Practice Questions

Q77.Let y = y(x) be the solution of the differential equation (1 + x2) dxdy + y = etan−1 x , y(1) = 0. Then y(0) is (1) 2 1 (eπ/2 −1) (2) 21 (1 −eπ/2) (3) 4 1 (1 −eπ/2) (4) 14 (eπ/2 −1)

Q77.Let →a = 2^i + ^j −^k, b = ((→a× (^i + ^j)) ×^i) ×^i. Then the square of the projection of →a on b is : (1) 1 (2) 2 3 3 (3) 2 (4) 1 5 →

Q77.Let OA→ =→a, OB→ = 12→a+ 4→b and OC→ = →b, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then area of the areaquadrilateralof S OABC is equal to _____ (1) 6 (2) 10 (3) 7 (4) 8

Q77.Let →𝑎= ^𝑖+ 𝛼 ^𝑗+ 𝛽 ^𝑘 , 𝛼, 𝛽∈𝑅. Let a vector →𝑏 be such that the angle between →𝑎 and →𝑏 is 𝜋 and →𝑏 = 6, If 4 →𝑎· →𝑏= 3√2, then the value of 𝛼2 + 𝛽2 | →𝑎× →𝑏|2 is equal to (1) 90 (2) 75 (3) 95 (4) 85 2 is equal to

Q77.Let A(2, 3, 5) and C(−3, 4, −2) be opposite vertices of a parallelogram ABCD if the diagonal −→ BD = ˆi + 2ˆj + 3ˆk then the area of the parallelogram is equal to (1) 1 2 √410 (2) 21 √474 (3) 1 2 √586 (4) 21 √306 → → →

Q77.Let three vectors →a = α^i + 4^j + 2^k, b = 5^i + 3^j + 4^k,→c= x^i + y^j + z^k form a triangle such that →c = →a −→b and the area of the triangle is 5√6. If α is a positive real number, then |→c|2 is equal to: (1) 16 (2) 14 (3) 12 (4) 10 → →−−−

Q77.The temperature 𝑇𝑡 of a body at time 𝑡= 0 is 160° 𝐹 and it decreases continuously as per the differential 𝑑𝑇 equation 𝑑𝑡= −𝐾𝑇−80, where 𝐾 is positive constant. If 𝑇15 = 120° 𝐹, then 𝑇45 is equal to (1) 85° 𝐹 (2) 95° 𝐹 (3) 90° 𝐹 (4) 80° 𝐹

Q77.Let →a = 4^i −^j + ^k,→b = 11^i −^j + ^k and →c be a vector such that (→a + →b) × →c = →c × (−2→a + 3→b). If (2→a + 3→b) ⋅→c = 1670, then |→c|2 is equal to : (1) 1609 (2) 1618 (3) 1600 (4) 1627 →

Q77.Let →𝑎= 3 ^𝑖+ ^𝑗−2 ^𝑘, 𝑏= 4 ^𝑖+ ^𝑗+ 7 ^𝑘 and →𝑐= ^𝑖−3 ^𝑗+ 4 ^𝑘 be three vectors. If a vectors →𝑝 satisfies →𝑝× →𝑏= →𝑐× →𝑏 and →𝑝⋅→𝑎= 0, then →𝑝⋅ ^𝑖− ^𝑗− ^𝑘 is equal to (1) 24 (2) 36 (3) 28 (4) 32

Q77.Let 𝑦= 𝑦𝑥 be the solution of the differential equation 𝑑𝑦 2𝑥𝑥+ 𝑦3 −𝑥𝑥+ 𝑦−1, 𝑦0 = 1. Then, 1 + 𝑦1 𝑑𝑥= √2 √2 equals: (1) 4 (2) 3 4 + √𝑒 3 −√𝑒 2 1 (3) (4) 1 + √𝑒 2 −√𝑒

Q77.Consider three vectors →a,→b, →c. Let |→a| = 2, |→b| = 3 and →a = →b × →c. If α ∈[0, 3 ] is the angle between the vectors →b and →c, then the minimum value of 27|→c −→a|2 is equal to: (1) 110 (2) 124 (3) 121 (4) 105

Q77.Let x = x(t) and y = y(t) be solutions of the differential equations dxdt + ax = 0 and dydt + by = 0 respectively, a, b ∈R. Given that x(0) = 2 ; y(0) = 1 and 3 y(1) = 2 x(1), the value of t, for which x(t) = y(t), is : (1) log 2 2 (2) log4 3 3 4 2 (3) log3 4 (4) log 3 → → and→cbe the vector such that →a×→c= b and →a⋅→c= 3, then

Q77.Consider a 𝛥𝐴𝐵𝐶 where 𝐴1, 3, 2, 𝐵−2, 8, 0 and 𝐶3, 6, 7. If the angle bisector of ∠𝐵𝐴𝐶 meets the line 𝐵𝐶 at 𝐷, then the length of the projection of the vector →𝐴𝐷 on the vector →𝐴𝐶 is: (1) 37 (2) √38 2√38 2 39 (3) (4) √19 2√38

Q77.The position vectors of the vertices A, B and C of a triangle are 2 ^i - 3 ^j + 3 ^k, 2 ^i + 2 ^j + 3 ^k and - ^i + ^j + 3 ^k respectively. Let 𝑙 denotes the length of the angle bisector AD of ∠BAC where D is on the line segment BC, then 2𝑙2 equals : (1) 49 (2) 42 (3) 50 (4) 45

Q78.If the mirror image of the point 𝑃( 3, 4, 9 ) in the line 𝑥−1 = 𝑦+ 1 = 𝑧−2 is 𝛼, 𝛽, 𝛾, then 14𝛼+ 𝛽+ 𝛾 is: 3 2 1 (1) 102 (2) 138 (3) 108 (4) 132 𝑥+ 3 𝑦−4 𝑧+ 1

Q78.Let y = y(x) be the solution of the differential equation (2x loge x) dxdy + 2y = x3 loge x, x > 0 and y (e−1) = 0. Then, y(e) is equal to (1) −3e (2) −32e (3) −23e (4) −2e

Q78.Let OA→ = 2→a, OB = 6→a + 5→b and OC = 3→b, where O is the origin. If the area of the parallelogram with −−→ → adjacent sides OA and OC is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to : (1) 32 (2) 40 (3) 38 (4) 35

Q78.The distance of the point 𝑄( 0, 2, – 2 ) form the line passing through the point 𝑃( 5, – 4, 3 ) and perpendicular to the lines →𝑟= −3 ^𝑖+ 2 ^𝑘+ 𝜆2 ^𝑖+ 3 ^𝑗+ 5 ^𝑘, 𝜆∈ℝ and →𝑟= ^𝑖−2 ^𝑗+ ^𝑘+ 𝜇− ^𝑖+ 3 ^𝑗+ 2 ^𝑘, 𝜇∈ℝ is (1) √86 (2) √20 (3) √54 (4) √74

Q78.Let 𝛼, 𝛽, 𝛾 be mirror image of the point 2, 3, 5 in the line 𝑥−1 = 𝑦−2 = 𝑧−3 . Then 2𝛼+ 3𝛽+ 4𝛾 is equal to 2 3 4 (1) 32 (2) 33 (3) 31 (4) 34 𝑥−1 𝑦+ 1 𝑧+ 4

Q78.If →a = ˆi + 2ˆj + ˆk, b = 3(ˆi −ˆj + ˆk) is equal to × − b →a⋅((→c → → b) −→c) (1) 32 (2) 24 (3) 20 (4) 36

Q78.Let a unit vector which makes an angle of 60∘ with 2^i + 2^j −^k and angle 45∘ with ^i −^k be C. Then → is : C + + (−12^i 1 ^j −√23 ^k) 3√2 (1) √2 + − + 3 + 21 )^i 1 )^j + √23 )^k ^i −12 ^k (2) ( √31 ( √31 3√2 ( √31 2√2 (3) √2 ^i + + 3 3√2 1 ^j −12 ^k (4) −√23 ^i + √23 ^j + ( 21 3 )^k

Q78.Let →a = 2^i + 5^j −^k,→b = 2^i −2^j + 2^k and→cbe three vectors such that (→c +^i) × (→a + →b +^i) = →a × (→c +^i). If →a ⋅→c = −29, then →c ⋅(−2^i + ^j + ^k) is equal to: (1) 15 (2) 12 (3) 10 (4) 5

Q78.Let →a = ^i + ^j + ^k,→b = 2^i + 4^j −5^k and →c = x^i + 2^j + 3^k, x ∈R. If →d is the unit vector in the direction of →b + →c such that →a ⋅→d = 1, then (→a × →b) ⋅→c is equal to (1) 11 (2) 3 (3) 9 (4) 6

Q78.Let →a = aiˆi + a2ˆj + a3ˆk and b = b1ˆi + b2ˆj + b3ˆk be two vectors such that →a = 1;→a⋅ b = 2 and b = 4. If × −3b, then the angle between b and →cis equal to : →c= 2(→a → → → b) JEE Main 2024 (30 Jan Shift 1) JEE Main Previous Year Paper (1) cos−1( √32 ) (2) cos−1(−1√3 ) 2 ) 3 (3) cos−1(−√32 ) (4) cos−1(

Q78.If the shortest distance between the lines is √n L2 : →r = 2(1 + μ)^i + 3(1 + μ)^j + (5 + μ)^k, μ ∈R , where gcd(m, n) = 1, then the value of m + n equals (1) 390 (2) 384 (3) 377 (4) 387