WBJEE 2015 Solved Mathematics Question Paper – Part 2
Get free access to WBJEE Solved Mathematics Question Paper for the year 2015. This solved paper will help students in their final level of preparation for WBJEE Exam.
Find WBJEE 2015 Solved Mathematics Question Paper – Part 2 in this article. This paper consists of 10 questions (#11 to #20) from WBJEE 2015 Mathematics paper. Detailed solution of these questions has been provided so that students can match their solutions.
Importance of Previous Years’ Paper:
Previous year 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 WBJEE engineering entrance exam has two sections – Mathematics, Physics and Chemistry. The Mathematics section of WBJEE 2015 engineering entrance exam consists of 80 questions.
12. Let a, b, c, d be any four real numbers. Then an + bn = cn + dn holds for any natural number n if
(A) a + b = c + d
(B) a – b = c – d
(C) a + b = c + d, a2 + b2 = c2 + d2
(D) a – b = c – d, a2 – b2 = c2 – d2
Put n = 1, a + b = c + d ...... (1)
Put n = 3, a3 + b3 = c3 + d3 ........... (2)
from (1) and (2) ab = cd
Consider a quadratic with roots (a3, b3)
x2 – (a3 + b3)x + a3b3 = 0 ........ (3)
Consider another quadratic with roots (c3, d3)
x2 – (c3 + d3)x + (cd)3 = 0 ................. (4)
Since a3 + b3 = c3 + d3 and (cd)3 = (ab)3
Both quadratic are same and quadratic cannot have more than 2 roots.
Here, a = c and b = d or a = d, b = c
(C) – 1
Ans. (No option is correct)
We have the binomial expression,
Except, r = 0, 15, 30, 45, 60, 75, 90 each term will give irrational terms.
So, there are 7 rational terms
Total number of terms in the given binomial expression = 101
No. of irrational terms
= 101 – 7 = 94
(A) p2 – 16p – 8q < 0
(B) p2 – 8p + 16q < 0
(C) p2 – 8p – 16q < 0
(D) p2 – 16p + 8q < 0
(A) 0 (zero)
(B) – 1
We know that by the property of cube root of unity:
By using the property of modulus, we have
Now, when the lines are concurrent then they satisfy all the equations simultaneously at certain point.
From (2) and (3)
Minimum value of t is obtained when a = 4
So, the least possible value of t = 8
Ans : (B)