# WBJEE 2015 Solved Mathematics Question Paper Part - 4

Find WBJEE Solved Mathematics Question Paper 2015. This solved paper will help students in their final level of preparation of WBJEE Exam.

Find **WBJEE Solved Mathematics Question Paper **for the year 2015. This article contains ten questions from the paper with their detailed solution. 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.

The Mathematics section of WBJEE engineering exam has 80 multiple choice questions.

**31.** The value of λ, such that the following system of equations has no solution, is

2x ─ y ─ 2z = 2

x ─ 2y + z = ─ 4

x + y + λz = 4

(A) 3 (B) 1 (C) 0(zero) (D) ─3

**Ans: (D)**

**Sol:**

We have,

2x ─ y ─ 2z = 2

x ─ 2y + z = ─ 4

x + y + λz = 4

For no solution,

Then *f*(100) is equal to

(A) 0(zero)

(B) 1

(C) 100

(D) 10

** **

**Ans: (A)**

**Sol:**

We have,

**33.**

**Ans: (A)**

**Sol.**

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

**34.** The area of the region bounded by the curve y = x^{3}, its tangent at (1, 1) and x-axis is

(A) 1/12

(B) 1/6

(C) 2/17

(D) 2/15

**Ans: (A)**

**Sol.**

**35.** If log_{0.2}(x─1)> log_{0}._{04}(x+5) then

(A) ─1 < x < 4 (B) 2 < x < 3

(C) 1 < x < 4 (D) 1 < x < 3

**Ans: (C)**

**Sol.**

We have,

log_{0.2}(x─1)> log_{0}._{04}(x+5)

**36.** The number of real roots of equation log_{e}x + ex = 0

(A) 0(zero) (B) 1

(C) 2 (D) 3

**Ans: (B)**

**Sol.**

The graph of both functions can be drawn as follows:

It can be seen from the graph that there is clearly one root of the given equation.

**WBJEE Sample Question Paper Set-II**

**37.** If the vertex of conic y^{2} ─ 4y = 4x ─ 4a always lies between the straight lines x + y = 3 and

2x + 2y ─ 1 = 0 then

**Ans: (B)**

**Sol.**

We have the cone:

y^{2} ─ 4y = 4x ─ 4a

y^{2} ─ 4y + 4 - 4 = 4x ─ 4a

(y -2)^{2} = 4(x – a + 1)

On comparing the above equation with the standard equation we get the vertex of conics as

(a ─1, 2)

Since, vertex lies between straight lines x + y = 3 and 2x + 2y ─ 1 = 0

So, (a ─ 1 + 2 ─ 3) (2a ─ 2 + 4 ─ 1) < 0

(a ─ 2) (2a + 1) < 0

-1/2 < a < 2

**38.** Number of intersecting points of the conic 4x^{2} + 9y^{2} = 1 and 4x^{2} + y^{2} = 4 is

(A) 1 (B) 2

(C) 3 (D) 0(zero)

**Ans: (D)**

**Sol.**

We have curves,

C_{1}: 4x^{2} + 9y^{2} = 1

C_{2}: 4x^{2} + y^{2} = 4

**39.**

**Ans: (C)**

**Sol.**

When the line lies on the plane then, the direction ratios of the line must be perpendicular to the direction ratios of normal to the plane.

The line must be perpendicular to the normal to plane.

3.1 + (2 + λ) (─2) + (─1).0 = 0

**40.**

**Ans: (D)**

**Sol.**