CBSE 2014 Question Paper for Class 12 Mathematics - Outside Delhi
March 18, 2017 | Author: aglasem | Category: N/A
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Description
AglaSem Schools
H$moS> Z§.
Series OSR
Code No.
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65/1
narjmWu H$moS >H$mo CÎma-nwpñVH$m Ho$ _wI-n¥ð >na Adí` {bIo§ &
Roll No.
Candidates must write the Code on the title page of the answer-book.
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Please check that this question paper contains 11 printed pages. Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate. Please check that this question paper contains 29 questions. Please write down the Serial Number of the question before attempting it. 15 minutes time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will not write any answer on the answer-book during this period.
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H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _o§ _w{ÐV n¥ð> 11 h¢ & àíZ-nÌ _| Xm{hZo hmW H$s Amoa {XE JE H$moS >Zå~a H$mo N>mÌ CÎma -nwpñVH$m Ho$ _wI-n¥ð> na {bI| & H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _| >29 àíZ h¢ & H¥$n`m àíZ H$m CÎma {bIZm ewê$ H$aZo go nhbo, àíZ H$m H«$_m§H$ Adí` {bI| & Bg àíZ-nÌ H$mo n‹T>Zo Ho$ {bE 15 {_ZQ >H$m g_` {X`m J`m h¡ & àíZ-nÌ H$m {dVaU nydm©• _| 10.15 ~Oo {H$`m OmEJm & 10.15 ~Oo go 10.30 ~Oo VH$ N>mÌ Ho$db àíZ-nÌ H$mo n‹T>|Jo Am¡a Bg Ad{Y Ho$ Xm¡amZ do CÎma-nwpñVH$m na H$moB© CÎma Zht {bI|Jo &
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J{UV MATHEMATICS
{ZYm©[aV g_` : 3 KÊQ>o
A{YH$V_ A§H$ : 100
Time allowed : 3 hours 65/1
Maximum Marks : 100 1
P.T.O.
AglaSem Schools
gm_mÝ` {ZX}e : g^r àíZ A{Zdm`© h¢ &
(ii)
Bg àíZ nÌ _| 29 àíZ h¢ Omo VrZ IÊS>m| _| {d^m{OV h¢ : A, ~ VWm g & IÊS> A _| 10 àíZ h¢ {OZ_| go àË`oH$ EH$ A§H$ H$m h¡ & IÊS> ~ _| 12 àíZ h¢ {OZ_| go àË`oH$ Mma A§H$ H$m h¡ & IÊS> g _| 7 àíZ h¢ {OZ_| go àË`oH$ N>… A§H$ H$m h¡ &
(iii)
IÊS> A _| g^r àíZm| Ho$ CÎma EH$ eãX, EH$ dmŠ` AWdm àíZ H$s Amdí`H$Vm AZwgma {XE Om gH$Vo h¢ &
(iv)
nyU© àíZ nÌ _| {dH$ën Zht h¢ & {\$a ^r Mma A§H$m| dmbo 4 àíZm| _| VWm N>… A§H$m| dmbo 2 àíZm| _| AmÝV[aH$ {dH$ën h¡ & Eogo g^r àíZm| _| go AmnH$mo EH$ hr {dH$ën hb H$aZm h¡ &
(v)
H¡$bHw$boQ>a Ho$ à`moJ H$s AZw_{V Zht h¡ & `{X Amdí`H$ hmo Vmo Amn bKwJUH$s` gma{U`m± _m±J gH$Vo h¢ &
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(i)
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General Instructions :
All questions are compulsory.
(ii)
The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each, Section B
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(i)
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comprises of 12 questions of four marks each and Section C comprises of 7 questions of six marks each. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(iv)
There is no overall choice. However, internal choice has been provided in
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(iii)
4 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions. (v)
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Use of calculators is not permitted. You may ask for logarithmic tables, if required. 2
AglaSem Schools
IÊS> A SECTION A
àíZ g§»`m 1 go 10 VH$ àË`oH$ àíZ 1 A§H$ H$m h¡ & Question numbers 1 to 10 carry 1 mark each. 1.
`{X N na
R = {(x, y) : x + 2y = 8}
EH$ g§~§Y h¡, Vmo
R
H$m n[aga {b{IE &
tan–1 x + tan–1 y =
If tan–1 x + tan–1 y =
h¡, Vmo
x + y + xy
H$m _mZ {b{IE &
, xy < 1, then write the value of x + y + xy. 4
A2 = A
`{X A EH$ Eogm dJ© Amì`yh h¡ {H$ Ohm± I EH$ VËg_H$ Amì`yh h¡ &
h¡, Vmo
7A – (I + A)3
H$m _mZ {b{IE,
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3.
, xy < 1 4
as
`{X
gl
2.
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If R = {(x, y) : x + 2y = 8} is a relation on N, write the range of R.
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If A is a square matrix such that A2 = A, then write the value of
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x–y If 2x – y
z w
3x 5.
`{X
=
z – 1 = 0 w
7
8
–2
4
6
7
8
4
–2
4
h¡, Vmo
x+y
H$m _mZ kmV H$s{OE &
4 , find the value of x + y. 5
h¡, Vmo
x
H$m _mZ kmV H$s{OE &
7
If
4 5
7
3x
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– 1 0
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x–y 2x – y
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`{X
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7A – (I + A)3, where I is an identity matrix.
, find the value of x.
6
4 3
P.T.O.
AglaSem Schools x
6.
`{X
f (x )
t sin t dt
h¡, Vmo
f(x)
H$m _mZ kmV H$s{OE &
0
x
If f (x)
t sin t dt , then write the value of f(x).
0
_mZ kmV H$s{OE 4
2
x 2
x 1
:
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7.
dx
Evaluate :
2
dx
‘p’
^ ^ ^ 3i + 2 j + 9k
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H$m dh _mZ kmV H$s{OE {OgHo$ {bE g{Xe g_m§Va h¢ &
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8.
x 1
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x 2
as
4
^ ^ ^ ^ ^ ^ a = 2i + j + 3k, b = – i + 2 j + k a . ( b c ) kmV H$s{OE & ( b c ), if ^ ^ j + 2k.
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Find a . ^ c =3i + 10.
^ ^ ^ i – 2p j + 3 k
^ ^ ^ 3 i + 2 j + 9k
and
^ ^ ^ c = 3i + j + 2k
h¢, Vmo
^ ^ ^ ^ ^ ^ a = 2i + j + 3k, b = – i + 2 j + k
and
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Find the value of ‘p’ for which the vectors ^ ^ ^ i – 2p j + 3 k are parallel.
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`{X EH$ aoIm Ho$ H$mVu` g_rH$aU
VWm
3–x y4 2z – 6 5 7 4
h¢, Vmo Cg aoIm H$m g{Xe
g_rH$aU {b{IE & If the cartesian equations of a line are vector equation for the line. 65/1
4
3–x y4 2z – 6 , write the 5 7 4
AglaSem Schools
IÊS> ~ SECTION B
àíZ g§»`m 11 go 22 VH$ àË`oH$ àíZ 4 A§H$ H$m h¡ & Question numbers 11 to 22 carry 4 marks each. 11.
`{X \$bZ
f : R R, f(x) = x2 + 2
àXÎm h¢, Vmo
fog VWm gof
VWm
x , x1 x –1 gof (– 3) kmV H$s{OE
g : R R, g(x) =
kmV H$s{OE Am¡a AV…
fog (2) VWm
Ûmam &
{gÕ H$s{OE {H$ 1– x 1 cos–1 x, – = 4 2 1 – x
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x 2 = 4 x 4
h¡, Vmo x H$m _mZ kmV H$s{OE &
Prove that
1– x –1 1 cos–1 x, x1 – = 4 2 2 1 – x
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1x – tan–1 1 x
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x – 2 + tan1 tan1 x – 4
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AWdm `{X
–1 x1 2
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1x – tan–1 1 x
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12.
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If the function f : R R be given by f(x) = x2 + 2 and g : R R be x given by g(x) = , x 1, find fog and gof and hence find fog (2) x –1 and gof (– 3).
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OR
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x – 2 + tan1 If tan1 x – 4 13.
65/1
x 2 = , find the value of x. 4 x 4
gma{UH$m| Ho$ JwUY_m] H$m à`moJ H$aHo$, {gÕ H$s{OE {H$ xy
x
x
5x 4 y
4x
2x
10x 8y
8x
3x
x3
5
P.T.O.
AglaSem Schools Using properties of determinants, prove that
14.
`{X
xy
x
x
5x 4 y
4x
2x
10x 8y
8x
3x
x = ae (sin – cos )
x3
y = ae (sin + cos )
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Find the value of
dx 2
dy + aby = 0. dx
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dh _mZ kmV H$s{OE {OgHo$ {bE
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x Ho$
dy + aby = 0. dx
x2
a2
b2
–
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dH«$
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16.
– (a + b)
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dx
2
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If y = P eax + Q ebx, show that
d 2y
as
– (a + b)
h¡, Vmo Xem©BE {H$
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y = P eax + Q ebx
d 2y
dy dx
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`{X
na
dy at = , if x = ae (sin – cos ) and dx 4
y = ae (sin + cos ). 15.
4
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H$m _mZ kmV H$s{OE &
h¡, Vmo =
y2
1
y = [x (x – 2)]2
EH$ dY©_mZ \$bZ h¡ &
AWdm Ho$ {~ÝXþ
( 2 a, b)
na ñne© aoIm VWm A{^b§~ Ho$ g_rH$aU kmV
H$s{OE &
Find the value(s) of x for which y = [x (x – 2)]2 is an increasing function. OR Find the equations of the tangent and normal to the curve at the point ( 2 a, b). 65/1
6
x2 a2
–
y2 b2
1
AglaSem Schools 17.
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:
0
4 x sin x 1 cos 2 x
dx
AWdm
:
x2
dx
x 2 5x 6
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_mZ kmV H$s{OE
Evaluate :
dx
as
0
1 cos 2 x
gl
4 x sin x
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OR
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dx
x 2 5x 6
AdH$b g_rH$aU
dy = 1 + x + y + xy dx y = 0 h¡ &
H$m {d{eï> hb kmV H$s{OE, {X`m J`m h¡ {H$
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18.
x2
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Evaluate :
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O~ x = 1 h¡, Vmo
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Find the particular solution of the differential dy = 1 + x + y + xy, given that y = 0 when x = 1. dx 19.
AdH$b g_rH$aU
(1 + x2)
–1 dy + y = e tan x dx
Solve the differential equation (1 + x2)
65/1
7
equation
H$mo hb H$s{OE &
–1 dy + y = e tan x . dx
P.T.O.
AglaSem Schools 20.
A, B, C VWm D, {OZHo$ pñW{V ^ ^ ^ ^ ^ ^ ^ ^ – j – k , 3 i + 9 j + 4 k VWm 4 (– i + j + k ) h¢,
Xem©BE {H$ Mma {~ÝXþ
g{Xe H«$_e… g_Vbr` h¢ &
^ ^ ^ 4i + 5j + k,
AWdm ^ ^ ^ H$m a = i + j + k ^ ^ ^ c = i + 2 j + 3 k Ho$ `moJ\$b
g{Xe
JwUZ\$b
1
Ho$ ~am~a h¡ &
^ ^ ^ b = 2i + 4 j – 5k
g{Xem|
VWm
H$s {Xem _| EH$ _mÌH$ g{Xe Ho$ gmW A{Xe
H$m _mZ kmV H$s{OE Am¡a AV…
H$s {Xem _| EH$
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_mÌH$ g{Xe kmV H$s{OE &
b + c
Show that the four points A, B, C and D with position vectors ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 4 i + 5 j + k , – j – k , 3 i + 9 j + 4 k and 4 (– i + j + k ) respectively are coplanar.
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OR
21.
EH$ aoIm {~ÝXþ
(2, –1, 3) go
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^ ^ ^ The scalar product of the vector a = i + j + k with a unit vector along ^ ^ ^ ^ ^ ^ the sum of vectors b = 2 i + 4 j – 5 k and c = i + 2 j + 3 k is equal to one. Find the value of and hence find the unit vector along b + c .
hmoH$a OmVr h¡ VWm aoImAm|
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^ ^ ^ ^ ^ ^ r = ( i + j – k ) + (2 i – 2 j + k ) VWm ^ ^ ^ ^ ^ ^ r = (2 i – j – 3 k ) + ( i + 2 j + 2 k ) na
b§~dV² h¡ & CgH$m g_rH$aU, g{Xe
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VWm H$mVu` ê$n _| kmV H$s{OE &
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A line passes through (2, –1, 3) and is perpendicular to the lines ^ ^ ^ ^ ^ ^ r = ( i + j – k ) + (2 i – 2 j + k ) and ^ ^ ^ ^ ^ ^ r = (2 i – j – 3 k ) + ( i + 2 j + 2 k ). Obtain its equation in vector and cartesian form.
22.
EH$ à`moJ Ho$ g\$b hmoZo H$m g§`moJ CgHo$ Ag\$b hmoZo go VrZ JwZm h¡ & àm{`H$Vm kmV H$s{OE {H$ AJbo nm±M narjUm| _| go H$_-go-H$_ 3 g\$b hm|Jo & An experiment succeeds thrice as often as it fails. Find the probability that in the next five trials, there will be at least 3 successes.
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AglaSem Schools
IÊS> g SECTION C
àíZ g§»`m 23 go 29 VH$ àË`oH$ àíZ Ho$ 6 A§H$ h¢ & Question numbers 23 to 29 carry 6 marks each.
Xmo {dÚmb` A VWm B AnZo MwZo hþE {dÚm{W©`m| H$mo {ZîH$nQ>Vm, gË`dm{XVm VWm ghm`H$Vm Ho$ _yë`m| na nwañH$ma XoZm MmhVo h¢ & {dÚmb` A AnZo H«$_e… 3, 2 VWm 1 {dÚm{W©`m| H$mo BZ VrZ _yë`m| Ho$ {bE àË`oH$ H$mo H«$_e… < x, < y VWm < z XoZm MmhVm h¡ O~{H$ BZ nwañH$mam| H$m Hw$b _yë` < 1,600 h¡ & {dÚmb` B AnZo H«$_e… 4, 1 VWm 3 {dÚm{W©`m| H$mo BZ _yë`m| Ho$ {bE Hw$b < 2,300 nwañH$ma ñdê$n XoZm MmhVm h¡ (VWm nhbo {dÚmb` O¡go hr VrZ _yë`m| na dhr nwañH$ma am{e XoZm MmhVm h¡) & `{X BZ VrZm| _yë`m| na {XE JE EH$-EH$ nwañH$ma H$s Hw$b am{e < 900 h¡, Vmo Amì`yhm| H$m à`moJ H$aHo$ àË`oH$ _yë` Ho$ {bE Xr JB© nwañH$ma am{e kmV H$s{OE & Cn`w©º$ VrZ _yë`m| Ho$ A{V[aº$ EH$ AÝ` _yë` gwPmBE, Omo nwañH$ma XoZo Ho$ {bE em{_b H$aZm Mm{hE &
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23.
Xem©BE {H$ EH$ r {ÌÁ`m Ho$ Jmobo Ho$ AÝVJ©V A{YH$V_ Am`VZ dmbo b§~-d¥Îmr` e§Hw$ H$s 4r 3
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Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award < x each, < y each and < z each for the three respective values to 3, 2 and 1 students respectively with a total award money of < 1,600. School B wants to spend < 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is < 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
D±$MmB© 8 27
h¡ & `h ^r Xem©BE {H$ Bg e§Hw$ H$m A{YH$V_ Am`VZ Jmobo Ho$ Am`VZ H$m
hmoVm h¡ &
Show that the altitude of the right circular cone of maximum volume that 4r can be inscribed in a sphere of radius r is . Also show that the 3 8 maximum volume of the cone is of the volume of the sphere. 27 65/1
9
P.T.O.
AglaSem Schools 25.
_mZ kmV H$s{OE
:
1 4
cos x sin4 x
dx
Evaluate :
26.
cos 4 x sin4 x
dx
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g_mH$bZ H$m à`moJ H$aHo$ EH$ Eogo {ÌH$moUr` joÌ H$m joÌ\$b kmV H$s{OE Omo erfm] (– 1, 2), (1, 5) VWm (3, 4) Ûmam {Kam h¡ &
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g_Vbm| x + y + z = 1 VWm 2x + 3y + 4z = 5 H$s à{VÀN>oXZ aoIm H$mo AÝV{d©ï> H$aZo dmbo VWm g_Vb x – y + z = 0 Ho$ b§~dV² g_Vb H$m g_rH$aU kmV H$s{OE & Cn`w©º$ kmV {H$E JE g_Vb H$s _yb-{~ÝXþ go Xÿar ^r kmV H$s{OE &
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Using integration, find the area of the region bounded by the triangle whose vertices are (– 1, 2), (1, 5) and (3, 4).
^ k) = 0
^ + 2k)
VWm g_Vb
Ho$ à{VÀN>oXZ {~ÝXþ H$s {~ÝXþ
(2, 12, 5)
go Xÿar kmV
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H$s{OE &
+
+ 4 ^j
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^ ^ r . (i – 2j
ch
^ ^ ^ ^ r = 2 i – 4 j + 2 k + (3 i
.s
aoIm
oo
AWdm
Find the equation of the plane through the line of intersection of the
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planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. Also find the distance of the plane obtained above, from the origin. OR Find the distance of the point (2, 12, 5) from the point of intersection of ^ ^ ^ ^ ^ ^ the line r = 2 i – 4 j + 2 k + (3 i + 4 j + 2 k ) and the plane ^ ^ ^ r . ( i – 2 j + k ) = 0. 65/1
10
AglaSem Schools 28.
EH$ {Z_m©UH$Vm© H§$nZr H$jm XII Ho$ {bE J{UV _| ghm`H$ {ejU gm_J«r Ho$ Xmo Z_yZo A VWm B ~ZmVr h¡ & Z_yZo A H$m àË`oH$ ZJ ~ZmZo Ho$ {bE 9 l_ K§Q>o Am¡a 1 l_ K§Q>m nm°{be H$aZo _| bJmVr h¡, O~{H$ Z_yZo B Ho$ àË`oH$ ZJ ~ZmZo _| 12 l_ K§Q>o VWm nm°{be H$aZo _| 3 l_ K§Q>o bJmVr h¡ & ~ZmZo VWm nm°{be H$aZo Ho$ {bE à{V gámh CnbãY A{YH$V_ l_ K§Q>o H«$_e… 180 VWm 30 h¢ & H§$nZr Z_yZo A Ho$ àË`oH$ ZJ na < 80 VWm Z_yZo B Ho$ àË`oH$ ZJ na < 120 H$_mVr h¡ & Z_yZo A VWm Z_yZo B Ho$ {H$VZo-{H$VZo ZJm| H$m {Z_m©U à{V gámh {H$`m OmE {H$ A{YH$V_ bm^ hmo ? Bg àíZ H$mo a¡{IH$ àmoJ«m_Z g_ñ`m ~ZmH$a J«m\$ Ûmam hb H$s{OE & à{V gámh A{YH$V_ bm^ Š`m h¡ ?
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VrZ {g¸o$ h¢ & EH$ {g¸o$ Ho$ XmoZm| Amoa {MV hr h¡, Xÿgam {g¸$m A{^ZV h¡ {Og_| {MV 75% ~ma àH$Q> hmoVm h¡ VWm Vrgam {g¸$m ^r A{^ZV h¡ {Og_| nQ>> 40% ~ma àH$Q> hmoVm h¡ & VrZ {g¸$m| _| go `mÑÀN>`m EH$ {g¸$m MwZH$a CN>mbm J`m & `{X {g¸o$ na {MV àH$Q> hmo, Vmo Š`m àm{`H$Vm h¡ {H$ dh XmoZm| Amoa {MV dmbm {g¸$m h¡ ? AWdm àW_ N>:$ YZ nyUmªH$m| _| go Xmo g§»`mE± `mÑÀN>`m ({~Zm à{VñWmnZ) MwZr JBª & _mZm X XmoZm| g§»`mAm| _| go ~‹S>r g§»`m ì`º$ H$aVm h¡ & `mÑpÀN>H$ Ma X H$m àm{`H$Vm ~§Q>Z kmV H$s{OE VWm Bg ~§Q>Z H$m _mÜ` ^r kmV H$s{OE &
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29.
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A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of < 80 on each piece of type A and < 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit ? Make it as an LPP and solve graphically. What is the maximum profit per week ?
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There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin ? OR Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X, and hence find the mean of the distribution. 65/1
11
P.T.O.
AglaSem Schools
QUESTION PAPER CODE 65/1 EXPECTED ANSWERS/VALUE POINTS SECTION - A Marks
1 3
4. 3
6. x sin x
7.
9. – 10
10. r = (3ˆi − 4ˆj + 3kˆ) + λ (–5ˆi + 7ˆj + 2kˆ)
1 1 17 (log 17 – log 5) or log 2 2 5
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8. p = −
3. – I
SECTION - B 2
as
x x getting fog (x) = f = +2 x –1 x –1
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11.
)
x2 + 2 x2 +1
½m 1½ m
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(
2 getting g of (x) = g x + 2 =
1½ m
.a
fog(2) = 6
1×10 = 10 m
.c
5. – 2
2. 1
em
1-10. 1. {1, 2, 3}
ch
11 10
½m
Putting x = cos θ in LHS, We get
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12.
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.s
g of (– 3) =
1m
2 cos θ – 2 sin θ 2 2 = tan–1 θ θ 2 cos + 2 sin 2 2
1m
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1 + cos θ – 1 – cos θ LHS = tan–1 1 + cos θ + 1 – cos θ
1 – tan θ 2 = tan −1 tan π − θ = tan–1 2 1 + tan θ 4 2
2 PDF created with pdfFactory trial version www.pdffactory.com
½+1 m
AglaSem Schools
=
π 1 π 1 − θ = − cos −1 x = R.H.S 4 2 4 2
½m
OR Given equation can be written as
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om ½m
.a
gl
x–2 1 = x–4 x+3
x 2 + x – 6 = x – 4 or x 2 = 2 ∴ x = ± 2
½+1 m
Operating R 2 → R 2 − 4R 1 and R 3 → R 3 − 8R 1 , we get
ch
13.
1+½ m
oo
⇒
2 2x + 6
ls
∴
x+2 x+4 –1 x + 2 = tan x + 4
½m
as
= tan–1
1– 1 +
x + 2 x + 4
em
x – 2 = tan–1 1 – tan–1 tan–1 x – 4
x 2x
x
0 − 2x 0 − 5x
2m
w
w
w
LHS =
.s
x+y x
Expanding along C2, we get – x (– 5x 2 + 4x 2 ) = x 3
14.
1+1 m
dx = a e θ (sin θ – cos θ) + a e θ (cos θ + sin θ) dθ
= 2 a eθ sin θ
1m ½m
dy = a e θ (sin θ + cos θ) + a e θ (cos θ – sin θ) = 2 a e θ cos θ dθ
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1m
AglaSem Schools
1m
dy dx
½m
at θ = π
= cot π 4
4
y = P e ax + Q e bx ⇒
= 1
dy = a P e ax + b Q e bx dx
1m
om
15.
dy 2 a e θ cos θ = = cot θ dx 2 a e θ sin θ
em as
d2y dy ∴ LHS = 2 – (a + b) + aby dx dx
{
}
{
{
.a
}
gl
= a 2 P e ax + b 2 Q e bx – (a + b) a P e ax + b Q e bx + ab P e ax + Q e bx
{
1m
.c
d2y = a 2 P e ax + b 2 Q e bx 2 dx
}
1m 1m
oo
ls
= P e ax a 2 – a 2 – ab + ab + Q e bx b 2 – ab – b 2 + ab
}
[
y = [x (x – 2) ] = x 2 – 2x 2
.s
16.
ch
= 0 + 0 = 0. = R.H.S.
]
2
∴
(
)
dy = 2 x 2 – 2x (2x – 2 ) dx
w
dy = 4 x (x – 1) (x – 2 ) dx
1m
w
w
⇒
dy = 0 dx ∴
x = 0, x = 1, x = 2
½m
Intervals are (– ∞ , 0), (0, 1), (1, 2), (2, ∞ )
½m
since ∴
⇒
1m
dy > 0 in (0, 1) or (2, ∞ ) dx
f(x) is increasing in (0, 1) U (2, ∞ ) OR
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1m
AglaSem Schools
x2 y2 2x 2y dy dy b2 x – 2 =1 ⇒ 2 – 2 = 0 ⇒ = 2 a2 b a b dx dx a y
slope of tangent at
(
2a , b =
)
slope of normal at
(
2a , b = –
2b a
)
½m
a 2b
½m
4x sin x dx 2 x
ls
π
∫
ch
0
π
2I = 4 π
sin x
∫ 1 + cos
.s
∴
2
x
dx
w
w
Put cos x = t ∴ sin x dx = – dt
∴
I = 2π
1
∫ 1
– dt 1+ t2
[
]
= 2 π tan –1 t
½m
or 2 π
1m
½m
w
0
–1
½m
π
4 (π – x) sin (π – x) 4 (π – x) sin x dx = ∫ dx 2 2 1 + cos (π – x) 1 + cos x 0
oo
x → (π – x) gives I =
½m
.a
∫ 1 + cos 0
as
π
I =
Let
2 (a2 + b2)
2 by =
½m
gl
i.e. ax +
)
em
(
a x– 2a 2b
.c
2 bx – ay = ab
and equation of normal is y – b = –
17.
)
om
(
2b x– 2a a
Equation of tangent is y – b = i.e.
1m
dt
∫ 1+ t
2
½m
1m
–1
1 –1
π π = 2 π – – = π 2 4 4
1m
OR I=∫
x +2 x 2 + 5x + 6
dx =
∫
1 1 (2x + 5) – 2 2 dx 2 x + 5x + 6 5
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1m
AglaSem Schools
=
=
∫
2x + 5 x + 5x + 6 2
dx
∫
(
x+ 5
1 log x + 2
x 2 + 5x + 6 –
)
1 – 2 2 2
2
½+½ m
5 2 + x + 5x + 6 + c 2
1+1 m
3 2
ls
c = –
log 1 + y = x +
oo
solution is
½m
1m
½+1 m
½m
x2 3 – 2 2
½m
ch
∴
x2 +c 2
⇒
x = 1, y = 0
em
log 1 + y = x +
.c
∫ (1 + x) dx
as
=
gl
dy
∫ 1+ y
om
dy = 1 + x + y + xy = (1 + x) (1+y) dx
∴
Given differential equation can be written as
.s
19.
1 2
dx –
.a
18.
1 2
–1 dy 1 1 + ⋅y = ⋅ e tan x 2 2 dx 1 + x 1+ x
w
w
1m
1
w
Integrating factor = e
∴ solution is, y ⋅ e tan
⇒
y ⋅ e tan
or
–1
x
y=
–1
x
=
=
∫ 1+ x 2 dx 1
∫ 1+ x
2
= e tan e 2 tan
–1
–1
x
x
1m dx
1 2 tan –1x e +c 2
1m
1m
–1 1 tan –1x e + c e – tan x 2
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AglaSem Schools
20.
A, B, C, D are coplaner, if AB ⋅ AC × AD = 0
1m
AB = – 4ˆi – 6ˆj – 2kˆ, AC = − ˆi + 4ˆj + 3kˆ, AD = − 8ˆi – ˆj + 3kˆ
1½ m
–4 –6 –2 –1 –8
4 –1
3 3
½m
om
AB ⋅ AC × AD =
= – 4 (15) + 6 ( 21) – 2 (33) = 0
em
b+c
=1
b+c
.a
gl
a ⋅ b + a ⋅c = b + c
or
oo
ls
(ˆi + ˆj + kˆ )⋅ (2ˆi + 4ˆj – 5kˆ ) + (ˆi + ˆj + kˆ )⋅ (λˆi + 2ˆj + 3kˆ ) = (2 + 4 – 5) + (λ + 2 + 3) =
∴
(λ + 6 )2
= (λ + 2 ) + 40
(λ + 2)ˆi + 6ˆj – 2kˆ
(λ + 2)2 + 36 + 4
⇒ λ =1
½m 1m ½m
=
3ˆi + 6ˆj – 2kˆ 7
or
3ˆ 6ˆ 2 ˆ i+ j– k 7 7 7
1m
w
w
b+c
½m
.s
2
w
b+c
ch
⇒
Hence
½m
as
Given that a ⋅
.c
OR
1m
21.
The direction perpendicular to the given lines is given by
(2ˆi − 2ˆj + kˆ ) × (ˆi + 2ˆj + 2kˆ ) ˆi ˆj kˆ = 2 –2 1 1 2 2
1m
= – 6ˆi − 3ˆj + 6kˆ or 2ˆi + ˆj – 2kˆ
∴ Vector equation of required line is
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1m
AglaSem Schools
(
)
(
r = 2ˆi − ˆj + 3kˆ + λ 2ˆi + ˆj − 2kˆ
)
1m
and the cartesian form is x –2 y +1 z–3 = = 2 1 –2
∴
p = 3 q, and p + q = 1
∴
p=
and q =
1 4
.c
3 4
om
Let probability of success be p and that of failure be q
P (atleast 3 successes) = P(r > 3) = P(3) + P(4) + P(5) 1
4
1 3 1 3 3 C3 ⋅ + 5 C 4 ⋅ + 5 C5 4 4 4 4 4
5
1m ½m
1½ m
10.27 5.81 243 918 459 + + = or 1024 1024 1024 1024 512
1m
oo
ls
=
.a
gl
=
3
as
2
5
em
22.
1m
4x + x +
w
Here
w
w
23.
+ 2y +
.s
3x
ch
SECTION - C
∴
3 2 1 4 1 3 1 1 1
y y
z
= 1600
+ 3z = 2300 + z = 900
1½
x 1600 y = 2300 or AX = B z 900
| A | = 3(–2) –2 (1) + 1 (3) = – 5 ≠ 0 ∴ X = A –1 B
½m
Cofactors are : A11 = – 2, A12 = – 1, A13 = 3 A 21 = – 1, A 22 = 2, A 23 = –1 A 31 = 5, A 32 = −5, A 33 = –5
1½ m
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AglaSem Schools
– 2 − 1 5 1600 x 1 ∴ y = – – 1 2 − 5 2300 5 z 3 – 1 – 5 900 ∴ x = 200, y = 300, z = 400
1½ m
i.e. Rs 200 for sincerity, Rs 300 for truthfulness and
om
Rs 400 for helpfulness
For correct fiqure
em
24.
.c
One more value like, honesty, kindness etc.
1m
½m
as
let radius of cone be x and its height be h.
gl
∴ OD = (h – r)
ls
.a
Volume of cone (v) =
½m
1 π x2h .............(i) 3
½m
oo
In ∆ OCD, x2 + (h – r) = r2 or x2 = r2 – (h – r)2 V=
ch
∴
2
1 1 2 3 π h {r2 – (h –r) } = π (– h + 2 h2r) 3 3
dv π = (– 3h2 + 4 hr) dn 3
1m
.s w w w
∴
dv = 0 dh
⇒ h=
4r 3
π 4π r 4r d2v π 0 y>0 For correct graph :
2m
2m
om
Vertices of feasible region are
.c
A (0, 10), B (12, 6), C (20, 0)
em
P(A) = 1200, P(B) = 1680, P(C) = 1600 ∴
For Max. P, No. of type A = 12 1m
as
No. of type B = 6
.a
Let event E1 : choosing first (two headed) coin
ls
29.
gl
Maximum Profit = Rs. 1680
oo
E2 : choosing 2nd (biased) coin
P(E1) = P(E2) = P(E3) =
1 3
½m
1m
.s
∴
ch
E3 : choosing 3rd (biased) coin
½m
P(A/E1) = 1, P(A/E2) =
w
∴
w
w
A : The coin showing heads. 75 3 60 3 = , P(A/E3) = = 100 4 100 5
1 ⋅1 3 P(E1/A) = 1 1 3 1 3 ⋅1 + ⋅ + ⋅ 3 3 4 3 5
=
1½ m
1+1m
20 47
1m OR
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AglaSem Schools
Total number of ways of selecting two numbers = 6 C 2 = 15
½m
Values of x (larger of the two) can be 2, 3, 4, 5, 6
1m
P(x = 2) =
1 2 3 , P(x = 3) = , P(x = 4) = 15 15 15
4 P(x = 5) = 15
.c
6 5 15 30 15
70 14 = 15 3
w
w
w
.s
ch
oo
ls
.a
Mean = ∑ x P(x) =
5 4 15 20 15
em
2 3 4 1 2 3 P(x) : 15 15 15 2 6 12 x P(x) : 15 15 15
as
x:
om
Distribution can be written as
gl
∴
2½
5 and P(x = 6) = 15
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1m
1m
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