WBJEE 2015 Maths Paper with Solutions

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WBJEE 2015 - Maths

Question 1. The value of limx22x3t2x2dt\lim_{x\rightarrow 2}\int_{2}^{x}\frac{3t^2}{x-2}dt is

  1. a. 10
  2. b. 12
  3. c. 8
  4. d. 16

Solution:

  1. Answer: (b)

    limx22x3t2x2\lim_{x\rightarrow 2}\int_{2}^{x}\frac{3t^2}{x-2}

    limx22x3t2dtx2\lim_{x\rightarrow 2}\frac{\int_{2}^{x}3t^2dt}{x-2}

    Which is 0/0 form

    Applying L’ hospital rule

    = limx23x23(2)2×01\lim_{x\rightarrow 2}\frac{3x^2-3(2)^2 \times 0}{1}

    {In numerator we will use Newton Leibnitz’s theorem}

    = 3 x 4 = 12


Question 2. If cot (2x/3) + tan(x/3) = cosec (kx/3) then the value of k is

  1. a. 1
  2. b. 2
  3. c. 3
  4. d. –1

Solution:

  1. Answer: (b)

    Solved WBJEE Maths 2015 Paper


Question 3. If θ ∈ (π/2, 3 π/2), then the value of

Solved WBJEE Maths 2015 Papers

  1. a. -2cotθ
  2. b. 2cotθ
  3. c. 2cosθ
  4. d. 2sinθ

Solution:

  1. Answer: (b)

    Solution for WBJEE Maths 2015 Paper

    {i.e. in IInd&IIIrd Quadrant where cos θ is negative.}

    Therefore, |2cos θ| = –2cos θ}

    = –2cos θ + 2cot θ + 2cos θ

    = 2cot θ


Question 4. The number of real solutions of the equation (sin x – x) (cos x – x2) = 0 is

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

Solution:

  1. Answer: (c)

    (sin x – x) (cos x – x2) = 0

    i.e. sin x = x OR cos x = x2

    Now let’s use graphs to get their solutions.

    Solution for WBJEE Maths 2015 Papers

    Therefore, we have total 3 solutions.


Question 5. Which of the following is not always true?

Solution for WBJEE 2015 Maths Paper

    Solution:

    1. Answer: (d)

      Let’s check for all the options:

      Solution for WBJEE 2015 Maths Papers

      Solution for Maths WBJEE 2015 Paper


    Question 6: If the four points with position vectors 2i^+j^+k^,i^+j^+k^,j^k^  and  λj^+k^-2 \hat i + \hat j + \hat k, \hat i + \hat j + \hat k, \hat j - \hat k \; and \; \lambda \hat j + \hat k are coplanar, then λ =

    1. a. 1
    2. b. 2
    3. c. –1
    4. d. 0

    Solution:

    1. Answer: (a)

      If four points A(-2, 1, 1), B(1,1,1), C(0, 1, –1) & D(0, λ, 1) are coplanar or lies in the same plane then vectors determined by them will also be coplanar.

      Consider,

      Solution for Maths WBJEE 2015 Papers

      Now applying the condition of co-planarity of three vectors

      i.e. their scalar triple product equals zero.

      Therefore, 3002022λ10=0\begin{vmatrix} 3 & 0 &0 \\ 2 & 0 &-2 \\ 2 & \lambda - 1 &0 \end{vmatrix}=0

      {Expanding above determinant along R1}

      3(0 + 2(λ–1)) + 0 + 0 = 0

      Or 6(λ–1) = 0

      Or λ = 1


    Question 7. For all real values of a0, a1, a2, a3 satisfying a0 + a1/2, a2/3, a3/4 = 0, the equation a0 + a1x + a2x2 + a3x3 = 0 has a real root in the interval

    1. a. [0, 1]
    2. b. [–1, 0]
    3. c. [1, 2]
    4. d. [–2, –1]

    Solution:

    1. Answer: (a)

      Here we are talking about roots of an equation function in some interval so think about Rolle’s theorem.

      Now, let f(x) = a0x + a1x2/2 + a2x3/3 + a3x4/4 + c

      (i) f(0) = c

      f(1) = a0 + a1/2, a2/3, a3/4 + c = c

      As, a0 + a1/2, a2/3, a3/4 = 0 (given)

      f(0) = f(1)

      (ii) Also, f(x) is a polynomial so continuous and differential everywhere.

      Therefore, By Rolle’s theorem f’(x) i.e. a0 + a1x + a2x2 + a3x2

      Will have at least one real root in [0,1]


    Question 8. Let f: R -> R be defined as

    Maths WBJEE 2015 Solved Paper

    Then which of the following is true?

    1. a. f is discontinuous for all x
    2. b. f is continuous for all x
    3. c. f is discontinuous at x = k π, where k is an integer
    4. d. f is continuous at x = k π, where k is an integer

    Solution:

    1. Answer: (d)

      We know on the real line ’x’ takes rational and irrational values in the very close neighbourhood.

      Therefore, the values of given function f(x) will oscillate to ‘0’ or sin|x|, corresponding to irrational and rational values of ‘x’ and so will be discontinuous at infinite points.

      The f(x) will be continuous at points where y = 0 & y = sin|x| meets (or equal)

      i.e. sin|x| = 0

      ⇒ x = k π, k ∈ I


    Question 9. If f: [0, π/2]-> R is defined as f(θ)=1tan  θ1tan  θ1tan  θ1tan  θ1f(\theta) = \begin{vmatrix} 1 & tan \; \theta &1 \\ -tan \; \theta & 1 & tan \; \theta\\ -1 & -tan \; \theta & 1 \end{vmatrix}. Then the range of f is

    1. a. (2, ∞)
    2. b. (–∞, –2)
    3. c. (2, ∞)
    4. d. (– ∞, 2]

    Solution:

    1. Answer: (c)

      Expanding given determinant along R1, we have

      = 1(1+ tan2θ) – tanθ(- tanθ + tanθ) + 1 (tan2θ + 1)

      = 2sec2θ

      Now, for θ ∈ [0, π/2]

      ⇒ sec2θ ∈ [1, ∞)

      ⇒ 2 sec2θ ∈ [2, ∞)

      i.e. Range of f(θ) -> [2, ∞)


    Question 10. If A and B are two matrices such that AB = B and BA = A, then A2 + B2 equals

    1. a. 2AB
    2. b. 2BA
    3. c. A+B
    4. d. AB

    Solution:

    1. Answer: (c)

      A2 + B2 = A. A + B. B

      = A. (BA) + B(AB)

      {Given AB = B & BA = A}

      = (AB) A + (BA) B

      =B. A + AB

      = A + B


    Question 11. If ω is an imaginary cube root of unity, then the value of the determinant

    Maths WBJEE 2015 Solved Papers
    is

    1. a. -2 ω
    2. b. -3 ω2
    3. c. -1
    4. d. 0

    Solution:

    1. Answer: (b)

      Applying operations: C1 -> C1 + C2, C2 -> C2 – C3

      Solved paper for Maths WBJEE 2015

      Expanding along R3

      = 0 + 0 + (ω–1) (ω2–ω)

      = ω3 – ω2 – ω2 + ω

      = 1–2ω2 + ω

      = (1+ω) - 2 ω^2

      {Because, 1 + ω + ω2 = 0, 1 + ω = – ω2}

      = –3ω2


    Question 12. Let a, b, c, d be any four real numbers. Then an + bn = cn + dn holds for any natural number n if

    1. a.a + b = c + d
    2. b.a – b = c – d
    3. c.a + b = c + d, a2 + b2 = c2 + d2
    4. d.a – b = c – d, a2 – b2 = c2 – d2

    Solution:

    1. Answer: (d)

      a, b, c, d ∈ R

      an + bn = cn + dn for all n ∈ N

      For n = 1, a +b = c + d …..(i)

      n = 2, a2 + b2 = c2 + d2 …..(ii)

      n = 3, a3 + b3 = c3 + d3 …..(iii)

      ⇒ (a + b) (a2 + b2 – ab) = (c + d) (c2 + d2 – cd)

      ⇒ a2 + b2 – ab = c2 + d2 – cd (using (i) & (ii))

      ⇒ ab = cd …..(iv)

      Now consider a quadratic equation with roots a3& b3

      x2– (a3 + b3) x + a2b3 = 0 …..(v)

      Similarly, equation with roots c3 & d3 will be

      x2 – (c3 + d3) x + c3d3 = 0 …..(vi)

      both equation (v) & (vi) are same with four roots a3, b3, c3, d3 which is not possible since a quadratic equation cannot have more than two roots.

      Therefore, it is possible only when a = c & b = d OR a = d & b = c.


    Question 13. If α, β are the roots of x2 – px + 1 = 0 and γ is root of x2 + px + 1 = 0, then (α + γ) (β + γ) is

    1. a. 0 (zero)
    2. b. 1
    3. c. –1
    4. d. p

    Solution:

    1. Answer: (a)

      α, β all roots of x2 – px + 1 = 0 ….(i)

      If we replace x by –x

      (i) becomes x2 + px + 1 = 0 …..(ii)

      This implies, (ii) will have roots –α and -β

      (transformation of equations)

      But we are given root of (ii) γ

      Therefore, Either γ = – α or – β

      So, (α + γ) (β + γ) = 0


    Question 14. Number of irrational terms in the binomial expansion of (31/2 + 71/3)100 is

    1. a. 90
    2. b. 88
    3. c. 93
    4. d. 95

    Solution:

    1. Answer: (G)

      (31/5 + 71/3)100

      Writing general term of given binomial

      i.e. Tr+1 = 100Cr (31/5)100–r (71/3)r

      Now the term will be rational when exponents of 3 and 7 both become a whole number

      i.e. (100-r)/5 & r/3

      r = 0, 15, 30, 45, 60, 75 & 90 (7 rational terms)

      Since total no. of terms are 101.

      Therefore, No. of irrational terms = 101–7 = 94.


    Question 15. The quadratic expression (2x + 1)2 – px + q ≠ 0 for any real x if

    1. a. p2 – 16p – 8q < 0
    2. b. p2 – 8p + 16q < 0
    3. c. p2 – 8p – 16q < 0
    4. d. p2 – 16p + 8q < 0

    Solution:

    1. Answer: (c)

      (2x + 1)2 – px + q ≠ 0

      or 4x2 + (4–p) x + 1 + q ≠ 0 for any real ’x’

      ⇒ Equation, 4x2 + (4–p) x +1 + q = 0 has no real roots

      Therefore, D < 0

      (4 – p)2 – 4. 4 (1 + q) < 0

      ⇒ 16 + p2 – 8p – 16 – 19q < 0

      ⇒ p2 – 8p – 16q < 0


    Question 16. The value of

    Solved papers for Maths WBJEE 2015

    1. a. 0 (zero)
    2. b. –1
    3. c. 1
    4. d. i

    Solution:

    1. Answer: (b)

      Solved Maths 2015 WBJEE Paper

      = ω + 1/ω

      = ω + ω2 = -1

      {As, ω3 = 1, ω3 = 1/ω, 1 + ω + ω2 = 0, ω + ω2 = -1}


    Question 17. Find the maximum value of |z| when |z - 3/2| = 2, z being a complex number.

    1. a. 1 + √3
    2. b. 3
    3. c. 1 + √2
    4. d. 1

    Solution:

    1. Answer: (b)

      Given, |z - 3/2| = 2

      From triangle inequality

      Solved Maths 2015 WBJEE Papers

      From right inequality

      |z| – 3/|z| ≤ 2

      ⇒ |z|2 – 2|z| – 3 ≤ 0

      ⇒ (|z| – 3)(|z| + 1) ≤ 0

      ⇒ –1 ≤ |z| ≤ 3

      ⇒ |z|max = 3


    Question 18. Given that x is a real number satisfying [5x2 – 26x + 5]/[3x2 – 10x + 3] < 0, then

    1. a. x < 1/5
    2. b. 1/5 < x < 3
    3. c. x > 5
    4. d. 1/5 < x < 1/3 or 3 < x < 5

    Solution:

    1. Answer: (d)

      Paper Solutions for Maths WBJEE 2015


    Question 19. The least positive value of t so that the lines x = t + α, y + 16 = 0 and y = αx are concurrent is

    1. a. 2
    2. b. 4
    3. c. 16
    4. d. 8

    Solution:

    1. Answer: (d)

      y = –16 ...(i)

      y = αx ...(ii)

      x = t + α ...(iii)

      From (i) and (ii)

      x = -16/α, now put in (iii)

      -16/α = t + α

      t = -(α + 16/α)

      Now, α + 16/α ≥ -8 for α > 0 and α + 16/α ≤ –8 for α < 0 {using A.M. ≥ G.M.}

      t = -(α + 16/α) ≤ -8; α > 0

      t = -(α + 16/α) ≥ 8; α < 0

      Therefore, least positive value of ‘t’ is 8

      Note: You can also use concept of maxima & minima.


    Question 20. If in a triangle, ΔABC, a2 cos2 A – b2 – C2 = 0, then

    1. a. π/4 < A <π/2
    2. b. π/2 < A <π
    3. c. A = π/2
    4. d. A <π/2

    Solution:

    1. Answer: (b)

      a2 cos2 A – b2 – C2 = 0 (given) cos2 A = (b2 + c2)/a2< 1

      {cos A ≤ 1 but A ≠0,so, cos A < 1 }

      Now from cosine rule

      cos A = (b2 + c2 - a2 )/ 2bc < 0

      {Because, b2 + c2< a2 }

      So, cos A < 0

      A ∈ (π/2, π)


    Question 21.

    Papers Solutions for Maths WBJEE 2015

      Solution:

      1. Answer: (a)

        For x ∈ [0, 3π/2]

        Question Paper Solutions for Maths WBJEE 2015

        From above graph

        |cos x| ≥ sin x

        For x ∈ [0, π/4] U [3π/4, 3π/2]


      Question 22. A particle starts moving from rest from a fixed point in a fixed direction. The distance s from the fixed point at a time t is given by x = t2 + at – b + 17, where a, b are real numbers. If the particle comes to rest after 5 sec at a distance of s = 5 units from the fixed point, then values of a and b are respectively

      1. a. 10, –33
      2. b. –10, –33
      3. c. –8, 33
      4. d. –10, 33

      Solution:

      1. Answer: (b)

        s = t2 + at – b + 17

        v = ds/dt|t=5 = 2t +a|t=5 = 0

        10 + a = 0

        a = – 10

        Also, at t = 5, s = 25

        Therefore, 25 = 52 + 5(–10) – b + 17

        or b = –33


      Question 23.

      Question Paper Solutions for Maths 2015 WBJEE

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

      Solution:

      1. Answer: (c)

        WBJEE 2015 Solved Math Paper


      Question 24. If

      WBJEE 2015 Solved Math Papers

      then the values of a, b are respectively

      1. a. 2, 2
      2. b. 1, 2
      3. c. 2, 1
      4. d. 2, 0

      Solution:

      1. Answer: (a)

        WBJEE 2015 Solution for Math Paper

        For above limit to be defined coefficient of x i.e. a- b must be zero

        ⇒ a = b

        Again, the value of limit will be the coefficient of x2

        i.e. a + b/2 = 3

        ⇒ a + a/2 = 3

        ⇒ a = 2 = b


      Question 25. Let P(x) be a polynomial, which when divided by x – 3 and x – 5 leaves remainders 10 and 6 respectively. If the polynomial is divided by (x – 3) (x – 5) then the remainder is

      1. a. –2x + 16
      2. b. 16
      3. c. 2x – 16
      4. d. 60

      Solution:

      1. Answer: (a)

        P(x) = (x–3) Q1(x) + 10

        ⇒ P(3) = 10

        P(x) = (x–5) Q2(x) + 6

        ⇒ P(5) = 6

        P(x) = (x–3) (x–5) Q3(x) + ax + b

        Put x = 3 & 5

        P(3) = 0 + 3a + b = 10

        P(5) = 0 + 5a + b = 6

        Solving above equations

        a = –2 & b = 16

        Remainder = –2x + 16


      Question 26. The integrating factor of the differential equation dy/dx + (3x2 tan-1y – x3)(1 + y2) = 0 is

      1. a. ex^2
      2. b. ex^3
      3. c. e3x^2
      4. d. e3x^3

      Solution:

      1. Answer: (b)

        dy/dx + (3x2 tan-1y – x3)(1 + y2) = 0

        WBJEE 2015 Solution for Math Papers

        Now, put tan–1y = z

        Differentiate with respect to x

        (1/(1+y2)dy/dx = dz/dx

        Therefore, dz/dx + z. 3x2 = x3

        P(x) = 3x2; Q(x) = x3

        WBJEE 2015 Solution for Math Questions


      Question 27. If y = e–xcos2x then which of the following differential equations is satisfied?

      WBJEE Solutions for Maths paper 2015

        Solution:

        1. Answer: (a)

          y = e–x. cos2x

          ⇒ y1 = e–x(–1) cos2x + e–x (–sin2x) (2)

          ⇒ y1 = –y – 2e–x sin2x

          ⇒ y2 = –y1 – 2(–e–x sin2x + e–x cos2x. 2)

          y2 = –y1 + 2e–x sin2x – 4e–x cos2x

          y2 = – y1 + (–y–y1) – 4y

          y2 + 2y1 + 5y = 0


        Question 28. In a certain town, 60% of the families own a car, 30% own a house and 20% own both a car and a house. If a family is randomly chosen, what is the probability that this family owns a car or a house but not both?

        1. a. 0.5
        2. b. 0.7
        3. c. 0.1
        4. d. 0.9

        Solution:

        1. Answer: (a)

          A -> Car, B -> House

          n(A) = 60%, n(B) = 30%, n(A n B) = 20%

          No. of families which owns car or house but not both

          = n(A U B) –n(A n B)

          = n(A) + n(B) – n(A n B) –n(A U B)

          = 70 + 30 – 2 × 20 = 50%

          So, Probability = 50/100 = 0.5


        Question 29. The letters of the word COCHIN are permuted and all the permutations are arranged in alphabetical order as in English dictionary. The number of words that appear before the word COCHIN is

        1. a. 360
        2. b. 192
        3. c. 96
        4. d. 48

        Solution:

        1. Answer: (c)

          WBJEE Solutions for Maths papers 2015

          Rank of COCHIN = 4! + 4! + 4! + 4! + 1

          = 4 × 4! + 1 = 97

          No. of words before COCHIN = 97–1 = 96


        Question 30. Let f : R -> R a continuous function which satisfies f(x)=0xf(t)dtf(x) = \int_{0}^{x} f(t)dt. Then the value of f(loge5) is

        1. a. 0 (zero)
        2. b. 2
        3. c. 5
        4. d. 3

        Solution:

        1. Answer: (a)

          f(x)=0xf(t)dtf(x) = \int_{0}^{x} f(t)dt

          Differentiate {using Newton Leibnitz theorem on R.H.S.}

          f’(x) = f(x)

          f'(x)/f(x) = 1

          Integrate w.r.t. x

          WBJEE Solution for Maths paper 2015

          f(x) = ex+c = ec . ex

          Now, f(0) = f(x)=00f(t)dtf(x) = \int_{0}^{0} f(t)dt = 0

          Therefore, k must be 0

          f(x) = 0 (const.)

          f(loge5) = 0


        Question 31. The value λ, is such that the following system of equations has no solution, is

        2x – y – 2z = 2

        x – 2y + z = –4

        x + y + λz = 4

        1. a. 3
        2. b. 1
        3. c. 0 (zero)
        4. d. –3

        Solution:

        1. Answer: (d)

          For number solution

          Δ = 0 and at least one of Δx, Δy, Δz must be non-zero (Cramer’s rule)

          WBJEE Solutions for Maths paper 2015 Question 31

          ⇒ 2(–2λ–1) + 1 (λ–1) – 2 (1+2) = 0

          ⇒ –3λ – 9 = 0

          ⇒ λ = – 3


        Question 32.

        WBJEE Solutions for Maths paper 2015 Question 32

        Then f(100) is equal to

        1. a. 0 (zero)
        2. b. 1
        3. c. 100
        4. d. 10

        Solution:

        1. Answer: (a)

          WBJEE Solutions for Maths paper 2015 Solution 32

          = x(x2–1) {1.(-2x + 2x)}

          = 0

          f(100) = 0


        Question 33.If sin-1(x – x2/4 + x3/4 – x4/8 + …) = π/6 where |x| < 2 then the value of x is

        1. a. 2/3
        2. b. 3/2
        3. c. -2/3<.p>d. -3/2

        Solution:

        1. Answer: (a)

          sin-1(x – x2/4 + x3/4 – x4/8 + …infinity) = π/6

          x – x2/4 + x3/4 – x4/8 + …infinity = ½

          Infinite G.P.; Common ratio = -x/2

          x/(1+x/2) = ½

          2x/(2+x) = ½

          x = 2/3


        Question 34.The area of the region bounded by the curve y = x3, its tangent at (1, 1) and x-axis is

        1. a. 1/12
        2. b. 1/6
        3. c. 2/17
        4. d. 2/15

        Solution:

        1. Answer: (a)

          y = x3

          dy/dx = 3x2

          At A(1,1), dy/dx = 3

          Equation of tangent at A (1, 1)

          y – 1 = 3(x – 1)

          y = 3x – 2

          Now,

          WBJEE Solutions for Maths paper 2015 Question 34


        Question 35. If log0.2(x – 1)> log0.04(x + 5) then

        1. a.–1 < x < 4
        2. b.2 < x < 3
        3. c.1 < x < 4
        4. d.1 < x < 3

        Solution:

        1. Answer: (c)

          (I) For log0.2(x – 1) and log0.04(x + 5) to be defined

          x –1 > 0 and x + 5 >0

          x >1 ....(i)

          (II) Now, log0.2(x – 1) > log0.04(x + 5)

          log0.2(x – 1) > log0.2 (x + 5)

          log0.2 (x – 1) > (1/2) log0.2 (x + 5)

          2log0.2 (x – 1) > log0.2 (x + 5)

          log0.2 (x – 1)2> log0.2 (x + 5)

          Taking antilog {note the inequality will change because base < 1}

          (x – 1)2< x + 5

          x2 – 3x – 4 < 0

          (x – 4)(x + 2) < 0

          –1 < x < 4 {but x > 1, (i)}

          Therefore, 1 < x < 4


        Question 36. The number of real roots of equation logex + ex = 0

        1. a. 0 (zero)
        2. b. 1
        3. c. 2
        4. d. 3

        Solution:

        1. Answer: (b)

          logex + ex = 0

          logex = - ex

          WBJEE Solutions for Maths paper 2015 Question 36

          The two curves meet at one point

          so, One solution


        Question 37.If the vertex of the conic y2 – 4y = 4x – 4a always lies between the straight lines x + y = 3 and 2x + 2y – 1 = 0 then

        1. a. 2 < a < 4
        2. b. -1/2< a < 2
        3. c. 0 < a < 2
        4. d. -1/2 < a < 3/2

        Solution:

        1. Answer: (b)

          y2 – 4y = 4 x – 4a

          ⇒ (y–2)2 = 4 (x – a + 1) {Parabola}

          ⇒ Vertex (a – 1, 2)

          Now, (a-1, 2) always lies between lines x + y – 3 =0 & 2x + 2y–1=0

          ⇒ (a – 1 + 2 – 3) (2(a–1)+2(2) –1) < 0 {Location of point with respect to line}

          ⇒ (a – 2) (2a + 1) < 0

          ⇒ -(1/2) < a < 2


        Question 38. Number of intersecting points of the conic 4x2 + 9y2 = 1 and 4x2 + y2 = 4 is

        1. a. 1
        2. b. 2
        3. c. 3
        4. d. 0 (zero)

        Solution:

        1. Answer: (d)

          C1: x2/(1/4) + y2/(1/9) = 1 and C2: x2/1 + y2/4 = 1

          WBJEE Solutions for Maths paper 2015 Question 38

          Clearly no point of intersection.


        Question 39. The value of λ for which the straight line (x-λ)/3 = (y-1)/(2+λ) = (z-3)/-1 may lie on the plane x – 2y = 0 is

        1. a. 2
        2. b. 0
        3. c. -1/2
        4. d. There is no such λ

        Solution:

        1. Answer: (c)

          If given line lies on the plane, the normal of plane must also be normal (perpendicular) to the line.

          Direction ratios of line (3, 2+λ, –1)

          Direction ratios Normal (1, –2, 0)

          (3)(1) + (2+λ)(–2) + (–1)(0) = 0

          – 1 – 2λ = 0

          λ=-1/2


        Question 40. The value of 2 cot-1(1/2) – cot-1(4/3) is

        1. a. – π/8
        2. b. 3 π/2
        3. c.π/4
        4. d.π/2

        Solution:

        1. Answer: (d)

          WBJEE Solutions for Maths paper 2015 Question 40


        Question 41. If the point (2cos θ, 2sin θ), for θ ∈ (0, 2 π) lies in the region between the lines x + y = 2 and x – y = 2 containing the origin, then θ lies in

        1. a. (0, π/2)
        2. b. [0, π]
        3. c. (π/2, 3 π/2)
        4. d. [π/4, π/2]

        Solution:

        1. Answer: (c)

          (2cosθ, 2sinθ) lies on the circle x2 + y2 = 4

          WBJEE Solved Paper Maths 2015

          The given point will lie in shaded region for θ∈(π/2, 3 π/2).


        Question 42. Number of points having distance √5 from the straight line x – 2y + 1 = 0 and a distance √13 from the lie 2x + 3y – 1 = 0 is

        1. a. 1
        2. b. 2
        3. c. 4
        4. d. 5

        Solution:

        1. Answer: (c)

          Let the point be (a, b)

          Now, |(a-2b+1)/√5| = √5 => a – 2b + 1 = ±5 …(1)

          Also |(2a+3b-1)/√13| = √13 => 2a + 3b – 1 = ±13 …(2)

          Anyone of equations (1) will give a point of intersection with anyone of equations (2).

          So total 4 points.


        Question 43. Let f : R-> R be defined as (x2-x+4)/( x2-x+4). Then the range of the function f(x) is

        1. a. [3/5, 5/3]
        2. b. (3/5, 5/3)
        3. c. (-∞, 3/5) ⋃(5/3, ∞)
        4. d. [-5/3, -3/5]

        Solution:

        1. Answer: (a)

          f(x) = y = (x2-x+4)/( x2-x+4). ; x belongs to R

          => y (x2 + x + 4) = x2 – x + 4

          => (y – 1) x2 + (y + 1) x + 4y – 4 = 0

          For x to be real, D ≥ 0

          => (y + 1)2 – 4×4× (y – 1)2 ≥ 0

          => (5y – 3)(–3y + 5) ≥0

          => (5y – 3)(3y – 5) ≤ 0

          => 3/5 ≤ y ≤5/3

          Also, check for y = 1; x = 0, which is valid.


        Question 44. The least value of 2x2 + y2 + 2xy + 2x – 3y + 8 for real numbers x and y is

        1. a. 2
        2. b. 8
        3. c. 3
        4. d. 1

        Solution:

        1. Answer: Bonus (G)

          z = 2x2 + y2 + 2xy + 2x – 3y + 8

          = {4x2 + 2y2 + 4xy + 4x – 6y + 16}

          = {(2x + y + 1)2 + (y – 4)2 – 1}

          = {(2x + y – 1)2 + (y – 4)2} –1/2 ≥ – 1/2

          So the least value is –1/2 at x = –5/2 and y = 4.


        Question 45. Let f : [–2, 2] -> R be a continuous function such that f(x) assumes only irrational values. If f(√2) = √2, then

        1. a. f(0) = 0
        2. b. f(√2-1) = √2-1
        3. c. f(√2-1 ) = √2+1
        4. d. f(√2-1 ) = √2

        Solution:

        1. Answer: (d)

          A continuous function assuming only irrational (or rational) value must be a constant function.

          So f(x) = √2, for all x belongs to R


        Question 46. The minimum value of cos θ+sin θ+2/sin θ for θ belongs to (0, π/2) is

        1. a. 2+√2
        2. b. 2
        3. c. 1+√2
        4. d. 2√2

        Solution:

        1. Answer: (a)

          y = sin θ + cos θ +2/sin θ ; θ belongs to (0, π/2)

          = sin θ + cos θ + 1/sin θcos θ

          As the expression remains unchanged by interchanging sin θ and cos θ so the minimum is achieved for sin θ = cos θ

          i.e. for θ = π/4 in (0, π/2)

          So, minimum value = 2 + √2.


        Question 47. Let xn = (1-1/3)2(1-1/6)2(1-1/10)2….(1-1/n(n+1)/2), n ≥2. Then the value of limnxn\lim_{n\to\infty }x_{n} is:

        1. a. 1/3
        2. b. 1/9
        3. c. 1/81
        4. d. 0 (zero)

        Solution:

        1. Answer: (b)

          WBJEE Solved Paper 2015 Maths


        Question 48. The variance of the first 20 natural numbers is

        1. a. 133/4
        2. b. 279/12
        3. c. 133/2
        4. d. 399/4

        Solution:

        1. Answer: (a)

          The variance of first n natural numbers is (n2-1)/12.

          WBJEE Maths Solved Paper 2015


        Question 49: A fair coin is tossed a fixed number of times. If the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads, then the probability of getting exactly one head is

        1. a. 1/64
        2. b. 1/32
        3. c. 1/16
        4. d. 1/8

        Solution:

        1. Answer: (b)

          Let, coin be tossed n times,

          Probability of getting exactly 3 heads = nC3(1/2)3(1/2)n-3

          Probability for exactly 5 heads = nC5 (1/2)5(1/2)n-5

          (Using Binomial probability distribution)

          Now, nC3(1/2)n = nC5 (1/2)n

          => n = 8

          Thus, required probability = 8C1(1/2)8 = 1/32


        Question 50. If the letters of the word PROBABILITY are written down at random in a row, the probability that two B-s are together is

        1. a. 2/11
        2. b. 10/11
        3. c. 3/11
        4. d. 6/11

        Solution:

        1. Answer: (a)

          Required probability alignments = (10!/2!)/(11!/2!2!)

          = 2/11


        Question 51. The number of distinct real roots of

        WBJEE Solutions Paper 2015 Maths

        1. a. 0 (zero)
        2. b. 2
        3. c. 1
        4. d. > 2

        Solution:

        1. Answer: (c)

          WBJEE 2015 Solutions Paper Maths


        Question 52. Let x1, x2, .........., x15 be 15 distinct numbers chosen from 1, 2, 3, ........., 15. Then the value of (x1 – 1) (x2 – 1), (x3 – 1)..........(x15 – 1) is

        1. a. always ≤ 0
        2. b. 0 (zero)
        3. c. always even
        4. d. always odd

        Solution:

        1. Answer: (b)

          Among x1, x2, …., x15; one of them, say xk is 1.

          xk–1 = 0

          So the given product is ‘0’.


        Question 53. Let [x] denote the greatest integer less than or equal to x. Then the value of  for which the function,

        WBJEE 2015 Maths Solutions Paper

        1. a. α = 0
        2. b. α = sin(–1)
        3. c. α = sin(1)
        4. d. α = 1

        Solution:

        1. Answer: (c)

          WBJEE 2015 Maths Paper Solutions


        Question 54. Let f(x) denotes the fractional part of a real number x. Then the value of 03f(x)2dx\int_{0}^{\sqrt{3}}f(x)^{2}dx

        1. a. 2√3-√2-1
        2. b. 0 (zero)
        3. c. √2-√3+1
        4. d. √3-√2+1

        Solution:

        1. Answer: (c)

          2015 WBJEE Solved Paper Maths


        Question 55. Let S = {(a, b, c) belongs to N × N × N : a + b + c = 21, a ≤ b ≤ c } and T = {(a, b, c) belongs to N× N × N : a, b, d are in A.P.}, where N is the set of all-natural numbers. Then the number of elements in the set S ⋂ T is

        1. a. 6
        2. b. 7
        3. c. 13
        4. d. 14

        Solution:

        1. Answer: (b)

          S : a + b + c = 21 and T: a + c = 2b

          => a + c = 14 and b = 7

          (1, 13), (2, 12), …….(6, 8) or a = b = c = 7

          Hence 7 triplets.


        Question 56: Let y = ex2e^{x^{2}} and y = ex2sinxe^{x^{2}}\sin x be two given curves. Then the angle between the tangent to the curves at any point of their intersection is

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

        Solution:

        1. Answer: (a)

          2015 Solved Paper WBJEE Maths


        Question 57. Area of the region bounded by y = |x| and y = –|x| + 2 is

        1. a. 4 sq. units
        2. b. 3 sq. units
        3. c. 2 sq. units
        4. d. 1 sq. units

        Solution:

        1. Answer: (c)

          2015 Solved Paper Maths WBJEE

          In figure; A(-1,1), B(0,2), C(1,1), 0(0,0)

          OABC is a square with side length √2

          So Area = √22 = 2 sq. units.


        Question 58. Let d(n) denote the number of divisors of n including 1 and itself. Then d(225), d(1125) and d(640) are

        1. a. in AP
        2. b. in HP
        3. c. in GP<.p>d. consecutive integers

        Solution:

        1. Answer: (c)

          (i) 225 = 32×52

          => d(225) = (2+1) (2+1) = 9

          (ii) 1125 = 32×53

          => d(1125) = 3 × 4 =12

          (iii) 640 = 27×5

          => d(640) = 8 × 2 =16

          9, 12, 16 are in GP


        Question 59. The trigonometric equation sin–1x = 2sin–12a has a real solution if

        1. a. |a|>1/√2
        2. b. 1/2 √2<|a|>1/√2
        3. c. |a|>1/2√2
        4. d. |a|≤1/2√2

        Solution:

        1. Answer: (d)

          2015 Maths WBJEE Solved Paper


        Question 60. If 2 + i and √5-2i are the roots of the equation (x2 + ax + b) (x2 + cx + d) = 0, where a, b, c, d are real constants, then product of all roots of the equation is

        1. a. 40
        2. b. 9√5
        3. c. 45
        4. d. 35

        Solution:

        1. Answer: (c)

          If 2–i and √5+2i are two roots then other roots must be 2 + i and √5-2i

          So, Product of roots = (2–i)(√5-2i )(2+i)( 5+2i) = 5×9 = 45.


        Question 61. In a triangle ABC, ∠C = 900, r and R are the in-radius and circum-radius of the triangle ABC respectively, then 2(r + R) is equal to

        1. a. b + c
        2. b. c + a
        3. c. a + b
        4. d. a + b + c

        Solution:

        1. Answer: (c)

          2015 WBJEE Maths Solved Paper

          Now, AB = 2R = b – r + a – r

          => 2 (R + r) = b + a


        Question 62. Let α, β be two distinct roots of a cos θ + b sin θ = c, where a, b and c are three real constants and q belongs to [0, 2 π]. Then α+ β is also a root of the same equation if

        1. a. a + b = c
        2. b. b + c = a
        3. c. c + a = b
        4. d. c = a

        Solution:

        1. Answer: (d)

          a cos θ + b sin θ = c

          2015 Maths Solved Paper WBJEE

          (c+a)b2-2b2a+(c-a)a2 = 0

          b2(c-a)+a2(c-a) = 0

          (c-a)(b2+a2) = 0

          c – a = 0

          c = a


        Question 63. For a matrix A = [100210321]\begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ 3& 2& 1 \end{bmatrix} if U1, U2 and U3 are 3 × 1 column matrices satisfying

        1. a. 6
        2. b. 0 (zero)
        3. c. 1
        4. d. 2/3
        5. a.sub>3 = 2, b3 = 1, c3 = –3

          2015 Solved Papers WBJEE Maths

          Sum of elements of U-1 = 0

        Solution:

        1. Answer: (b)

          Let Ui = [aibici]\begin{bmatrix} a_{i}\\ b_{i}\\ c_{i} \end{bmatrix}

          i = 1,2,3

          2015 WBJEE Solved Papers Maths

          a.sub>3 = 2, b3 = 1, c3 = –3

          2015 Solved Papers WBJEE Maths

          Sum of elements of U-1 = 0


        Question 64. Let f : N -> R be such that f(1) = 1 and

        f(1) + 2f(2) + 3f(3) + .......... + nf(n) = n(n + 1)f(n), for all n belongs to N, n ≥ 2, where N is the set of natural numbers and R is the set of real numbers. Then the value of f(500) is

        1. a. 1000
        2. b. 500
        3. c. 1/500
        4. d. 1/1000

        Solution:

        1. Answer: (d)

          f(1) = 1

          n = 2; f(1) + 2f(2) = 2.3 f(2)

          f(2) = 1/4

          n = 3; f(1) + 2f(2) +3f(3) = 12f(3)

          f(3) = 1/6

          => f(n) = 1/2h

          So f(500) = 1/1000.


        Question 65. If 5 distinct balls are placed at random into 5 cells, then the probability that exactly one cell remains empty is

        1. a. 48/125
        2. b. 12/125
        3. c. 8/125
        4. d. 1/125

        Solution:

        1. Answer: (a)

          Total no. of ways to put 5 balls in 5 cells = 55

          No. of ways so that exactly one cell remains empty

          (i) Select one cell to be left empty = 5C1

          (ii) Distribute 5 balls to 4 cells so that no cell will be left empty.

          Distribution will be of 1, 1, 1 and 2

          2015 Solved Papers Maths WBJEE


        Question 66. A survey of people in a given region showed that 20% were smokers. The probability of death due to lung cancer, given that a person smoked, was 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is 0.006, what is the probability of death due to lung cancer given that a person is a smoker?

        1. a. 1/140
        2. b. 1/70
        3. c. 3/140
        4. d. 1/10

        Solution:

        1. Answer: (c)

          E -> death due to lung cancer

          E1 -> Person is a smoker

          E2 -> Person is a non-smoker

          P(E1) = 1/5

          P(E2) = 4/5

          P(E) = 0.006

          P(E/E1) = 10P(E/E2)

          Now, P(E) = P(E1).P(E/E1)+P(E2).P(E/E2)

          0.006 = (1/5)P(E/E1)+(4/5)(1/10)P(E/E1)

          => P(E/E1) = 3/140


        Question 67. A person goes to the office by car or scooter or bus or train, the probability of which are 1/7, 3/7, 2/7 and 1/7 respectively. The probability that he reaches office late, if he takes a car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, the probability that he travelled by car is

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

        Solution:

        1. Answer: (a)

          Using Baye's Theorem

          2015 Maths WBJEE Solved Papers

          Let Ei -> Probability of travelling by car, scooter, bus, train

          P(E/E1)-> Probability of reaching on time

          = (1/7)(7/8) ÷(1/7)(7/9)+(3/7)(8/9)+(2/7)(5/9)+(1/7)(8/9)

          = 1/7


        Question 68. The value of (x2)dx{(x2)2(x+3)7}13\int \frac{(x-2)dx}{\left \{ (x-2)^{2}(x+3)^{7} \right \}^{\frac{1}{3}}} is

        1. a. (3/20)((x-2)/(x+3))4/3+c
        2. b. (3/20)((x-2)/(x+3))3/4+c
        3. c. (5/12)((x-2)/(x+3))4/3+c
        4. d. (3/20)((x-2)/(x+3))5/3+c

        Solution:

        1. Answer: (a)

          2015 WBJEE Maths Solved Papers


        Question 69. Let f : R ->R be differentiable at x = 0. If f(0) = 0 and f’(0) = 2, then the value of lim x->0 (1/x)[f(x)+f(2x)f(3x)+..+f(2015x)

        1. a. 2015
        2. b. 0 (zero)
        3. c. 2015 × 2016
        4. d. 2015 × 2014

        Solution:

        1. Answer: (c)

          Applying L’ hospital rule

          2015 Maths Solved Papers WBJEE

          = 2×2015×2016/2

          = 2015×2016


        Question 70. If x and y are digits such that 17! = 3556xy428096000, then x + y equals

        1. a. 15
        2. b. 6
        3. c. 12
        4. d. 13

        Solution:

        1. Answer: (a)

          Since 17!is divisible by 9,

          So sum of digits must be divisible by 9

          48 + x + y is divisible by 9

          x + y = 6 or 15 ; x, y belongs to {0,1,2,..a}

          Again,17! is divisible by 11,

          |Sum of digits at odd places – sum of digits at even places|

          Must be divisible by 11

          i.e. |10 + x - y| is divisible by 11

          => |x - y| = 1 (only possible value)

          Hence x + y = 15, (x, y) = (8,7) or (7,8)


        Question 71. Which of the following is/are always false?

        1. a. A quadratic equation with rational coefficients has zero or two irrational roots
        2. b. A quadratic equation with real coefficients has zero or two non-real roots
        3. c. A quadratic equation with irrational has zero or two rational roots
        4. d. A quadratic equation with integer coefficients has zero or two irrational roots

        Solution:

        1. Answer: (c)

          A -> A quadratic equation with rational coefficient can have any type of roots.

          Ex:

          (i) x2 = 0 -> zero roots

          (ii) x2 – x – 1 = 0 -> Irrational roots

          so can be true

          B -> Again a quadratic equation with real coefficients can have any type of roots.

          So, can be true

          C -> A quadratic equation with irrational distinct coefficients can never have zero or rational roots.

          So, always false

          D -> It can have any roots

          So, can be true.


        Question 72. If the straight line (a – 1)x – by + 4 = 0 is normal to the hyperbola xy = 1 the which of the followings does not hold?

        1. a. a > 1, b > 0
        2. b. a > 1, b < 0
        3. c. a < 1, b < 0<.p>d. a < 1, b > 0

        Solution:

        1. Answer: (a,c)

          x.y = 1

          Slope of normal = -dx/dy = x2 or 1/y2>0

          So given line is normal if its slope,

          i.e. (a-1)/b >0

          A and C are possible.


        Question 73. Suppose a machine produces metal parts that contain some defective parts with probability 0.05. How many parts should be produced in order that probability of at least one part defective is ½ or more? (Given log1095 = 1.977 and log102 = 0.3)

        1. a. 11
        2. b. 12
        3. c. 15
        4. d. 14

        Solution:

        1. Answer: (c,d)

          Probability of at least one defective part

          =1- non defective part

          = 1 – (0.95)n ≥1/2

          => (0.95)n ≤1/2

          Taking log

          WBJEE 2015 Solutions Paper Maths


        Question 74. Let f: R-> R be such that f(2x – 1) = f(x) for all x belongs to R. If f is continuous at x = 1 and f(1) = 1, then

        1. a. f(2) = 1
        2. b. f(2) = 2
        3. c. f is continuous only at x = 1
        4. d. f is continuous at all points.

        Solution:

        1. Answer: (a,d)

          f(2x-1) = f(x) for all x belongs to R

          Replacing x -> (x+1)/2

          f(x) = f((x+1)/2)

          WBJEE Solutions Paper Maths 2015


        Question 75. If cos x and sin x are solutions of the differential equation a0(d2y/dx2)+a1(dy/dx)+a2y = 0, where a0, a1, a2 are real constants then which of the followings is/are always true?

        1. a. A cos x + B sin x is a solution, where A and B are real constants
        2. b. A cos(x+ π/4) is solution, where A is real constant
        3. c. A cos x sin x is a solution, where A is real constant
        4. d. A cos(x+ π/4) +B sin(x- π/4) is a solution, where A and B are real constants

        Solution:

        1. Answer: (a,d)

          y = a sin x+b cos x

          dy/dx = a cos x-b sin x

          d2y/dx2 = -a sin x-b cos x = -y

          d2y/dx2+y = 0

          The given difference = n is general solution of y = a sin x + b cos x

          So a and d are correct options.


        Question 76. Which of the following statements is/are correct for 0 < θ < π/2?

        Solved WBJEE 2015 Maths Paper

          Solution:

          1. Answer: (a, c)

            0 < θ < π/2 ;0 < n θ ≤nπ/2

            {0<n<1, nπ/2 <π/2}

            So, 0 <cos θ <1

            Or 0 <cosnθ<1 …(1)

            K < cos nθ< 1

            Where 0 < k < 1

            Previous Year Solved Questions WBJEE 2015 Maths

            Therefore, cosnθ≤ cos n θ


          Question 77. Let 16x2 – 3y2 – 32x – 12y = 44 represent a hyperbola. Then

          1. a. Length of the transverse axis is 2√3
          2. b. Length of each latus rectum is 32√3
          3. c. Eccentricity is √ (19/3)
          4. d. Equation of a directrix is x = √19/3

          Solution:

          1. Answer: (a, b, c)

            16x2 – 3y2 – 32x – 12y = 44

            ⇒ 16(x2-2x) – 3(y2+4y) = 44

            ⇒ 16(x2-2x+1) – 3(y2+4y+4) = 44+4

            ⇒ 16(x-1)2 –3(y+2)2 = 48

            ⇒ (x-1)2/3 – (y+2)2/16 = 1

            Length of transverse axis = 2a = 2√3

            Length of Latus Rectum = 2b2/a = 32/√3

            Previous Year Questions Solutions WBJEE 2015 Maths


          Question 78. For the function f(x) = [1/[x]], where [x] denotes the greatest integer less than or equal to x, which of the following statements are true?

          1. a. The domains is (–∞, ∞)
          2. b. The range is {0} U {–1} U {1}
          3. c. The domain is {–∞, 0} U [1, ∞)
          4. d. The range is {0} U {1}

          Solution:

          1. Answer: (b, c)

            f(x) = [1/[x]]

            For Domain: [x] ≠ 0

            x ≠ [0, 1)

            Therefore, Df = R – [0, 1)

            Range:

            1/[x] -> 1, ½, 1/3, ¼, -1, -1/2, -1/3, -1/4

            [1/[x]] -> 1, 0, -1

            Therefore, Rf = {-1, 0, 1}


          Question 79. Let f be any continuously differentiable function on [a, b] and twice differentiable on (a, b) such that f(a) = f’(a) = 0 and f(b) = 0. Then

          1. a. f'(a) = 0
          2. b. f'(x) = 0 for some x ∈ (a, b)
          3. c. f''(x) = 0 for some x ∈ (a, b)
          4. d. f'''(x) = 0 for some x ∈ (a, b)

          Solution:

          1. Answer: (b, c)

            Applying Rolle's Theorem on f(x) in [a, b]

            f(a) = f(b) = 0

            f(x) is constant and differential function

            f'(c) = 0 for c ∈ (a, b)

            Now, again applying Rolle’s Theorem on f’(x)

            f'(a) = f'(c) = 0

            Therefore, In (a, c), f"(x) will be zero at some point.


          Question 80. A relation ρ on the set of real number R is defined as follows:

          1. a. ρ is reflexive and symmetric
          2. b. ρ is symmetric but not reflexive
          3. c. ρ is symmetric and transitive
          4. d. ρ is an equivalence relation

          Solution:

          1. Answer: (b, c)

            x ρ y ⇒x. y > 0

            Reflexive: x ρ x ⇒x . x > 0

            x2> 0

            Not true for x = 0

            Therefore, not reflexive

            Symmetric: x ρ y ⇒x. y > 0

            ⇒y. x > 0

            ⇒y ρ x

            Therefore, symmetric

            Transitive: x ρ y ⇒ x . y > 0 ...(i)

            y ρ z ⇒ y. z > 0 ....(ii)

            From (i) and (ii)

            x. z. y2> 0

            x . z > 0

            x ρ z

            Therefore, Transitive.


          Video Lessons - WBJEE 2015 - Maths

          WBJEE 2015 Maths Question Paper with Solutions

          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions
          WBJEE 2015 Maths Question Paper with Solutions

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