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.

**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.

**27. Let α,β be the roots of x ^{2}–x–1 = 0 and S_{n} = α^{n}+β^{n}, for all integers n ≥ 1. Then for every integer n≥2**

(A) S_{n}+S_{n–1} = S_{n+1}

(B) S_{n}–S_{n–1} = S_{n}

(C) S_{n–1} = S_{n+1}

(D) S_{n}+S_{n–1} = 2S_{n+1}

Ans : (A)

Sol:

We have,

x^{2}–x–1 = 0

So,

α+β =1

Also,

α^{2}–α–1 = 0

α^{2} = α + 1 …(1)

And, β^{2}–β–1=0

β^{2} = β + 1 …(2)

Now,

S_{n} +S_{n–1} = α^{n}+β^{n} + α^{n-1}+β^{n-1} = (α^{n}+α^{n}–1)+( β^{n}+β^{n–1}) = α^{n–1} (α+1) + β^{n–1} (β+1)

From (1) and (2)

S_{n} +S_{n–1} = α^{n–1} (α^{2}) + β^{n–1} (β^{2}) = α^{n+1} + β^{n+1} = S_{n+1}

So, S_{n} +S_{n–1} = S_{n+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 a ^{3}sin(B–C) + b^{3} sin(C–A) + c^{3}sin(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 .

a^{3}sin(B–C) + b^{3} sin(C–A) + c^{3}sin(A–B) = 0

**WBJEE 2016 Solved Physics and Chemistry Question Paper**

**29. In the Argand plane, the distinct roots of 1+z+z ^{3}+z^{4} = 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+z^{3}+z^{4} = 0,

(1+z) (1+z^{3})= 0,

z = –1, –1, –ω, –ω^{2}

**30. The number of digits in 20 ^{301} (given log_{10}2 = 0.3010) is**

(A) 602

(B) 301

(C) 392

(D) 391

**Ans: (C)**

**Sol: **

log 20^{301} = 301 × log 20 = 301 × 1.3010 = 391.6010,

So, the number of digits in 20^{301 }is 392 digits.

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