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WBJEE 2014 Solved Mathematics Question Paper Part 13

Apr 6, 2017 19:04 IST

    Find WBJEE 2014 Solved Mathematics Question Paper – Part 13 in this article. This paper consists of 5 questions (#61 to #65) from WBJEE 2014 Mathematics paper. Detailed solution of these questions has been provided so that students can match their solutions.

    Importance of Previous Years’ Paper:

    Previous years’ 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. So, this paper will certainly boost your confidence.

    About WBJEE Exam

    WBJEE is a common entrance examinations held at state level for admission to the undergraduate level engineering and medical courses in the state of West Bengal. The Mathematics section of WBJEE 2014 engineering entrance exam consists of 80 questions.

    61. We define a binary relation on the set of all 3 × 3 real matrices as A B if and only if there exist invertible matrices P and Q such that B = PAQ–1. The binary relation is

    (A) Neither reflexive nor symmetric

    (B) Reflexive and symmetric but not transitive

    (C) Symmetric and transitive but not reflexive

    (D) An equivalence relation

    Ans : (D)

    Hints :

    For Reflexive,

    A.I = IA,

    A = AAA-1

    A = AI = A

    so reflexive.

    For Symmetric,

    B = PAQ–1,

    BQ = PA,

    P -1BQ = A or A = (P–1) B. (Q–1)–1,so symmetric.

    For Transitive,

    B = PAQ–1,

    C = PBQ 1=P.PAQ–1.Q–1= (PP)A(QQ)–1, so transitive

     wbjee maths questions

    63. For any two real numbers θ and ϕ, we define θRφ if and only if sec2θ – tan2φ = 1. The relation R is

    (A) Reflexive but not transitive

    (B) Symmetric but not reflexive

    (C) Both reflexive and symmetric but not transitive

    (D) An equivalence relation

    Ans : (D)


    For reflexive, θ = φ

    so sec2θ – tan2θ = 1, Hence Reflexive

    For symmetric, sec2θ – tan2φ = 1

    So, (1 + tan2θ) – (sec2φ – 1) = 1 so, sec2φ – tan2θ = 1. Hence symmetric

    For Transitive,

    Let sec2θ – tan2φ = 1 and sec2φ – tan2γ = 1

    so, 1 + tan2φ – tan2γ = 1 or,

    sec2θ – tan2γ = 1.

    Hence it is transitive

    64. A particle starting from a point A and moving with a positive constant acceleration along a straight line reaches another point B in time T. Suppose that the initial velocity of the particle is u > 0 and P is the midpoint of the line AB. If the velocity of the particle at point P is v1 and if the velocity at time T/2 is v2 , then

    (A) v1> v2


    (B) v1> v2

    (C) v1< v2

    (D) v1=1/2 v2

    Ans: (B)

    Since the particle is moving with a positive constant acceleration hence it’s velocity should increase, when it moves from point A to B. Also, P is the mid-point of line AB.

    maths questions paper 2014 for wbjee

    WBJEE 2014 Solved Mathematics Question Paper – Part 12


    So the time taken to travel AP is more than that of the time taken for PB. So the instant T/2 is before P. Hence, v1 > v2 since velocity increases from A to B.



    If the velocity of the particle at point P is v1 and if the velocity at time  is v2 , then

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