Practice Questions

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

Q77.Let y(x) be the solution of the differential equation 2x2 dy + (ey −2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to: (1) loge(2e) (2) loge 2 (3) 2 (4) 0

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

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 −−−−−

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)

Q78.The equation of the plane passing through the line of intersection of the planes →r⋅(ˆi + ˆj + ˆk) + 4 = 0 and parallel to the x-axis, is →r⋅(2ˆi + 3ˆj −ˆk) + + 6 = 0 (1) →r⋅(ˆi 3ˆk) + 6 = 0 (2) →r⋅(ˆi −3ˆk) + 6 = 0 (3) →r⋅(ˆj −3ˆk) −6 = 0 (4) →r⋅(ˆj −3ˆk)

Q78.Let O be the origin. Let OP→ = xˆi + yˆj −ˆk and OQ→ = −ˆi + 2ˆj + 3xˆk, x, y ∈R, x > 0, be such that −−−−−→ → → → → PQ = √20 and the vector OP is perpendicular to OQ. If OR = 3ˆi + zˆj −7ˆk, z ∈R, is coplanar with OP −→ and OQ, then the value of x2 + y2 + z2 is equal to (1) 7 (2) 9 (3) 2 (4) 1

Q78.The distance of the point 1, 1, 9 from the point of intersection of the line = = and the plane 1 2 2 𝑥+ 𝑦+ 𝑧= 17 is: (1) 19√2 (2) 2√19 (3) √38 (4) 38

Q78.The lines x = ay −1 = z −2 and x = 3y −2 = bz −2, (ab ≠0) are coplanar, if: (1) b = 1, a ∈R −{0} (2) a = 1, b ∈R −{0} (3) a = 2, b = 2 (4) a = 2, b = 3