Practice Questions

Q76.Let y = y(x) be the solution of the differential equation x(1 −x2) dxdy + (3x2y −y −4x3) = 0, x > 1 with y(2) = −2. Then y(3) is equal to (1) −18 (2) −12 (3) −6 (4) −3

Q76.Let the solution curve y = y(x) of the differential equation (1 + e2x)( dxdy y) (0, π2 ). Then, x→∞exy(x)lim is equal to JEE Main 2022 (29 Jul Shift 1) JEE Main Previous Year Paper (1) π (2) 3π 4 4 (3) π (4) 3π 2 2 → b = b + λ→c. If→b and →care non-

Q76.If 𝑏𝑛= ∫02 cos2𝑛𝑥sin𝑥𝑑𝑥, 1 1 1 (1) 𝑏3 - 𝑏2, 𝑏4 - 𝑏3, 𝑏5 - 𝑏4 are in an A.P. with (2) 𝑏3 - 𝑏2, 𝑏4 - 𝑏3, 𝑏5 - 𝑏4 are in an A.P. with common common difference-2 difference 2 (3) 𝑏3 - 𝑏2, 𝑏4 - 𝑏3, 𝑏5 - 𝑏4 are in a G.P. (4) 1 1 1 are in an A.P. with common 𝑏3 - 𝑏2, 𝑏4 - 𝑏3, 𝑏5 - 𝑏4 difference -2

Q76.If dx + 2x−1 = 0, x, y > 0, y(1) = 1 , then y(2) is equal to (1) 2 + log2 3 (2) 2 + log2 2 (3) 2 −log−2 3 (4) 2 −log2 3 → →

Q76.If dx dy + 2y tan x = sin x, 0 < x < π2 and y( π3 ) = 0 , then the maximum value of y(x) is JEE Main 2022 (26 Jul Shift 1) JEE Main Previous Year Paper (1) 1 (2) 3 8 4 (3) 1 (4) 3 4 8 → →

Q76.If y = y(x) is the solution of the differential equation (1 + e2x) dxdy + 2(1 + y2)ex = 0 and y(0) = 0, then 2 + (y(logc √3)) is equal to: 6(y′(0) ) (1) 2 (2) −2 (3) −4 (4) −1

Q76.The differential equation of the family of circles passing through the points (0, 2) and (0, −2) is (1) 2xy dxdy + (x2 −y2 + 4) = 0 (2) 2xy dxdy + (x2 + y2 −4) = 0 (3) 2xy dxdy + (y2 −x2 + 4) = 0 (4) 2xy dxdy −(x2 −y2 + 4) = 0 →

Q76.The area bounded by the curve y = x2 −9 and the line y = 3 is (1) 8√6 −16√12 −72 (2) 8√6 + 8√12 −72 (3) 16√6 + 16√12 −72 (4) 16√6 −16√12 −64 → → → → → is b b × b × × (→c×→a) →c

Q76.If the solution curve of the differential equation ((tan−1 y) −x)dy = (1 + y2)dx passes through the point (1, 0) then the abscissa of the point on the curve whose ordinate is tan(1) is (1) 2 (2) 2e (3) 3 (4) 2e e →

Q76.The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is (1) 9 (2) 7 (3) 5 (4) 3

Q76.The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by x2 . If the curve xy−x2y2−1 passes through the point (1, 1), then e ⋅y(e) is equal to (1) 1−tan(1) (2) tan(1) 1+tan(1) (3) 1 (4) 1+tan(1) 1−tan(1)

Q76.Let a smooth curve y = f(x) be such that the slope of the tangent at any point (x, y) on it is directly proportional to ( −yx ). If the curve passes through the points (1, 2) and (8, 1), then y( 81 ) is equal to (1) 2 loge 2 (2) 4 (3) 1 (4) 4 loge 2 → → → →

Q76.Let x = x(y) be the solution of the differential equation 2ye y2 dx + (y2 )dy Then, x(e) is equal to (1) e loge(2) (2) −e loge(2) (3) e2 loge(2) (4) −e2 loge(2)

Q76.Let y = y1(x) and y = y2(x) be two distinct solutions of the differential equation dxdy = x + y, with y1(0) = 0 and y2(0) = 1 respectively. Then, the number of points of intersection of y = y1(x) and y = y2(x) is (1) 0 (2) 1 (3) 2 (4) 3 → →

Q76.The general solution of the differential equation 𝑥- 𝑦2𝑑𝑥+ 𝑦5𝑥+ 𝑦2𝑑𝑦= 0 is 4 3 4 3 (1) 𝑦2 + 𝑥 = 𝐶𝑦2 + 2𝑥 (2) 𝑦2 + 2𝑥 = 𝐶𝑦2 + 𝑥 3 4 3 4 (3) 𝑦2 + 𝑥 = 𝐶2𝑦2 + 𝑥 (4) 𝑦2 + 2𝑥 = 𝐶2𝑦2 + 𝑥 → → → → → →

Q77.If 2, 3, 9, 5, 2, 1, 1, 𝜆, 8 and 𝜆, 2, 3 are coplanar, then the product of all possible values of 𝜆 is (1) 21 (2) 59 2 8 57 95 (3) (4) 8 8

Q77.Let a and b be two unit vectors such that |(a + b) + 2(a × b)| = 2. If θ ∈(0, π) is the angle between ˆa and ˆb , then among the statements: (S1) : 2 ˆa × ˆb = ˆa −ˆb is 1 + (S2) : The projection of ˆa on 2 (ˆa ˆb) (1) Only (S1) is true. (2) Only (S2) is true. (3) Both (S1) and (S2) are true. (4) Both (S1) and (S2) are false. JEE Main 2022 (24 Jun Shift 2) JEE Main Previous Year Paper

Q77.Let →a and b be the vectors along the diagonal of a parallelogram having area 2√2. Let the angle between →a and → → → → → → × −2b, then an angle between b and →cis b be acute. →a = 1 and →a. b = →a× b . If →c= 2√2(→a b) (1) −π (2) 5π 4 6 (3) π (4) 3π 3 4 P . Then the

Q77.Let the vectors →𝑎= 1 + 𝑡 ^𝑖+ 1 - 𝑡 ^𝑗+ ^𝑘, →𝑏= 1 - 𝑡 ^𝑖+ 1 + t ^𝑗+ 2 ^𝑘 and →𝑐= 𝑡 ^𝑖- 𝑡 ^𝑗+ ^𝑘, 𝑡∈𝑅 be such that for 𝛼, 𝛽, 𝛾∈𝑅, 𝛼 →𝑎+ 𝛽 →𝑏+ 𝛾 →𝑐= →0 ⇒𝛼= 𝛽= 𝛾= 0. Then, the set of all values of 𝑡 is (1) a non-empty finite set (2) equal to 𝑁 (3) equal to 𝑅- 0 (4) equal to 𝑅

Q77.Let →a = 3ˆi + ˆj and→b = ˆi + 2ˆj + ˆk. Let →cbe a vector satisfying →a× (→ ×→c) parallel, then the value of λ is (1) −5 (2) 5 (3) 1 (4) −1 θ is the angle between the vectors

Q77.If dy + ex(x2 −2)y = (x2 −2x)(x2 −2)e2x and y(0) = 0 , then the value of y(2) is dx (1) −1 (2) 1 (3) 0 (4) e →

Q77.Let 𝐴𝐵𝐶 be a triangle such that 𝐵𝐶= →𝑎, 𝐶𝐴= 𝑏, 𝐴𝐵= →𝑐, →𝑎= 6√2, 𝑏= 2√3 and 𝑏· →𝑐= 12 Consider the statements : 𝑆1: →𝑎× →𝑏+ →𝑐× →𝑏- →𝑐= 62√2 - 1 𝑆2: ∠𝐴𝐵𝐶= cos-1√ 23. Then (1) both 𝑆1 and 𝑆2are true (2) only 𝑆1 is true (3) only 𝑆2 is true (4) both 𝑆1 and 𝑆2 are false 𝑥- 3 𝑦+ 4 𝑧- 7

Q77.Let →a = αˆi + ˆj + βˆk and b = 3ˆi −5ˆj + 4ˆk be two vectors, such that →a× b = −ˆi + 9ˆi + 12ˆk. Then the → → projection of b −2→a on b +→a is equal to (1) 2 (2) 395 (3) 9 (4) 465 → → → 23 × b × 2ˆj is equal to ⋅ˆk = 2 , then

Q77.Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 tan x(cos x −y). if the curve passes π through the point ( π4 , 0), then the value of ∫ 0 2 ydx is equal to (1) (2 −√2) + √2π (2) 2 − √2π (3) (2 + √2) + √2π (4) 2 + √2π →

Q77.The area enclosed by y2 = 8x and y = √2x that lies outside the triangle formed by y = √2x, x = 1, y = 2√2 , is equal to (1) 16√2 (2) 11√2 6 6 (3) 13√2 (4) 5√2 6 6