Practice Questions

Q77.Let y = y(x) be the solution of the differential equation x tan( xy )dy = (y tan( xy ) −x)dx, −1 ≤x ≤1, y( 12 ) = π6 . Then the area of the region bounded by the curves x = 0, x = √21 and y = y(x) in the upper half plane is: (1) 1 8 (π −1) (2) 121 (π −3) (3) 4 1 (π −2) (4) 16 (π −1)

Q77.Let α be the angle between the lines whose direction cosines satisfy the equations l + m −n = 0 and l2 + m2 −n2 = 0. Then the value of sin4 α + cos4 α is : (1) 5 (2) 1 8 2 (3) 3 (4) 3 8 4

Q77.Let y = y(x) be the solution of the differential equation (x −x3)dy = (y + yx2 −3x4)dx, x > 2 If y(3) = 3, then y(4) is equal to: (1) 4 (2) 12 (3) 8 (4) 16 b If magnitudes of the vectors →a, b and →care √2, 1 and

Q77.Let three vectors →a, b and →cbe such that →a× b =→c, b ×→c=→a and →a = 2. Then which one of the following is not true? b b b × is 2 (1) →a× ((→ → → (2) → +→c) ( −→c)) = 0 Projection of →a on ( ×→c) + = 8 (4) 3→a+→b −2→c 2 = 51 (3) [→a →b →c] [→c →a →b ] JEE Main 2021 (22 Jul Shift 1) JEE Main Previous Year Paper = 2. If P(α, β, γ) is the

Q77.Let →a = ˆi + 2ˆj −3ˆk and b = 2ˆi −3ˆj + 5ˆk. If →r×→a = b ×→r,→r⋅(αˆi + 2ˆj + ˆk) 2 is equal to : = −1, α ∈R, then the value of α + →r →r⋅(2ˆi + 5ˆj −αˆk) (1) 9 (2) 15 (3) 13 (4) 11

Q77.Let y = y(x) be a solution curve of the differential equation (y + 1) tan2 xdx + tan xdy + ydx = 0, x ∈(0, π2 ). If lim xy(x) = 1, then the value of y( π4 ) is: x→0+ (1) π 4 + 1 (2) π4 −1 (3) π 4 (4) −π4 is equal b

Q77.If for a > 0, the feet of perpendiculars from the points A(a, −2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, −a, −1) and D respectively, then the length of line segment CD is equal to : (1) √31 (2) √41 (3) √55 (4) √66

Q77.In a triangle ABC, if BC→ = 3, CA→ = 5 and BA→ = 7, then the projection of the vector BA→ on BC→ is equal to JEE Main 2021 (20 Jul Shift 2) JEE Main Previous Year Paper (1) 19 (2) 13 2 2 (3) 11 (4) 15 2 2

Q77.The population 𝑃= 𝑃𝑡 at time 𝑡 of a certain species follows the differential equation 𝑑𝑃 0 . 5𝑃- 450. If 𝑑𝑡= 𝑃0 = 850, then the time at which population becomes zero is: (1) log𝑒9 (2) 2log𝑒18 1 (3) log𝑒18 (4) 2log𝑒18 𝑥- 3 𝑦- 4 𝑧- 5

Q77.Let 𝑦= 𝑦( 𝑥) be the solution of the differential equation 𝑑𝑦 1 + 𝑥𝑒𝑦- 𝑥, - √2 < 𝑥< √2, 𝑦0 = 0 𝑑𝑥= , then the minimum value of 𝑦𝑥, 𝑥∈-√2, √2 is equal to : (1) 2 - √3 - loge2 (2) 2 + √3 + loge2 (3) 1 + √3 - loge√3 - 1 (4) 1 - √3 - loge√3 - 1

Q77.If →a and→b are perpendicular, then →a× (→a (→a (→a →b))) 4→ (1) →a b (2) →0 → 4→ 1 (3) →a× b (4) 2 →a b

Q77.If 𝑦= 𝑦( 𝑥) is the solution curve of the differential equation 𝑥2 d𝑦+ 𝑦- 1 0; 𝑥> 0 and 𝑦( 1 ) = 1, 𝑥d𝑥= 1 then 𝑦 is equal to : 2 (1) 3 + e (2) 3 - e 3 1 1 (3) - (4) 3 + 2 √e √e

Q77.A differential equation representing the family of parabolas with axis parallel to y−axis and whose length of latus rectum is the distance of the point (2, −3) from the line 3x + 4y = 5, is given by: (1) 11 d2x dy2 = 10 (2) 11 dx2d2y = 10 d2y (3) 10 = 11 (4) 10 d2xdy2 = 11 dx2 = 1 and

Q77.If f(x) = {ax2 + b ; |x| < 1 respectively: (1) 1 2 , 12 (2) 12 , −32 (3) 2 5 , −32 (4) −12 , 32

Q77.If 𝑦0 = 0, then for 𝑦= 1, the value of 𝑥 lies in the interval : 𝑑𝑥= 2𝑥+ 2𝑥+ 𝑦log𝑒2, 1 (1) 1, 2 (2) 2, 1 (3) 2, 3 (4) 0, 1 2

Q77.If the curve y = y(x) is the solution of the differential equation 2(x2 + x5/4)dy −y(x + x1/4)dx = 2x9/4dx, x > 0 which passes through the point (1, 1 −43 loge 2), then the value of y(16) is equal to (1) 4( 313 + 38 loge 3) (2) ( 313 + 38 loge 3) (3) 4( 313 −83 loge 3) (4) ( 313 −83 loge 3) −−

Q77.If vectors →a1 = xˆi −ˆj + ˆk and →a2 = ˆi + yˆj + zˆk are collinear, then a possible unit vector parallel to the vector xˆi + yˆj + zˆk is: (1) + 1 (−ˆj √2 ˆk) (2) √31 (ˆi +ˆj −ˆk) (3) + ˆk) √2 1 (ˆi −ˆj) (4) √31 (ˆi −ˆj JEE Main 2021 (26 Feb Shift 2) JEE Main Previous Year Paper

Q77.Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If dt, 0 ≤x ≤1 and f(0) = 0, then : lim x21 ∫x0 x→0 ∫x0 √1 −(f ′(t))2 dt = ∫x0 f(t) f(t)dt (1) does not exist (2) equals 0 (3) equals 1 (4) equals 21 2x+y−2x

Q77.Let f(x) be a differentiable function defined on [0, 2] such that f ′(x) = f ′(2 −x) for all x ∈(0, 2), f(0) = 1 and f(2) = e2. Then the value of ∫20 f(x)dx is (1) 2(1 + e2) (2) 1 + e2 (3) 1 −e2 (4) 2(1 −e2) = 1 and

Q77.Let y = y(x) be the solution of the differential equation xdy = (y + x3 cos x)dx with y(π) = 0, then y( π2 ) is equal to: (1) π2 4 + π2 (2) π22 + π4 (3) π2 2 −π4 (4) π24 −π2

Q77. n→∞[ (1) 1 (2) 1 2 4 (3) 1 (4) 1 3

Q77.Which of the following is true for y(x) that satisfies the differential equation dy = xy −1 + x −y; y(0) = 0 dx (1) y(1) = e−12 −1 (2) y(1) = e 12 −e−12 (3) y(1) = 1 (4) y(1) = e 21 −1 → → + 2ˆj + = −3, then →r⋅(2ˆi −3ˆj + ˆk) is

Q77.Let y = y(x) be the solution of the differential equation dydx = 2(y + 2 sin x −5)x −2 cos x such that y(0) = 7. Then y(π) is equal to (1) 7eπ2 + 5 (2) eπ2 + 5 (3) 2eπ2 + 5 (4) 3eπ2 + 5

Q77.Let y = y(x) be solution of the differential equation loge( dxdy ) y(−23 loge 2) = α loge 2 , then the value of α is equal to: JEE Main 2021 (27 Jul Shift 1) JEE Main Previous Year Paper (1) −14 (2) 41 (3) 2 (4) −12 →

Q77.Let y = y(x) be the solution of the differential equation dxdy = (y + 1)((y + 1)ex2/2 −x), y(2) = 0. Then the value of dxdy at x = 1 is equal to (1) −e3/2 (2) − 2e2 (e2+1)2 (1+e2)2 (3) e5/2 (4) 5e1/2 (1+e2)2 (e2+1)2 −−−−−