CBSE 2015 FOREIGN EXAMINATION

Similar documents
INTEGRALS. Chapter 7. d dx. 7.1 Overview Let d dx F (x) = f (x). Then, we write f ( x)

Chapter 16. 1) is a particular point on the graph of the function. 1. y, where x y 1

SOLVED EXAMPLES. be the foci of an ellipse with eccentricity e. For any point P on the ellipse, prove that. tan

Instructions for Section 1

TOPIC 5: INTEGRATION

CONIC SECTIONS. MODULE-IV Co-ordinate Geometry OBJECTIVES. Conic Sections

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS SEMESTER TWO 2014 WEEK 11 WRITTEN EXAMINATION 1 SOLUTIONS

Answers & Solutions. for MHT CET-2018 Paper-I (Mathematics) Instruction for Candidates

CBSE 2013 ALL INDIA EXAMINATION [Set 1 With Solutions]

CIVL 8/ D Boundary Value Problems - Rectangular Elements 1/7

Lecture 4. Conic section

SCIENCE ENTRANCE ACADEMY PREPARATORY EXAMINATION-3 (II P.U.C) SCHEME 0F EVALUATION Marks:150 Date: duration:4hours MATHEMATICS-35

Functions and Graphs 1. (a) (b) (c) (f) (e) (d) 2. (a) (b) (c) (d)

Continuous Random Variables: Basics

Integration Continued. Integration by Parts Solving Definite Integrals: Area Under a Curve Improper Integrals

Lecture 11 Waves in Periodic Potentials Today: Questions you should be able to address after today s lecture:

COHORT MBA. Exponential function. MATH review (part2) by Lucian Mitroiu. The LOG and EXP functions. Properties: e e. lim.

MAXIMA-MINIMA EXERCISE - 01 CHECK YOUR GRASP

MATHEMATICS PAPER IB COORDINATE GEOMETRY(2D &3D) AND CALCULUS. Note: This question paper consists of three sections A,B and C.

CONTINUITY AND DIFFERENTIABILITY

CBSE-XII-2015 EXAMINATION. Section A. 1. Find the sum of the order and the degree of the following differential equation : = 0

Ch 1.2: Solutions of Some Differential Equations

Board Answer Paper: October 2014

π = tanc 1 + tan x ...(i)

, between the vertical lines x a and x b. Given a demand curve, having price as a function of quantity, p f (x) at height k is the curve f ( x,

1 are perpendicular to each other then, find. Q06. If the lines x 1 z 3 and x 2 y 5 z

(2) If we multiplied a row of B by λ, then the value is also multiplied by λ(here lambda could be 0). namely

KCET 2016 TEST PAPER WITH ANSWER KEY (HELD ON WEDNESDAY 4 th MAY, 2016)

EXPECTED ANSWERS/VALUE POINTS SECTION - A

Lecture contents. Bloch theorem k-vector Brillouin zone Almost free-electron model Bands Effective mass Holes. NNSE 508 EM Lecture #9

The Z transform techniques

Combinatorial Networks Week 1, March 11-12

CSE303 - Introduction to the Theory of Computing Sample Solutions for Exercises on Finite Automata

MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. Note: This question paper consists of three sections A, B and C.

This Week. Computer Graphics. Introduction. Introduction. Graphics Maths by Example. Graphics Maths by Example

HIGHER ORDER DIFFERENTIAL EQUATIONS

5. B To determine all the holes and asymptotes of the equation: y = bdc dced f gbd

The Derivative of the Natural Logarithmic Function. Derivative of the Natural Exponential Function. Let u be a differentiable function of x.

MATHEMATICS PART A. 1. ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC

2008 AP Calculus BC Multiple Choice Exam

MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. Note: This question paper consists of three sections A,B and C. SECTION A

1. If * is the operation defined by a*b = a b for a, b N, then (2 * 3) * 2 is equal to (A) 81 (B) 512 (C) 216 (D) 64 (E) 243 ANSWER : D

12/3/12. Outline. Part 10. Graphs. Circuits. Euler paths/circuits. Euler s bridge problem (Bridges of Konigsberg Problem)

Winter 2016 COMP-250: Introduction to Computer Science. Lecture 23, April 5, 2016

5/9/13. Part 10. Graphs. Outline. Circuits. Introduction Terminology Implementing Graphs

MATHEMATICS (B) 2 log (D) ( 1) = where z =

Mathematics. Mathematics 3. hsn.uk.net. Higher HSN23000

Math 61 : Discrete Structures Final Exam Instructor: Ciprian Manolescu. You have 180 minutes.

SUBJECT: MATHEMATICS ANSWERS: COMMON ENTRANCE TEST 2012

Construction 11: Book I, Proposition 42

KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a

1997 AP Calculus AB: Section I, Part A

FORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81

( ) Geometric Operations and Morphing. Geometric Transformation. Forward v.s. Inverse Mapping. I (x,y ) Image Processing - Lesson 4 IDC-CG 1

Multi-Section Coupled Line Couplers

Self-Adjointness and Its Relationship to Quantum Mechanics. Ronald I. Frank 2016

The Matrix Exponential

b. How many ternary words of length 23 with eight 0 s, nine 1 s and six 2 s?

CSE 373: More on graphs; DFS and BFS. Michael Lee Wednesday, Feb 14, 2018

SAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS

Last time: introduced our first computational model the DFA.

DIFFERENTIAL EQUATION

FSA. CmSc 365 Theory of Computation. Finite State Automata and Regular Expressions (Chapter 2, Section 2.3) ALPHABET operations: U, concatenation, *

NORMALS. a y a y. Therefore, the slope of the normal is. a y1. b x1. b x. a b. x y a b. x y

MASSACHUSETTS INSTITUTE OF TECHNOLOGY HAYSTACK OBSERVATORY WESTFORD, MASSACHUSETTS

MATHEMATICS PAPER IA. Note: This question paper consists of three sections A,B and C. SECTION A VERY SHORT ANSWER TYPE QUESTIONS.

UNIT # 08 (PART - I)

The Matrix Exponential

Solution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers:

Single Correct Type. cos z + k, then the value of k equals. dx = 2 dz. (a) 1 (b) 0 (c)1 (d) 2 (code-v2t3paq10) l (c) ( l ) x.

TOPPER SAMPLE PAPER - 5 CLASS XI MATHEMATICS. Questions. Time Allowed : 3 Hrs Maximum Marks: 100

Polygons POLYGONS.

CBSE 2018 ANNUAL EXAMINATION DELHI

QUESTION PAPER CODE 65/2/2/F EXPECTED ANSWER/VALUE POINTS

Minimum Spanning Trees

(HELD ON 21st MAY SUNDAY 2017) MATHEMATICS CODE - 1 [PAPER-1]

Section 3: Antiderivatives of Formulas

1 Introduction to Modulo 7 Arithmetic

Sundials and Linear Algebra

Supplementary Materials

Binomials and Pascal s Triangle

Floating Point Number System -(1.3)

Floating Point Number System -(1.3)

ME 522 PRINCIPLES OF ROBOTICS. FIRST MIDTERM EXAMINATION April 19, M. Kemal Özgören

Linear Algebra Existence of the determinant. Expansion according to a row.

Lesson-5 ELLIPSE 2 1 = 0

Elliptical motion, gravity, etc

10. The Discrete-Time Fourier Transform (DTFT)

MATHEMATICS PAPER & SOLUTION

# 1 ' 10 ' 100. Decimal point = 4 hundred. = 6 tens (or sixty) = 5 ones (or five) = 2 tenths. = 7 hundredths.

Chem 104A, Fall 2016, Midterm 1 Key

SOLUTION CLASS-XII / (CBSE)

( β ) touches the x-axis if = 1

National Quali cations

Marking Scheme. Section A 3. 2 [1] l m n 1 n 1 cos [1] Direction ratios of the given line are 2, 1, 2.

Objective Mathematics

Derangements and Applications

Drill Exercise Find the coordinates of the vertices, foci, eccentricity and the equations of the directrix of the hyperbola 4x 2 25y 2 = 100.

KENDRIYA VIDYALAY SANGATHAN: CHENNAI REGION CLASS XII PRE-BOARD EXAMINATION Q.No. Value points Marks 1 0 ={0,2,4} 1.

Paths. Connectivity. Euler and Hamilton Paths. Planar graphs.

Transcription:

CBSE 05 FOREIGN EXAMINATION (Sris SSO Cod No 65//F, 65//F, 65//F : Forign Rgion) Not tht ll th sts hv sm qustions Onl thir squnc of pprnc is diffrnt M Mrks : 00 Tim Allowd : Hours SECTION A Q0 Find th r of prlllogrm whos djcnt sids r rprsntd th vctors iˆ kˆ nd 4j ˆ kˆ ˆi ˆj kˆ Lt i ˆ kˆ nd 4j ˆ kˆ Now, 0 i ˆ 4ˆj 8kˆ 0 4 rquird r i ˆ 4ˆj 8kˆ 44 6 64 4 4 4 squnits Q0 Find th sum of th intrcpts cut off th pln z 5 on th coordint s z Givn pln z 5 i, 5/ 5 5 On compring to z, w hv, nd z-intrcpts s 5/, 5, 5 rspctivl c 5 5 sum of th intrcpts 5 ( 5) Q0 Find th unit vctor in th dirction of th sum of th vctors iˆ j ˆ kˆ nd 4iˆ j ˆ kˆ Lt p i ˆ j ˆ kˆ 4i ˆ j ˆ kˆ 6i ˆ kˆ p 6i ˆ kˆ rquird unit vctor, p ˆ [6i ˆ k] ˆ p 6 7 d d 4 Q04 Writ th sum of th ordr nd dgr of th diffrntil qution 0 d d 4 Ordr nd dgr of th diffrntil qution 0 is nd rpctivl rquird sum of ordr nd dgr 4 d Q05 Writ th solution of th diffrntil qution d d Givn d C or, log C log 5 6 Q06 If A 4, thn writ th cofctor of th lmnt of its nd row 4 7 Cofctor of lmnt, C [6 ( 7)( )] SECTION B Q07 Find th point on th curv 9, whr th norml to th curv mks qul intrcpts on th s W hv 9 Lt th rquird point P (, ) So, 9 (i) d On diffrntiting th q of curv wrt, 8 d 6 For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) d 6 mt mn t P 6 6 Sinc th norml mks qul intrcpts with th s, mn Tht is, 6, 6, (ii) 6 6 Solving (i) nd (ii) simultnousl, w gt : 9, 9 6 6 4 4 4 0, 4 0 ( 4) 0, ( 4) 0 0, 4 8 8 So, 0, Hnc th rquird points r 4, nd (0,0) n d d Q08 If [ ], thn show tht ( ) n n W v n d n n d n d n d n d d d d d d n ( ) n d d ( ) n n d d ( ) n Q09 Find whthr th following function is diffrntil t nd or not :, f (),, f () f () ( ) W hv Lf () lim lim lim, f () f () ( ) ( ) Rf () lim lim lim Lf () Hnc f () isn t diffrntil t f () f () ( ) ( ) Lf () lim lim lim, f () f () ( ) ( )( ) Rf () lim lim lim lim ( ) ( ) Lf () Hnc f () is diffrntil t Q0 In prlimnt lction, politicl prt hird pulic rltion firm to promot its cndidts in ws tlphon, hous clls nd lttrs Th cost pr contct (in pis) is givn in mtri A s 40 Tlphon A 00 Hous Cll 50 Lttrs Th numr of contcts of ch tp md in two citis X nd Y is givn in th mtri B s For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) Tlphon Hous Cll Lttrs 000 500 5000 Cit X B 000 000 0000 Cit Y Find th totl mount spnt th prt in th two citis Wht should on considr for csting his/hr vot prt s promotionl ctivit or thir socil ctivitis? 40 Tlphon Th cost pr contct (in pis) is givn in mtri A s A 00 Hous Cll 50 Lttrs Th numr of contcts of ch tp md in two citis X nd Y is givn in th mtri B s Tlphon Hous Cll Lttrs 000 500 5000 Cit X B 000 000 0000 Cit Y Th totl mount spnt th prt in th two citis BA 40 4 000 500 5000 0 Tht is, 00 500 0 0 000 000 0000 6 0 50 5 4 0 5000 0 8 0 50 98 6 0 5000 5000 84 40 00 44 5 990000 Amount spnt in Cit X (in pis) 0000 Amount spnt in Cit Y (in pis) Hnc, th prt spnt 990000 pis (or, `9900) in th Cit X nd, 0000 pis (or, `00) in th Cit Y On should considr prt s socil ctivitis instd of promotionl ctivitis of th prt for csting his/hr vot Q Evlut : sin( ) Lt I sin( ) (i) Appling intgrl B Prts, w gt d I sin( ) sin( ) I sin( ) cos( ) I sin( ) cos( ) d I sin( ) cos( ) cos( ) I sin( ) cos( ) sin( ) I sin( ) cos( ) sin( ) 9 I sin( ) cos( ) I [B (i) 4 4 9 I I [sin( ) cos( )] I [sin( ) cos( )] C 4 4 / cos Q Evlut : / For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) Considr I / / cos (i) / cos / cos I / / / / On dding (i) & (ii), w gt : I / / / 0 I cos / / For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg 4 Using f () f ( ) I / / / / cos cos I cos / / cos (ii) I cos f () cos cos( ) f ( ), i f is vn function nd, f () f (), if f is n vn function 0 I sin / sin sin 0 0 Q Thr mchins E, E nd E in crtin fctor producing lctric uls, producs 50%, 5% nd 5% rspctivl, of th totl dil output of lctric uls It is known tht 4% of th uls producd ch of mchins E nd E r dfctiv nd tht 5% of thos producd mchin E r dfctiv If on ul is pickd up t rndom from d s production, clcult th proilit tht it is dfctiv OR Two numrs r slctd t rndom (without rplcmnt) from positiv intgrs,, 4, 5, 6, nd 7 Lt X dnot th lrgr of th two numrs otind Find th mn nd vrinc of th proilit distriution of X Lt E, E nd E dnot th vnts tht olts producd mchins E, E nd E rspctivl Lt A th vnt tht th slctd ul is dfctiv 50 5 4 5 P(E ), P(E ) P(E ), P(A E ) P(A E ), P(A E ) 00 00 4 00 5 00 0 7 Thrfor, P(A) P(E )P(A E ) P(E )P(A E ) P(E )P(A E ) 5 4 5 4 0 400 OR Sinc X dnots th lrgr of th two numrs otind from,, 4, 5, 6 nd 7 So vlus of X :, 4, 5, 6, 7 X P(X) 6 5 6 5 5 4 6 5 6 5 5 5 6 5 6 5 6 5 5 6 4 6 5 6 5 6 5 6 5 5 7 5 6 5 6 5 6 5 6 5 6 5 5 4 5 85 Now, mn X P(X) 4 5 6 7 5 5 5 5 5 5 And vrinc 4 5 4 X P(X) Mn 4 5 6 7 5 5 5 5 5 9 Q4 Th two vctors ˆ j k ˆ nd i ˆ ˆj 4kˆ rprsnt th two sids vctors AB nd AC rspctivl of tringl ABC Find th lngth of th mdin through A

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) Hr AB ˆj k, ˆ AC i ˆ ˆj 4kˆ AB BC AC BC (i ˆ ˆj 4k) ˆ ( ˆj k) ˆ i ˆ ˆj kˆ Sinc mdin through A mts th sid BC t th midpoint of BC If M is th mid-point of BC thn, BM ˆi ˆj kˆ 5 Thrfor AM AB BM ˆj kˆ iˆ ˆj kˆ ˆi kˆ 5 4 Lngth of mdin, AM units Q5 Find th qution of pln which psss through th point (,, 0) nd contins th lin 6 z 4 5 4 Th qution of pln through (,, 0) is A( ) B( ) C(z 0) 0(i) whr A, B, C r th dr s of th norml to th rquird pln As pln (i) contins th lin 6 z 4 with dr s s, 5, 4 so, A 5B 4C 0(ii) 5 4 Also (, 6, 4) lis on th givn lin nd pln (i) s wll so, A( ) B(6 ) C(4 0) 0 i, 0A 4B 4C 0 0A B C 0(iii) Solving (ii) nd (iii), w gt A B C i, dr s of th norml to pln (i) r,, Hnc qution of pln is : ( ) ( ) z 0 i, z Q6 If tn (cos ) tn (cos c ), ( 0), thn find th vlu of OR If tn tn tn tn, thn find th vlu of n(n ) cos Givn tn (cos ) tn ( cos c ) tn tn ( cos c ) cos cos cos tn tn tn tn sin cos sin 0 cos sin sin sin sin [cos sin ] 0 sin 0 or cos sin 0 0 or cot 0 or But 0 4 4 OR W hv tn tn tn tn n(n ) (n ) n tn tn tn tn n(n ) tn tn tn tn tn n tn (n ) tn (n ) tn n tn (n ) tn (n ) tn tn tn tn (n ) n n tn tn tn tn n n Q7 If A nd I is th idntit mtri of ordr, thn show tht A 4A I Hnc find A OR If A nd B nd (A B) A B, thn find th vlus of nd 5 4 W hv A A AA 4 5 (i) For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg 5

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) 0 8 4 0 5 4 Also 4A I 4 (ii) 0 4 8 0 4 5 B (i) nd (ii), w gt : A 4A I Pr-multipling oth sids A w gt : A AA 4A A A I IA 4I A 0 A 4I A 4 0 A OR W v (A B) A B (A B)(A B) A B AA AB BA BB AA BB AB BA So, B qulit of mtrics, w gt :,,, On solving ths qutions, w gt :, 4 Q8 Using proprtis of dtrminnts, prov th following : ( ) LHS : Lt B R R R, R R R 0 0 0 0 Tking 0 0 ( ) 0 0 Epnding long R ( ) 0 0 0 ( ) RHS Q9 Evlut : Lt I I sin( ) OR Evlut : sin( ) sin( ) sin( ) sin( ) cos cos( ) sin sin( ) common from R nd R ( 4)( 9) I sin ( ) sin( ) oth I cos sin cot( ) I cos sin log sin( ) C SECTION C Q0 Solv th givn diffrntil qution : cos d cos sin 0 OR Solv th givn diffrntil qution : Givn cos d cos sin 0 d 0 For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg 6

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) sin cos d cos d dv Put v v dv sin v vcos v v v cos v v cos v dv sin v v log sin v v log log C sin cos d cos dv sin v vcos v v v cos v dv sin v vcos v v vcos v v cos v cos v v dv sin v v log log C sin v v log C sin sin C OR Givn 0 d d 0 Considr I I t t t t t I t log t dt I C sin C or, sin whr ( )( ) 0 d I d 0(i) tdt Put t t t dt t I log I dt t Sustituting th vlu of I in (i), w gt : log C log C is th rquird solution Q Find th proilit distriution of th numr of doults in four throws of pir of dic Also find th mn nd vrinc of this distriution 5 Lt E : gtting doult on th pir of dic P(E), P(E) 6 6 Lt X :Numr of doults in four throws of pir of dic So vlus of X r 0,,,, 4 X 0 4 P(X) 4 5 6 65 96 4 C 6 6 5 500 96 4 5 C 6 6 50 96 4 5 C 6 6 0 96 4 6 96 For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg 7

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) Mn 65 500 50 0 864 X P(X) 0 4, 96 96 96 96 96 96 65 500 50 0 Vrinc X P(X) (Mn) 0 4 96 96 96 96 96 5 Vrinc 9 Q Lt f : N R function dfind s f () 4 5 Show tht f : N S, whr S is th rng of f, is invrtil Also find th invrs of f Hr f : N R, f () 4 5 Lt n ritrr lmnt of rng S of function f Thn = 4 + + 5, for som in N, which implis tht = ( + ) + 6 6 This givs s 6 6 Lt us dfin g : S N g() Now, gof () g f () g(4 5) g ( ) 6 ( ) 6 6 6 6 And, fog() f g() f 6 Hnc, gof = I N nd fog = I S This implis tht f is invrtil with f = g 6 6 So, f i, f () Q Using intgrtion, find th r of th rgion oundd th lin 0, th curv nd -is W hv 0 (i) nd (ii) Solving (i) & (ii), 0 ( )( ) 0, Rquird r (i) (ii) 0 0 0 8 0 4 0 squnits Q4 Find th distnc of th point P(,, ) from th pln z 5 msurd prlll to th lin whos dirction cosins r proportionl to,, 6 Eqution of lin through P(,,) nd prlll to lin whos dr s r proportionl to,, 6 is : z (i) 6 An rndom point on lin (i) is Q(,, 6 ) If Q lis on th givn qution of pln z 5 thn, ( ) ( ) ( 6 ) 5 7 For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg 8

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) So, coordints of th point Q r 6 Q,, 7 7 7 i, 9 5 Q,, 7 7 7 9 5 Rquird distnc, PQ unit 7 7 7 Q5 Mimis z 8 9, sujct to th constrints givn low : 6, 6,,, 0 To Mimis z 8 9, Sujct to th constrints givn low : 6 6, 6, ;, 0 6 Cornr points Vlu of z A(0,) 9 B, 0 6 94 8 C, Mvlu D(,0) 6 O(0, 0) 0 So mimum vlu of z is ttind t 0 6, nd mimum vlu is 8 Q6 Find th minimum vlu of ( ), whr c OR Find th coordints of point of th prol stright lin Givn c (i) 7 which is closst to th c ds c Lt S ( ) S nd, d S c ds c For locl points of mim nd/or minim, 0 c d S c 0 / t c c Also, minimum vlu of S c c Rplcing vlu of c in (i), w gt Tht is, S c S c S is minimum t c OR Givn lin is i, 0 (i) Also lt th rquird point on th prol 7 P(h, k) h k Distnc of P from lin = 0 is, s ( ) k h 7h (i) For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg 9

CBSE 05 Annul Em Ppr (Forign) Compild B O P Gupt (+9-965050480) h h 7h h 4h 5 h 4h 5 s, (i) s 0 0 0 (h ) ds (h ) 0 (h ) d s s & 0 0 dh 0 0 dh 0 So, s is lst for ll vlu of h ds (h ) For locl points of mim nd/or minim, 0 h dh 0 B (i), k ( ) 7( ) 8 Hnc th coordints of th rquird points on th givn prol r P(, 8) All Rights Rsrvd with O P Gupt Disclimr : All cr hs n tkn whil prpring this solution drft Solutions hv n vrifid prominnt cdmicins hving vst knowldg nd princ in tching of Mths Still if n rror is found, pls ring it to our notic Kindl forwrd our concrns/fdcks through mssg or WhtsApp @ +9965050480 or mil t thopgupt@gmilcom Lt s lrn Mths with smil:-) For vrious stuffs on Mths, pls visit t : wwwthopguptcom Pg 0