# WBJEE 2014 Solved Mathematics Question Paper – Part 5

Find WBJEE Solved Mathematics Question Paper for the year 2014. This solved paper will help students in their final level of preparation for WBJEE Exam.

Find **WBJEE 2015 Solved Mathematics Question Paper – Part 5** in this article. This paper consists of 5 questions (#21 to #25) 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.

**21. Let p, q be real numbers. If α is the root of x^{2}+ 3p^{2 }x + 5q^{2} = 0, β is a root of **

** x ^{2 }+ 9p^{2}x + 15q^{2} = 0 and 0 < α < β, then the equation x^{2 }+ 6p^{2}x + 10q^{2} = 0 has a root γ that always satisfies **

(A) γ = α/4 + β

(B) β < γ

(C) γ = α/2 + β

(D) α < γ < β

**Ans: (D)**

**Sol: **

Let us assume that,

f(x) = x^{2 }+ 6p^{2}x + 10q^{2}

As α is the root of *x*^{2}+ 3*p*^{2 }*x* + 5*q*^{2} = 0

So, α^{2 }+ 3p^{2}α + 5q^{2} = 0

Also, β is a root of x^{2 }+ 9p^{2}x + 15q^{2} = 0

So, β^{2}+9p^{2}β+15q^{2 }= 0

Now,

f(α) = α^{2 }+ 6p^{2}α + 10q^{2} = (α^{2}+3p^{2}α+5q^{2}) + (3p^{2}α+5q^{2}) = 0 + 3p^{2}α+5q^{2} > 0

Again, f(β) = β^{2 }+ 6p^{2}β + 10q^{2} = (β^{2}+9p^{2}β+15q^{2}) – (3p^{2}β+5q^{2}) = 0 – (3p^{2}β + 5q^{2}) < 0

So, there is one root γ such that, α < γ < β.

(1+x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + ---------+C_{n}x^{n} ,

(x+1)^{n} = C_{0}x^{n} + C_{1}x^{n–1}+C_{2}x^{n–2} + ------+C_{n}

So, coefficient of x^{n} in [(1+x)^{n} × (x+1)^{n}] i.e. (1+x)^{2n} is ( C_{0}^{2} +C_{1}^{2}+ -------+C_{n}^{2}), which is ^{2n}C_{n}

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

**23. Ram is visiting a friend. Ram knows that his friend has 2 children and 1 of them is a boy. Assuming that a child is equally likely to be a boy or a girl, then the probability that the other child is a girl, is**

(A) 1/2

(B) 1/3

(C) 2/3

(D) 7/10

**Ans : (C)**

**Sol: **

Let us denote the boy with *B* and the girl with *G*. Let S be the sample space for two children and one of them is boy.

S = {BG, BB,}

The probability that one of the children is boy and the other is girl = 1/2

(A) A^{n} = I and A^{n - 1} ≠ I

(B) A^{m} ≠ I for any positive integer m

(C) A is not invertible

(D) A^{m} = 0 for a positive integer m

**Ans: (A)**

**Sol:**

**Also Get:**

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