⃗=
Unit vector in the direction of𝑎𝑎⃗ + 𝑏𝑏
1
𝜃𝜃 = Cos-1 ( )
3
(2𝑖𝑖⃗ +2𝑗𝑗 )
√8
�⃗ = 1
Projection of 𝑎𝑎⃗ on 𝑏𝑏
3
√3
ℷ = −15
��⃗
𝑎𝑎 X 𝑏𝑏�⃗ + ��⃗
𝑎𝑎 X 𝑐𝑐⃗ +��⃗
𝑏𝑏 X 𝑐𝑐⃗ + ��⃗
𝑏𝑏 X 𝑎𝑎⃗ +��⃗𝑐𝑐 X
𝑎𝑎⃗ + ��⃗𝑐𝑐 X 𝑏𝑏�⃗
=0
5
1
2
2
5
1
2
1
1
5

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