Govt. of NCT of Delhi: Directorate of EducationExamination Cell. Room No. 222-A

Old Secretariat. Delhi-110054

No. DE. 5 (31)/ Pt-IV/Exam/ 2016/60'\- (, I0 Date: o~hl/.(.II CIRCULAR II

Sub: The Question Paper Design in the Subject of Mathematics(Class XII) for CBSE Exam-2017

Please find enclosed herewith a copy of CSSE/ACAD/DD(SC)/2016Circular No. Acad-32/2016 dated 26th July. 2016 on the subject cited above. Thecircular highlights the changes in the question paper design of Mathematics (classXII) which will be applicable for the academic session 2016-17.

In its circular, CSSE has provided question-wise break up, difficultylevel as well as choice(s) in the question paper and also elucidated that there willbe no overall choice in the paper. However, 30% internal choices will be given in 4marks and 6 marks questions.

Circular is very important for all the class XII Maths (041) students.Therefore, all Heads of Schools are directed to bring this to the notice of concernedteachers along with the Sample Question Paper enclosed herewith. Thisinformation may also be disseminated to the students concerned and their parents.

Encl: As above~\"\ ,\, -."-(DR. SUNIT A s. KAUSHIK)Addl. DE (Exam)

All Heads of Govt.lGovt. Aided/Unaided Recognized Schools through DEL E

Copy to:

1. PS to Secretary2. P.S. to Director3. All ROEs/DOEs (District & Zonal)/DEOs through Del E.4. ADE (IT) to get it placed on website of Directorate of Education.5. Guard file.

~

~ ~ 1".

(SHARDA NEJA)050 (Exam)

Em(lil: ~Clg(lndh.cbse@live.comWebsite: www.cbse(lc(ldemicnl(.in

Tel: 011-23220155011-23220083

~

~

CENTRAL BOARD OF SECONDARY EDUCATION

(Etl.uol I l/u Iodtl ell1y.1 HUjr I jdlj] dl v/tu . d lot. Rr I «Bu)f'I(~ I nul 171 bll NV;lluy (1)11JOnt ,0i;1 fnYyl&110002

(An Autonomous Orgonilotion under the Union Ministry of Human Resource Development, GOVI.of India)"Shiksha Sedan", 17, Institutional Area, Rouse Avenue, New Delhi-HOOOZ

CBSEIACAOJl)O(SC)120 16 26.07.2016Circular No. Acad-32/2016

All IIH' Heads uf Schools affltlnted to CHSE

Subject: The Question Paper Design in the subject of Mathematics (Class XII) for the BoardExamination 2017

Dear Principal

The following will be applicable in the subject Mathematics (041) for class XII for the academicsession 2016-17 and Board examination 2017:

I. Question Paper Design

S, Typology of VSA SA LA-I LA-II Marks wcightageNo, Questions (I Mark) (2 Marks) (4 Marks) (6 Marks)I Remembering 2 2 2 I 20 20%2 Understanding I 3 4 2 35 35%J Application I - 3 2 25 25%4 HOTS - 3 I - 10 10%5 Evaluation I (VBQ) - I 10 10%- -

Total Ix4-4 2x8-16 4xll-44 6x6-36 100 100%Value based question of 04 marks

2. Quustion-wlsc break-up

Type of Marks per Total no. of Total MarksQuestion Question QuestionsVSA I 4 4SA 2 8 16

LA-I 4 II 44LA-II 6 6 36Total 29 100

3. Difficulty levelDifflcultv level welahtaseEasy 20%Averaae 60%Difficult 20%

4. Choicc(s):There will be no overall choice in the question paper. However, 30% internal choices will be given in4 marks and 6 marks questions.

,, .It is requested that the above information be brought to the notice of concerned teachers and studentsalcngwith the Sample Question Paper enclosed with the Circular. 0') C---"'

For <lily query or clarification, you may also contact Sh. Subhnsh Chand, Deputy Director,at SCh'IIlU.cbsc((llgmai l.colll

With warm regards

Sugandh SharmaAdditional Director and In-charge (Research and Innovation)

Copy with a request to respective Heads of Dircctoratcs/KVS/NVS/CTSA as indicated below to alsodlssemtnate the information to all concerned schools under their jurisdiction:J. The Commissioner, Kendriyu V idynluya Sangathan, IS-Institutional Area, Shaheed Jeet Singh Marg,

New Delhi-! 10016.2. The Commissioner, Navodaya Vidyalaya Sam ill, B-15, Sector-62, lnsuturional Area, NOlda-20 1309,3 The Director of Education, Directorate of Education, Govr. of NCT of DeIhl, Old Secretariat, Delhi-

I 10054.4. The Director of Public Instructions (Schools), Union Territory Secretariat, Sector 9, Chaudigarh-l ou

017,5. The Director of Education, Govt. of Sikkim, Gangtok, Sikkim - 73710 I6. The Director of Schoo! Education, Govt. of Arunachal Pradesh, ltanagar - 791 III7. The Director of Education, Govt. of A&N Islands, Port Blair - 7441018. The Director of Education, S.I.E., CBSE Cell, VIP Road, Junglee Ghat, P,O. 744103, A&N Islands.9 The Secretary, Central Tibetan School Administration, ESS Plaza, Community Centre, Sector 3

Rohini,Oelhl-11008510. The Additional Director General of Army Education, A -Wing, Sena Bhawan, DHQ, PO, New Delhi-

I 1000 I.II. The Secretary AWES, Integrated Headquarters of MoD (Army), FDRe Building No, 202, Shankar

Vihar (Near APS), Delhi Cantl-IIOOIO.12 All Regional Directors! Regional Officers of CBSE with the request to send this circular to all the

Heads of the affiliated schools ofthe Board III their respective regions,13. All ASSOCiate Professor & Additional Directors/ Advisors/ Consulrams/Educat ion Officers,14, All Additional DIrector! JOint Oil-ector! Deputy Director! ASSIstant DIrector, Vocational Cell.IS. All Assistant Professor & Joint Directors, CBSEHi. All ASSIstant Professor & Deputy Directors, CBSE17. Tile Officer In charge of LT_ with the request to put this Circular on the CBSE websites.18. The ASSIstant Librarian, (BSE19. The Publtc Relations Officer, CBSE20_ Tile Hindi Officer, CBSE21. SPS to Chairman, CBSE22 SPS to Secretary, CBSE23. SI)S to Controller ofExarninations, CBSE24, SP$ 10 Director (Sp~CIClIExams and CTET), CBSE25 PA to Professor & Oil-ector (Academics, Research, Training & Innovation), CBSE26, PA to Director (Information Technology)

Suguudh SharmaAdditional Dtrecror and lncharge (Research nnd Innovation)

SAMPLE QUESTION PAPER

MATHEMATICS (041)

CLASS XII - 2016-17

Time allowed : 3 hours Maximum Marks: 100

General Instructions:

(i) All questions are compulsory.(ii) This question paper contains 29 questions.(iii) Question 1- 4 in Section A are very short-answer type questions carrying 1 mark each.(iv) Question 5-12 in Section B are short-answer type questions carrying 2 marks each.(v) Question t 3-23 in Section Care long-answer-I type questions carrying 4 marks each.(vi) Question 24-29 in Section Dare long-answer-It type questions carrying 6 marks each.

Section-A

Questions 1 to 4 carry 1 mark each.

1. State the reason why the Relation R={(a,b):aSb1}on the set R of real numbers is not

reflexive.

2. If A is a square matrix of order 3 and 12.111= klAI, then find the value of k.

3. If (/ and iJ are two nonzero vectors such that In x iJl = a .b, then find the angle between a and b.

4. If * is a binary operation on the set R of real numbers defined by a * b = a +b - 2, then find the

identity element for the binary operation * .

Section-B

Questions 5 to 12 carry 2 marks each.

5. ISimplify cot" ~ for x <-I.

x--I

6. Prove that the diagonal elements of a skew symmetric matrix are all zeros.

7._, 5x I I dy 2 3

If y = Ian ~ ,- r::< x < tz > then prove that - ~ +I-6x- -.../6 ,,6 dx I +4X2 I+9x2'

8. If x changes from 4 to 4.01, then find the approximate change in loge x.

9. Find J( I-X,)' e'dx.l+x

10. Obtain the differentia! equation of the family of circles passing through the points (a,O)anct

(-a,O)

7 I I12. If peA) ~ =, PCB) ~ -, peA n B) ~ -c, then find P(A! B).

5 3 )

Section-C

Questions 13 to 23 carry 4 marks each.

13. IfA~[~ ~2lthen using A-', solve the following system of equations: x- 2y ~ -1,2x+ Y ~ 2.

14.

1I2x-1 x<-

Discuss the differentiability of the function lex) = ' 2 at x =..!..I 23-6x,x:2':-2

OR

"For what value of k is the following function continuous at x = --?6

Ilx) ~

J3sinx+cos.x 1rf[ ,x;;t:.-6

x+-6

"k,x=--6

15.. ') ~ d2 Y ')

If x = «sin pt,y = bcca pt, then show that (a- -x- )y------z +6- = O.dx

16. Find the equation of the normal to the curve 2y = X2, which passes 'through the point (2, 1).

OR

Separate the interval [0, ~ }nto subintervals in which the function I(x) = sin" X+C054 x is

strictly increasing or strictly decreasing.

17. A magazine seller has 500 subscribers and collects annual subscription charges of Rs.300 per

subscriber She proposes to increase the annual subscription charges and it is believed that for

every increase of Re I, one subscriber will discontinue. What increase \\'i11 bring maximum

income to her? Make appropriate assumptions In order to apply derivatives to reach the

solution. Write one important role of magazines in our lives.

. f Sin x18. Pind dx.[cos' x+l)(cos2 x+4)

19. Find the general solution of the differential equation (I + tan y)( dx -dy) +2xdy = O.

OR

20. Prove that o.{(b+c)x(a+2b+Jc)}=[a b -l21.

. .. x-I y-2 Z-Q x-4 y-IFmd the values of a so that the following lines are skew: -- = -' - =--,-- =-- = z.

2 J 4 5 2

22. A bag contains 4 green and 6 'white balls. Two balls are drawn one by one without replacement.

If the second ball drawn is white, what is the probability that the first ball drawn is also white?

23. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find

the probability distribution of the number of diamond cards drawn. Also, find the mean and the

variance of the distribution.

Section-D

Questions 24 to 29 carry 6 marks each.

24. Let f:[O,OCl)-')-Rbe a function defined by f(x)=9x2 +6x-5. Prove that f is not invertible.

Modify, only the codomain of f to make f invertible and then find its inverse.

OR

Let * be a bina ry operation defined on Qx Q by (a, b) * (c, d) = (ac,b +ad), where Q is the set

of rational numbers. Determine, whether * is commutative and associative. Find the identity

element for * and the invertible elements of Qx Q.

25. Using properties of determinants, prove that a

(a +b)'c

c c

(b+c)'a =2(a+b+c)'.

a

b(c + a)'

bb

OR

r

pa+q

qa + r = 0, then, uSing properties of determinants, proveIf p*O,q*O and

p

qq

pa+q qa+r 0

that at least one of the following statements is true: (a) P. q, r are in G P. (b) a is a root of the

equation px" + 2qx+ r = O.

26. Using integration, find the area of the region bounded by the curves y =)5 _X2 and y = IX-II.

27_

,,r f xsmxcosx dx

Evaluate the allowing: . ~ 4 •. \.o sm x-i-cos x

OR

"Evaluate J(x + e2-')dx as the limit of a sum.o

28. Find the equation of the plane through the point (4, ·3, 2) and perpendicular to the line of

intersection of the planes x - y + 2z - 3 = 0 and 2x - y -Sz = O.Find the point of intersection of

the line r = 1+2) -I: + ;{.(1 +3) -9£) and the plane obtained above.

29. In a mid-day meal programme, an NCO wants to provide vitamin rich diet to the students of an

Mea school .The dietician of the NGO wishes to mix two types of food in such a way that

vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C.

Food 1 contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C. Food 2 contains 1

unit per Kg of vitamin A and 2 units per kg of vitamin C. It costs Rs 50 per kg to purchase Food

'l and Rs 70 per kg to purchase Food 2. Formulate the problem <.IS LPP and solve it graphically

for the minimum cost of such a mixture?

-·0--0--0--

MATHEMATICS (041)CLASS XII - 2016-17

Marking SchemeSection A

1. 1 (I)' 'I I):2 >"2 ~ l"2'2" ~ R. Hence, R is not reflexive. [1]

2. k = 23 :::: 8 III

[11

[1]

3. smB=cos8=>B=4Y

4. eE R is the identity element for * if {/ *e=e*a =a'ia E R :::::>e=2

Section B

5. Let sec" x = B. Then x = secB and for x < -1, ;rr< e < n2

[1/2[

Given expression = cot" (-cot 8) [1/2[

_I Ll _I 1r== cot (cot(7r-o»=;r-s~c x as O<7r-B<-

2[1]

6. Let A be a skew-symmetric matrix. Then by definition A' =-A

=> the (i, j)!h element of A'::::the (i, [sth element oj( - /f)

=> the (j,i)!h element of A::::-the (i, j)lh element of A

[1/2[

[1/2[

[1/2]

For the diagonal elements i:::: j => the (l,i)th element of A = -the (i,i)th element of A

=> the (i,Olh element oj A = 0 Hence, the diagonal elements are all zero. [1/2]

7. _I 3x+2x _I Iy = tan = tan 3x + tan - 2x. 1-3x2x

[1[

dy 3 2-= ,+ 2dx 1+9x- 1+4x

8. Let y=logex,x=4,6x=.01

[1[

[1/2[

dy-=-dx x

[1/2[

dy=(dY] xSx=~I~=.0025dx .\"~4 400

[11

Jl 1 Zx J9. Given integral = --7 ~ , c'itsI+x- (I+x'r

[1[

=_I~, e'+c as !!.......(_I~,)= -2xl+x- dx l+w ' (l+x2/ [1[

,()2"" 210."- + y-b ~ 0" +b"or,x" +Y -2by ~ a (1) [1]

11.

12.

13.

dy dy2x+2y--2b-::::O~2bdx dx

ely2x+2y-

dx [1/2]dydx

Substituting in (1), (x2_y2_a2)dy -2xy=Odx

[1/2]

Iii + EI' + 10- EI' ~ 2 (Iiil' + lEI')=>IEI'~2116=>IEI= 46

[1]

[1/2]

[1/2]

peA / 8) = P(AnB)PCB)

[1/2]

1- PIA uB)=

1- PCB)

1-[ PCA) + PCB) - peA n B) 11- PCB)

[1/2]

[1/2]

7=

10[1/2]

Section C

IAI ~5

adjA =[~2 ~]

A"' = ndjA =.!.[ 1 21

]IAI 5-2

[1/2]

[1+1/2]

[1/2]

[Xl [-I]Given system of equations is AX:::: B, where X:::: y ,B = 2 [1/2]

[l/2]

3 4=> x:::: -:-,y =";

) )[1/2]

14.

I I I(--h)- f(-) 2(--h)-1-0LI'(~)= 11m 2 2 = 11m 2 =22 '•...•u- -h h__rr -II [1+1/2[

I I Ij(-+h)- f(-) 3-6(-+h)-0

Rj'(~)=lllTI 2 2 -elim ? =-62 11--->0' h IHlr h

[1+1/2]

LI'(~);t: RI'( ~) I f is not differentiable at x = ~2 2 2

OR

[lJ

2sin(x+ n)lim I(x) = lim 6

r. ;r 7[.f->-'(, '<-"-'(, X + _

6

[2J

=2 rn;r

[(--)=k. 6 [1/21

For the continuity of f(x) at x = - 7r J(- .7)= lim f(x) ~ k = 26 6 H-~

6

[1/2J

15. dx dy I .- = apcos pt,- = - ,psm pIdl dt

[lJ

dy -bp sin pt b= =--tanpt

dx apcos pt a[1/2J

eI"y -bpsec' pt dt=

dx2 a dx[1]

-bpsec2 pi _b12 2 el2y 2

= X = 2 2 c> (a -x )Y-2 +b = 0a pacospt (a -x)y dx

dv-' = x The slope of the normal atdx

[1+1/2J

16. Let the normal beat (X"Y,) to the curve 2y=x",

-f -[(x"y,) = 7( '-'-IY')~= x,

dx (",,\',)

rn

-IThe equation of the normal is Y- Y, =-(x-xl)

x,[1/2J

. -[The point (2, 1) satisfies it 1- Yl = -(2-x,) => X,YI = 2 .... (1)

x,[1/2]

Also, 2YI = x/ .....(2) [1/2]

Solving (1) and (2), vveget x, = 23 'YI = 23 [1/2J

2 2

The reg u ired equation of the normal is x + 23 y = 2 + 23 [1J

OR

I'(x) = 4sin3 .rcosx-c qcos ' xsin x = -sin4x II]

, ITJ (x) ~ 0 => x ~-4

[1]

In the interval Sign of f'(x) Conclusion Marks

(0,: )-ve as

f ' 'I d ,,[0, : ) II]0<4x<1< IS stnct y ecreasmg In

IT ff +ve as IT IT('4''2) f' 'I' ,,['4''2) [1]

;r < 4x < 2ff IS strict Y increasIng In

17. Increase in subscription charges = Rs X, Decrease in the number of subscriber= x. Obviously, x

is a whole number. 11(2]

Income is given by y = (500 - x)(300 + x). Let us assume for the time being

O:':;x<500,xeR [1]

dy dy- ~200-2x,- ~O=> x ~ IOOdx ell:

dl~; =_2,(dl~J =-2<0dx dx

.'=11)0

[1(2]

[1(2]

Y is maximum when x = 100, which is a whole number. Therefore, she must increase the

subscription charges by Rs 100 to have maximum income. [1(2]

Magazinescontribute,a great deal to the development of our knowledge. Through valuable and

subtle critical and commentary articles all culture, social civilization, new life style we learn

a lot of interesting things. Through reading magazines, our mind and point of view are

consolidated and enriched.[1]

18. cosx=t~-sinxdx=dtThegiveninlegral =-J ~ dt,(r +I)(r +4)

[I]

, -I A 8Putr=y, =--+--

(y+I)(y+4) y+1 y+4[1(2]

-I 1-I ~ (y+4)A + B(y+ I),O~ A + 8,-1 ~4A+ B:, A ~ 3,8 ~ ')

h '. 1f dl 1f dt 1 -, 1 _,1T e givenintegral =-- +- =--Ian t+-tan -+cJ (1'+1) J (1'+4) J 6 2

1 _, ( ) 1 I cos x=--tan cosx +c-tan" --+c362

[I]

[1]

[1(2]

19.dx ?

(l+tany)dx=(I+tany-2x)dy=>-+ - x=1dy l+tany

f3,~[I]

J2".' Jt'in""CO$y)+("".y .•."n")d

I.F. ::::e 140";0." :::: e . "";.1''''';'')' y ::::el"g,,(o~S."HiI1Yl+!' :::: (COS Y + sin y)e!' [2]

[I]._r(cosy+siny)eJ" = J (cos y+sin y)eJ'dy::::::>x(cosy+siny)e-" ::::e' sin y+c

OR

( )'x I :;;r , -- e:... r :... dx I >;

We have (I +e")dx = ('--I)e'dY => -' =) .' = le'-), hence homogeneous [1/2]y dy (l+eY) y

dx dvx=vy,-=v+y-

dy dy[1/2]

f l:e" dv=-J(~Ve + \I Y

[I]

log, Ie' + 0 = -log, Lvi + log, C

::::::> log, ICe" + v)yl:::: loge C

[I]

[1/2]

,<> (e' + \I)Y:::: ±c:::: A::::::> (e" + X)y = A, the general solution

y[1/2]

20.- - -

LHS = ii . (b xii + 2b xb +bb xc +cx ii + 2i5xb + 3cxi5)- - - -::::ii . (b x 0) + 3i'i· (b x c) + ii· (ex 1i) + 20 (cxb)as b xb =cxc =0

::::3[0 b CJ + 2[ a c EJ::::3[ii b (,J-2[ii b +r- b cJ

[I]

[I]

[I]

[I]

21. As 2: 3: 4 *" 5; 2: 1, the lines are not parallel

Any point on the first line is (2,,1, + 1,3/L+ 2, 4,1 + a)

Any point on the second line is (5,L1+4,2j1+ I,p)

[1(2]

[I]

[I]

lines will be skew, if apart from being non parallel, they do not intersect. There must not exist

a pair of values of A,;"I, which satisfy the three equations simultaneously:

2/1.+ 1= 5jJ +4, 31t+ 2 = 2;1 + 1,41t+ a = pSolving the first two equations, we get A = -1, f.i =-1

These values will not satisfy the third equation if a#. 3

[1]

[1/2]

22. Let £1 = First ball drawn is white, £2 = First ball drawn is green, A =Second ball drawn is

white [1]

The required probability, by Bayes' Theorem, =

prE IA)= P(E,)xP(AIE,), P(E,)xP(AI E,)+P(E,)xP(AI E,)

6 5-x-10 9 5== =6 5 4 6-x-+-x- 9

10 9 10 9

23. Let X denote the random variable. X=O, 1, 2 n = 2, P = %, q = 3/4

[1]

[2]

[1(2]

o 2 Total Marks1Xi

6(16 2(16 1(2

[l(2]

pi

o

[1+1(2]

o 6(16 4(16

1Mean = Lx.p. =-

" 2

Variance = IX;2p;-(Ixip;)2

5 1 3

8 4 8

Section D

24. 'ix E [0,00),y = 9x' +6x-5 = (3x+ I)' -6;'-5 .(1) Range f = [-5,00) e codomain f, hence, f

is not onto and hence, not invertible

Let us take the modified codomain f = [-S,w)

Let us now check whether f is one-one. Let Xl' X2 E [0,(0) such that

5(8

[l(2]

[1(2]

[l(2]

[2]

[1(21

Since, with the modified codomain = the Range f f is both one-one and onto, hence invertible.

[ )" JY+6-1From (1) above, for any y E -5,00 ,(.)x+I)-=y+6~x= ...•,-t [_ ) [ ., JY+6 -II :-),00 -> O,oo),f (y) = 3

OR

Let (G,b ),( c,d) E Qx Q Then b + ad may not be equal to d + cb. We find that

(1,2) * (2,3) = (2,5), (2,3) * (1,2) = (2,7) " (2,5) Hence, + is not commutative.

[1+1(2]

[1]

[1]

[1]

Let

(a,b ),( c,d),( e,f) E QxQ,«(a,b) .(c,d)) • (e,f) ~(ace,b+ad +acf) ~(a,b )*«(c,d))*(e,f))

Hence, '" is associative. [lJ

(x,y) E QxQ is the identity element for * if

(x,y) *(a.b) ~ (a,b )*(x,y) ~ (a,b )'i(a,b) E QxQ;e"

(xa,y+xb):= (ax,b+ay) = (a,b )i.e.,xa =a,y+xb =b+oy =b, (x, y) = (1, 0) satisfies these

equations. Hence, (1, 0) is the identity element for '"

(C,d)EQXQ is the inverseof (a,b)EQxQ if

[2]

I -b(c,d).( a,b) ~ (a,b)' (c,d) ~ (I,O),Le.,(ac,b + ad) ~ (ca,d +cb) ~ (1,0) => c ~ -,d ~-. The

a a

I -binverse of (a,b)EQxQ,a"O;s (-,-) [2J

a a

(a + b)' , c'c-I ,

(b + c)',

LH5~- a- a-abc

b' b' (c+(I)'

(a+b+c)(a+b-c)~ 0

abc(b+c+a)(b-c-(l)

o(b+c+a)(b+c-a)

(b+c+a)(b-c-a)

c'c/ (CI~CI-C3,C2.....-tC2-C3)

(c +a )'

[IJ

((I+b+c)'((I+b-c) 0 c'

0 (b+c-o),~ a-

abc(b-c-a) (b-c-a) (c+a )'

(o+b-c) 0,

(a+b+c)'c-

~ 0 (b+c-a) (I (R, --> R, -(R, + R,))abc

(-2a) (-2c) 2ea

[1/2J

[1J

, (ac+bc-c2) a c2

(a+b+c)'= 0 (ba+C(I-02) 02

abcca(-20c) (-2ca) 2ca

[1/2J

( )' (ac+bc)

a+b+c ,= 0'

abcca o

c' c'a2 (CI~CI+C3,C2--7C2+Cl)2ea

[IJ(ba+ca)o

, , (a+b)(a+b+ct2c'a'

= uabcca 0

c c

(b+c)o

a (C, ~ C1 +C),C2 ---7 C2 +C3)

1

[l[

( )'a+b+c 2, 3= (ab+ac+b-+bc-ac)=2(a+b+c)b .

[1(2]

OR

pq q'Given equation ~ - pq pr

pqpa+q qa+r

,pqa+o:

pqa + pr = 0o

[1]

0, q2 _ pro: - pr

1=0 (R, ->R,-R,)=>- pq pr pqa+ pr

pqpa+o qa+r 0

[1]

, 0q- - PI'

pqa+ pr =0=> pq prpq

pa+q qa+r 0

, 0(( -pr

qa+r =0=> p q rpq

pa+q qa+r 0

, 0 0C( - PI'

qa+r = O(C, -) C, -CJ)=> q -qaq

0pa +q qa +r

[1]

[1]

[l[

( 2 _ ')r:::::> I I (q2a+rq+pqa2+q2a)=O:::::>(q2-pr)(2qa+r+pa2)=O~q2-pr=O(i.e.,p,q,T

q

are in GP) or 2qa + r + pa2 =Ofi.e.. a is a root of the equation 2qx + r+ px2 =0 [1[

26.

~ /-,

/., .,

Figure [J Marks]

. C->II ' 11Solving y = \l5-x· ,Y= x-I we get (x-I)- =5-x eer -x-2 = 0:::) X= 2,-1 [1]

, ,The required area = the shaded area= J (../5- x? -(\- x»)dx + J (../5_x2

- (x-l»)dx~ ,

2l?}v[2]

, , , [ ]' [ ']' [' ]'_ .1 . .. I.!. _I X X x.=J(~d.'-J(I-Xfl.'+J(-(x-I»dx=-'~+5S," - - x-, - --x_I _I I 2 J5 _I - _I 2 I

[1+ 'h]

[1/2]

27.

r. "ff J[ 1r "If .2 • '2(--x)s1Il(--r)cos(--x) "2 (--x)COSXSlIlX

1 = f ,\Sin ,\ cos x I = f 2 2 2 f 2 ' d4 ./ as dx , I = 4 4 X

o Sin x+cos x o sin4('r -x)+cos\l1' -x) u COS x-r sm x2 2

[1]

.~ 7r .-f2 (:;)cosxsmx

21 = - 4 . 4 dxc cos x-t-sm x

[1/2],

If ''E" r.- - -21=(7r)[J C~SXSil~~ dx+f c~sxsi~,:dx]=(7r)[Jtanxse~2\Lt+Jcose~2xcotxdx] [2]

2 oeos X+51O x 2.cos x-r-sm x 2 0 I+tan x ::: cot x+l, ,ff I I 0 I

21 = (-)[ f ~(- f ~p Isubstituting4 01+( ,i+p

, ~ 2 2tan- .r = t .cot" x = p r» 2 tanxsec xdx = dt,-2cOI xcosec xdx = dp [1]

[1]

,I=~

16[1/2]

OR

, "I(x)=x+e",fI(x)dx= lim hIf(rh),nh=4

,,_"".h ....•Uo 1"=1

[1)

" ""I(rh) = rh +e"", II(rh) = hII'+ Ie""~=l I I

[1]

I n(1I + 1) 1h e2""_1== I +e2 e21,_1 [2)

f" . 1111+11 e8 -I If(x)dx= lim [nh __ +e2h

21 x-]o ,,-.-.o,h_O 2 e r -\ 2 [1)

2h

I' [44+11 'h e8_1 I e8_1;;:::;im --+e- Jh x-];;:::;8+--h40 2 e- -I 2 2

21t

[11

28,J

i1=b xb = I -\, ,k2 =sT +7)+/~-3

[2]

2 -I

The equation of the plane is r.ii=(S/+7)+£) (4i-3)+2k),i.e.,r.(S?+7}+k)=1 11]

The position vector of any point on the given line is (I + A)T + (2 +3.tl)} + (-1-9A)k [1]

Wehave (l+.:t)5+(2+H)7+(-I-9,1.)I~1 [1]

J, ~ -I [1/2]

The position vector of the required point is -) + 8/~ [1/2]

29. Let x kg of Pood 1 be mixed with y kg of Food 2. Then to minimize the cost, C = SOx + 70y

subject to the following constraints: 2x + y 2. 8, x + 2y:2:1 O,x 2. 0, Y 2. 0 [2]

\

" "

Graph [2]

At C Marks

(a, 8) Rs 560

(2,4) Rs 380 [1].

(10.0) Rs 500

In the half plane SOx + 70y < 380, there is no point common with the feasible region. Hence, theminimur» cost is Rs 380. [1]