mathematics set 4, model papers of madhya pradesh board of secondary education, xiith class

7/29/2019 Mathematics set 4, Model papers of Madhya Pradesh Board of Secondary Education, XIIth Class

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gk;j lsd.Mjh Ldwy ijh{kk& 2012&13HIGHER 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 ik esa n [k.M v ,oa c fn, x, gSaA(4) [k.M v esa fn, x, 'u - 1 ls 5 rd oLrqfu"B 'u gSaA(5) [k.M c esa 'u - 6 ls 21 esa vkarfjd fodYi fn, x, gSaA

Instructions(1) All questions are compulsory to solve.(2) Marks have been indicated against each question(3) There are two sections 'A' and 'B' are given in the question paper.(4) In section A question No. 1 to 5 are of objective type questions.

(5) Internal options are given in question No. 6 to 21 of Section-B

[k.M&v oLrqfu"B 'u (Objective Type Questions)'u&1 R;sd oLrqfu"B 'u esa fn, x, fodYia esa ls lgh mkj pqfu,&

5 1 vad

(i)1

(x 4) (x 6)+ + ds vkaf'kd f gksxh&

(a)1 1

2 (x 4) 2 (x 6)

+ + (b)1 1

5 (x 4) 2 (x 6)

+ +

(c) 1 53 (x 4) 2 (x 6)++ + (d) 2 1x 4 x 6+ +

(ii) 2 tan1 xdk eku gS&

(a) tan1 22x

1 x+(b) tan1 2

2x

1 x

(d) tan121 x

2x

+(d) tan1

21 x

2x

(iii) fcUnq (2, 1, 4) dh y v{k ls nwjh gxh&

(a) 20 (b) 1


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(c) 12 (d) 10

(iv) lerYk 2x y + 2z + 1 = 0 dh ewYk fcUnq ls nwjh gxh&

(a) 1 (b)1

3

(c) 3 (d) 3

(v) Aur

vjBur

ds fLkfr lfn'k e'k% 2i 9j 4k $ $ vj 6i 3j 8k +$ $ gSA ABuuur

dk ifj.kke gS&(a) 11 (b) 12

(c) 13 (d) 14

Choose and Write the correct answer from the given options provided in everyobjective type question

(i) Partial fraction of

1

(x 4) (x 6)+ + is&

(a)1 1

2 (x 4) 2 (x 6)

+ + (b)1 1

5 (x 4) 2 (x 6)

+ +

(c)1 5

3 (x 4) 2 (x 6)+

+ + (d)2 1

x 4 x 6

+ +(ii) The value of 2 tan1 x is&

(a) tan12

2x

1 x+(b) tan1

2

2x

1 x

(d) tan12

1 x

2x

+(d) tan1

21 x

2x

(iii) Distance of the points (2, 1, 4) from y axis is&

(a) 20 (b) 1

(c) 12 (d) 10

(iv) Distance from origin to plance 2x y + 2z + 1 = 0 is&

(a) 1 (b) 13

(c) 3 (d) 3

(v) The position vactors of vactor Aur

andBur

are 2i 9j 4k $ $ and 6i 3j 8k +$ $

respectively then the magnitude of ABuuur

is&

(a) 11 (b) 12

(c) 13 (d) 14


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'u&2- fuEufYkf[kr dkua esa lR;/vlR; dku NkVdj viuh mkjiqfLrdk esafYkf[k,A 5 1 vad

(i)logx

x dx dk eku1

2(log x)2 + c gSA

(ii) xY dk

r (2i j k) + +$ $ $

= 5 dsU (2, 1, 1) gSA(iii) lg&laca/k r rkk lekJ;.k xq.kkd abxy rkk byx esa laca/k r = bxy , byx grk

gSA

(iv) fkqt dh rhu a ef/;dkv a }kjk fu/kkZfjr lfn'k a dk ;x 'kwU; grk gSA

(v) n pj a ds e/; lg&laca/k xq.kkad lnSo r < 1 lUrq"V djrk gSA

Write True / False in the following statements

(i) The value oflogx

xdx is

1

2(log x)2 + c

(ii) The centre of the sphere r (2i j k) + +$ $ $ = 5 is (2, 1, 1)

(iii) The relation between correlation cofficient r and the regretion coefficient

bxy and byx is r = bxy . byx

(iv) The vector's sum of the median of triangle directed from the vertex is

zero.

(v) The co-relation cofficient of two variables always satisfied the relation r< 1.

'u&3- lgh tM+h cukb,A 5 1 vad[k.M&v [k.M&c

(i)dx

tan x cot x+ (a) 3 sin2 x cos x

(ii)d

dx(tan x) (b) cot2x

(iii) ;fn a 2i 3j 5k = +r

$ $ (c)17

7

rkk b 2i 2j 2k = + +r $ $

r a . br r

dk eku

(iv) 0.4396E05 + 0.3512E02 dk eku (d) cos 2x

4(v) ewYk fcUnq ls lerYk (e) 0.1251E08

3x 2y + 6z = 17 dh nwjh (f) 0(g) sec2 x


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Make the correct pairsSection-A Section-B

(i)dx

tan x cot x+ (a) 3 sin2 x cos x

(ii)

d

dx (tan x) (b) cot2x

(iii) If a 2i 3j 5k = +r

$ $ (c)17

7

and b 2i 2j 2k = + +r

$ $

then value of a . br r

(iv) Value of 0.4396E05 + 0.3512E02 (d) cos 2x

4

(v) The length of perpendicular (e) 0.1251E08from the origin to the plane (f) 0

3x 2y + 6z = 17 (g) sec2 x

'u&4- fj Lkkua dh iwfrZ dhft,& 5 1 vad

(i) dbZ QYku f (x) fcUnq x] ij mfPp"B gS r f"(x) dk eku ---------------- gxkA

(ii) U;wVu jsQlu fof/k ls 10 dk ?kuewYk ke iqujkofk i'pkr~ -----------------------gxkA

(iii) ;fn e0 = 1, e1 = 2.72 , e2 = 7.39 r leYkEc prqqZt fu;e ls 3 x0

e dx--------------------A

(iv) 2

dx

x a2

dk eku ------------------------ gSA

(v) fcUnq ar

ls g dj tkus okY rkk xr

ij YkEcor~ lerYk dk lehdj.k ---------------- gSA

Fill in the Blanks(i) If any function f (x) is maxima at x] then value of f"(x) is ....................

(ii) Cube root of 10 by Nuwtion Raphson's method after first interaction is

..........................

(iii) If e0 = 1, e1 = 2.72 , e2 = 7.39 then by trapezoidal rule integral3 x

0e dx

--------------------

(iv) The value 2

dx

x a2 is------------------------


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(v) Equation of the plance passing through the point ar

and normal to xr

is........................

'u&5- fuEufYkf[kr 'ua ds mkj ,d 'kCn/okD; esa fYkf[k,& 51 vad

(i) flEilu ds ,d frgkbZ fu;e dk lwk fYkf[k,A

(ii) yz&lerYk dk lehdj.k fYkf[k,A

(iii) sec x dx dk eku D;k gxkA(iv) ax dk x ds lkis{k vodYku xq.kkad fYkf[k,A

(v) 0.65731E05 + 0.58918E05 = .........................

Write the answer of each question in one word / sentence of the following

(i) Write the fomula of Simpsons' one third ruly.

(ii) What is the equation of yz&plane

(iii) What is the value of sec x dx(iv) Write differential cofficient of ax with rest to x(v) 0.65731E05 + 0.58918E05 = .........................

[k.M&c(Section-B)

vfrYk?kq mkjh; 'u (Very Short Answer Type Questions)

'u&6- f 2 22x 3(x 1) (x 1)

+ d vkaf'kd f esa O; dhft,A 4 vad

Resolve 2 22x 3

(x 1) (x 1)

+ into partial fraction

vkok (or)

f2

2

x 7x

x 2x 8

++

d vkaf'kd f esa fo dhft,A

Resove2

2

x 7x

x 2x 8

++

into partial fraction

'u&7- fl) dhft, fd& 4 vad

sin14

5+ sin1

5

13+ sin1

16

65 = 2

Prove that

sin14

5+ sin1

5

13+ sin1

16

65 = 2


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vkok (or)

fl) dhft,&

sin1 x + sin1 x 1 =

2

Prove thatsin1 x + sin

1 x 1 =2

'u&8- ;fn y =1 x

1 x

+

g r fl) dhft, fd (1 x2)dy

dx+ y = 0

4 vad

If y =1 x

1 x

+

then prove that (1 x2)dy

dx+ y = 0

vkok (or)

;fn y sin x sin x sin x ........= + + + g r fl) dhft, fd&

(2y 1)dy

dx= cos x

If y sin x sin x sin x ........= + + + , then prove that

(2y 1)

dy

dx = cos x

'u&9-x x

x x

e e

e e

+

dk vdoYk xq.kkad Kkr dhft,A 4 vad

Find the differential coefficient of

x x

x x

e e

e e

+

vkok (or)

;fn y = log1 cos mx1 cos mx

+ g r

dy

dxdk eku Kkr dhft,A

If y = log1 cos mx

1 cos mx

+ , then find the value of

dy

dx

'u&10- ,d okh; IYV dh fkT;k 0.2 lseh fr lsds.M dh nj ls c


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vkok (or)

QYku f (x) = x3 6x2 + 11x 6 dh vUrjkYk [1, 3] esa jY es; dh tkpdhft,A

Verify Rolle's threorem for the function f (x) = x3 6x2 + 11x 6 in the

interval [1, 3]'u&11- fuEufYkf[kr vkdM+ a lsx rkky ds e/; lg&laca/k xq.kkad dh x.kuk dhft,A

4 vad

x 5 9 13 17 21

y 12 20 25 33 35

Calculate the coefficient of corelation between x and y on the basis of thefollowing data when x and y are

x 5 9 13 17 21

y 12 20 25 33 35

vkok (or)

;fn cov (x, y) = 2.25 , var (x) = 6.25 , var (y) = 20.25 g] r lg&laca/k xq.kkad (x, y) Kkr dhft, rkk lg&laca/kh dh fr h crkb,A

If cov (x, y) = 2.25 , var (x) = 6.25 , var (y) = 20.25 then find the coefficient

of correlation (x, y) and also discribe the nature of correlation.

'u&12- ;fn n lekJ;.k js[kkv a ds chp dk d.k gS] lekJ;.k xq.kkad byx =1.6 rkk bxy = 0.4r tan dk eku Kkr dhft,A 4 vad

If the angle between two regression lines of , regression coefficient byx =1.6 and bxy = 0.4, then calculate the value of tan

vkok (or)

n pj jkf'k; a x vkSj y ds e/; lg&laca/k xq.kkad gS] r fl) dhft, fd2 2 2

x y (x y)

x y2

+ = tgk

2 2x y, rkk

2(x y) e'k% x , y vj (x y)

ds lkj.k xq.kkad gSaA

If the correlation coefficient between two variables x andy is, then show

that

2 2 2x y (x y)

x y2

+ = where

2 2x y, and

2(x y) the variance of x ,

y and (x y) respectively.

Yk?kq mkjh; 'u (Short Answer Type Questions)

'u&13- 52 ik a dh QsaVh gqbZ rk'k dh xh esa ls 2 ;knPN;k fudkY tkrs gSa] nu a

YkkYk ;k nu a bs gus dh D;k kf;drk gS\ 5 vad


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Two cards are drawn at rendom from a well shuffled pack of 52 cards, what

is the probability that either both are red or both are aces?

vkok (or)

n ikl a dh ,d Qsad esa ;x 9 ;k 11 kIr u djus dh kf;drk Kkr dhft,A

In the single through of two dice what is the probability of not obtaining atotal of 9 and 11.

'u&14- fcUnq &1] 3] 2 ls xqtjus okY ml lerYk dk lehdj.k Kkr dhft, tlerYk x + 2y + 2z = 5 rkk 3x + 3y + 2z = 8 ij YkEck gSA 5 vad

Find the equation of the plance which passes through the point (1, 3, 2) and

prependicular to the planes x + 2y 3z = 5 and 3x + 3y + 2z = 8

vkok (or)

,d pj lerYk ewYk fcUnq ls P nwjh ij jgrk gS rkk v{k a d A, B vj C

fcUnqv a ij dkVrk gSA fl) dhft, fd prq"QYkdOABC ds dsUd dk fcUnq ik

2 2 2 2

1 1 1 16

x y z p+ + = gS

A variable plane is at a constant distance P from a origin and meet the axisat points A, B and C. Show that the locus of the centroid of tetrahodron

OABC is 2 2 2 21 1 1 16

x y z p+ + =

'u&15- lfn'k fof/k ls fl) dhft, fd& sin ( + ) = sin cos + cos

sin 5 vadProve by vector method sin ( + ) = sin cos + cos sin

vkok (or)

fl) dhft, fd&

( ) ( ) ( )a b c b c a c a b 0 + + + =r r r r r r r r r

Prove that

( ) ( ) ( )a b c b c a c a b 0 + + + =r r r r r r r r r

'u&16- ;fn f(x) =2

2

x 1

x 1

+

g r fl) dhft, fd f1

x

= (f) x 5 vad

If f(x) =

2

2

x 1

x 1

+

, then prove that f1

x

= (f) x

vkok (or)fuEufYkf[kr QYku dh x = 0 ij lkarR; dh foospuk dhft,&


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f (x) =2

1 cos x; x 0

x1

; x 02

Discss the continuity of the following at x = 0

f (x) =2

1 cos x; x 0

x1

; x 02

'u&17- fuEu dk eku Kkr dhft,& 5 vad

dx

5 4 sin x+Evaluate the followingdx

5 4 sin x+vkok (or)

fl) dhft,/ 2

0

tan xdx

41 tan x

=

+

Prove that/ 2

0

tan xdx

41 tan x

=

+'u&18- o x2 = 4y rkk js[kk x = 4y 2 ls f?kjs {k dk {kQYk Kkr dhft,A

Find the area bounded by the curve x2 = 4y and the line x = 4y 2.

5 vadvkok (or)

fuEufYkf[kr dk eku Kkr dhft,&

dx

a cos x bsin x+Evaluate the following

'u&19- fuEu vodYk lehdj.k d gYk dhft,& 5 vad

x + ydy

dx= 2y

Solve the following differential equation

x + y

dy

dx = 2y


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vkok (or)fuEu vodYk lehdj.k d gYk dhft,&

(1 + x2)dy

dx+ 2xy 4x2 = 0

Solve the following differential equation

(1 + x2)dy

dx+ 2xy 4x2 = 0

'u&20- mu js[kkv a ds chp dk d.k Kkr dhft, ftudh fnd~&dT;k, fuEufYkf[krlehdj.k a }kjk fu/kkZfjr gSa& 6 vad

2l + 2n m = 0 ,oa ml + mn + nl = 0Find the angle between those lines, where direction cosines are recognised bythe following equation

2l + 2n m = 0 and ml + mn + nl = 0

vkok (or)

fl) dhft, fd js[kk,x y 2 z 3

1 2 3

+= = rkk

x 2 y 6 z 3

2 3 4

= = lerYkh;

gSaA budk frPNsn fcUnq h Kkr dhft,A

Prove that the linesx y 2 z 3

1 2 3

+= = and

x 2 y 6 z 3

2 3 4

= = are coplaner.

Also find the point of intersection of these lines

'u&21- mu js[kkv a ds chp dh U;wure nwjh Kkr dhft,] ftudh lfn'k lehdj.kgS& 6 vad

r (1 2 ) i (2 3 ) j (3 4 ) k = + + + + + $ $

r (2 3 ) i (3 4 ) j (4 5 ) k = + + + + + $ $

Find the shortest distance between the two lines, whose vector equationsare

r (1 2 ) i (2 3 ) j (3 4 ) k = + + + + + $ $

r (2 3 ) i (3 4 ) j (4 5 ) k = + + + + + $ $

vkok (or)

ml xksys dk lehkj.k Kkr dhft, t fcUnqv a 0] &2] &4 rkk 2] &1] &1ls xqtjrk gS rkk ftldk dsU js[kk 5y + 2z = 0 = 2x 3y ij fLkr gSA

Find the equation of the sphere which posses through points (0, 2, 4) and

(2, 1, 1) and 5y + 2z = 0 = 2x y.


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vkn'kZ mkj (Model Answer)[k.M&v

mkj&1(i) (a) (ii) (b)

(iii) (a) (iv) (b)

(v) (d)

mkj&2(i) lR; (ii) vlR;(iii) vlR; (iv) lR;(v) vlR;

mkj&3-(i) (d) (ii) (g)

(iii) (f) (iv) (e)

(v) (c)mkj&4-

(i) _.kkRed (ii) 2.1667

(iii) 6.915 (iv) log2 2x x a + + c

(v) ( ) r a n 0 =mkj&5-

(i)

b

a f(x) dx =h

3 [yo + yn + 4 (y1 + y3 ......... yn1) + 2 (y2 + y4 .......yn2)]

(ii) x = 0 (iii) log (sec x + tan x)

(iv) ax loge a

(v) 0.124649E06

[k.M&c (Section-B)mkj&6-

2 2 2 2 2

2x 3 A Bx C Dx E

(x 1)(x 1) (x 1) (x 1) (x 1)

+ += + +

+ + + 1 vad

2x 3 = A (x2 + 1) + (Bx + c) (x 1) (x2 + 1) + (Dx + E) (x 1)

x = 1 1 = 4A A = 1

41 vad

vc 2x 3 = A (x4 + 2x3 + 1) + (Bx + C) (x3 x2 + x 1)+ D (x2 x) + E (x 1)

2x 3 = A (x4 + 2x3 + 1) + B (x4 x3 + x2 x) + C (x3 x2 + x 1)

+ D (x2 x) + E (x 1)

ltkrh; in a dh rqYkuk djus ij&


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0 = A + B B =1

4

0 = B + C C =1

4

0 = 2A + B C + D

=2 1 1

4 4 4 + + D D =

1

2

2 = B + C + D + E

2 = 1 1 1

4 4 2+ + E E =

5

21 vad

lehdj.k 1 ls

2 22x 3(x 1) (x 1)

+ = 2 2 21 (x 1) (x 5)

4(x 1) 4(x 1) 2(x 1) + + +

+ + 1 vad

'u&6 dk vkok dk mkj&

2

2

x 7x

x 2x 8

++

= 1 + 25x 8

x 2x 8

++

------(i)

ekuk 25x 8

x 2x 8

++

=5x 8

(x 4) (x 2)

++

=A B

(x 4) (x 2)+

+ -------(ii) 1 vad

5x + 8 = A (x 2) + B (x + 4)

x ds xq.kkad a dh rqYkuk djus ij 5 = A + B --------(iii) 1 vad

vpj in a dh rqYkuk djus ij 8 = 2A + 4B -----------(iv)

(iii) o (iv) d gYk djus ij A = 2 , B = 3 1 vadvc lehdj.k (i) ls

2

5x 8

x 2x 8

++

=2 3

x 4 x 2+

+ lehdj.k (ii) ls

2

2

x 7x

x 2x 8

++

= 1 +2 3

x 4 x 2+

+ 1 vad


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mkj&7

ck;k i{k Yus ij&

sin14

5+ sin1

5

13+ sin1

16

65

=1 14 5sin sin

5 13

+ + sin1 16

65

= sin1

224 5 5 4

1 15 13 13 5

+ + sin1

16

65

1 vad

= sin

1

4 12 5 3

5 13 13 5

+ + sin

1

16

65

= sin163

65+ sin1

16

65 1 vad

= sin1

2 263 16 16 63

1 165 65 65 65

+

1 vad

=63 63 16 16

65 65 65 65 +

= sin14225

4225

= sin1 1 =2

nk;k i{k 1 vad


ck;k i{k sin1 x + sin1 1 x sin1 x = cos1 1 x 1 vad

sin1 x = cos1 1 x

cos1 1 x + sin1 1 x 1 vad

=2

= nk;k i{k[ cos1 x + sin1 x =

2

]

2 vad

mkj&8

x x

x xd e edx e e

+


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=

x x x x x x x x

x x 2

d d(e e ) (e e ) (e e ) (e e )

dx dx

(e e )

+ +

1 vad

=

x x x x x x x x

x x 2

(e e ) (e e ) (e e ) (e e )

(e e )

+ +

1 vad

2x 2x 2x 2x

x x 2

(e e 2) 2(e e 2)

(e e )

+ + +

= x x 24

(e e ) 2 vad


y = log1 cos mx

1 cos mx+

dy

dx=

d

dxlog

2

2

mx2sin

2mx

2 cos2

=

d

dx log tan

mx

2 1 vad

tanmx

2= t j[kus ij

dy

dx=

d

dxlog t

=d

dt. log t

dt

dx1 vad

= ddt

. log t . ddx

. tan mx2

=1

t.

m

2sec2

mx

2

=1

mxtan

2

m

2. sec2

mx

21 vad


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=2

mxcos

m 12. .mx mx2

sin cos2 2

= mmx mx2 sin .cos

2 2

= msin mx

1 vad

dy

dx= m cosec (mx)

mkj&9

y =1 x

1 x

+

y =

(1 x)

(1 x)

+log Yus ij log y = log (1 x) log (1 + x)

log y = log1

2(1 x)

1

2log (1 + x) 1 vad

x ds lkis{k vodYku djus ij&

1 dy

y dx=

1 1 1 1( 1) 1

2 (1 x) 2 (1 x)

+

= 1 1 12 1 x x 1

+ + 1 vad

= 1 1 x 1 x

2 (1 x) (1 x)

+ + +

1 dy

y dx= 2

1

1 x

1 vad

(1 x2)dy

dx = y

(1 x2)dy

dx+ y = 0 ;gh fl) djuk kk 1 vad


y = sin x sin x sin x .........+ +

y = sin x y+ 1 vad

nu a i{k a dk oxZ djus ij&


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y2 = sin x + y 1 vad

x ds lkis{k vodYku djus ij&

2ydy

dx= cos x

dy

dx1 vad

(2y 1)dydx

= cos x 1 vad

;gh fl) djuk kkA

mkj&10

{kQYk A = r2

dA

dr= 2 r 1 vad

fkT;k ifjorZu dh nj = drdt

= [email protected]

vc {kQYk dh njdA

dt=

dA dr.

dr dt1 vad

= 2 r (0.2) 1 vad r = 10lseh

{kQYk ifjorZu dh nj = 2 . 10 (0.2)= 4 [email protected] 1 vad

'u&10 dk vkok dk mkj

f (x) = x3 6x2 + 11 x 6

(i) f (x), varjkYk [1, 3] esa lrr~ gSA 1 vad

(ii) f' (x) = 3x2 12x + 11 ftldk vfLrRox [1, 3] ds lh eku a ds fYk, gSA1 vad

blfYk, f (x), varjkYk (1, 3) ds fYk, vodYkuh; gSA

(iii) f (1) = 0 rkk f (3) = 0

f (1) = f (x)vr% jY es; dh lh fLkfr;k larq"V grh gSaA 1 vad

vr% c (1, 3) dk vfLrRo bl dkj gxk fd&f' (c) = 0 3c2 12 c + 11 = 0

c =12 144 132

6

c =

1

2 3


17/31

(17)

G

c ds nu a eku varjkYk (1, 3) esa fLkr gSA 1 vadvr% jY es; fl) gqbZA

mkj&11x y u v u2 v2 u.v

5 12 8 13 64 169 1049 20 4 5 16 25 20

13 25 0 0 0 0 0

17 33 4 8 16 64 32

21 25 8 10 64 100 80

65 125 u = 0 v=0 u2=160 v2=358 u.v=236

2 vad

u = 0 , v = 0, u2 = 160, v2 = 358, u . v = 236

lwk =2 2 2 2

n u.v u. v

[n u ( u) ][n v ( v) ]

=5 236 0 0

[5 160 0][5 358 0]

1 vad

=236 236

160 358 57280=

=236

239.33= 0.98 1 vad


cov (x . y) = 2.25 , var (x) = 6.25 , var (y) = 20.25

(x, y) = cov (x, y) 1 vad

(x, y) =cov (x,b)

var (x) . var (y)

=2.25

6.25 20.25

1 vad

=2.25

2.5 4.5

=225

25 45

1 vad

= 0.2

pwfd dk eku _.kkRed gS vr% lg&laca/k iw.kZ _.kkRed gSA 1 vadmkj&12

lekJ;.k xq.kkad byx = 1.6 1 vad


18/31

(18)

G

bxy = 0.4

vr% nu a js[kkv a dh o.krk m1 = 1.6 , m2 = 0.4

chp dk d.k gSA

tan =1 2

1 2

m m

1 m m

+ =

1.6 0.4

1 1.6 0.4

+ 2 vad

tan =1.2

1.64= 0.73

tan = 0.73 1 vad'u&12 dk vkok dk mkj&

2x y =

21 [x y) (x y)]n

=21 [x x) (y y)]

n 1 vad

=21

[x x) (y y) 2(x x) (y y)]n

+

= 2 21 1 2

[x x) (y y) (x x) (y y)]n n n

+

1 vad

2x y = x2 + y2 2 x y 1 vad

2 xy= x2 + y

2 2x y

;k =2 2 2

x y x y

x y2

+

1 vad

Yk?kq mkjh; 'u (Short Answer Type Question)

mkj&13

n(S) = 52C2 = 1326

eku Yk E = nu a i YkkYk gus dh ?kVuk

F = nu a i bs gus dh ?kVuk

rc E F = n YkkYk bs gus dh ?kVuk 1 vad

n (E) = 26C2= 325

n (F) = 4C2= 6 1 vad

rkk n (E F) =2

C2 = 1


19/31

(19)

G

P (E) =n(E) 325

n(S) 1326=

P (F) =n(F) 6 1

n(S) 1325 221= =

P (E F)=n (E F) 1

n(S) 1326

= 1 vad

P nu a YkkYk ;k bs gus ij= P (E) + P (F) P (E F)

=325 6 1

1326 1325 1326+ 1 vad

=

330 55

1326 221= 1 vad'u&13 dk vkok dk mkj&

;gk n (S) = 6 6 = 36ekuk A = nu a ikl a ij 9 vkus dh ?kVuk

= {(4, 5), (5, 4), (3, 6), (6, 3)} 1 vad

B = nu a ikl a ij 11 vkus dh ?kVuk

= [(5, 6) , (6, 5)]

P (A) = 436

P (B) =2

361 vad

pwfd nu a ?kVuk, viot gSaA

P (A B)= P (A) + P (B) 1 vad

=4 2

36 6+

=6 1

36 6

1 vad

ijarq ;x 9 o 11 u vkus dh kf;drk

= 1 P (A B)

= 1 1 5

6 6= 1 vad

mkj&14

fcUnq &1] 3] 2 ls xqtjus okY lerYk dk lehdj.k gS&

a (x+1) + b (y3) + c (z2) = 0----------1 1 vad;g fn, gq, lerYk a ij YkEc gxk&


20/31

(20)

G

a + 2b + 2c = 0

3a + 3b + 2c = 0 1 vad

a

4 6=

b c

6 2 3 6=

a2

=b b c4 4 3

= =

= k

a = 2k , b = 4k , c = 3k 1 vadeku 1 esa j[kus ij&

2k (x+1) 4k (y3) + 3k (z2) = 0 1 vad

2x 4y + 3z + 8 = 0 1 vad


ekuk lerYk dk lehdj.kx y z

a b c+ + = 1 ---------1

dsU ls bldh nwjh p =2 2 2

1

1 1 1

a b c

+ + 1 vad

21

p = 2 2 2

1 1 1

a b c+ + -------2 1 vadA, B, C ds funsZ'kkad e'k% (a, o, o)] (0, b, 0) rkk (0, 0, c) gS vr% prq"QYkd

OABC ds dsUd ds funsZ'kkad

x =a

4, y =

b

4, z =

c

41 vad

a = 4x , b = 4y , c = 4z;s eku lehdj.k 2 esa j[kus ij dsUd dk fcUnqik

21

p= 2 2 2

1 1 1

16x 16y 16z+ + 1 vad

;k 216

p= 2 2 2

1 1 1

x y z+ + 1 vad

mkj&15

x&v{k ds lkkOP rkkOQ e'k% o d.k cukrs gSaA OCuuur

o ODuuur

e'k%

OPuuur

o OQ

uuur

dh vj ,dkad lfn'k gSA rcC

oD

ds funsZ'kkad e'k%(cos A, sin

A) rkk (cos B, sin B) g asxsA


21/31

(21)

G

1 vad

OCuuur

= (cos A) i (sin A) j+$ $

ODuuur

= (cos B) i (sin B) j+$ $ 1 vad

OD OCuuur uuur

= (cos B) i (sin B) j cos A) i (sin A) i + $ $ $ $

1 vad

|OD| |OC| sin COD esa = (cos B sin A) i j (sin B cos A) j i $ $ $ $

;k 1.1 sin (A + B) esa 1 vad

=

(sin A cos B) n (cos A sin B) ( n) sin (A + B) = sin A cos B + cos A sin B 1 vad


ck;k i{k ( ) ( ) ( )a b c b c a c a b + + r r r r r r r r r

= ( ) ( ) ( ) ( ) ( ) ( )a.c b a.b c b.a c b.c a c.b a c.a b + + r r r r r r r r r r r r r r r r r r

2 vad

= ( ) ( ) ( ) ( ) ( ) ( )c.a b a.b c a.b c b.c a b.c a c.a b + + r r r r r r r r r r r r r r r r r r

2 vad

= 0 nk;k i{k

vfn'k xq.kuQYk e fofues; fu;e dk ikYku djrk gS a.b b.a=r r r r

1 vad

mkj&16

f (x) =

2

2

x 1

x 1

+


22/31

(22)

G

f1

x

=

2

2

11

x1

1x

+1 vad

f1

x =

2

2

1 x

1 x

+

= 2

2

x 1

x 1

+

= f(x) 2 vad

f1

x

= f(x) 2 vad


L.H.L.

x o

lim

f(x) =

h 0

lim

f (0 h) 1 vad

= 21 cos( h)

( h)

h 0lim 2

1 cos h

h

=

h 0lim

2

2

h2sin

2

h1 vad

= 2h 0lim

2h

sin 2h

2

1

41 vad

=2

4 1 =

1

2

R.H.L.

x 0

lim+

f (x) =h 0lim f (0 + h)

= 2h 0

1 cos hlim

h

=

2

2h 0

h2 sin

2limh

= 2 1

4 1 =

1

2

f (0) =1

2


23/31

(23)

G

x olim f(x) = f (0) =

x 0

lim+

+ f(x) 1 vad

vr% QYku x = 0 ij larr gS

mkj&17

ekuk tan x2= t

1

2sec2

x

2dx = dt 1 vad

;k dx =2

2dt

xsec

2

dx = 2

2dt

x1 tan

2+

= 22dt

1 t+

vc I =dx

5 4 sin x+ =( )2 2

2 dt

4 t1 t 5

1 t

+ + +

1 vad

2

x2tan

2

sin x x1 tan2

= +

Q

= 22 dt

5 5t 8t+ + = 22 dt

85t t 1

5+ +

=2 2

2 dt

5 4 3t

5 5

+ +

1 vad

= 1

4t

2 1 5tan3 35

5 5

+

=12 5t 4

tan3 3

+ 1 vad


24/31


25/31

(25)

G

vh"V { skQYk (OAB) =leYkac prqqZt BNMA dk {kQYk & [{kQYk BNO + {kQYk OMA]

1 vad

1 vad

= 21

1 11 3 y dx2 4

+

=2

2

1

15 x

8 4 dx

=

23

1

15 x

8 12

1 vad

= 15 18 12

(8 + 1

=15 3

8 4

=9

8oxZ bdkbZ 1 vad


I = dxa cos x bsin x+

Put a = r sin , = r cos

r = 2 2a b+ , = tan1a

b 1 vad

I =dx

r sin cos x r cos . sin x +

= dxr sin (x )+ 1 vad


26/31

(26)

G

=1

r cosec (x + ) dx 1 vad

=1

rlog

xtan

2

+ + c

=1

rlog

xtan

2 2

+ + c 1 vad

= 2 2

1

a b+log

1x 1 atan tan

2 2 b

+ + c

1 vad

mkj&19

x + y dydx

= 2y

dy

dx=

2y x

y

1 vad

ekuk y = vx dy

dx= v + x

dv

dx

v + xdv

dx

=2v 1

v

xdv

dx=

2v 1

v

v 1 vad

lekdYku djus ijdx

x = 2v dv

(v 1)

log x = log (v 1) +1

v 1+ c 1 vad

log x + log (v 1) = 1v 1

+ c

logy

x 1x

=

1

y1

x

+ c 1 vad

log (y x) =x

y x + c 1 vad


27/31

(27)

G


( 1 + x2)dy

dx+ 2 xy 4x2 = 0

;k 2dy 2x

dx 1 x+ + y =

2

2

4x

1 x+ 1 vad

P = 22x

1 x+Q =

2

2

4x

1 x+I.F. = ePdx

=

2xdx

21 xe +

= elog(1+x2)

= 1 + x2 1 vad

vh"V gYk gxk&y I.F. = (I.F.) dx + c

y (1 + x2) =

2

2

4x

1 x+ (1 + x2) dx + c

;k y (1 + x2) = 4x2 dx + c 1 vad

=24x

3

+ c

3y (1 + x2) = 4x3 + 3c 1 vad

nh?kZ mkjh; 'u (Long Anwer Question)mkj&20

2l + 2n m = 0 ----------1

ml + mn + nl = 0 ---------2

lehdj.k 1 ls m = 2 (l + n) 1 vad

-;g eku 2 esa j[kus ij

2 (l + n) n + nl + 2l (l + n) = 0

;k 2n2 + 2nl + nl + 2l2 + 2ln = 0;k 2l2 + 5nl + 2n2 = 0 1 vad

(2l + n) (2n + l) = 0

2l + n = 0 ------3 1 vad

;k l + 2n = 0 ----------4lehdj.k 3 ls 2l + 0 . m + n = 0

lehdj.k 1 ls 2l m + 2n = 0


28/31

(28)

G

otz xq.ku ls0 1+

l=

m

2 4=

n

2 0

;k1

l=

m

2=

n

2

1l

=m

2=

n

21 vad

ke js[kk ds fnd~ vuqikr &1] 2] 2lehdj.k 4 ls l + 0.m + 2.n = 0lehdj.k 1 ls 2l m + 2n = 0

0 2+l

=m

4 2=

n

1

;k2l = m

2= n

1 nwljh js[kk ds fnd~ vuqikr 2] 2] &1 ;fn nu a js[kkv a ds chp dk d.k

gS r 1 vad

cos =( 1) (2) (2) (2) 2 ( 1)

1 4 4 4 4 1

+ + + + + +

=2 4 2

9

+ = 0

= 90 1 vad'u&20 dk vkok dk mkj

js[kk,1

1

x xl

=1

1

y y

m

=

1

1

z z

n

rkk2

2

x xl

=2

2

y y

m

=

2

2

z z

n

lerYkh; g axh ;fn&

2 1 2 1 2 1

1 1 1

2 2 2

x x y y z zm n

m n

l

l= 0 1 vad

;gk (x1 y1 z1) = (0, 2, 3) , (x2 y2 z2) = (2, 6, 3)

l1 = 1 , m1 = +2, n1 = 3 , l2

= 2 , m2 = 3, n2 = 4

2 0 6 2 3 3

1 2 3

2 3 4

+

=

2 4 5

1 2 3

2 3 4

1 vad


29/31

(29)

G

2

1 2 3

1 2 3

2 3 4= 0 vr% js[kk, lerYkh; gSa 1 vad

ke js[kk ds fdlh fcanqds funsZ'kkad (r , 2r + 2, 3r 3) gSA ;fn og fcUnqf}rh;

js[kk ij h fLkr gS r&r 2

2

=

2r 2 6

3

+ =

3r 3 3

4

1 vad

gYk djus ij r = 2 1 vadvr% frPNsn fcUnq gxk (2, 2 2 + 2 , 3 2 3) vkkZr~ (2, 6, 3)

1 vad

mkj&21

fn;k gS rr

=

i 2 j 3k + +$ $ +

( )2i 3j 4k + +$ $ -------1vj r

r= 2i 3j 4k + +$ $ + ( )3i 4j 5k + +$ $ -------2

1ar

= 3i 2j 3k + +$ $ , b1 = 2i 3j 4k + +$ $ 1 vad

vj 2ar

= 2i 3j 4k + +$ $ , b2 =3i 4 j 5k + +$ $

vc 2 1a ar r

= 2i 3j 4k i 2j 3k + + $ $ $ $ 1 vad

=

i j k+ +$ $

vj 1 2b br r

=

i j k

2 3 4

3 4 5

$ $

1 vad

= i (15 16) j (12 10) k (8 9) + + $ $

= i 2 j k + $ $

1 2| b b |r r = 1 4 1 6+ + = 1 vad%

U;wure nwjh =( ) ( )2 1 1 2

1 2

a a . b b

| b b |

r r r r

r r

( ) ( )i j k . i 2 j k 6

+ $ $ $ $= 0 2 vad


30/31

(30)

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ekuk xksys dk lehdj.k

x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 ------1

xY dk dsU (u, v, w) 1 vad

js[kk 5y + 2z = 0 = 2x 3y ij fLkr gS&

5(v) + 2 (w) = 0 = 2 (u) 3 (v)

5v + 2w = 0 --------2

rkk 2u 3v = 0 ------3 1 vad

iqu% fcUnq (0, 2, 4) rkk (2, 1, 1) xY ij (1) ij gS

0 + 4 + 16 4v 8w + d = 0

20 4v 8w + d = 0 1 vadblh dkj 6 + 4u 2v 2w + d = 0 ---------4 1 vad

lehdj.k 2] 3] 4] 5 d gYk djus ij

u = 3 , v = 2 , w = 5, d = 12 1 vad

lehdj.k 1 esa ;s eku j[kus ij

x2 + y2 +z2 + 2 (3) x + 2 (2) y + 2.5.z + 12 = 0

vr% x2 + y2 + z2 6x 4y + 10z + 12 = 0 1 vad


31/31

'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

mathematics set 4, model papers of madhya pradesh board of secondary education, xiith class

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