Find **WBJEE 2014 Solved Mathematics Question Paper – Part 16** in this article. This paper consists of 5 questions (#76 to #80) 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.

(A) f(x) is continuous at x = 1

(B) f(x) is not continuous at x = 1

(C) f(x) is differentiable at x = 1

(D) f(x) is not differentiable at x = 1

**Ans: (A,D)**

**Sol: **

We have,

Clearly f(x) is continuous but not differentiable at x = 1.

**78.** If u(x) and v(x) are two independent solutions of the differential equation

then additional solution (s) of the given differential equation is (are)

(A) y = 5 u(x) + 8 v(x)

(B) y = c_{1}{u(x) – v(x)} + c_{2} v(x), c_{1} and c_{2} are arbitrary constants

(C) y = c_{1} u(x) v(x) + c_{2} u(x)/v(x), c_{1} and c_{2} are arbitrary constants

(D) y = u(x) v(x)

**Ans : (A,B)**

**Sol: **

Any linear combination of u(x) and v(x) will also be a solution.

So, the correct options are:

y = 5 u(x) + 8 v(x)

y = c_{1}{u(x) – v(x)} + c_{2} v(x)

**79.** For two events A and B, let P(A) = 0.7 and P(B) = 0.6. The necessarily false statements(s) is/are

(A) P(A ∩B) = 0.35

(B) P(A ∩B) = 0.45

(C) P(A ∩B) = 0.65

(D) P(A ∩B) = 0.28

**Ans: (C,D)**

**Sol:**

P(A U B) = P(A) + P(B) - P(A ∩B)

P(A U B) = 0.7 + 0.6 - P(A ∩B)= 1.3 - P(A ∩B)

P(A) ≤ P(A U B) ≤ 1

0.7 ≤ P(A U B) ≤ 1

0.7 ≤1.3 - P(A ∩B) ≤ 1

0.3 ≤ P(A ∩B) ≤ 0.6

**80.** If the circle x^{2 }+ y^{2} + 2gx + 2fy + c = 0 cuts the three circles x^{2} + y^{2} – 5 = 0, x^{2} + y^{2} – 8x – 6y + 10 = 0 and x^{2} + y^{2} – 4x + 2y – 2 = 0 at the extremities of their diameters, then

(A) C = – 5

(B) fg = 147/25

(C) g + 2f = c + 2

(D) 4f = 3g

**Ans : (A,B,D)**

**Sol: **

Common chords of the circle will pass through the centres.

c = -5, 8g + 6f = -35, 4g – 2f = -7

g = -14/5

f = -21/10

**Also Get:**

**WBJEE Previous Years' Question Papers**