JEE Main 2020 Maths Paper With Solutions Shift 1 Sept 4

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September 4 Shift 1 - Maths

Question 1.Let y = y(x) be the solution of the differential equation, xy'-y = x2(xcosx+sinx),x > 0. If y(π) = π, then y’’(π/2)+y(π/2) is equal to:

  1. a) 2+(π/2)+(π2/4)
  2. b) 2+(π/2)
  3. c) 1+(π/2)
  4. d) 1+(π/2)+ (π2/4)

Solution:

  1. xy’-y = x2(x cosx + sinx) x > 0, y(π) =π

    y'-(1/x)y = x(x cosx+sin x)

    I.F = e1xdxe^{-\int\frac{1}{x}dx }

    = e- ln x

    = 1/x

    y(1/x) = ∫(1/x)x(x cos x+sin x) dx

    (y/x) = ∫(x cos x+ sin x) dx

    (y/x) = ∫ (d/dx)(x sin x) dx

    (y/x) = x sin x+C

    y = x2 sin x+Cx

    x = π, y = π

    π = πC

    C = 1

    y = x2 sin x+x

    y(π/2) = (π2/4)+(π/2)

    y' = 2x sinx+x2cosx+1

    y’’ = 2sinx+2xcosx+2x cosx-x2sinx

    y’’(/2) = 2-(2/4)

    y(π/2)+ y’’(π/2) = 2+π/2

    Answer:(b)


Question 2. The value of r=02050rC6\sum_{r=0}^{20} \: ^{50-r}C_{6} is equal to:

  1. a) 51C7-30C7
  2. b) 51C7+30C7
  3. c) 50C7-30C7
  4. d) 50C6-30C6

Solution:

  1. r=02050rC6\sum_{r=0}^{20} \: ^{50-r}C_{6}

    = 50C6+49C6+48C6+…+31C6+30C6

    Add and subtract 30C7

    Using nCr+ nCr-1 = n+1Cr

    30C6+30C7 = 31C7

    31C6+31C7 = 32C7

    Similarly solving

    51C7-30C7

    Answer: (a)


Question 3. Let [t] denote the greatest integer ≤ t. Then the equation in x, [x]2+2[x+2]-7 = 0 has:

  1. a) exactly four integral solutions.
  2. b) infinitely many solutions.
  3. c) no integral solution.
  4. d) exactly two solutions.

Solution:

  1. [x]2+2[x+2]-7 = 0

    [x]2 +2[x]-3 = 0

    let [x] = y

    y2+3y-y -3 = 0

    (y -1)(y+3) = 0

    [x] = 1 or [x] = -3

    x ∈ [1,2) or x ∈ [-3,-2)

    Answer: (b)


Question 4. Let P(3, 3) be a point on the hyperbola, (x2/a2)-(y2/b2) = 1. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair (a2, e2) is equal to:

  1. a) (9, 3)
  2. b) (9/2, 2)
  3. c) (9/2, 3)
  4. d) (3/2, 2)

Solution:

  1. (x2/a2)-(y2/b2) = 1

    Point P(3, 3) on hyperbola.

    (9/a2)-(9/b2) = 1 ..(i)

    Equation of normal (a2x/3)+(b2y)/3 = a2

    At x axis y = 0

    a2x/3 = a2e2

    x = 3e2 = 9

    3e2 = 9

    e2 = 3

    e = √3

    e2 = 1+b2/a2 = 3

    b2 = 2a2 ..(ii)

    Put in equation 1

    (9/a2)-(9/2a2) = 1

    (9/2a2) = 1

    a2 = 9/2

    (a2, e2) = (9/2,3)

    Answer: (c)


Question 5. Let (x2/a2)+(y2/b2) = 1 (a>b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, Φ(t) = (5/12)+t-t2, then a2+b2 is equal to:

  1. a) 135
  2. b) 116
  3. c) 126
  4. d) 145

Solution:

  1. L.R = 2b2/a = 10 ..(i)

    Φ(t) = (5/12)-(t-1/2)2+(1/4)

    = (8/12)-(t-1/2)2

    Φ(t)max = 2/3 = e

    e2 = 1-(b2/a2)

    = 4/9

    b2/a2 = 5/9

    From (i)

    b2/a.a = 5/9

    5/a = 5/9

    a = 9

    b2 = 45

    a2+b2 = 81+45 = 126

    Answer: (c)


Question 6. Let f(x)=x(1+x)2dxf(x) = \int \frac{\sqrt{x}}{(1+x)^{2}}dx , x≥0. Then f(3)-f(1) is equal to:

  1. a) (-π/6)+(1/2)+(√3/4)
  2. b) (π/6)+(1/2)-(√3/4)
  3. c) (-π/12)+(1/2)+(√3/4)
  4. d) (π/12)+(1/2)-(√3/4)

Solution:

  1. f(x)=x(1+x)2dxf(x) = \int \frac{\sqrt{x}}{(1+x)^{2}}dx

    Substituting x = tan2t

    dx = 2tan t sec2t dt

    f(x) = tant.2tantsec2tsec4tdt\int \frac{\tan t. 2\tan t\sec ^{2}t}{\sec ^{4}t}dt

    f(x) = 2∫sin2t dt

    x = 3 t = π/3

    x = 1 t = π/4

    JEE Main 2020 Paper With Solution Maths Sept 4 Shift 1

    Answer: (d)


Question 7. If 1+(1-22×1)+(1-42×3)+(1-62×5)+......+(1-202×19) = α - 220β, then an ordered pair (α, β) is equal to:

  1. a) (10,97)
  2. b) (11,103)
  3. c) (11,97)
  4. d) (10,103)

Solution:

  1. Tn = 1-(2n)2(2n-1)

    = 1-4n2(2n-1)

    = 1-8n3+4n2

    Sn = n=110Tn\sum_{n=1}^{10}T_{n}

    = n-8n3+4n2

    = n-8×(n2(n+1)2/4)+4n(n+1)(2n+1)/6

    S10 = 10-2×100×121 +(2/3)×10×11×21

    = 10-24200+1540

    = 10-22660

    Sum of series = 11-220×103

    = α - 220β

    α = 11

    β = 103

    Answer: (b)


Question 8. The integral (xxsinx+cosx)2dx\int \left ( \frac{x}{x\sin x+\cos x} \right )^{2}dx is equal to: (where C is a constant of integration)

  1. a) tan x-(x sec x/(x sin x+cos x))+C
  2. b) sec x-(x tan x/(x sin x+cos x))+C
  3. c) sec x+(x tan x/(x sin x+cos x))+C
  4. d) tan x+(x sec x/(x sin x+cos x))+C

Solution:

  1. (xxsinx+cosx)2dx\int \left ( \frac{x}{x\sin x+\cos x} \right )^{2}dx

    JEE Main 2020 Papers With Solutions Sept 4 Maths Shift 1

    = tan x-(x sec x/(x sin x+cos x))+C

    Answer: (a)


Question 9. Let f(x)= |x-2| and g(x) = f(f(x)) ,x∈0,4]. Then 03(g(x)f(x))dx\int_{0}^{3}(g(x)-f(x))dx is equal to:

  1. a) 1/2
  2. b) 0
  3. c) 1
  4. d) 3/2

Solution:

  1. f(x) = |x-2| = {x2,x22x,x<2\left\{\begin{matrix} x-2, & x\geq 2\\ 2-x, & x< 2 \end{matrix}\right.

    g(x) = ||x-2|-2| = {x4,x2x,x<2\left\{\begin{matrix} \left | x-4 \right |, & x\geq 2\\ \left | x \right |, & x< 2 \end{matrix}\right.

    = {4x,x[2,4)x4,x4x,x[0,2)\left\{\begin{matrix} 4-x, & x\in [2,4)\\ x-4 ,&x\geq 4 \\ x, & x\in [0,2) \end{matrix}\right.

    03(g(x)f(x))dx\int_{0}^{3}(g(x)-f(x))dx

    = 03g(x)dx03f(x)dx\int_{0}^{3}g(x)dx-\int_{0}^{3}f(x)dx

    = [(1/2)×2×2+1+(1/2)×1×1]-[(1/2)×2×2+(1/2)×1×1)]

    = (7/2)-(5/2)

    = 1

    JEE Main 2020 Maths Papers With Solutions Sept 4 Shift 1

    Answer: (c)


Question 10. Let x0 be the point of local maxima of f(x) = a.(b×c)\vec{a}.(\vec{b}\times \vec{c}) where a=xi^2j^+3k^\vec{a}= x\hat{i}-2\hat{j}+3\hat{k}, b=2i^+xj^k^\vec{b}= -2\hat{i}+x\hat{j}-\hat{k} and c=7i^2j^+xk^\vec{c}= 7\hat{i}-2\hat{j}+x\hat{k}. Then the value of a.b+b.c+c.a\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a} at x = x0 is:

  1. a) -22
  2. b) -4
  3. c) -30
  4. d) 14

Solution:

  1. a.(b×c)\vec{a}.(\vec{b}\times \vec{c})

    = x232x172x\begin{vmatrix} x & -2 & 3\\ -2 & x & -1\\ 7 & -2 & x \end{vmatrix}

    = x(x2-2)+2(-2x+7)+ 3(4-7x)

    = x3-2x- 4x+14+12-21x

    f(x) = x3- 27x+26

    f’(x) = 3x2-27 = 0

    x = 3

    f’’(x) = 6x

    at x = 3, f” (3) = 6×3= 18 > 0

    at x = -3, f” (-3) = 6×3 = -18 < 0

    Max at x0 = -3

    a=(3,2,3)\vec{a}= (-3,-2,3), b=(2,3,1)\vec{b}= (-2,-3,-1), c=(7,2,3)\vec{c}= (7,-2,-3)

    So a.b+b.c+c.a\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}

    = 6+6-3-14+6+3-21+4-9

    = 25-47

    = -22

    Answer: (a)


Question 11.A triangle ABC lying in the first quadrant has two vertices as A(1,2) and B(3,1). If BAC = 900 and ar(ΔABC) = 5√5 s units, then the abscissa of the vertex C is:

  1. a) 1+√5
  2. b) 1+2√ 5
  3. c) 2√5-1
  4. d) 2+√5

Solution:

  1. JEE Main 2020 Maths Papers Sept 4 Shift 1 With Solutions

    AB = √(4+1) = √5

    (1/2)×√5×x = 5√5

    x = 10

    mAB = -1/2

    mAC = 2 = tanθ

    Since sinθ = 2/√5

    cosθ = 1/√5

    by parametric co-ordinates

    a = 1+ xcosθ = 1+10×1/√5

    = 1+2√5

    Answer: (b)


Question 12.Let f be a twice differentiable function on (1,6). If f(2) = 8, f’(2) = 5, f’(x) ≥1 and f’’(x) ≥4, for all x ∈ (1,6), then:

  1. a) f(5)+f’(5) ≥28
  2. b) f’(5)+f’’(5) ≤20
  3. c) f(5) ≤10
  4. d) f(5)+f’(5) ≤26

Solution:

  1. f(2) = 8, f’(2) = 5, f’(x) ≥1 and f’’(x) ≥4,

    x∈(1,6)

    JEE Main Solution 2020 Maths Papers Sept 4 Shift 1

    adding (1) and (2) we get,

    f(5)+f’(5) ≥28

    Answer: (a)


Question 13.Let α and β be the roots of x2-3x+p = 0 and 1/γ and 1/δ be the roots of x2-6x+q = 0. If α, β, γ, δ form a geometric progression. Then ratio (2q+p): (2q-p) is:

  1. a) 33 :31
  2. b) 9 : 7
  3. c) 3 : 1
  4. d) 5 : 3

Solution:

  1. Roots of x2-3x+p = 0 are α and β.

    The roots of x2-6x+q = 0 are γ and δ.

    α + β = 3

    γ + δ = 6

    α = a , β = ar, γ = ar2, δ = ar3

    a(1+r) = 3 ...(i)

    ar2(1+r) = 6 ...(ii)

    Divide (ii) by (i)

    r2 = 2

    α.β = p = a2r

    γ.δ = q = a2r5

    (2q+p)/(2q-p) = (2r4+1)/( 2r4-1)

    = (2×22+1)/( 2×22-1) = 9/7

    Answer: (b)


Question 14. Let u = (2z+i)/(z-ki), z = x+iy and k>0. If the curve represented by Re(u) +Im(u) =1 intersects the y-axis at the points P and Q where PQ =5, then the value of k is:

  1. a) 4
  2. b) 1/2
  3. c) 2
  4. d) 3/2

Solution:

  1. u = (2z+i)/(z-ki)

    z = x+iy

    JEE Main Solution Sept 4 Shift 1 2020 Maths Papers

    Re(u)+Img(u) = 1

    2x2+(2y+1)(y -k )+x+2xk = x2 + (y-k)2

    at y - axis, x = 0

    (2y +1)(y- k) = (y- k )2

    2y2 +y- 2yk- k = y2+k2-2yk

    Roots of y2+y-(k +k2) = 0 are y1 and y2

    Diff. of roots = 5

    √(1+4k+4k2) = 5

    4k2+4k = 24

    k2+k-6 = 0

    (k +3)(k-2) = 0

    k = 2

    Answer: (c)


Question 15. If A = [cosθisinθisinθcosθ]\begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}, (θ = π/24) and A5 = [abcd]\begin{bmatrix} a & b\\ c& d \end{bmatrix}, where i = √-1, then which one of the following is not true?

  1. a) a2-d2 = 0
  2. b) a2-c2 = 0
  3. c) 0 ≤a2+b2 ≤1
  4. d) a2-b2 = 1/2

Solution:

  1. A = [cosθisinθisinθcosθ]\begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}

    A2 = [cosθisinθisinθcosθ][cosθisinθisinθcosθ]\begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}

    JEE Main Sept 4 Shift 1 2020 Solved Maths Papers

    a = d = cos (5θ)

    b = c = i sin (5θ)

    a2-b2 = cos25θ + sin2

    = 1

    Answer: (d)


Question 16. The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is:

  1. a) 3
  2. b) 9
  3. c) 7
  4. d) 5

Solution:

  1. (5+7+10+12+14+15+x+y)/8 = 10

    x+y = 17 ..(i)

    Variance = (52+72+102+122+142+152+x2+y2)/8 -100 = 13.5

    (739+x2+y2)/8 - 100 = 13.5

    x2+y2 = 169 ..(ii)

    ∴ x = 12, y = 5

    |x-y| = 7

    Answer: (c)


Question 17. A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:

  1. a) 37
  2. b) 29
  3. c) 65
  4. d) 55

Solution:

  1. Shift 1 JEE Main Sept 4 2020 Solved Maths Papers

    n(B) ≤n(A∪B) ≤n(U)

    76 ≤76+63-x ≤100

    -63 ≤-x ≤-39

    63 ≥x ≥ 39

    Answer (d)


Question 18.Given the following two statements

(S1): (q˅p)→(p ↔ ~q) is a tautology.

(S2): ~q ˄ (~p ↔ q) is a fallacy. Then:

  1. a) only (S1) is correct.
  2. b) both (S1) and (S2) are correct.
  3. c) only (S2) is correct
  4. d) both (S1) and (S2) are not correct.

Solution:

  1. Shift 1 2020 JEE Main Sept 4 Solved Maths Papers

    Answer (d)


Question 19. Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is:

  1. a) 5
  2. b) 20/3
  3. c) 10/3
  4. d) 6

Solution:

  1. Shift 1 Sept 4 JEE Main 2020 Solved Maths Papers

    Using similar triangle concept

    tan θ1 = 10/x = H/x1

    x1 = Hx/10

    tan θ2 = 15/x = H/x2

    x2 = Hx/15

    Since x1+x2 = x

    (Hx/10)+( Hx/15) = x

    15H+10H = 150

    H = 150/25 = 6 m

    Answer: (d)


Question 20. If (a+√2b cos x)( a-√2b cos y) = a2-b2, where a>b>0, then dy/dx at (π/4, π/4) is:

  1. a) (a+b)/(a-b)
  2. b) (a-2b)/(a+2b)
  3. c) (a-b)/(a+b)
  4. d) (2a+b)/(2a-b)

Solution:

  1. (a+√2b cos x)( a-√2b cos y) = a2-b2

    Differentiating both sides w.r.t.y

    -√2b sin x (dy/dx)(a-√2b cos y)+ (a+√2b cos x)(√2b sin y) = 0

    x = y = π/4

    -b(dx/dy) (a-b)+(a+b)b = 0

    dx/dy = (a+b)/(a-b)

    Answer: (a)


Question 21. Suppose a differentiable function f(x) satisfies the identity f(x+y) = f(x)+f(y)+xy2+x2y,for all real x and y. If limx0f(x)x=1\lim_{x\to0}\frac{f(x)}{x}=1 then f’(3) is equal to:

    Solution:

    1. f(x+y) = f(x)+f(y)+xy2+x2y

      x = y = 0

      f(0) = 2f(0)

      f(0) = 0

      Now,

      Maths Shift 1 JEE Main Sept 4 2020 Solved Papers

      f’(x) = 1+0+x2

      f’(x) = 1+x2

      f’(3) = 10

      Answer: (10)


    Question 22.If the equation of a plane P, passing through the intersection of the planes, x+4y-z+7 = 0 and 3x+y+5z = 8 is ax+by+6z = 15 for some a, b∈R, then the distance of the point (3,2,-1) from the plane P is ….. units.

      Solution:

      1. p1+λp2 = 0

        (x +4y-z+7)+λ(3x+y+5z-8) = ax+by+6z-15

        (1+3λ)/a = (4+λ)/b = (-1+5λ)/6 = (-7-8λ)/-15

        ∴ 15-75λ = 42-48λ

        -27 = 27λ

        = -1

        ∴ plane is -2x+3y-6z+15 = 0

        d = |-6+6+6+15)/√(4+9+36)|

        = 3 units

        Answer: (3)


      Question 23. If the system of equations

      x-2y+3z = 9

      2x+y+z = b

      x-7y+az = 24, has infinitely many solutions, then a-b is equal to:

        Solution:

        1. D = 0

          12321117a\begin{vmatrix} 1 & -2& 3\\ 2 & 1 & 1\\ 1& -7 & a \end{vmatrix} = 0

          1(a +7)+2(2a -1)+3(-14-1) = 0

          a+7+4a-2-45 = 0

          5a = 40

          a = 8

          D1 = 923b112478=0\begin{vmatrix} 9 & -2& 3\\ b & 1 & 1\\ 24& -7 & 8 \end{vmatrix} = 0

          9(8+7)+2(8b-24)+3(-7b-24) = 0

          135+16b-48-21b-72 = 0

          15 = 5b

          b = 3

          Hence a - b = 8-3 = 5

          Answer: (5)


        Question 24.Let (2x2+3x+4)10 = r=020arxr\sum_{r =0}^{20}a_{r}x^{r}. Then a7/a13 is equal to:

          Solution:

          1. Given (2x2+3x+4)10 = r=020arxr\sum_{r =0}^{20}a_{r}x^{r}

            Replace x by 2/x in above identity:

            Maths Shift 1 Sept 4 2020 JEE Main Solved Papers

            Now, comparing coefficient of x7 from both sides

            (take r = 7 in L.H.S. and r = 13 in R.H.S)

            210a7 = a13213

            a7/a13 = 23 = 8

            Answer: (8)


          Question 25.The probability of a man hitting a target is 1/10. The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4 is:

            Solution:

            1. Probability of hitting, P(H) = 1/10

              probability of missing, P(M) = 9/10

              We have, 1-(probability of all shots result in failure ) >1/4

              = 1-P(M)n > 1/4

              = 1-(9/10)n > 1/4

              (9/10)n <3/4

              n≥3

              Answer (3)


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            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1

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            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1
            JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1

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