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WBJEE 2015 Solved Mathematics Question Paper Part - 4

Jan 30, 2017 18:45 IST

    Find WBJEE Solved Mathematics Question Paper for the year 2015. This article contains ten questions from the paper with their detailed solution. Previous year question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam.

    The Mathematics section of WBJEE engineering exam has 80 multiple choice questions.

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

    2x ─ y ─ 2z = 2

    x ─ 2y + z = ─ 4

    x + y + λz = 4

    (A) 3               (B) 1         (C) 0(zero)        (D) ─3

    Ans: (D)

    Sol:

    We have,

    2x ─ y ─ 2z = 2

    x ─ 2y + z = ─ 4

    x + y + λz = 4

    For no solution,

    WBJEE 2015

    WBJEE 2015

    Then f(100) is equal to

    (A) 0(zero)             

    (B) 1

    (C) 100                   

    (D) 10

     

    Ans: (A)

    Sol:

    We have,

    WBJEE 2015

    33.

    WBJEE 2015

    Ans: (A)

    Sol.

    WBJEE 2015

    WBJEE 2016 Solved Physics and Chemistry Question Paper

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

    (A) 1/12

    (B) 1/6

    (C) 2/17

    (D) 2/15

    Ans: (A)

    Sol.

    WBJEE 2015

    35. If log0.2(x─1)> log.04(x+5) then

    (A) ─1 < x < 4             (B) 2 < x < 3

    (C) 1 < x < 4               (D) 1 < x < 3

    Ans: (C)

    Sol.

    We have,

    log0.2(x─1)> log.04(x+5)

    WBJEE 2015

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

    (A) 0(zero)           (B) 1

    (C) 2                   (D) 3

    Ans: (B)

    Sol.

    The graph of both functions can be drawn as follows:

    WBJEE 2015

    It can be seen from the graph that there is clearly one root of the given equation.

    WBJEE Sample Question Paper Set-II

    37. If the vertex of conic y2 ─ 4y = 4x ─ 4a always lies between the straight lines x + y = 3 and

    2x + 2y ─ 1 = 0 then

    WBJEE 2015

    Ans: (B)

    Sol.

    We have the cone:

    y2 ─ 4y = 4x ─ 4a

    y2 ─ 4y + 4 - 4 = 4x ─ 4a

    (y -2)2 = 4(x – a + 1)

    On comparing the above equation with the standard equation we get the vertex of conics as

    (a ─1, 2)

    Since, vertex lies between straight lines x + y = 3 and 2x + 2y ─ 1 = 0

    So, (a ─ 1 + 2 ─ 3) (2a ─ 2 + 4 ─ 1) < 0

    (a ─ 2) (2a + 1) < 0

    -1/2  < a < 2

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

     (A) 1              (B) 2

    (C) 3               (D) 0(zero)

    Ans: (D)

    Sol.

    We have curves,

    C1: 4x2 + 9y2 = 1

    C2: 4x2 + y2 = 4

    WBJEE 2015

    39.

    WBJEE 2015

    Ans: (C)

    Sol.

    When the line lies on the plane then, the direction ratios of the line must be perpendicular to the direction ratios of normal to the plane.

    The line must be perpendicular to the normal to plane.

    3.1 + (2 + λ) (─2) + (─1).0 = 0

    WBJEE 2015

    40.

    WBJEE 2015

     Ans: (D)

    Sol.

    WBJEE 2015

    WBJEE 2016 Solved Mathematics Question Paper

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