12 Maths Model Paper PDF

Maheshwari Public School Vijay Path, Tilak Nagar, Jaipur MODEL PAPER-1/Class-12/Subject-Mathematics Time Allowed: 3:15 Hrs. MM: 80 ijh{kkfFkZ;ksa ds fy, lkekU; funsZ’k:(General I...

Maheshwari Public School Vijay Path, Tilak Nagar, Jaipur MODEL PAPER-1/Class-12/Subject-Mathematics Time Allowed: 3:15 Hrs. MM: 80 ijh{kkfFkZ;ksa ds fy, lkekU; funsZ’k:(General Instruction to the examinees:) 1- ijh{kkFkhZ loZizFke vius iz’u i= ij ukekadu vfuok;Zr% fy[ksA Candidate must write first his/her Roll No. on the question peper compulsorily. 2- lHkh iz’u djus vfuok;Z gSA All the questions are compulsory. 3- iz’u i= ds fgUnh o vaxzsth :ikUrj eas fdlh izdkj dh =qfV@vUrj@fojks/kkHkkl gksus ij fgUnh Hkk"kk ds iz’u dks gh lgh ekusA If there is any error/difference/contradiction in Hindi & English versions of the question paper, the question of Hindi version should be treated valid 4. ftu ÁÜu¨a esa vkarfjd [k.M gSa mu lÒh ds mÙkj ,d lkFk gh fy[ksa For questions having more than one part the answers to those parts are to be written together in continuity 5. Marks Distribution Table:- Section Q. Nos. Types of questions Marks per question A 1 (i) to (xviii) MCQ’s 1 (1x 18)=18 2(i) to (vi) Fill in the blanks 1 (1 x 6)=6 3(i) to (xii) VSA 1(1 x 12)=12 B 4 to13 SA-1 2(2 x 10)=20 C 14-17 SA-2 3(3 x 4)=12 D 18 to 20 LA 4(4 x 3)=12 6. iz’u la[;k 18,19 vkSj 20 esa vkUrfjd fodYi gSaA bu iz’uksa esa ls vkidks ,d gh fodYi djuk gSA There are internal choices in Q. No. 18,19 and 20. You have to attempt only one of the alternatives in these questions. Section-A Q.1 (i) ;fn f: R → R ,d Qyu bl Ádkj gS fd f(x) = 5x – 4 rks f-1(6)Kkr dhft;sA If f: R → R is a function such that f(x) = 5x - 4 then find f-1(6) (A) 1 (B) 2 (C) 3 (D) 4 1 𝜋 (ii) ;fn cot −1 𝑥 + tan−1 (3) = 2 rks x dk eku Kkr dhft,A 1 𝜋 If cot −1 𝑥 + tan−1 (3) = then value of x is. 2 1 (a) -1 (b) 3 (c) 1 (d) 3 𝑘+4 −1 𝑎 −1 (iii) ;fn [ ]=[ ]rks a dk eku Kkr dhft;sA 3 𝑘−6 3 −4  𝑘+4 −1 𝑎 −1 If [ ]=[ ], then find the value of a. 3 𝑘−6 3 −4 (a) 9 (b) 6 (c) 3 (d) 2 𝑘 8 (iv) ;fn lkjf.kd | | = 4 gks rks 𝑘 dk eku gksxk & 2 4 𝑘 8 If determinant | | = 4, then value of k will be 2 4 (a) −3 (b) 3 (c) −5 (d) 5 𝑠𝑒𝑐𝑥−1 𝑑𝑦 (v) ;fn 𝑦=√ rks dk eku Kkr dhft,A 𝑠𝑒𝑐𝑥+1 𝑑𝑥 𝑠𝑒𝑐𝑥−1 𝑑𝑦 If 𝑦 =√ then find the value of 𝑠𝑒𝑐𝑥+1 𝑑𝑥 1 𝑥 1 𝑥 (a) 𝑠𝑒𝑐 2 𝑥 (b) 𝑠𝑒𝑐 2 2 (c) 2 𝑐𝑜𝑠𝑒𝑐 2 2 (d) None of these 2 (vi) ,d o`Ùk dh f=T;k 0.7 cm/s dh nj ls c< jgh gSA bldh ifjf/k dh o`f) dh nj D;k gS tc r = 4.9 cm gS\ The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference, if r = 4.9 cm? (a) 1.4𝜋 (b) 6𝜋 (c) 6.1𝜋 (d) None of these 1 1 (vii) ∫0 √1−𝑥 2 𝑑𝑥 dk eku gksxk\ 1 1 Value of ∫0 √1−𝑥2 𝑑𝑥 will be- (a) 2 (b) -1 (c) -2 (d) 1 𝑑𝑦 (viii) vody lehdj.k + 𝑦 = 𝑒 𝑥 dh ?kkr gksxh& 𝑑𝑥 𝑑𝑦 The degree of the differential equation 𝑑𝑥 + 𝑦 = 𝑒 𝑥 will be- (a) 1 (b) 2 (c) 0 (d) 3 (ix) nks lfn’kksa 𝑎⃗ o 𝑏⃗⃗ ds ifjek.k Øe’k% √3 o 2gSa vkSj 𝑎⃗. 𝑏⃗⃗ = √6 gS rks 𝑎⃗ rFkk 𝑏⃗⃗ ds chp dk dks.k Kkr dhft;sA Find the angle between two vectors𝑎⃗and 𝑏⃗⃗ with magnitudes √3 and 2, respectively having 𝑎⃗. 𝑏⃗⃗ = √6. 𝜋 𝜋 𝜋 𝜋 (a) 4 (b) 3 (c) 2 (d) − 4 (x) fcUnqvks (2,1,2) vkSj (4,2,0) ls xqtjus okyh js[kk dh fnd~~&dkslkbu D;k gksxh What is direction cosine of a line which passes through the points (2,1,2) and (4,2,0). 2 1 −2 −2 1 2 2 −2 1 1 2 −2 (a) 3 , 3 , (b) ,3,3 (c) 3 , 3 , 3 (d) 3 , 3 , 3 3 3 𝑥−5 𝑦+2 𝑧 𝑥 𝑦 𝑧 (xi) ;fn js[kk;sa = = 1 rFkk = 𝑝 = 3 ijLij yEc gksrh gks rks p dk eku gksxk& 7 −5 1 𝑥−5 𝑦+2 𝑧 𝑥 𝑦 𝑧 If the lines = = 1 and 1 = 𝑝 = 3 are perpendicular, then value of p will be- 7 −5 (A) −2 (B) 2 (C) 1 (D) −1 (xii) 𝑦 = √𝑥 rFkk y = x ls ifjc) {ks= dk {ks=Qy Kkr dhft;sA Find the area enclosed by curve 𝑦 = √𝑥 and y = x. 1 1 2 (a) 1 (b) 9 (c) 6 (d) 3 (xiii) oØ y2=4x, y &v{k ,oa js[kk y=3 ls f?kjs {ks= dk {ks=Qy gS & Area of the region bounded by the curve y2 = 4x; y-axis and the line y=3 is 9 9 9 (A) 2 (B) (C) (D) 4 3 2 𝐵 (xiv) ;fn P(A) =0.8, P(B) = 0.5 vkSj P(𝐴) = 0.4 rks P(A∩B) Kkr dhft;sA 𝐵 If P(A) =0.8, P(B) = 0.5 and P(𝐴) = 0.4. Find P(A∩B). (a) 0.32 (b) 0.64 (c) 0.98 (d) 0.9 (xv) ;fn iklksa dk ,d tksM+k mNkyk tkrk gS] rks izR;sd ikls ij le vHkkT; la[;k izkIr djus dh izkf;drk D;k gksxh& The probability of obtaining an even prime number on each die, when a pair of dice is rolled, will be- 1 1 1 (a) 0 (b) 3 (c) 12 (d) 36 (xvi) nks ikls ,d lkFk mNkyus ij dqy ;ksx 5 vkus dh Ákf;drk gS& 1 1 1 (a) 1/18 (b) 12 (c) 9 (d) 6 Two dices are drawn simultaneously. Probability to get total of 5 is- 1 1 1 (a) 1/18 (b) 12 (c) 9 (d) 6 (xvii) ,d ifjokj esa nks cPps gSA ;fn ;g Kkr gks fd cPpksa esa ls de ls de ,d cPpk yM+dk gS] rks nksuksa cPpksa ds yM+dk gksus dh Ákf;drk gS & A family has two children. Probability that both the children are boys, given that at least one of them is a boy, is – (xviii) ,d o`Ùk dh f=T;k r=6 ij x ds lkis{k {ks=Qy esa ifjorZu dh nj gS & (a) 10𝜋 (b) 12𝜋 (c) 11𝜋 (d) 8𝜋 The rate of change of the area of a circle with respect to its radius at r cm = 6 is – (a) 10𝜋 (b) 12𝜋 (c) 11𝜋 (d) 8𝜋 2𝑥 Q.2 (i) ;fn 𝑦 = sin−1 (1+ 𝑥 2 ) rks __________ 2𝑥 𝑑𝑦 If 𝑦 = sin−1 (1+ 𝑥 2 ) then 𝑑𝑥 = __________ 1 (ii) ;fn tan−1(1) + cos −1 ( ) =__________ √2 1 If tan−1(1) + cos −1 ( ) = __________ √2 𝑑𝑦 (iii) ;fn 3x + 2y =sin 𝑥 rks = __________ 𝑑𝑥 𝑑𝑦 If 3x + 2y =sin 𝑥 then 𝑑𝑥 = __________ (iv) ,slh 2 /kukRed la[;k,¡ ftudk ;ksx 60 rFkk xy3 vf/kdre gks_______________ Two positive number whose sum is 60 and xy3 is maximum will be_______________ 𝑑𝑦 (v) vody lehdj.k 𝑑𝑥 + 𝑦 = 𝑒 𝑥 dh ?kkr gksxh_______ 𝑑𝑦 The degree of the differential equation 𝑑𝑥 + 𝑦 = 𝑒 𝑥 will be_______ (vi) ;fn lfn’k 𝑎⃗ vkSj 𝑏⃗⃗ n¨ gS r¨ lfn’k 𝑎⃗ vkSj 𝑏⃗⃗ ds chp dks.k 𝑠𝑖𝑛 𝜃 =____________ g¨xkA If 𝑎⃗ and 𝑏⃗⃗ are two vectors, then the angle between 𝑎⃗ and 𝑏⃗⃗ will be 𝑠𝑖𝑛 𝜃 =____________. 1 2 3 Q.3(i) fuEu lehdj.k dks gy dhft;s |4 𝑥 6| = 0 7 8 9  1 2 3 Solve the following equation: |4 𝑥 6| = 0 7 8 9 3 −11 1 (ii) lkjf.kd | 5 0 0| dk eku Kkr dhft;sA −10 3 0 3 −11 1 Find the value of determinant | 5 0 0| −10 3 0 (iii) og vUrjky Kkr dhft, ftleas Qyu 𝑓(𝑥) = sin 𝑥 − cos 𝑥 o/kZeku ;k ákleku gks tcfd 𝑥 ∈ (0, 𝜋) Find the interval in which function 𝑓(𝑥) = sin 𝑥 − cos 𝑥 is increasing or decreasing, 𝑥 ∈ (0, 𝜋) 𝑑𝑦 (iv) oØ 𝑦 = 𝑥 3 − 𝑥 + 1 dh fcanq 𝑥 = 2 ij Kkr dhft, & 𝑑𝑥 Find the value of dy/dx of curve 𝑦 = 𝑥 3 − 𝑥 + 1 at point 𝑥 = 2 (v) ∫ log 𝑥 𝑑𝑥 dk eku Kkr dhft;sA Evaluate ∫ log 𝑥 𝑑𝑥. (vi) ∫ 𝑒 𝑥 sec(1 + tan 𝑥) 𝑑𝑥 dk eku Kkr dhft;sA Evaluate ∫ 𝑒 𝑥 sec(1 + tan 𝑥) 𝑑𝑥 𝑑𝑦 1+𝑦 2 (vii) vody lehdj.k = 1+𝑥 2 dks gy dhft;sA 𝑑𝑥 𝑑𝑦 1+𝑦 2 Find the general solution of the differential equation 𝑑𝑥 = 1+𝑥 2 (viii) (3𝑎⃗ − 5𝑏⃗⃗). (2𝑎⃗ + 7𝑏⃗⃗)dk eku Kkr dhft;sA Evaluate the product (3𝑎⃗ − 5𝑏⃗⃗ ). (2𝑎⃗ + 7𝑏⃗⃗). (ix) ;fn 𝑎⃗ = 2𝑖̂ + 𝑗̂ + 3𝑘̂ vkSj 𝑏⃗⃗ = 3𝑖̂ + 5𝑗̂ − 2𝑘̂ rks 𝑎⃗ × 𝑏⃗⃗ Kkr dhft,A If 𝑎⃗ = 2𝑖̂ + 𝑗̂ + 3𝑘̂ and 𝑏⃗⃗ = 3𝑖̂ + 5𝑗̂ − 2𝑘̂ then find 𝑎⃗ × 𝑏⃗⃗. (x) ;fn 𝑎⃗ = 𝑖̂ − 2𝑗̂ + 3𝑘̂ vkSj 𝑏⃗⃗ = 3𝑖̂ − 2𝑗̂ + 𝑘̂ rks 𝑎⃗. 𝑏⃗⃗ Kkr dhft;sA Find 𝑎⃗. 𝑏⃗⃗ if 𝑎⃗ = 𝑖̂ − 2𝑗̂ + 3𝑘̂ and 𝑏⃗⃗ = 3𝑖̂ − 2𝑗̂ + 𝑘̂. (xi) fcUnq ftldk fLFkfr lfn'k 2i- j+4 k + ls xqtjus okyh js[kk dk lfn'k lehdj.k Kkr dhft, tks lfn'k i+ 2j- k ds lekUrj gSA Find the equation of the line passing through the point with position vector 2i- j+4 k and its parallel to the vector i+ 2j- k. (xii) fl) dhft, fd Qyu 𝑦 = 𝑥 2 + 2𝑥 + 𝑐 vody lehdj.k 𝑦 2 − 2𝑥 − 2 = 0 dk gy gSA Prove that the function 𝑦 = 𝑥 2 + 2𝑥 + 𝑐 is a solution of the differential equation 𝑦 2 − 2𝑥 − 2 = 0 Section-B Q.4 fl) dhft, fd Z esa ifjÒkf"kr ,d lac/a k R, 𝑎𝑅𝑏 ⟺ 𝑎 − 𝑏] 3 ls foÒkT; ,d rqY;rk laca/k gSA Prove that a Relation R, 𝑎𝑅𝑏 ⟺ 𝑎 − 𝑏, 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 3 defined in Z is an Equivalence Relation.  0  tan  / 2  Q.5 ;fn A    rFkk I d¨fV 2 dk rRled vkO;wg gS r¨ fl) dhft, fd%  tan  / 2 0   Cos   Sin   I  A   I  A    Sin  Cos     0  tan  / 2  If A    and I is Identity matrix of order 2 then prove that:  tan  / 2 0   Cos   Sin   I  A   I  A    Sin  Cos   cos 𝜃 sin 𝜃 Q.6 ;fn vkO;wg 𝐴 = [ ] gks rks 𝐴𝐴𝑇 Kkr dhft;sA − sin 𝜃 cos 𝜃 cos 𝜃 sin 𝜃 If matrix 𝐴 = [ ], then find 𝐴𝐴𝑇. − sin 𝜃 cos 𝜃 1 Q.7 ;fn 𝐴 = [1 2 3] vkSj 𝐵 = rks (𝐴𝐵)−1 Kkr dhft;sA 3 1 If 𝐴 = [1 2 3] and 𝐵 = then find (𝐴𝐵)−1. 3 1−cos(𝑐𝑥) ; 𝑥≠0 Q.8 ;fn Qyu 𝑓(𝑥) = { 𝑥 sin 𝑥 fcUnq 𝑥 = 0 ij larr gS rks 𝑐 dk eku Kkr dhft,A 1⁄ ; 𝑥=0 2 1−cos(𝑐𝑥) ; 𝑥≠0 If following function is continuous at x= 0, then find c 𝑓(𝑥) = { 𝑥 sin 𝑥 1⁄ ; 𝑥=0 2 Q.9 Qyu 𝑓(𝑥) = |𝑥| + |𝑥 − 1| dk 𝑥 = 0 rFkk 𝑥 = 1 ij lkarR; dk ijh{k.k dhft,A If 𝑓(𝑥) = |𝑥| + |𝑥 − 1| then test for the continuity at 𝑥 = 0 and 𝑥 = 1 𝑑𝑦 Q.10 ;fn 𝑦 = tan−1(𝑡𝑎𝑛𝑥 + 𝑠𝑒𝑐𝑥) rks dk eku Kkr dhft, 𝑑𝑥 𝑑𝑦 If 𝑦 = tan−1(𝑡𝑎𝑛𝑥 + 𝑠𝑒𝑐𝑥) then find the value of 𝑑𝑥 Q.11 vUrjky Kkr dhft, ftuesa 𝑓(𝑥) = 𝑥 2 − 6𝑥 + 5 ls iznÙk Qyu 𝑓 − (𝑎) fujarj o/kZeku gSA (𝑏) fujarj àkleku gSA Find the interval in which the function 𝑓(𝑥) = 𝑥 2 − 6𝑥 + 5 is (a) increasing (b) decreasing 𝑐𝑜𝑠 2𝑥 Q.12 gy dhft;sA Solve: ∫ (𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥)2 𝑑𝑥 Q.13 o`Ùk 𝑥 2 + 𝑦 2 = 𝑎2 ls f?kjs {ks= dk {ks=Qy Kkr dhft;sA Find the area enclosed by curve 𝑥 2 + 𝑦 2 = 𝑎2. Section-C Q.14 ;fn 𝑎⃗, 𝑏⃗⃗ vkSj 𝑐⃗ ek=d lfn’k bl izdkj gSa fd 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗ = 0 rks 𝑎⃗.⃗⃗⃗⃗ ⃗⃗⃗ 𝑐⃗ + 𝑐. 𝑏 + 𝑏. ⃗⃗⃗ 𝑎⃗ dk eku Kkr dhft;sA If 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are unit vectors such that 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗ = 0. Then find the value of 𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗.⃗⃗⃗⃗ 𝑎. Q.15 ,d ifjokj eas nks cPpsa gSaA ;fn ;g Kkr gS fd nksuksa cPpksa eas ls de ls de ,d cPpk yM+dk gS rks nksuksa cPpksa ds yM+dk gksus dh izkf;drk Kkr dhft;sA A family has two children. What is the probability that both the children are boys given that at least one of them is a boy. 1 Q.16 Evaluate eku Kkr dhft, ∫ √7−6𝑥−𝑥 2 𝑑𝑥 Q.17 ,d xksy xqCckjs ds ry dk {ks=Qy 2 lseh2/lsd.M dh nj ls c< jgk gS] rks bldk vk;ru fdl nj ls c< jgk gS] tcfd f=T;k 6 lseh gS\  The radius of the spherical ballon is increasing at the rate of 2𝑐𝑚2 /𝑠𝑒𝑐.At what rate is the volume of the ballon increasing when radius is 6 cm. Section-D Q.18 leryksa 𝑟⃗. (𝑖̂ + 2𝑗̂ + 3𝑘̂) − 4 = 0 vkSj 𝑟⃗. (2𝑖̂ + 𝑗̂ − 𝑘̂) + 5 = 0 ds izfrPNsnu ls g¨dj tkus okys rFkk lery 𝑟⃗. (5𝑖̂ + 3𝑗̂ − 6𝑘̂) + 8 = 0 ds yEcor~ lery dk lehdj.k Kkr dhft,A Find the equcation of the plane which passes through the intersection of two given planes 𝑟⃗. (𝑖̂ + 2𝑗̂ + 3𝑘̂) − 4 = 0 and 𝑟⃗. (2𝑖̂ + 𝑗̂ − 𝑘̂) + 5 = 0 and it perpendicular to the plane 𝑟⃗. (5𝑖̂ + 3𝑗̂ − 6𝑘̂) + 8 = 0 OR nh x;h js[kkvksa 𝑟⃗ = (1 + 𝑡)𝑖̂ + (𝑡 + 2)𝑗̂ + (3 + 2𝑡)𝑘̂ vkSj 𝑟⃗ = (𝑠 − 1)𝑖̂ + (2𝑠 + 1)𝑗̂ − (𝑠 + 1)𝑘̂ ds chp dh U;wure nwjh Kkr dhft;sA Find the shortest distance between the given lines. 𝑟⃗ = (1 + 𝑡)𝑖̂ + (𝑡 + 2)𝑗̂ + (3 + 2𝑡)𝑘̂ and 𝑟⃗ = (𝑠 − 1)𝑖̂ + (2𝑠 + 1)𝑗̂ − (𝑠 + 1)𝑘̂ 𝜋 𝑥 sin 𝑥 Q.19 Evaluate eku Kkr dhft, : ∫0 𝑑𝑥 1+cos2 𝑥 OR 1 Solve gy dhft, : ∫ 𝑑𝑥 √sin3 𝑥 sin(𝑥+𝑎) Q.20 xzkQh; fof/k ls fuEu jSf[kd izkx s zkeu leL;k dks gy dhft,A Solve the following linear programming problem graphically Maximise Z = 3x +9y Constraints (vojks/k) 𝑥 + 3𝑦 ≤ 60, 𝑥 + 𝑦 ≥ 10, 𝑥 ≤ 𝑦, 𝑥𝑦 ≥ 0 OR xzkQh; fof/k ls fuEu jSf[kd izksxzkeu leL;k dks gy dhft,A Solve the following linear programming problem graphically Manimise Z = x +2y Constraints (vojks/k) 2𝑥 + 𝑦 ≥ 3, 𝑥 + 2𝑦 ≥ 6, 𝑥𝑦 ≥ 0