Download - Mathematics, set 2, Model papers of Madhya Pradesh Board of Secondary Education, XIIth Class
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7/29/2019 Mathematics, set 2, Model papers of Madhya Pradesh Board of Secondary Education, XIIth Class
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'u&ik CYkw fUV
ijh{kk & gk;j lsds.Mjhd{kk&12 iw .kkd&100fo"k;& mPp xf.kr le;&3 ?k.Vk
- bdkbZ bdkbZ ij oLrq fu"B dq Ykfu/kkZfjr 'u vadokj 'u a dh la[;k 'u vad
1 vad 4 vad 5 vad 6 vad
1- vkaf'kd f 05 01 01 & & 01
2- izfrYke QYku 05 01 01 & & 01
3- lerYk T;kferh;
4- lerYk 15 04 & 01 01 025- ljYk js[kk ,oa xYkk
6- lfn'k
7- lfn'k a dk xq.kuQYk 15 04 & 01 01 02
8- lfn'k a dk fkfoeh;T;kferh; esa vuq;x
9- QYku] lhek rkk 05 & & 01 & 01
lkarR;10- vodYku 10 02 02 & & 02
11- dfBu vodYku
12- vodYku dk vuq;x 05 01 01 & & 01
13- lekdYku
14- dfBu lekdYku 15 05 & 02 & 02
15- fuf'pr lehdYku
16- vodYk lehdj.k 05 & & 01 & 01
17- lglaca/k 05 01 01 & & 01
18- lekJ;.k 05 01 01 & & 01
19- kf;drk 05 & & 01 & 01
20- vkafdd fof/k;k 05 05 & & & &
dqYk 100 25 07 07 02 16
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gk;j lsd.Mjh Ldwy ijh{kk&2012&2013HIGHER SECONDARY SCHOOL EXAMINATION
kn'kZ 'u&i=
Model Question Paper
mPp xf.krHIGHER MATHEMATICS
(Hindi and English Versions)
Time& 3 aVs Maximum Marks100funsZ'k&
(1) lh 'u gy djuk vfuok;Z gSA(2) 'u a ij vk/kkfjr vad muds lEeq[k n'kkZ, x, gSaA(3) 'u - 1 ls 5 rd oLrqfu"B 'u gSaA(4) 'u 6 ls 21 rd R;sd 'u esa vkarfjd fodYi fn, x, gSaAInstructions(1) All questions are compulsory to solve.(2) Marks have been indicated against each question(3) From Question No. 1 to five are objective type questions
(4) Internal options are given in question No. 6 to 21
[k.M&v Section-A oLrqfu"B 'u (Objective Type Questions)
'u&1 R;sd oLrqfu"B 'u esa fn, x, fodYia esa ls lgh mkj pqfu,&
(i)1
( 2)x x + ds vkaf'kd f gSa&
(a)1 1
1x x
+(b)
1 1
2 1x x+
+ +
(c)1 1 1
2 2x x
+ (d)
1 1
2 1x x
+ +
(ii) tan11
2
+ tan11
3 dk eku gS&
(a) tan11
6(b)
3
(d)4
(d)
6
(iii) fcUnq (4, 3, 5) dh Y v{k ls nwjh gS&
(a) 34 (b) 5
(c) 41 (d) 15
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(iv) ml lerYk dk lehdj.k t v{k a ls bdkbZ YkackbZ ds vUr% [k.M dkVrk gS]gS&(a) x + y + z = 0 (b) x + y + z = 1
(c) x + y + z = 3 (d) x + y + z = 1
(v) ;fn fdlh js[kk ds fnd~vuqikr 1, 3, 2 g r ml js[kk dh fnddT;k,
gSa&
(a)1 3 2
, ,14 14 14
(b)
1 2 3, ,
14 14 14
(c)1 3 2
, ,14 14 14
(d)
1 2 3, ,
14 14 14
Write the correct anwer from the given options provided in every objectivetype question
(i) Partial fractions of
1
( 2)x x + are
(a)1 1
1x x
+(b)
1 1
2 1x x+
+ +
(c)1 1 1
2 2x x
+ (d)
1 1
2 1x x
+ +
(ii) The value of tan11
2+ tan1
1
3 is
(a) tan1 16
(b)3
(d)4
(d)
6
(iii) Distance of the point (4, 3, 5) from Y axis is
(a) 34 (b) 5
(c) 41 (d) 15
(iv) Equation of a plane which cuts the unit intercepts with the coordinate axis
is(a) x + y + z = 0 (b) x + y + z = 1
(c) x + y + z = 3 (d) x + y + z = 1
(iv) If direction ratios of a line are 1, 3, 2 then direction consines of line
are
(a)1 3 2
, ,14 14 14
(b)
1 2 3, ,
14 14 14
(c)
1 3 2
, ,14 14 14
(d)
1 2 3
, ,14 14 14
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'u&2- fuEufYkf[kr dkua esa lR;/vlR; dku NkVdj viuh mkjiqfLrdk esafYkf[k,A
(i) fcUnqv a (1, 2, 3) vj (4, 5, 6) d feYkkus okYkh js[kk ds fnd~vuqikr 5,3, 9 gSaA
(ii) rhu vlekUrj] v'kwU; lfn'k lerYkh; gus ds fYk, mudk vfn'k fkd
xq.kuQYk 'kwU; grk gSA(iii) sin (cos1x) dk vodYku xq.kkad 'kwU; grk gSA(iv) sin x + cos x dk egke eku 2 gSA
(v) . .ii j j k k + + dk eku 3 gSAWrite True / False in the following statement
(i) The direction ratios of the line joining the points (1, 2, 3) and (4, 5, 6)
are 5, 3, 9
(ii) If three non parallel non zero vectos are coplaner than the scalar triple
product of them will be zero.
(iii) The differential Coefficient of sin (cos1x) is zero.
(iv) The maximum value of sin x + cos x is 2
(v) The value of . .ii j j k k + + is 3.'u&3- fj Lkkua dh iwfrZ dhft,&
(i) sin x dk nok vodYkt --------------- ------- grk gSA
(ii) Yxzkt loZlfedk ls ( ) ( ). .a b c d H H H KH
-----------------------
(iii) x
y d -------------------- lekJ;.k xq.kkad dgrs gSaA
(iv) ;fn 0.75 r < 1 g r pj a esa ----------------- lg&laca/k grk gSA
(v) nks v'kwU; lfn'k aH
vkSj bH
lekarj gksrs gSa ;fn vkSj ;fn---------------- -Fill in the Blanks
(i) The nth derivative of sin x is ----------------------
(ii) By Lagrang's inequality ( ) ( ). .a b c d H H H KH
= -----------------------
(iii) xy
is -------------------- regression Coefficient
(iv) If 0.75 r < 1 then ----------------- Co-relation in variables.
(v) Two non zero vectors aH and b
H are parallel if and only if -----------------
'u&4- [k.M v ds fYk, [k.M c esa ls lgh mkj pqudj tM+h cukb,AMatch the Column by choosing from section (B) for section (A)
[k.M v [k.M c(Section-A) (Section-B)
(i) tanx dx (a) sin1x
a+ c
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(ii) 21
1 x+ dx (b)1
asec1
x
a
+ c
(iii) 2 2
1
a x dx (c) cosec1 x
a+ c
(iv) 2 2
1
a x
(d) log (cos x) + c
(v) 2 21
a x
+ dx (e) tan
1x + c
(f) cos1x
a
+ c
(g)1
a cot1
x
a
+ c
'u&5- fuEufYkf[kr 'ua ds mkj ,d 'kCn/okD; esa fYkf[k,&(i) U;wVu jSQlu fof/k ls fdlh la[;k dk oxZewYk~ Kkr djus dk lwk fYkf[k,A(ii) vkafdd fofk;ksa ls lacaf/kr leyEc prHkZqt fu;e gsrq lw+= fyf[k,A(iii) vkafdd fofk;ksa ls lacaf/kr flEilu dk ,d frgkbZ fu;e fyf[k,A(iv) U;wVu jSQlu fof/k ls fdlh la[;ky dk ?kuewYk Kkr djus dh fof/k dk lwk
fYkf[k,A(v) 0.3542E05 + 0.2681 E05 dk eku fYkf[k,A
Write the anwer of each question in one word/sentence of the following(i) Write the formula for square root of a number by Newton Reiphsons
method.
(ii) Write the formula Simpson's rule related by numerical method.
(iii) Write one thrid rule of Simson's related to numerical method.
(iv) Write the formula for cube root of a number by Newtan's reiphsons
method.
(v) Write the value of 0.3542E05 + 0.2681 E05.
[k.M&c(Section-B)
vfrYk?kq mkjh; 'u (Very Short Answer Type Questions)'u&6- fuEu O;atda d vkaf'kd f esa O; dhft,A 4 vad
2
1
5 6x x +Solve following expression into partial Fractions
2
1
5 6x x +vkok (or)
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2 5
( 1) ( 2)
x
x x
+ d vkaf'kd f a esa O; dj&
2 5
( 1) ( 2)
x
x x
+ Solve in partial fractions
'u&7- fl) dhft, fd 4 vad
tan11
a b
ab
+ + tan
11
b c
bc
+ + tan
1
1
c a
ca
+ = 0
Prove that- tan11
a b
ab
+ + tan
11
b c
bc
+ + tan
1
1
c a
ca
+ = 0
vkok (or)
fl) dj fd sin13
5
+ cos112
13
= sin156
65
Prove that sin13
5+ cos1
12
13= sin1
56
65'u&8- ;fn y = log x + log x + log x .......... g r 4 vad
fl) dhft, fd1
(2 1)
dy
dx x y=
If y = log x + log x + log x .......... then prove
that
1
(2 1)
dy
dx x y=
vkok (or)
y =1 x
1 x
+
dk x ds lkis{k vodYku dhft,A
Differentiate y =1 x
1 x
+
with respect to x.
'u&9- log tan4 2
x + dkxds lkis{k vodYku djA 4 vad
Differentiate log tan4 2
x + with respect to x
vkok (or)
tanx dk ke fl)kar ds }kjk vodYk xq.kkad Kkr djA
Find differential coefficient of tanx by the first principle
'u&10- ,d d.k /okZ/kj ij dh vj Qsadk tkrk gSA xfr dk lehdj.kS = ut 4.9 t2 gSA 20 ehVj dh pkbZ ij igqpus ds fYk, d.k dk kjafd
osx Kkr djA 4 vad
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A particle is thrown vertically upwards. The law of motion is S = ut 4.9
t2. Find the initial velocity of the particle to reach the height of 20 metres
vkok (or)fl) dhft, QYku x3 3x2 + 3x + 7 dk eku fcUnq x = 1 ij u r mfPp"Bvj u gh fufEu"B gSA
Prove that function x3 3x2 + 3x + 7 neither have a maxima nor minima atx = 1
'u&11- fuEu vkdM+a ls lekJ;.k js[kkva ds lehdj.k Kkr djA 4 vad
x 2 4 6 8 10
y 6 5 4 3 2
Find the equation of regression of lines from the following data
x 2 4 6 8 10
y 6 5 4 3 2
vkok (or)fuEu vkdM+ a ds vk/kkj ijx dhy ij lekJ;.k js[kk dk lehdj.k Kkr djA;fn y = 90 r x dk eku Kkr djAJs.kh x ylekUrj ek/; 18 100ekud fopyu 14 20
x vj y esa lg laca/k xq.kkad = 0.8From the followoing data. Find the line of regression ofx on y and estimate
the value of x, if y = 90.Series x y
Arithmetic Mean 18 100
Standard Deviation 14 20
Coefficient of coreelation between x and y = 0.8'u&12- dkYkZ fi;lZu fof/k dk ;x dj fuEu vkdM+a ls lg&laca/k xq.kkadKkr dhft,A 4 vad
x 3 4 6 8 9
y 90 100 130 160 170
Find the coefficient of correlation between x and y by using Karl Pearson'smethod from following data
x 3 4 6 8 9
y 90 100 130 160 170
vkok (or)fl) dhft, fd lg&laca/k xq.kkad r dk eku 1 ls +1 ds chp grk gSAProve that the value of coefficent of correlation lines between 1 and +1
y?kq mkjh; 'u (Short Answer Type Questions)'u&13- ,d lerYk v{k a d fcUnqABC ij feYkrk gS] blls cus ABC dk
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dsUd (a, b, c) gSA fl) dhft, fd lerYk dk lehj.k 3x y z
a b c+ + = gSA
5 vad
A plane intercepts the coordinate axis at ABC respectively the centroid of
ABC is (a, b, c) then prove that the equation of plane is 3x y za b c
+ + = -
vkok (or)ml lerYk dk lehdj.k Kkr dj t ewYk fcUnq ls gdj tkrk gS vj lerYk ax + 2y z = 1 rkk 3x 4y + 2 = 5 ij Ykac gAFind the equation of a plane which passes through the origin and perpendicular
to the planes x + 2y z = 1 and 3x 4y + 2 = 5.
'u&14- ;fn ABC dk dsUd G g r fl) dhft, fd 5 vad
GA GB GC O+ + =
KKKH KKKH KKKH KH
If G be the centriod of a triangle ABC then prove that GA GB GC O+ + =KKKH KKKH KKKH KH
vkok (or)lfn'k a ds ;x dk lkgp;Z fu;e fYkf[k, ,oa mls fl) dhft,State and prove Associative law of vector addition
'u&15- ;fn f(x) = log1
1
x
x
+
g r fl) dj fd& 5 vad
f (a) + f (b) = f 1
a b
ab
+
+
If f(x) = log1
1
x
x
+
then prove that
f (a) + f (b) = f1
a b
ab
+ +
vkok (or)
30
tan sinlim
x
x x
x
dk eku Kkr djA
Evaluate 30
tan sinlim
x
x x
x
'u&16-5 4sin
dx
x+ dk eku Kkr dj& 5 vad
Evaluate5 4sin
dx
x+vkok (or)
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22 6 8
dx
x x+ + dk eku Kkr dj
Evaluate 22 6 8
dx
x x+ +
'u&17- nh?kZo`k2 2
2 2
x y
a b+ = 1 dk {kQYk Kkr djA 5 vad
Find the area of ellipse
2 2
2 2
x y
a b+ = 1
vkok (or)
fl) dhft, fd/ 2
0
sin
sin cos
x
x x
+ dx = 4
Prove that / 20
sinsin cos
xx x
+ dx = 4
'u&18- vodYk lehdj.k gYk dhft,A
sec2x tan y dx + sec2 y . tan x dy = 0
Solve the following differential equation
sec2x tan y dx + sec2 y . tan x dy = 0
vkok (or)
fuEu vodYk lehdj.k d gYk djA(1 + y2) dx = (tan1 y x) dy
Solve the following differential equation
(1 + y2) dx = (tan1 y x) dy
'u&19- fdlh 'u d gYk djus ds A ds frdwYk la;xkuqikr 4 : 3 rkkmlh 'u d gYk djus ds B ds vuwdwYk la;xkuqikr 7 : 5 gSaA ;fn nuagYk djus dh df'k'k djrs gSa 'u ds gYk gus dh kf;drk Kkr djA
5 vad
The odds against A solving a problem are 4 : 3 and odds in favour of Bsolving that problem are 7 : 5. What is the probability that the problem will
be solved if they both try.
vkok (or),d ikls d n ckj mNkYkk tkrk gSA 4 ls vf/kd vad vkuk lQYkrk ekuk tkrkgSA lQYkrv a dh la[;k dk kf;drk caVu Kkr dhft,AA die is thrown twice A number greater than 4 is taken success find the
probablity distribution of number of successes.
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nh?kZ mkjh; 'u (Long Answer Type Question)
'u&20- fl) dhft, fd js[kk,1 2 3
2 3 4
x y z = = vj.
2 3 4
3 4 5
x y z = =
lerYkh; gSaA bu js[kkva ds frPNsn fcUnq Kkr djA
6 vad
Prove that the lines1 2 3
2 3 4
x y z = = and
2 3 4
3 4 5
x y z = = are coplaner
also find the point of intersection of lines.
vkok (or)
ml xY dk lehdj.k Kkr dj t fcUnqv a (1, 3, 4)] (1, 5, 2) vj (1, 3, 0) ls gdj tkrk gS rkk ftldk dsU lerYk x + y + z = 0 ij fLkr gSAFind the equation of sphere which passes through the points (1, 3, 4),
(1 5, 2) and (1, 3, 0) whose centre lines on the plane x + y + z = 0
'u&21- fuEu js[kkva ds chp dh U;wure nwjh Kkr djA 6 vad
( )2= + + + +H
r i j k i j k
( )2 2 2= + + +H
r i j k i j k
Find the shortest distance between two following lines
( )2= + + + +H
r i j k i j k
( ) ( )2 2 2= + + +H
r i j k i j k
vkok (or)fuEu fcUnqv a ls gdj tkus okY lerYk dk lehj.k Kkr djA
2 6 6 , 3 10 9 + + i j k i j k vj 5 6 i kFind the equation of plane passing through the points.
2 6 6 , 3 10 9 + + i j k i j k and 5 6 i kvkn'kZ mkj
mkj&1
(i) (c)1 1 1
2 2x x
+ (ii) (c)
4
(iii) (c) 41 (iv) (b) x + y + z = 1
(v) (a)1 3 2
, ,14 14 14
mkj&2
(i) vlR; (ii) lR;(iii) vlR; (iv) vlR;(v) lR;
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mkj&3
(i) sin2
nx
+
(ii)ac aa
bc bd
H H H H
HH HKH
(iii) x dky ij (iv) mPpLrjh; /kukRed
(v) 0a b =H H
mkj&4- lgh tfM+;k&[k.M v [k.M c
(i) tanx dx (d) log (cos x) + c
(ii) 21
1 x+ dx (e) tan1 x + c
(iii)2 2
1
a xdx (a) sin1
x
a+ c
(iv) 2 2
1
a x
(c) cosec1 + x
ac
(v) 2 21
a x
+ dx (g)
1
acot1
x
a
+ c
mkj&5-
(i) xn+1 =
1 Nx
n2 xn
+
(ii)b
a
hf(x)dx
2= [yo + yn + 2 {y1 + y2 + ...........yn1}] ;gk h =
b a
n
(iii)b
a
hf(x)dx
3= [yo + 4 (y1 + y3 + ........yn1) + 2 (y2 + ........yn1) + yn]
(iv) xn+1 = 2
12
3
+
n
n
Nx
x
mkj&6-gj ds xq.ku[kaM djus ij
x2 5x + 6 = x2 2x 3x + 6
= x (x 2) 3 (x 2)
= (x 2) (x 3) 1 vad
vr% 21
5 6x x +=
1
( 3) ( 2)x x
2
1
5 6x x + =1
( 3) ( 2)x x = ( 3) ( 2)A B
x x+ ekuk .........(I) 1 vad
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;k1
( 3) ( 2)x x =( 2) ( 3)
( 3) ( 2)
A x B x
x x
+
;k 1 = A (x 2) + B (x 3) ..........(II)lehdj.k II esa x 3 = 0 x = 3 j[kus ij
1 = A (3 2) + 0 1 vad A = 1blh dkj lehdj.k II esa x 2 = 0 x = 2 j[kus ij
1 = 0 + B (2 3)
;k 1 = B B = 1
vr% 21
5 6x x +=
1 1
3 2x x
1 vad
'u&6 dk vkok dk mkj
2 5
( 1) ( 2)
x
x x
+ = 1 2
A B
x x+
ekuk ........(I) 1 vad
;k2 5
( 1) ( 2)
x
x x
+ =
( 2) ( 1)
( 1) ( 2)
A x B x
x x
+
;k 2x + 5 = A (x 2) + B (x 1) .........(II) 1 vadlehdj.k II esa x 2 = 0 x = 2 j[kus ij
2 2 + 5 = 0 + B (2 1)
;k B = 9 1 vadlehdj.k II esa x 1 = 0 x = 1 j[kus ij2 1 + 5 = A (1 2) + 0
;k 7 = A A = 7vr% okafNr vkaf'kd f
2 5
( 1) ( 2)
x
x x
+ =
7 9
( 1) 2x x
+
1 vad
'u&7 dk mkjge tkurs gSa fd&
lwk tan1 x tan1 y = tan1( )
(1 )
x y
xy
+ .....(I) 1 vad
mi;qZ lwk esax = a ,oa y = b j[kus ij
tan1 a tan1 b = tan11
a b
ab
+ .....(II) 1 vad
blh dkj
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tan1 b tan1 c = tan11
b c
bc
+ .....(III)
,oa tan1 c tan1 a =1
c a
ca
+ ......(IV)
vr% tan11
a b
ab
+ + tan
11
b c
bc
+ + tan
11
c a
ca
+
= tan1 a tan1 b + tan1 b tan1 c
+ tan1 c tan1 a
= 0 1 vad
vr% Li"Vr% tan11
a b
ab
+ + tan
11
b c
bc
+ + tan
11
c a
ca
+ 1 vad
vkok
sin13
5+ cos1
12
13 = sin1
56
65 1 vad
ekuk cos112
13= ......(I)
rc cos =12
13led .k ABC esa
AC2 = AB2 + BC2
;k 132 = AB2 + (12)2
;k AB = 5
vr% sin =5
13 = sin1
5
13 ......(I)
Li"V% cos112
13= sin1
5
13lehdj.k (I) ls
vc L.H.S. sin13
5
+ cos112
13
= sin13
5+ sin1
5
13......(II)
ge tkurs gSa fd lwk&
sin1 x + sin1 y = sin12 2
1 1x y y x + 1 vad
vr% sin13
5+ cos1
12
13
= sin15
3+ sin1
513
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= sin1
2 23 5 5 3
1 15 13 13 5
+
1 vad
= sin1
3 12 5 4
5 13 13 5
+
= sin156
65
1 vad
Li"V% sin15
3+ cos1
12
13 = sin1
56
65
'u&8 dk mkj
fn;k gS& y = log log log .........x x x+ + ......(I)
;k y = logx y+ 1 vadnu a i{k a dk oxZ djus ij
y2 = log x + y
;k y2 y = log x 1 vadnu a i{k a dk x ds lkis{k vodYku djus ij
d
dx(y2 y) =
d
dx(log x) 1 vad
;k 2y dy dydx dx
= 1x
;k (2y 1)dy
dx=
1
x
dy
dx=
1
(2 1)x y 1 vad
vkok dk mkj&
fn;k gS y =1
1
x
x
+
nu a i{k a dk oxZ djus ij
y2 =1
1
x
x
+
........(I) 1 vad
leh- I dk nu a i{k a dk oxZ djus ij
d
dx(y2) =
1
1
d x
dx x
+ 1 vad
;k 2y dydx
= 21 1(1 )x x
x + + kxQYk lwk
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;k 2ydy
dx= 2
2
(1 )x
+
;kdy
dx=
1
(1 ) (1 )y x x
+ +
y dk va'k o gj esa xq.kk djus ij
;kdy
dx= 2(1 ) . (1 )
y
y x x
+ + 1 vad
;kdy
dx=
1. (1 ) . (1 )
1
y
xx x
x
+ + +
y2 =1
1
x
x
+
dy
dx = 2
1
1x 1 vad'u&9 dk mkj
gYk& ekuk fd y = log tan4 2
x + nu a i{k a dk x ds lkis{k vodYku djus ij
dy
dx=
d
dxlog tan
4 2
x +
tan4 2
x + = t ekuk 1 vad
dy
dx=
d
dx(log t)
=d
dtlog t
dt
dx .
dy dy dt
dx dt dx
= 3 1 vad
=1
tan
4 2
. +
d x
t dx
[
4 2
x+ = u Yus ij]
=1
.d
t dutan u
d
dx
4 2
x + 1 vad
=1
tsec2u .
d
dx 4 2
x +
=1
tan
4 2
x +
sec24 2
x + .1
2
u ,oa t ds eku okfil djus ij
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=
2sec1 4 2
2tan
4 2
+ +
x
x
vr% log tan4 2
d x
dx
+ =
2sec1 4 2
2tan
4 2
+ +
x
x1 vad
'u&9 dk vkok dk mkj
gYk& ekuk f (x) = tanx
f (x + h) = tan( )+x h
lwk ddx
f (x) =0
( ) ( )lim
+ h
f x h f xh
1 vad
tand
xdx
=0
limh
tan( ) tan+ x h xh
=0
limh
tan( ) tan tan( ) tan
tan( ) tan
x h x x h x
h x h x
+ + +
+ +1 vad
va'k o gj esa tan( )x h+ tanx dk xq.kk djus ij
=0
tan( ) tanlim
tan( ) tanh
x h x
x h x
+ + + 1 vad
=0
sinlim
2 cos ( ) cos .( tan ( ) tan ) + + +hh
h x h x x h x
= 0 0
1 sin 1lim lim
2 cos ( ) cos . tan ( ) tan
+ + +
h h
h
h x h x x h x
=1 1
2 cos . cos . tanx x x
=
21 sec
2 tan
x
x1 vad
'u&10 dk mkjgYk& fn;k gS fd S = ut 4.9 t2 .........(I)lehdj.k (I) d t ds lkis{k vodYku djus ij
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osxdS
dt= u 9.8 t ......(II) 1 vad
tc d.k egke pkbZ ij igqprk gS rc mldk osx 'kwU; gxkrc u 9.8 t = 0
;k t = 9.8u 1 vad
lehdj.k (I) ls 20 = u .9.8
u 4.9
2
9.8
u 1 vad
;k 20 =2 2
9.8 2 9.8
u u
;k u2 = 20 2 9.8;k u2 = 392
;k u = 392
u = 14 2 eh@ls- 1 vad'u&10 dk vkok dk mkj
ekuk y = x3 3x2 + 3x + 7 ......(I)x ds lkis{k vodYku djus ij&
rcdy
dx= 3x2 6x + 3 ......(II) 1 vad
iqu% x ds lkis{k vodYku djus ij2
2
d y
dx= 6x 6 .....(III) 1 vad
QYku ds mfPp"B ;k fufEu"B gus ds fYk,
dy
dx= 0
rc 3x2 6x + 3 = 0;k x2 2x + 1 = 0
;k (x 1)2 = 0x = 1
x = 1 ds fYk,2
2
d y
dx= 6 (1 1)
;k2
2
d y
dx= 0 vfuf'prrk 1 vad
lehdj.k (III) d x ds lkis{k vodYku djus ij
3
3d y
dx= 6 0 1 vad
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vr% QYku dk x = 1 ij u r mfPp"B ,oa u gh fufEu"B eku gSA'u&11 dk mkj
gYk&x y x2 y2 xy
2 6 4 36 12
4 5 16 25 20
6 4 36 16 24
8 3 64 9 24
10 2 100 4 20
x = 30 y = 20 x2 = 220 y2 = 90 xy = 100
2 vadekuk fd y dh x ij lekJ;.k js[kk dk lehdj.k fuEukuqlkj gS&
y = na + bx;k 20 = 5a + 30b .....(I)
mi;qZ lkj.kh ls eku j[kus ij,oa xy = ax + bx2
;k 100 = 30a + 220 b .....(II) 1 vadlehdj.k (I) o lehdj.k (II) d gYk djus ij
a = 7
b =
1
2vr% y dh x ij lekJ;.k js[kk dk lehdj.k
y = 7 0.5 x 1 vad'u&11 dk vkok dk mkj&
gYk& ge tkurs gSa fd lekJ;.k xq.kkad x dky ij
bxy = x
y
1 vad
= 0.8
14
20= 0.56
x dh y ij lekJ;.k js[kk dk lehdj.kx x = bxy (y y ) 1 vad
;k x 18 = 0.56 (y 100) (x = 18 , y = 100);k x = 38 + 0.56 y 1 vadvc pwfd y = 90rc x = 38 + 0.56 90
;k x = 12.4 1 vad
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'u& 12 dk mkj
gYk& X =x
n
=30
5
1 vad
= 6
blh dkj Y =650
5= 130
lg&laca/k xq.kkad gsrq lkj.kh
x y x x y y (xx ).(yy ) (xx )2 (y y )2
3 90 3 40 120 9 1600
4 100 2 30 60 4 900
6 130 0 0 0 0 0
8 160 2 30 60 4 900
9 170 3 40 120 9 1600
x = 30 y = 650 (xx ). (xx )2 (yy )2
(yy ) = 26 = 5000
= 360
1 vad
lg&laca/k xq.kkad r = 2 2( ) . ( )
( ) ( )
x x y y
x x y y
1 vad
=360
26 . 5000
=18
5 13
= 0.998
~ 1 ifjiw.kZ /kukRed lg&laca/k 1 vad'u&12 dk vkok dk mkj
ge tkurs gSa fd n pj a x oy ds chp lg&laca/k xq.kkad
r =2 2
( ) . ( )
( ) . ( )
x x y y
x x y y
1 vad
r =2 2
.
.
X Y
X Y
.......(I)
x x =X ,oa y y = Y fYk[kus ij
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fdUrq 'oktZ vlfedk (Schwart's Inequality) ls(XY)2 (X)2 (Y)2 1 vad
;k 2
2 2
( )
( ) ( )
XY
X Y
1 1 vad
;k2
.
XY
X Y
1
;k r2 1 lehdj.k (I) ls 1 vad 1 x + 1
Yk?kq mkjh; 'u'u&13 dk mkj
gYk& ekuk lerYk X ls ] Y v{k ls rkkZ v{k ls vUr%[k.M dkVrk gS rc lerYk dklehdj.k
1x y z
r+ + =
......(I)
lerYk X, Y rkkZv{k a d e'k% A, B rkkC fcUnqv a ij feYkrk gSA vr% fcUnqv a A, B, C dsfunsZ'kkad e'k% (, O, O)] (O O) rkk (O, O Y)g axsA
pwfd 'ukuqlkj ABC ds dsUd ds funsZ'kkad (a, b, c) gSaA
vr% x = 1 2 33
x x x+ +, y =
1 2 3
3
y y y+ +rkk z = 1 2 3
3
z z z+ +1 vad
a =0 0
3
+ +=
3
= 3a
b =0 0
3
+ +=
3
= 3b
c =0 0
3
+ + =
3
= 3c 1 vad
lehdj.k I esa , rkk ds eku j[kus ij
3 3 3
x y z
a b c+ + = 1
;kx y z
a b c+ + = 3
vr% okafNr lerYk dk lehdj.k
x y z
a b c+ + = 3
gxk 1 vad
1 vad
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vkok dk mkj&gYk& ewYk fcUnq ls gdj tkus okY lerYk dk lehdj.k
ax + by + cz = 0 ......(I) 1 vadlerYk (I) lerYk a ftuds lehdj.k uhps fn, x, gSa] ij Ykac gSA
x + 2y z = 1 ......(II)
,oa 3x 4y + 2 = 5 ......(III)vr% a + 2b c = 0 .....(IV)
3a 4b + c = 0 ......(V) 2 vadlehdj.k (IV) ,oa (V) d gYk djus ij
2 4 3 1 4
a b c
c= =
;k1 2 5
a b c= = = k ekuk 1 vad
vr% a = k , b = 2k ,oa c = 5klehdj.k (I) ls
kx + 2ky + 5kz = 0
;k x + 2y + 5z = 0;gh okafNr lerYk dk lehdj.k gSA 1 vad
'u&14 dk mkjgYk& ekuk ewYk fcUnqO ds lkis{kABC ds 'kh"kZA, B, C ds fLkfr lfn'k e'k%
, ,a b cH H H
gS rc , ,OA a OB b OC c= = =KKKH H KKKH H KKKH H
qtkv aBC, AC rkkAB ds e/; fcUnqe'k% D, E rkk F gSA ef/;dk,AD,BE rkk CF dk frPNsn fcUnq G ABC dk dsUd gSA 1 vad
D, BC dk e/; fcUnq gS2
b cOD
+=
H H
ABC dk dsUd G, AD d 2 : 1 ds vuqikr esa fokftr djrk gS 1 vad
OG =2( ) / 2 1
3
b c a+ +
=3+ +
b c a
=3
a b c+ +H H H
.....(I) 1 vad
GA = A fLkfr lfn'k & G dk fLkfr lfn'k
=2
3
a b c H H H
......(II)
blh dkj&
GBKH = B dk fLkfr lfn'k & G dk fLkfr lfn'k
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GBKKKH
=2
3
b a c .....(III)
,oa GCKH
=2
3
c a c H H H
......(IV) 1 vad
lehdj.k II, III ,oa IV ls
G A GB GC + +KH KKKH KKKH
=2 2 2
3 3 3
a b c b a c c a b + +
H H H H H H H H H
=2 2 2
3
a b c b a c c a b + + H
vr% GA GB GC + + = O 1 vadvkok dk mkj&
gYk& ;fn ,a b rkk c rhu v'kwU; lfn'k gS rclfn'k a ds ;x ds lkgp;Z fu;ekuqlkj
( )a b c+ +H H H
= ( )a b c+ +H H H
miifk& fpk ds vuqlkj eku dk lfn'k ,a bH H
rkk cH
dkxt ds rYk esa gSA tgk O ewYkfcUnq gS fpk
ds vuqlkj ,a OA b AB= =H KKKH H KKKH
rkk c BO=H KKKH
cuk;k rkkOC d feYkk;k fcUnq OB rkk AC d h feYkk;kA
AOB esa lnf'k ;x dk fkqt fu;e Ykxkus ij& 1 vad
OBKKKH
= OA AB+KKKH KKKH
1 vad
= a b+H H
vc fkqt OBC esa lfn'k ;x dk fkqt fu;e Ykxkus ij
OCKKKH
= OB BC +KKKKH KKKH
= ( )a b c+ +H H H
.....(1) 1 vadblh dkj ABC esa lfn'k ;x dk fkqt fu;e Ykxkus ij&
ACKKKH
= AB BC+KKKKH KKKH
= b c+H H
1 vadrkk OAC esa lkfn'k ;x dk fkqt fu;e Ykxkus ij&
OCKKKH
= OA AC +KKKH KKKH
= ( )a b c+ +H H H
......(2)
vr% lehdj.k 1 vj 2 ls
( )a b c
+ +
H H H
= ( )a b c
+ +
H H H
;gh lfn'k a ds ;x dk lkgp;Z fu;e gSA
1 vad
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'u 15 dk mkj&
gYk& fn;k gS f (x) = log1
1
x
x
+
......(1)
lehdj.k (1) esa x = a j[kus ij
f(a) = log 11
aa+ .....(2) 1 vad
lehdj.k (1) esa x = b j[kus ij
f (b) = log1
1
b
b
+
......(3) 1 vad
lehdj.k (2) vj (3) d tM+us ij
f (a) + f (b) = log1
1
+
a
a+ log
1
1
b
b
+
f (a) + f (b) = log (1 ) (1 )(1 ) (1 )
a ba b
+ +[logm + logn = log m n]
f (a) + f (b) = log1
1
a b ab
a b ab
++ + +
.......(4) 1 vad
lehdj.k (1) esa x =1
a b
ab
++
j[kus ij
f1a b
ab+ + = log
1 1
11
a b
aba b
ab
+ +
++
+
= log
1
11
1
+ +
+ + ++
ab a b
abab a b
ab
f 1
a b
ab
+ + = log
1
1
a a ab
a b ab
++ + + ......(5) 1 vad
lehdj.k (4) o (5) ls&
f (a) + f (b) = f1
a b
ab
+ + 1 vad
vkok dk gYk&
30
tan sinlim
x
x x
x
= 3
0
sinsin
coslim
x
xx
x
x
1 vad
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= 30
1sin 1
coslim
x
xx
x
=
( )30
sin 1 coslim
cosx
x x
x x
= 20
tan 1 coslim
x
x x
x x
1 vad
= 20 0
tan 1 coslim lim
x x
x x
x x
= 1 20
(1 cos ) (1 cos )lim
(1 cos )x
x x
x x
++ 0
tanlim 1
x
x
x
=
1 vad
=
2
20
1 coslim
(1 cos )x
x
x x
+
=
2
20
1 sinlim
1 cosx
x
xx +
=0
sin sinlim
. (1 cos )x
x x
x x x
+ 1 vad
=0
00
sin sinlim
limlim (1 cos )
x
xx
x x
x x
x
+ 0sin
lim 1x
x
x
= 3
=1 1
1cos
=1
1 1+
=1
21 vad
'u&16 dk mkj&
ekuk I =5 4 sin
dx
x+
I =
2
2 tan25 4
1 tan2
dx
x
x
+
+
2
tan2sin
1 tan2
x
xx
= +
3
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I =2
2
5 1 tan 8 tan2 2
1 tan
2
dx
x x
x
+ +
+
I =
2
2
1 tan2
5 5tan 8tan2 2
xdx
x x
+
+ + 1 vad
I =
2
2
sec2
5tan 8tan 5
2 2
+ +
xdx
x x tan2
x= tj[kus ij
sec22
x dx = 2dt
1 vad
I = 22
5 8 5
dt
t t+ +
=2
5
2 815
dt
t t+ +
=2
5 2
2 4 4 162( ) 15 5 25
dt
t t + + +
1 vad
=2
5
24 9
5 25
dt
t + +
j[kus ij 1 vad
t +4
5= u
dt = du
I =2
2
2
5 3
5
+
du
u
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=1
1
2 3 tan35 5
5
u
= 12 5tan3 3
u
=1
45( )
2 5tan3 3
+
t
=12 (5 )tan
3 3
+ t u
I =1 (5tan 4)2 2tan
3 3
x
+
1 vad
'u&16 dk vkok dk mkj&
22 6 8
dx
x x+ + = 21
2 3 4
dx
x x+ + 1 vad
=22
1
23 72 2
dx
x + +
1 vad
=1
31 2tan
7 72
2 2
x
+
2 vad
=11 2 3tan
7 7
x +
1 vad
'u&17 dk mkj&gYk& nh?kZ o`k dk lehdj.k
2 2
2 2
x y
a b+ = 1
2
2
y
b= 1
2
2
x
a
y2 = b22 2
2a x
a
1 vad
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y =2 2b
a xa
vc ewYk fcUnq O ds fYk, x = O rkk fcUnqA ds fYk, x = a gS vkkZr~ {kQYkAOB ds fYk, x dk eku O ls a rd fopfjr grk gSA vr% lekdYku lhek, x =0 ls x = a gxh
nh?kZok dk {kQYk = 4 {kQYk OAB dk {kQYk 1 vad
= 40
aydx
= 42 2
0
a ba x dx
a lehdj.k 1 ls
=
22 2 1
0
4sin
2 2
+
ab x a x
a xa a
=2 2 2 1
0
2sin
+
ab x
x a x aa a
1 vad
=2 2 2 1 2 12
sin 0 sin 0b a
a a a a aa a
+
=2 1 2 12 0 sin (1) sin 0
ba a
a
+ 1 vad
=22 0
2b aa
= ab oxZ bdkbZ 1 vad
'u&17 dk vkok dk mkj&
ekuk I = 20
sin
sin cos
xdx
x x
+
= 20
sin
2sin cos
2 2
x
dxx x
+ 1 vad
I = 20
cos
cos sin
xdx
x x
+lehdj.k 1 o 2 ls tM+us ij
I + I = 20
sin cos
sin cos sin cos
x xdx
x x x x
+
+ +
2 vad
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2 I = 20
sin cos
sin cos
x xdx
x x
++
2I = 20
dx
2I = [ ] / 20x
1 vad
2I =2
O
I =4
1 vad
'u&18 dk mkj&fn;k x;k lehdj.k gS&
sec2
x tan y dx + sec2
y tan x dy = 0 sec2 x tan y dx = sec2 x tan x dy
2sec
tan
xdx
x=
2sec
tan
ydy
y 1 vad
2sec
tan
xdx
x =2
sec
tan
ydy
y ...........(1) 1 vad
ekuk tan x = t 1 vadrkk tan y = 4rc sec2 x dx = dt 1 vadrkk sec2y dy = du
vr% lehdj.k 1 lsdt
t = 4du
log t = log u + log c log t + log u = log c log (t u) = log c t u = c 1 vad
tan x tan y = c'u&18 dk vkok dk mkj&
gYk& fn;k x;k vodYku lehdj.k gS& (1 + y2) dx = (tan1 y x) dy 1 vad
(1 + y2)dx
dy= tan1 y x
(1 + y2)dx
dy+ x = tan1 y
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21
1
dxx
dy y
+ + =
1
2
tan
1
y
y
+;g lehdj.k x esa js[kh; vodYk lehdj.k gSA
vr% bldh rqYkuk
dx
dy + Px = Q ls djus ij 1 vad
P = 21
1 y+Q =
1
2
tan
1
y
y
+ 1 vad
lekdYku xq.kkad (I.F.) =Pdy
e 1 vad
=
1
21dy
ye +
= etan1
y
vr% x (I.F.) = Q . (I.F.) dy + c ls vh"V gYk gxk&
x . etan1y =
1
2
tan
1
y
y
+ . etan1y dy + c
x . etan1y = . .tt e dt c+[tan1 y = tekuk 2
1
dy
y+
= dt rc]
nk, i{k esa t d ke QYku ekudj [k.M'k% lekdYku djus ij&
x . etan1y = t . et 1. et dt + C x . etan1y = t . et et + c x . etan1y = tan1 y etan1y etan1y + c 1 vad
x = tan1 y 1 + c etan1y
'u&19 dk mkj&?kVuk A dk frdwYk la;xkuqikr = 4 : 3
P (A) =3 3
4 3 7=
+1 vad
?kVuk B dk vuqdwYku la;xkuqikr = 7 : 5
P (B) =7 7
7 5 12=
+1 vad
pwfdA vjB Lorak ?kVuk, gSaP (A B) = P (A) . P (B)
=
3 7 1
7 12 4 = 1 vad
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'u gYk gus dh k;fdrkP (A B) = P (A) + P (B) P (A B) 1 vad
=3 1 1
7 12 4+
= 36 49 2184+ = 6484 1 vad
=16
21'u&19 dk vkok dk mkj&
n (S) = 4, 5, 6
4 ls vf/kd vad vkus dh kf;drk
P (A) =2 1
6 3
= 1 vad
4 ls vf/kd u vkus dh kf;drk
P (A) = 1 1 2
3 3= 1 vad
;fn X ;knfPNd pj gS rc x = 0, 1, 2 dbZ lQYkrk ug
P0 =P (x = 0) = P (AA ) =2 2 4
3 3 9 = 1 vad
P1 = P (x = 1) = P (AA ) + P (AA )
= 1 2 2 1 43 3 3 3 9
= 1 vad
P2 = P (x = 2) = P (AA) =1 1 1
3 3 9 = 1 vad
x 0 1 2
P14
9
4
9
1
91 vad
'u&20 dk mkj
gYk& js[kkv a ds lehdj.k&1 2 3
2 3 4
= =
x z z= r ekuk .......(I)
,oa2 3 4
3 4 5
= =
x y z= r1 ekuk .....(II)
lehdj.k (I) ls&x = 1 + 2r .......(a)
y = 2 + 3r .......(b)
z = 3 + 4r ........(c) 1 vad
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blh dkj lehdj.k (II) ls&x = 2 + 3r1 .......(a)
y = 3 + 4r1 .......(b)
z = 3 + 4r1 ........(c) 1 vadjs[kk (I) o (II) frPNsnh js[kk, gSa&
1 + 2r = 2 + 3r1;k 2r 3r1 1 = 0 .....(III)blh dkj 3r 4r1 1 = 0 ......(VI),oa 4r 5r1 1 = 0 .....(V) 1 vadlehdj.k III, IV o V d gYk djus ij&
3 4
r
=
1 1
3 2 8 9
r=
+ +
;k 1r
=1 1
1 1
r
= r = 1 ,oa r1 = 1 1 vadr o r1 ds eku lehdj.k (V) d larq"V djrs gSa vr% nh xbZ js[kk, lerYkh;
gSaA x = 1 + 2r leh- (a) esa r dk eku j[kus ij
x = 1 2
x = 1
blh dkj y = 1
,oa z = 1 2 vadvr% fPNsn fcUnq (1 , 1 , 1) gxkA'u&20 dk vkok dk mkj&
ekuk fd xY dk lehdj.kx2 + y2 + z2 + 2 ux + 2 vy + 2wz + d = 0 ......(1)
pwfd fcUnq (1, 3, 4) , (1, 5 , 2) rkk (1, 3 , 0) xY ij fLkr gS
vr% 12 + (3)2 + 42 + 2u 1 + 2 v (3) + 2 4 + d = 012 + (5)2 + 22 + 2u 1 + 2 v (5) + 2 . 2 + d = 0
12 + (3)2 + (O)2 + 2u 1 + 2 v (3) + 2 . 0 + d = 01 vad
gYk djus ij2u 6v + 8 + d = 26 .......(2)
2u 10 v + 4 + d = 30 ......(3)2u 6v + d = 10 ......(4) 1 vad
pwfd xY (1) dk dsU lerYk x + y + z = 0 ij fLkr gSA u + v + w = 0 ......(5) 1 vad
lehdj.k 4 & lehdj.k 2w = 2
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lehdj.k 5 & lehdj.k 34v + 4 w = 20
v = 3
lehdj.k (5) lsu = (v + w)
u = 1 2 vadu, v, ds eku a d lehdj.k 5 esa j[kus ij
2 18 + d = 10
d = 0
vr% lehdj.k I lsx2 + y2 + z2 2x + 6y 4z +10 = 10 1 vad
;gh xY dk lehdj.k gSA'u&21 dk mkj&
fn, x, lehdj.k a ls Li"V gS fd&1a =
2i j k+ +
1b = i j k+ + 1 vad
2a = 2 i j k
2b = 2 2i j k+ + 1 vad
2 1a aKKH KKH
= ( ) ( )2 2 + + i j k i j k
2 1a aKKH KKH
= 3 2i j k + 1 vad
1 2b bKH KKH
=
1 1 1
2 1 2
i j k
= (2 1) (2 2) (1 2)i j k + +
= 3 3i k+ 1 vad
U;wure nwjh =2 1 1 2
1 2
.| |
a a b bb b
1 vad
=
( ) ( )3 2 3 39 9
i j k i k + +
=3 6
3 2
=
3
2 1 vad
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dsoYk ifj.kke Yus ij nwjh _.kkRed ug grh'u&21 dk vkok dk mkj&
ekuk fd fn, x, fcUnq A , B o C rkk P dbZ vU; fcUnq gS ftldk fLkfrlfn'k r gSA 1 vad
AP = P dk fLkfr lfn'k & A dk fLkfr lfn'k
;k AP = ( )2 6 6r i j k + 1 vadAB = B dk fLkfr lfn'k & A dk fLkfr lnf'k
= ( ) ( )3 10 9 2 6 6i j k i j k + +
= 4 3i j k + 1 vad
ACKKKH
= C dk fLkfr lfn'k & A dk fLkfr lfn'k
=
( )
( )5 6 2 6 6i k i j k
+
= 3 6i j 1 vadpwfd , ,AP AC AB
KKKH KKKH KKKHlerYkh; gSa
, , KKKH KKKH KKKHAP AB AC = 0
{ } { }( 2 6 6 ) . ( 4 3 ) ( 3 6 )r i j k i j k i j + + = 0 ......(I)
fdUrq { }( 4 3 ) ( 3 6 )i j k i j +
=
1 4 3
3 6 0
i j k
= (0 18) (0 9) (6 12)i j k + +
= 18 9 18i j k + + 1 vadvr% lehdj.k (I) ls&
(2 6 6 ) .[ 9 (2 2 )]r i j k i j k + + + = 0
;k (2 2 )r i j k +4+612 = 0
;k (2 2 )r i j k = 2 1 vad;gh lerYk dk lehdj.k gSA