lkjf.kd
y?kq mÙkjh;
1 ;fn
2x 5
8 x
6 5
, rks x Kkr dhft,A
8 3
1 x
x2
2 ;fn Δ = 1 y
2
1 z
1
y , Δ1 = yz
x
z2
1
zx
y
1
xy , rks fl¼ dhft, fd Δ + Δ1 = 0
z
3 fcuk izlj.k fd,] fn[kkb, fd
cosec 2
2
cosec
42
40
cot
4 n'kkZb, fd
x
p
q
cot 2
1
2
1
2
=0
p q
x q = (x – p) (x2 + px – 2q2)
q x
nh?kZ mÙkjh;
5 ;fn
x 2 3
1 x 1 = 0, dk ,d ewy x = – 4 gks rks vU; nks ewyksa dks Kkr dhft,A
3 2 x
6 ,d f=kHkqt ABC esa ;fn
1
1 sin A
1
1 sin B
1
1 sin C
0
sinA +sin 2 A sinB+sin 2 B sinC+sin 2 C
rks fl¼ dhft, fd ΔABC ,d lef}ckgq f=kHkqt gSA
gy
2x 5
8 x
1 gesa fn;k gS
6 5
blfy,
8 3
⇒ x2 = 9 ⇒
2x2 – 40 = 18 – 40
2
gesa fn;k gS
1
1
yz
x
x =±3
1 1
zx xy
y z
iafDr;ksa vkSj LraHkksa dk ijLij ifjorZu djus ij gesa izkIr gksrk gS
1
1 yz
1 zx
1 xy
x
x
y
z
=
1
y
xyz
z
xyz
x2
xyz
y2
xyz
z2
x 1 x2
xyz
y 1 y2
= xyz
,
z 1 z2
=
⇒
C1 vkSj C2 dk ijLij ifjorZu djus ij
1 x
x2
(–1) 1 y
y2
1 z
2
z
–
Δ1 + Δ = 0
4
3 C1 → C1 – C2 – C3 dk iz;ksx djus ij ge ikrs gSa fd
cosec 2 – cot 2 – 1
cot 2 – cosec 2
0
cot 2
1 cosec 2
40
0
1
1
2
=
cot 2 θ
1
0 cosec 2 θ −1 = 0
0
40
2
C1 → C1 – C2 dk iz;ksx djus ij ge ikrs gSa fd
x p
p x
0
p q
x q
q x
1 p q
( x p) 1 x q
0 q x
0 p + x 2q
= ( x − p) −1
x
q , R1 → R1 + R2 dk iz;ksx djus ij
q
x
0
C1 osQ vuqfn'k izlj.k djus ij ge ikrs gSa
( x p) ( px x 2 2q 2 ) = ( x p) ( x 2
px 2q 2 )
R1 → (R1 + R2 + R3) dk iz;ksx djus ij ge ikrs gSa fd
5
C2 → C2 – C1, C3 → C3 – C1, osQ iz;ksx ls ge ikrs gSa
x 4 x 4 x 4
x
1
1 .
3
2
x
1
0
0
( x 4) 1 x 1
0 .
3
1 x 3
R1 ls mHk;fu"B (x + 4) ysus ij ge ikrs gSa
R1 osQ vuqfn'k izlj.k djus ij ge ikrs gSa
Δ = (x + 4) [(x – 1) (x – 3) – 0]. ijarq Δ = 0 fn;k gS blfy,
1 1 1
( x 4) 1 x 1
3 2 x
ekuk Δ =
6
1
1 sin A
1
1 sin C
1
1 sin B
1
1
1
1 sin A 1 sin B 1 sin C
cos 2 A
=
vr% x = – 4 osQ vfrfjDr vU; nks ewy 1 rFkk 3 gSaA
sinA +sin 2 A sinB+sin 2 B sinC+sin 2 C
=
1
1 sin A
x = – 4, 1, 3
cos 2 B
cos 2 C
0
0
sin B sin A
sin C sin B
cos 2 A cos 2 A cos 2 B
cos 2 B cos 2 C
, R3 → R3 – R2
, (C3 → C3 – C2 vkSj C2 → C2 – C1)
R1 osQ vuqfn'k izlj.k djus ij ge ikrs gSa
Δ = (sinB – sinA) (sin2C – sin2B) – (sinC – sin B) (sin2B – sin2A)
= (sinB – sinA) (sinC – sinB) (sinC – sin A) = 0
⇒
sinB – sinA = 0 ;k sinC – sinB ;k sinC – sinA = 0
⇒
A = B ;k B = C ;k C = A
vFkkZr~ f=kHkqt ABC lef}ckgq f=kHkqt gSA