anmol notes

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 )