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# WBJEE 2014 Solved Mathematics Question Paper – Part 6

Mar 22, 2017 17:37 IST

Find WBJEE 2015 Solved Mathematics Question Paper – Part 6 in this article. This paper consists of 5 questions (#26 to #30) 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.

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.

27. Let α,β be the roots of x2–x–1 = 0 and Sn = αnn, for all integers n ≥ 1. Then for every integer n≥2

(A) Sn+Sn–1 = Sn+1

(B) Sn–Sn–1 = Sn

(C) Sn–1 = Sn+1

(D) Sn+Sn–1 = 2Sn+1

Ans : (A)

Sol:

We have,

x2–x–1 = 0

So,

α+β =1

Also,

α2–α–1 = 0

α2 = α + 1                           …(1)

And, β2–β–1=0

β2 = β + 1                         …(2)

Now,

Sn +Sn–1 = αnn + αn-1n-1  = (αnn–1)+( βnn–1) = αn–1 (α+1) + βn–1 (β+1)

From (1) and (2)

Sn +Sn–1 = αn–12) + βn–12) = αn+1  + βn+1 = Sn+1

So, Sn +Sn–1 = Sn+1

28. In a ΔABC, a,b,c are the side of the triangle opposite to the angles A,B,C respectively. Then the value of a3sin(B–C) + b3 sin(C–A) + c3sin(A–B) is equal to

(A) 0

(B) 1

(C) 3

(D) 2

Ans : (A)

Sol.

By using sine law and 3A formula, the value of .

a3sin(B–C) + b3 sin(C–A) + c3sin(A–B) = 0

WBJEE 2016 Solved Physics and Chemistry Question Paper

29. In the Argand plane, the distinct roots of 1+z+z3+z4 = 0 (z is a complex number) represent vertices of

(A) a square

(B) an equilateral triangle

(C) a rhombus

(D) a rectangle

Ans : (B)

Sol:

We have,

1+z+z3+z4 = 0,

(1+z) (1+z3)= 0,

z = –1, –1, –ω, –ω2

30. The number of digits in 20301 (given log102 = 0.3010) is

(A) 602

(B) 301

(C) 392

(D) 391

Ans: (C)

Sol:

log 20301 = 301 × log 20 = 301 × 1.3010 = 391.6010,

So, the number of digits in 20301 is 392 digits.

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