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September 4 Shift 2 - Maths

1. Suppose the vectors x1, x2 and x3 are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. If x1 =

\(\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}\)
, x2 =
\(\begin{bmatrix} 0\\ 2\\ 1 \end{bmatrix}\)
, x3 =
\(\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}\)
, b1=
\(\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\)
, b2 =
\(\begin{bmatrix} 0\\ 2\\ 0 \end{bmatrix}\)
, b3 =
\(\begin{bmatrix} 0\\ 2\\ 2 \end{bmatrix}\)
, then the determinant of A is equal to

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

Solution:

  1. Using AX = B

    A =

    \(\begin{bmatrix} a_{1} & a_{2} &a_{3} \\ a_{4}& a_{5} & a_{6} \\ a_{7}& a_{8} & a_{9} \end{bmatrix}\)

    a1+a2+a3 = 1

    a4+a5+a6 = 0

    a7+a8+a9 = 0

    2a2+a3 = 0

    2a5+a6 = 2

    2a8+a9 = 0

    a3 = 0, a6 = 0, a9 = 2

    a8 = -1

    a5 = 1

    a2 = 0

    ⇒ a1 = 1

    a4 = -1

    a7 = -1

    A =

    \(\begin{bmatrix} 1 & 0 & 0\\ -1& 1 & 0\\ -1& -1 & 2 \end{bmatrix}\)

    A = 2×1 = 2

    Answer: a

2. If a and b are real numbers such that (2+α)4 = a+bα, where α = (-1+i√3)/2 then a+b is equal to:

  1. a) 33
  2. b) 57
  3. c) 9
  4. d) 24

Solution:

  1. (2+α)4 = a+bα

    (2+(-1+i√3)/2)4 = a+bα

    JEE Main 2020 Paper With Solution Maths Sept 4 Shift 2

    b√3/2 = 9√3/2

    ⇒ b = 9

    a = 0

    a+b = 9

    Answer: c

3. The distance of the point (1, -2, 3) from the plane x-y+z = 5 measured parallel to the line (x/2) = (y/3) = (z/-6) is:

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

Solution:

  1. Equation of line through (1,-2,3) whose d.r.s. are (2,3,-6)

    (x-1)/2 = (y+2)/3 = (z-3)/-6 = λ

    Any point on the line (2λ+1, 3λ-2, -6λ+3)

    Put in (x-y+z = 5)

    2λ+1- 3λ+2-6λ+3 = 5

    -7λ = -1

    λ = 1/7

    Distance = √((2λ)2+(3λ)2+(6λ)2)

    = √(4λ2+9λ2+36λ2)

    = 7λ

    = 1 unit

    Answer: d

4. Let f: (0, ∞)→ (0, ∞) be a differentiable function such that f(1) = e and

\(\lim_{t\to x}\frac{t^{2}f^{2}(x)-x^{2}f^{2}(t)}{t-x}=0\)
. If f(x) = 1, then x is equal to:

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

Solution:

  1. f(1) = e ..(i)

    \(\lim_{x\to0}\frac{t^{2}f^{2}(x)-x^{2}f^{2}(t)}{t-x}=0\)

    L’ Hospital rule

    ⇒ limt→x (2tf2(x)-2x2f(t).f’(t)) = 0

    ⇒ 2xf2(x)-2x2f(x).f’(x) = 0

    ⇒ 2xf(x){f(x)-xf’(x)} = 0

    ⇒ f’(x)/f(x) = 1/x

    ln f(x) = ln x+ln c

    ⇒ f(x) = cx ..(ii)

    If x = 1

    f(1) = c(1)

    f(1) = c

    From equation (i) and (ii)

    c = e ..(iii)

    From (iii)

    f(x) = ex

    ⇒ y = ex or y = cx

    if f(x) = 1

    ⇒ x = 1/e

    Answer: c

5. Contrapositive of the statement :

‘If a function f is differentiable at a, then it is also continuous at a’, is:

  1. a) If a function f is not continuous at a, then it is not differentiable at a.
  2. b) If a function f is continuous at a, then it is differentiable at a.
  3. c) If a function f is continuous at a, then it is not differentiable at a.
  4. d) If a function f is not continuous at a, then it is differentiable at a.

Solution:

  1. Contrapositive of p → q = ∼q → ∼p

    Answer: a

6. The minimum value of 2sinx+2cosx is:

  1. a) 21-√2
  2. b) 21-1/√2
  3. c) 2-1+√2
  4. d) 2-1+1/√2

Solution:

  1. Using A.M. ≥ G.M.

    y = 2sinx+2cosx

    (2sinx+2cosx)/2 ≥ √(2sinx+cosx)

    2sinx+cosx ≥ 21×2(sinx+cosx)/2

    2sinx+cosx ≥ 2(2+sinx+cosx)/2

    ⇒ (2sin x+2cos x)minimum = 2(2-√2)/2

    = 21-1/√2

    Answer: b

7. If the perpendicular bisector of the line segment joining the points P(1 ,4) and Q(k, 3) has y-intercept equal to -4, then a value of k is:

  1. a) -2
  2. b) √15
  3. c) √14
  4. d) -4

Solution:

  1. mPQ = (4-3)/(1-k)

    ⇒

    \(m_{\perp}\)
    = k-1

    Midpoint of PQ = ((k+1)/2, 7/2)

    equation of perpendicular bisector

    y-7/2 = (k-1)(x-(k+1)/2)

    for y intercept put x = 0

    y = (7/2)-(k2-1)/2 = -4

    (k2-1)/2 = 15/2

    ⇒ k = 4 and k = -4

    Answer: d

8. The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x2-1 below the x-axis, is:

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

Solution:

  1. JEE Main 2020 Papers With Solutions Sept 4 Maths Shift 2

    Area = 2a(a2-1)

    A = 2a3-2a

    dA/da = 6a2-2 = 0

    d2A/da2 = 12a

    at a = -1/√3

    d2A/da2 = -4√3 <0

    So area is maximum at a = -1/√3

    Amax = (-2/3√3)+(2/√3)

    = (-2+6)/3√3

    = 4/3√3 sq. units

    Answer: d

9. The integral

\(\int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\tan ^{3}x.\sin ^{2}3x(2\sec ^{2}x.\sin ^{2}3x+3\tan x.\sin 6x)dx\)
is equal to

  1. a) 9/2
  2. b) -1/18
  3. c) -1/9
  4. d) 7/18

Solution:

  1. JEE Main 2020 Maths Papers With Solutions Sept 4 Shift 2

    = (1/2)[9×0-(1/3)×(1/3)×1]

    = -1/18

    Answer: b

10. If the system of equations

x+y+z = 2

2x+4y–z = 6

3x+2y+ λz = μ

has infinitely many solutions, then

  1. a) λ-2μ = -5
  2. b) 2λ+μ = 14
  3. c) λ+2μ = 14
  4. d) 2λ-μ = 5

Solution:

  1. D = 0

    ⇒

    \(\begin{vmatrix} 1 & 1 & 1\\ 2& 4 & -1\\ 3 & 2 & \lambda \end{vmatrix}\)
    = 0

    (4λ+2)-1(2λ+3)+1(4-12) = 0

    4λ+2-2λ-3-8 = 0

    2λ = 9

    ⇒ λ = 9/2

    Dx =

    \(\begin{vmatrix} 2 & 1 & 1\\ 6& 4 & -1\\ \mu & 2 & \frac{9}{2} \end{vmatrix}\)
    = 0

    ⇒ μ = 5

    Now check option

    2λ+μ = 14

    Answer: b

11. In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six The game stops as soon as either of the players wins. The probability of A winning the game is :

  1. a) 5/31
  2. b) 31/61
  3. c) 30/61
  4. d) 5/6

Solution:

  1. sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}

    P(sum 7) = 6/36

    sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}

    P(sum 6) = 5/36

    P(Awin) =

    \(P(6)+P(\bar{6}).P(\bar{7}).P(6)+..\)

    = (5/36)+(31/36)×(30/36)×(5/36)+…

    = (5/36)÷(1-(31×30)/(36×36)

    = (5×36)/(36×36-31×30)

    = 5×36/366

    = 30/61

    Answer: c

12. If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1+x)n+5 are in the ratio 5:10:14, then the largest coefficient in this expansion is :

  1. a) 792
  2. b) 252
  3. c) 462
  4. d) 330

Solution:

  1. Tr:Tr+1:Tr+2

    n+5Cr-1: n+5Cr: n+5Cr+1 = 5:10:14

    (n+5)!/(r-1)!(n+6-r)! : (n+5)!/r!(n+5-r)! = 5/10

    r/(n+6-r) = 1/2

    (r+1)!(n+4-r)!/r!(n+5-r)! = 5/7

    2r = n+6-r

    3r = n+6 ..(i)

    (r+1)/(n+5-r) = 5/7

    7r+7 = 5n + 25-5r

    12r = 5n+18 ..(ii)

    From eq.(i) and (ii)

    4(n+6) = 5n+18

    n = 6, r = 4

    Then we can find (1+x)11

    Largest coefficient in the expansion (1+x)11 = 11C6 = 462

    Answer: c

13. The function JEE Main 2020 Maths Papers Sept 4 Shift 2 With Solutions

  1. a) both continuous and differentiable on R-{-1}
  2. b) continuous on R-{-1} and differentiable on R-{-1,1}
  3. c) continuous on R-{1} and differentiable on R-{-1,1}
  4. d) both continuous and differentiable on R-{1}

Solution:

  1. JEE Main Solution 2020 Maths Papers Sept 4 Shift 2

    At x = 1, f(1) = π/2

    f(1+) = 0

    discontinuous ⇒ non diff.

    At x = -1

    f(-1) = 0

    f(-1-) = (1/2){+1-1} = 0

    cont. at x = -1

    JEE Main Solution Sept 4 Shift 2 2020 Maths Papers

    At x = -1, f’(-1) = -1/2

    And f’(-1+) = 1/2

    So, at x = -1 it is non differentiable.

    Answer: c

14. The solution of the differential equation (dy/dx)-((y+3x)/loge(y+3x)) + 3 = 0 is: (where c is a constant of integration)

  1. a) x-loge(y+3x) = C
  2. b) x-(1/2)(loge(y+3x))2 = C
  3. c) x-2loge(y+3x) = C
  4. d) y+3x-(1/2)(logex)2 = C

Solution:

  1. (dy/dx)-((y+3x)/ln(y+3x))+3 = 0

    ⇒ (dy/dx)+3 = (y+3x)/ln(y+3x)

    Let ln(y + 3x) = t

    (1/y+3x)×((dy/dx)+3) = dt/dx

    (y+3x)dt/dx = (y+3x)/t

    ⇒ tdt = dx

    t2/2 = x+C

    (1/2)(ln(y+3x))2 = x+C

    Answer: b

15. Let λ ≠ 0 be in R. If α and β are the roots of the equation, x2-x+2λ = 0 and α and γ are the roots of the equation, 3x2-10x+27λ = 0, then βγ/λ is equal to:

  1. a) 27
  2. b) 9
  3. c) 18
  4. d) 36

Solution:

  1. x2-x+2λ = 0 (α, β) ..(i)

    3x2-10x+27λ = 0 (α, γ) ..(ii)

    Multiply (i) by 3

    3x2-3x+6λ = 0 ..(iii)

    Equation (ii)-(iii)

    -7x+21λ = 0

    α = 3λ

    Put in equation (i)

    9λ2-3λ+2λ = 0

    9λ2-λ = 0

    ⇒ λ = 1/9

    ⇒ α = 1/3

    αβ = 2/9

    ⇒ β = 2/3

    αγ = 1

    ⇒ γ = 3

    βγ/λ ⇒ (2/3)×3÷1/9

    = 18

    Answer: c

16. The angle of elevation of a cloud C from a point P, 200 m above a still lake is 300. If the angle of depression of the image of C in the lake from the point P is 600, then PC (in m) is equal to :

  1. a) 200√3
  2. b) 400√3
  3. c) 400
  4. d) 100

Solution:

  1. JEE Main Sept 4 Shift 2 2020 Solved Maths Papers

    (h-200)/x = tan 300

    (h+200)/x = tan 600

    (h+200)/(h-200) = 3

    h+200 = 3h-600

    2h = 800

    h = 400

    (h-200)/PC = sin 300

    PC = 400 m

    Answer: c

17. Let

\(\bigcup_{i=1}^{50}X_{i}= \bigcup_{i=1}^{n}Y_{i}= T\)
where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi ’s and exactly 6 of sets Yi’s, then n is equal to :

  1. a) 15
  2. b) 30
  3. c) 50
  4. d) 45

Solution:

  1. Number of elements of set T =

    \(\bigcup_{i=1}^{50}X_{i}= \bigcup_{i=1}^{n}Y_{i}\)

    ⇒ 50×10/20 = n×5/6

    ⇒ (50/2)×(6/5) = n

    ⇒ n = 30

    Answer: b

18. Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. If P(1, β ), β> 0 is a point on this ellipse, then the equation of the normal to it at P is :

  1. a) 8x-2y = 5
  2. b) 4x-2y = 1
  3. c) 7x-4y = 1
  4. d) 4x-3y = 2

Solution:

  1. e = 1/2

    x = a/e = 4

    ⇒ a = 2

    e2 = 1-b2/a2

    ⇒ 1/4 = 1-(b2/4)

    (b2/4) = 3/4

    ⇒ b2 = 3

    Elipse (x2/4)+(y2/3) =1

    P(1,β)

    x = 1

    (1/4)+(β2/3) = 1

    β2/3 = 3/4

    ⇒ β = 3/2

    ⇒ P(1, 3/2)

    Equation of normal (a2x/x1)-(b2y/y1) = a2-b2

    (4x/1)-(3y/3/2) = 4-3

    4x-2y = 1

    Answer: b

19. Let a1, a2, ..., an be a given A.P. whose common difference is an integer and Sn = a1+a2+ .... +an. If a1 = 1, an = 300 and 15 ≤ n ≤ 50, then the ordered pair (Sn-4, an-4) is equal to:

  1. a) (2480,248)
  2. b) (2480,249)
  3. c) (2490,249)
  4. d) (2490,248)

Solution:

  1. a1 = 1, an = 300, 15≤ n ≤50

    300 = 1+(n-1)d

    (n-1) = 299/d

    n-1 = integer, d has to be a factor of 299.

    So, d = 23 or 13

    if d = 23

    then n-1 = 13

    n = 14 (reject)

    or d = 13

    n-1 = 23

    n = 24 is possible (15 ≤ n <50)

    Or S20 = (20/2){2+19×13}

    = 10×249

    = 2490

    a20 = 1+19×13

    = 248

    (S20, a20) = (2490, 248)

    Answer: d

20. The circle passing through the intersection of the circles, x2+y2-6x = 0 and x2+y2-4y = 0, having its centre on the line, 2x-3y+12 = 0, also passes through the point:

  1. a) (-1,3)
  2. b) (1,-3)
  3. c) (-3,6)
  4. d) (-3,1)

Solution:

  1. S1+λ(S1-S2) = 0

    x2 + y2-6x+λ(4y-6x) = 0

    x2+y2-6x(1+ λ)+4λy = 0

    Centre (3(1+λ ), -2λ) put in 2x-3y+12 = 0

    6+6λ+6λ+12 = 0

    12λ = -18

    λ = -3/2

    Circle is x2+y2+3x-6y = 0

    Check options

    (-3,6) is the point.

    Answer: c

21. Let {x} and [x] denote the fractional part of x and the greatest integer ≤ x respectively of a real number x. If

\(\int_{0}^{n}{x}dx\)
,
\(\int_{0}^{n}[x]dx\)
and 10(n2-n), (nN, n>1) are three consecutive terms of a G.P., then n is equal to

    Solution:

    1. \(\int_{0}^{n}\left \{ x \right \}dx = n\int_{0}^{1}x dx\)

      =

      \(n\left ( \frac{x^{2}}{2}\right )_{0}^{1}\)

      = n/2

      \(\int_{0}^{n}[x ]dx = \int_{0}^{1}0dx+ \int_{1}^{2}1dx +\int_{2}^{3}2dx+..\int_{n-1}^{n}(n-1)dx\)

      ⇒ 1+2+…n-1 = n(n-1)/2

      ⇒ n/2, n(n-1)/2, 10(n2-n) → G.P

      ⇒ n2(n-1)2/4 = (n/2)×10×n×(n-1)

      ⇒ n-1 = 20

      ⇒ n = 21

      Answer: 21

    22. A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

      Solution:

      1. 6C4×1×32 = 15×9 = 135

        Answer: 135

      23. If

      \(\vec{a}= 2\hat{i}+\hat{j}+2\hat{k}\)
      , then the value of
      \(\left | \hat{i} \times (\vec{a}\times \hat{i})\right |^{2}+\left | \hat{j} \times (\vec{a}\times \hat{j})\right |^{2}+\left | \hat{k} \times (\vec{a}\times \hat{k})\right |^{2}\)
      is equal to:

        Solution:

        1. \(\left | \hat{i} \times (\vec{a}\times \hat{i})\right |^{2} = \left | \vec{a} -(a.\hat{i})\hat{i}\right |^{2}\)

          =

          \(\left | \hat{j}+2\hat{k} \right |^{2}\)

          = 1+4

          = 5

          Similarly

          \(\left | \hat{j} \times (\vec{a}\times \hat{j})\right |^{2} =\left | 2\hat{i}+2\hat{k} \right |^{2}\)
          = 4+4

          = 8

          \(\left | \hat{k} \times (\vec{a}\times \hat{k})\right |^{2} =\left | 2\hat{i}+\hat{j} \right |^{2}\)
          = 4+1

          = 5

          ⇒ 5+8+5 = 18

          Answer: 18

        24. Let PQ be a diameter of the circle x2+y2 = 9. If α and β are the lengths of the perpendiculars from P and Q on the straight line, x+y = 2 respectively, then the maximum value of αβ is:

          Solution:

          1. Shift 2 JEE Main Sept 4 2020 Solved Maths Papers

            α =

            \(\left | \frac{3\cos \theta +3\sin \theta -2}{\sqrt{2}} \right |\)

            β =

            \(\left | \frac{+3\cos \theta +3\sin \theta +2}{\sqrt{2}} \right |\)

            αβ =

            \(\left | \frac{(3\cos \theta +3\sin \theta)^{2} -4}{{2}} \right |\)

            ⇒ αβ =

            \(\left | \frac{9 +9\sin 2\theta -4}{{2}} \right |\)

            ⇒ αβ =

            \(\left | \frac{5 +9\sin 2\theta }{{2}} \right |\)

            αβmax = (9+5)/2

            = 7

            Answer: 7

          25. If the variance of the following frequency distribution :

          Class

          10-20

          20-30

          30-40

          Frequency

          2

          x

          2

          is 50, then x is equal to

            Solution:

            1. Shift 2 2020 JEE Main Sept 4 Solved Maths Papers

              \(\bar{x}\)
              = (100+25x)/(4+x)

              \(\bar{x}\)
              = 25

              400/(4+x) = 50

              x = 4

              Answer: 4

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            Video Lessons - September 4 Shift 2 Maths









            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4

            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4
            JEE Main 2020 Maths Paper With Solutions Shift 2 September 4