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WBJEE 2014 Solved Mathematics Question Paper – Part 7

Mar 23, 2017 18:24 IST

    Find WBJEE 2014 Solved Mathematics Question Paper – Part 7 in this article. This paper consists of 5 questions (#31 to #35) 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.

    WBJEE 2014

    WBJEE 2014

    WBJEE 2014

    WBJEE 2016 Solved Physics and Chemistry Question Paper

    34. If α, β are the roots of ax2 + bx + c = 0 (a≠0) and α + h, β + h are the roots of px2 +qx+r = 0 (p≠0) then the ratio of the squares of their discriminants is

    (A) a2:p2

     (B) a:p2

    (C) a2:p

    (D) a:2p

    Ans: (A)

    Sol.

    We have the quadratic equation,

    ax2 + bx + c = 0

    of which α and β are the roots

    So,

    WBJEE 2014

    35. Let f(x) = 2x2+5x+1. If we write f(x) as

    f(x) = a(x + 1)(x – 2) +b(x – 2)(x – 1)+c(x – 1)(x + 1) for real numbers a,b,c then

    (A) there are infinite number of choices for a,b,c

    (B) only one choice for a but infinite number of choices for b and c

    (C) exactly one choice for each of a,b,c

    (D) more than one but finite number of choices for a,b,c

    Ans: (C)

    Sol:

    We have,

    f(x) = 2x2+5x+1                      …(1)

    Given, f(x) can be written as:

    f(x) = a(x + 1)(x – 2) +b(x – 2)(x – 1)+c(x – 1)(x + 1)

    f(x) = a(x2 –x - 2) +b(x2 - 3x + 2) + c(x2 - 1)

    f(x) = (a + b + c)x2 - (a + 3b)x -2a + 2b – c             …(2)

    From (1) and (2)

    a + b + c = 2             …(3)

    a + 3b = - 5               …(4)

    -2a + 2b – c = 1        …(5)

    Adding (3), (4), (5)

    a + b + c + a + 3b -2a + 2b – c = 2 – 5 + 1

    6b = -2

    Putting value of b in (4) we get the value of a

    Again putting the value of a and b in equation (3) we get the value of c.

    So, we have exactly one choice for each of a,b,c.

    Also Get:

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