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WBJEE 2014 Solved Mathematics Question Paper – Part 16

Apr 25, 2017 17:44 IST

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.

WBJEE 2017

(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,

WBJEE 2014

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

WBJEE 2014

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

WBJEE 2014

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

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

(B) y = c1{u(x) – v(x)} + c2 v(x), c1 and c2 are arbitrary constants

(C)  y = c1 u(x) v(x) + c2 u(x)/v(x), c1 and c2 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 = c1{u(x) – v(x)} + c2 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 x2 + y2 + 2gx + 2fy + c = 0 cuts the three circles x2 + y2 – 5 = 0, x2 + y2 – 8x – 6y + 10 = 0 and x2 + y2 – 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 Sample Papers

WBJEE Previous Years' Question Papers

WBJEE Online Test

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