# anmol notes

(SESSION : 2014)

SUBJECT : MATHS( xf.kr )

SOLUTION ( gy )

19.

LHS

bc c a ab

qr

r p

pq

yz zx xy

C1 C1 + C2 + C3

2(a b c ) c a a b

=

2(p q r )

r p

pq

2( x y z) z x x y

abc c a ab

=2

pqr

r p

pq

xyz zx xy

C2 C2 – C1

C3 C3 – C1

abc

=2

b c

pqr

q r

xyz 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

x2

y 1

z3

=

=

a

b

c

It is perpendicular to the lines

a + 2b + 3c = 0

– 3a + 2b + 5c = 0

a

b

c

=

=

10 6

59

26

a

b

c

=

=

4

14

8

a

b

c

=

=

2

7

4

Equation of straight line is

x2

y 1

z3

=

=

2

7

4

cartesian equation

Vector equation

ˆ

ˆ

j

j

r = (2 ˆ + ˆ + 3 k...

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