QUESTION PAPER. Mathematics. 1. For the event to be completed is 5 th throw, 4 th and 5 th throw must be 4. Also 3 rd throw must be other cases are

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1 Mathematics QUESTION PAPER 1. For the event to be completed is 5 th throw, 4 th and 5 th throw must be 4. Also 3 rd throw must be other cases are = = = Hence (B) is the correct answer. Let the observations be x 1, x,..., x 5 Given (x 1 3) + (x 3) (x 5 3 ) = 5 x 1 + x x 5 = 155 Now Mean = x 1+x +...+x 5 5 = = Given I = cos(log x e )dx Let log x e = t 1 dx = dt dx = x et dt I = cost e t dt = 1 et ((cost + sint) + (cost sint))dt = 1 et (sint + cost) + C = x (sin(log e x ) + cos(log x e )) + C Hence (A) is the correct answer. 4. S K = K(K+1).K = K+1 S 1 + S S 1 = ( ) + ( 3 3 ) ( 11 ) = 1 4 [ ] = 1 4 [ = = 5 1 A A = ] = 33

2 cot 5. lim 3 x tanx x π cos(x+ π 4 4 ) = lim (1 tan 4 x) x π tan 3 x.( cosx 1 1 sinx) (1 tanx)(1+tanx) sec = lim x. x π tan 3 x.(cosx sinx) 4 (cosx sinx).(1+tanx).sec = lim x. x π cosx.tan 3 x.(cosx sinx) 4 = 8 Hence (C) is the correct answer. a 6. Let I = f(x)g(x)dx a = f(x)(4 g(x))dx a = 4 f(x)dx a I = 4 f(x)dx a I = 4 f(x)dx a I = f(x)dx a f(x)g(x)dx I a = f(a x)f(a x)dx 7. Since given vectors are coplanar, μ 1 1 Hence 1 μ 1 = 1 1 μ μ 3 3μ + = μ = 1, 1, Clearly (A) is the correct answer. 8. tan 1 ( x+3x 1 x 3x ) = π 4 5x 1 6x = 1 5x = 1 6x 6x + 5x 1 = x = 1, 1 6 But, since x >. Hence x = 1 6 R has only one element.

3 9. The given D. E can be written as dy dx + y x = ln x which is a linear D. E. Now I. F. = e 1 x dx = e ln x = x Hence solution of given D. E is y. x x. ln x dx y. x = x x ln x + C 4 Given y() = ln 1 C = y = x ln x x 4 y(e) = e 4 Hence (A) is the correct answer Area of region ABCDA is, = [(x + ) (x + 1)]dx 3 = (x x + 1)dx = = 1 9 = 15 = [ x3 x + x] The given ratio is 5th term from beginning 5 th term from the end T 5 T 7 = 4 1 C 4 ( 1 3 )1 4 ( 1 1) C 6 ( 1 3 )1 6 ( 1 1).33 = 4. (36) 1 3 Hence (A) is the correct answer

4 1. Corner points are 3π and π 4 4 of non-differentiable points { 3π 4, π 4 } 13. Equation of give circles are (x 1) + (y 1) = 4 (x 1) + (y 1) = 4 C 1 (1, 1) r 1 = (x 3) + (y 3) = 4 C (3, 3) r = PC 1 = PC =, C 1 C = 8 PC 1 + PC = C 1 C C 1 PC = π Hence area of quadrilateral PC 1 QC = 1 = 4 (D) is the correct option. 14. Solution: (B) p ~q a + b + d ~p q c (p ~q) (~p q) φ ((p ~q) (~p q)) (~p ~q) ~p ~q

5 Given P [ 3 1 ] P 5 = [ 15 1 ] Now Q P 5 = I 3 q 1 15 =, q =, q 3 15 = q 1 = 15, q 31 = 135, q 3 = 15 q 1+q 31 = = 1 q 3 15 Option (B) is the correct answer. 16. As we know, product of numbers is even when atleast one of the number must be even. Hence total subsets of A in which product of numbers is even = Total subsets total subsets in which all the elements are odd. = Since vertex of the hyperbola are (, ) and (, ), hence transverse axis of the hyperbola is x axis and centre is O(, ) Equation of hyperbola can be taken as x a + y b = 1 a = and ae = 3 e = 3 Since b = a (e 1) b = 9 4 = 5 Hence equation of hyperbola is x 4 y 5 = 1 Clearly point ( 6, 5 ) doesn t lies on the hyperbola.

6 18. If a complex number if purely imaginary, then it must be equal to minus times its conjugate. z α z+α = (z α z +α ) zz + αz αz α = (zz αz + αz α ) z = α α = 4 α = ± 19. Given f(θ) = 3cosθ + 5 sin (θ π 6 ) = 3cosθ + 5 (sinθ. 3 cosθ. 1 ) = 5 3 sinθ + 1 cosθ Maximum value of f(θ) is ( 5 3 ) + ( 1 ) = 19. Let α and β are the roots of given equation then α β = λ. Given λ + 1 λ = 1 α β + β α = 1 (α + β) = 3αβ ( n(m 4) 3m ) = 3. 3m ( n(m 4) 3m ) = 3. 3m m = 4 ± 3 Hence least value of m is Given (x) y = 4. e x y, taking log both the sides we, get y ln x = ln 4 + x y ln 4+x y = ( ) ln x+1 dy dx 1 (ln x+1). (ln 4+x) x = (ln x+1) (1 + ln x) dy dx x ln = (x.ln ) x

7 . To minimize the calculation, 3 numbers in G. P can be taken as a, a, ar. r Given product of 3 numbers is 51. a r. a. ar = 51 a = 8 Also, a + 4, a + 4, ar are in A.P r (a + 4) = a r ar r 5r + = (as a = 8) r = or 1 Hence numbers may be 4, 8, 16 or 16, 8, 4 In both the cases sum = 8 3. Each box contains 1 balls numbered from 1 to 1. n 1, n, n 3 are numbers on the balls drawn from the box B 1, B and B 3 respectively such that n 1 < n < n 3. i.e., all 3 numbers n 1, n, n 3 must be different and can be arranged only in one way (increasing). Now n 1, n, n 3 can be selected in 1 C 3 ways. Hence total number ways = 1 C 3. 1 = 1 C α β 1 4. For unique solution α 1 + β 1 α β C 1 C 1 + C + C 3 α + β + β 1 α + β β 1 α + β + β 1 β 1 (α + β + ) β 1 1 β α + β + Clearly point (, 4) satisfying the given condition.

8 5. Since given parabola is symmetric about the y axis, hence rectangle will also be symmetric about y axis. Let one vertex of the rectangle on the x axis be (α, ), then Area of rectangle A = α. (1 α ) da dα = 4 6α = α =, A = 3 6. Two different approaches we can use here. Approach 1: Let x be (t, t), then t t 1 Area of ΔPXQ = Δ = 1. 1(t t 6) Δ = t = 1 Hence area of ΔPXQ = 15 sq. units Approach : 4 For maximum area tangent to the parabola at X must be parallel to PQ. Let X(t, t), then dy dx = 4 (dy dx ) (t, t) = 1 t

9 1 t = = t = 1 X (1, 1) Area of ΔPXQ = 15 sq. units 4 7. Given, line perpendicular to given line passes through (7, 15) and (15, β) Hence 15 β = β = β = x + 4y λ = (7 λ)(31 λ) < {Since centres lie opposite side} λ (7, 31). (1) 7 λ 5 1 & 31 λ 5 7 λ 5 & 31 λ 1 λ or λ 1. () and λ 1 or λ 41. (3) (1) () (3) λ [1,1] 9. i j k Vector perpendicular to face OAB = 1 1 = 5i j 3k 1 3 i j k Vector perpendicular to face ABC = 1 1 = i 5j 3k 1 1 Angle between two faces cosθ = = θ = cos 1 ( )

10 Physics 1. By similar triangles ΔAEC~ΔIHA y L = d L Total distance (CD) = y + d + d + y = 3d. After along time So I = V R G = 15.5 = 6A 3. P 1 = ql P = ql Angle between them 6 Resultant dipole moment P = (ql) + (ql) + (ql) cos6 = 3 As P 1 = P direction of resultant is along -y axis this option is (A)

11 4. a rel = v rel = 1 1 v rel t t = 1 1 = 1s v bullet = = 9 m s v particle = 1 1 = 1 m s S = = 95 meter 5. a+1 Gm dm F = x a a+1 GM (A + Bx )dx x a a+1 a+1 = GM [ A dx + B dx] x a = GM[A[ 1 x ] a a+1 + B 1 ] = GM [A ( 1 a 1 a + )] + B 1 a

12 6. Torque on the rod about O, τ = kx. l = k. l. θ. l = k l. θ l = moment of inertia of rod about O, ω = l k M l f = 1 π M 6k = 6k M 7. R A = V = = P 5 R B = 1 = As resistance are in series current through them will be same. P A = i R A P B i = R B = 4 1 Possible option is C 8. d d cos6o = V s V a V a = V s = V

13 9. B = ( μ i 4πd ) 1 4 = 4π 1 7 i 4π ( 4 1 ) i = A 1. r = mqv qb = 1 r m q 11. v = ω k = 5 = 5 m s 1. Y = A. A. B. B. AB = (A. AB) + (B. AB) = A. (A. B) + B. (A + B) = A. B + B. A It is XOR gate 13. F r = du d = kr For circular motion F r = kr = mv r kr = mv (1) Bohr s quantization = mvr = nh π.... () From (1) & () m v m = kr 1 m ( nh πr ) = kr n h 4π mk = r4 r = ( h 4π mk ) 1 4 n 1 r n From equation (1) U n KE = 1 mv PE = 1 kr E = K + U = 1 mv + 1 kr = kr n

14 14. A C = 1 A C + A m = 16 A C A m = 4 A C = 1, A m = 6 μ = A m A c = Q i G Q C = i G I-case CQ = II-case C θ 5 v +R v 5 (+ 5R 5+R ) 5+R.() (1) From (1) and () R = Ω 16. dr = cdl l According to Questions l C dl = c dl l l l 1 Acording to Quation l C dl = c dl l l l 1 ( l) l = ( l)l 1 l = l 4 l = l = 1 =.5 m Let q 1 and q be the rate of flow of heat through inner part from outer part respecting net flow of heat q = q 1 + q

15 = K 1πR l πr. ΔT + K 1. π[(r) R ]. ΔT l l ΔT(K 1 + 3K ) If the cylinder is replaced by a single Material of thermal conductivity K then K. (R) q =. ΔT l On comparison K = K 1+3 K hc λ = (KE) Max + φ = φ 3.1 = φ φ =.76eV V + 1 = V = = 4 1 v 1 = 3 Distance of first image B = 3 = 14 3 cm 1 V 3 14 = = = Real image 7 cm right of B. 1. Energy dissipated when switch is thrown from 1 to.. mv x = mv

16 v x = v v y = v V net = ( v ) + ( v ) = v So path will be elliptical 3. t 1 = x = x (here total of two trains is x) v u 5 t = x v + u = x 11 t 1 = 11 t 5 4. Least count of screw gauge = 5 μm pitch L. C = no. of div on cirxular scale 5μm = 1mm N N = 5. I = 1 ε E. C I = 1 ε E V I =.96I 1 εe v =.96 1 ε E C E =.96 c v E =.96 μ r 1E = μ r ε r (For most of the transparent mediu μ r 1) Therefore.96 μ Hence E =.96 μ E = 4

17 6. u = gl(1 cos θ ) (i) V = velocity of ball after collision m M v = ( m + M ) u Since ball rises up to angle θ 1 m M v = gl(1 cos θ 1 ) = ( ) u (ii) m + M From (i) & (ii) m M m + m = 1 cos θ 1 = sin ( θ 1 ) 1 cos θ 7. U i + K i = U f + K f 8. KQ r + = KQ r v = KQ m ( 1 r 1 r ) v = KQ m ( 1 r 1 r ) + 1 mv sin ( θ ) M m = θ θ 1 θ + θ 1 M = ( θ θ 1 θ + θ 1 ) m m( + 1 ) = mk K = K = 5 K = 5 1 cm

18 9. Potential gradient = x = = ( 4 R+5 ) = R = R + 5 R = 395 Ω Chemistry 1. (ΔT f ) X = (ΔT f ) Y k f m x = k f m y 4 1 A 96 M = 3.7A 3A = 1 1 M 88. Clean water would have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more 3. Iodine gets oxidized to IO 3 when it reacts with an oxidizing agent (HNO 3 ). The oxidation number of I will be 1 = x + 3 ( ) 1 = x 6 x = = +5 Oxidation I IO 3 Reduction HNO 3 (dil) NO HNO 3 (conc) NO 4. ΔG = nfe cell = 965 = 386 kj ΔS = nf de dt = J o C = 96.5J at 98 K TΔS = 98 ( 96.5J) = 8.8 kj At constant T(= 48K) and pressure ΔG = ΔH TΔS ΔH = ΔG + TΔS = = kj

19 5. [Og 118 ]8s is configuration for Z = 1 it will belong to II nd group 6. Boiling point force of attraction in b, c, d -Hydrogen bonding takes place hence maximum force of attraction so high Boiling point. In (a) only vanderwall force of attraction (feable force) so having low Boiling point 7. a > b > c Acidic character % S character 1 +I Because acidic character stability of conjugate base (Anion) (a) CH C Anion ( ve) on Csp (b) CH 3 C C ve on Csp & + I of CH 3 (d) CH = C H( ve) on Csp Acidic character order CH C Csp > CH 3 C C Csp with+i of CH 3 > CH = C Csp So acidic character order = a > b > c 8. Cp is a molar heat capacity at constant pressure. It is a function of temperature it does not varies with pressure 9. PHBV is obtained by copolymerization of 3 Hydroxybutanoic acid & 3 hydroxypentanoic acid. 1.

20 11. A + B C + D Initially conc. a 1.5a At eq. a x 1.5 (a x) x x At equilibrium a x = 1.5a x.5 a = x a = x k C = (x) (x) (a x)(1.5a x) = 4x. x (x)(x) = 4 1. Lysine P I = 9.7 (basic) Aspargin Serine Neutral amino acid P I = 5.4 chain Alanine P I = 5.7 neutral with polar, but non ionisable side P I = 6 (neutral)

21 13. The value of adsorption is dependent on the inter molecular forces of attraction. Among the given options, H will have the weakest van der waal s force (London Dispersion), in which only electrons are involved. Hence, as force of attraction is weakest for H, it will have the lowest adsorption value. 14. If the complex is [M(H O) 6 ]Cl, M will have + oxidation state and hence, it has already lost electrons from s orbital since it has μ = 3.9 ( 3(3 + )) this tells us that there will be 3 unpaired so, the configuration can either be: 15. The electrolytic cell which is used for extraction of aluminum is a steel vessel. The vessel is lined with carbon, which acts as cathode and graphite is used at the anode. 16. K, Rb and Cs form super oxides on reaction with excess air Rb + O RbO (excess) RbO + H O RbOH = H O + O 17. A(s) B(g) + C(g)k P1 = x atm P 1 P 1 + P D(s) C(g) + E(g)k P = y atm P 1 + P P k p1 = P 1 (P 1 + P ) k p = P (P 1 + P ) k P1 + k P = (P 1 + P ) x + y = (P 1 + P ) P 1 + P = x + y (P 1 + P ) = x + y P Total = P B + P C + P E = (P 1 + P ) = x + y

22 18.. Ketone react with RMgx gives 3 o Alcohol 1. Nitriles are selectively reduced by DIBAL-H to imines followed by hydrolysis to aldehydes similarly, esters are also reduced to aldehyde with DlBAL-H. According to unit of rate constant it is a zero order reaction then half life of reaction will be t1 = C k = 5μg = 5year.5 μg year 3. PV = ZnRT P = ZnRT V At constant T and mol P Z V

23 4. As 1L solution have 1 3 mol CaSO 4 Eq. of CaSO 4 = eq. of CaCO 3 In 1L solution n CaSO4 v. f = n CaCO3 v. f. 1 3 = n CaCO3 n CaCO3 = 1 3 mol in 1L W CaCO3 = g in 1 L solution hardness in terms of CaCO 3 = W CaCO W total 1 6 = g 1g 1 6 = 1ppm Now n NaOH is 5 ml = M V = 5 Mass of NaOH is 5 ml = 4 g 1 =.1 mol 7. Give K 3 [Co(CN) 6 ] is inner orbital complex with hybridization d sp 3 and octahedral geometry. Ligands are approaching metal along the axes. Hence d x y,d z orbitals are directly in front of the ligands. 9. More nucleophilic nitrogen, more reactive with alkyl halide.

CHEMISTRY. JEE(Main) 2019 DATE : (SHIFT-1) PAPER 1 CONCEPT BASED

CHEMISTRY. JEE(Main) 2019 DATE : (SHIFT-1) PAPER 1 CONCEPT BASED CHEMISTRY JEE(Main) 2019 DATE :12-01-2019 (SHIFT-1) PAPER 1 CONCEPT BASED 1. [M(H 2 O) 6 ]Cl 2 is a coordination compound with spin only magnetic moment 3.9B.M. Find M? (a) Fe 2+, Co 2+ (b) V 2+, Co 2+