Find WBJEE 2014 Solved Mathematics Question Paper – Part 12 in this article. This paper consists of 5 questions (#56 to #60) 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.
56. If a, b and c are positive numbers in a G.P., then the roots of the quadratic equation
(logea)x2 – (2logeb)x + (logec) = 0 are
Since, a, b and c are in GP.
b2 = ac
Taking log both side:
log (b2) = log(ac)
2logeb = logea + logec
⇒logea – 2logeb + logec = 0
We have the quadratic equation,
(logea)x2 – (2logeb)x + (logec) = 0
Since, 1 satisfies the equation
Therefore 1 is one root and other root say β
(A) f satisfies the conditions of Rolle’s theorem on [–1, 1]
(B) f satisfies the conditions of Lagrange’s Mean Value Theorem on [–1, 1]
(C) f satisfies the conditions of Rolle’s theorem on [0, 1]
(D) f satisfies the conditions of Lagrange’s Mean Value Theorem on [0, 1]
Ans : (D)
f(x) is non differentiable at x = 0
Also, f(0) ≠ f(1)
So, f does not satisfies the conditions of Rolle’s theorem on [0, 1].
It satisfies the condition of Lagrange’s Mean Value Theorem on [0, 1].
58. Let z1 be a fixed point on the circle of radius 1 centered at the origin in the Argand plane and
z1 ≠ ± 1. Consider an equilateral triangle inscribed in the circle with z1, z2, z3 as the vertices taken in the counter clockwise direction. Then z1z2z3 is equal to
59. Suppose that f(x) is a differentiable function such tha f′(x) is continuous, f′(0) = 1 and f″(0) does not exist. Let g(x) = xf′(x). Then
(A) g′(0) does not exist
(B) g′(0) = 0
(C) g′(0) = 1
(D) g′(0) = 2
Ans : (C)
g(x) = xf′(x)
By using first principle of derivative