# anmol notes

Topics: Trigraph, Gh, Analytic geometry Pages: 3 (491 words) Published: August 25, 2014
12 th CBSE SAT (2)
(SESSION : 2014)
SUBJECT : MATHS( xf.kr )
SOLUTION ( gy )
19.

LHS
bc c a ab
qr

r p

pq

yz zx xy

C1  C1 + C2 + C3
2(a  b  c ) c  a a  b

=

2(p  q  r )

r p

pq

2( x  y  z) z  x x  y
abc c a ab

=2

pqr

r p

pq

xyz zx xy

C2  C2 – C1
C3  C3 – C1
abc

=2

b c

pqr

q r

xyz y z

C1  C1 + C2 + C3
a b c

=2

p q r
x y z
a b c

=2

p q r
x y z

Hence Proved
20.

x = a sin 2t (1 + cos2t)
dx
= 2a cos 2t (1 + cos 2t) – 2a sin22t
dt
= 2a cos 2t + 2a cos2 2t – 2a sin2 t
= 2a cos 2t + 2a cos 4t
y = b cos 2t (1 – cos 2t)
dy
= – 2b sin2t (1 – cos 2t) + b cos 2t (2 sin 2t)
dt

= – 2b sin 2t + 2b sin 2t cos 2t + 2b sin 2t cos 2t
= – 2b sin 2t + 2b sin 4t

dy
dy / dt
=
dx
dx / dt

12th CBSE SOLUTION_MATHS SAT (2)_PAGE # 1

=

2b sin 4t  2b sin 2t
2a cos 2t  2a cos 4t

=

b (sin 4t  2b sin 2t )
a cos 2t  2a cos 4t

(sin   sin )

 dy 
2
b
 
 dx t   = a cos   cos 
4

2

b (1)
b
=
=
a (1)
a

21.

hence proved

x (1 + y2) dx – y (1 + x2) dy = 0
y (1 + x2) dy = x (1 + y2) dx
ydy
1 y

2

x

=

1 x2

dx

Integrating both sides
dy
x
y
=
dx
I  y2
I  x2

1
2

2y

 1 y

2

dy =

1
2

2x

 1 x

2

dx

1
1
log |1 + y2| =
log |1 + x2| + logC
2
2

log |1 + y2| = log |1 + x2| + 2 logC
log |1 + y2| = log |C2 (1 + x2)|
when y = 1, x = 0
2 = C2
 1 + y2 = 2 (1 + x2)
y2 = 2 (1 + x2)
y2 = 2x2 + 1
y=
22.

2x2  1

Equation of staight line passing through (2, 1, 3) is
x2
y 1
z3
=
=
a
b
c

It is perpendicular to the lines
 a + 2b + 3c = 0
– 3a + 2b + 5c = 0
a
b
c
=
=
10  6
59
26
a
b
c
=
=
4
 14
8
a
b
c
=
=
2
7
4

 Equation of straight line is
x2
y 1
z3
=
=
2
7
4

cartesian equation
Vector equation

ˆ
ˆ
j
j
r = (2 ˆ + ˆ + 3 k...