 JEE Main Solved Mathematics Sample Paper Set-XII

Find JEE Main 2017 Solved Mathematics Sample Paper Set-XII in pdf format. This sample paper will be very useful for the IIT JEE Main examination to be held on 2 April, 8 April, 9 April 2017. This paper consists of 30 questions from Mathematics. Each question of this paper is very important from examination point of view and has been developed very carefully by Subject Experts at Jagranjosh.

IIT JEE Examination is held to give admission to various prestigious engineering colleges such as IITs, NITs and various others prestigious engineering institutions.

This examination is held at two levels: JEE Main and JEE Advanced

The JEE Main Solved Mathematics Sample Paper Set-XII in this article consists of 30 questions from Mathematics. The questions in this paper have been asked from the complete syllabus.  Each question of this paper is very important from examination point of view and has been developed very carefully by Subject Experts at Jagranjosh.

Importance of Sample Paper

Practicing sample papers and previous years’ question papers help in assessing your preparation and time management. It will also help in increasing your marks in examination.

Questions

1.         If a and b are roots of the equation x2 + x + 1 = 0. The equation whose roots are a19, b7 is

(A) x2 - x -1 = 0         (B) x2 - x + 1 = 0        (C) x2 + x -1 = 0         (D) x2 + x + 1 = 0 (A) cos x + i sin x        (B) m/2                     (C) 1                            (D) (m + 1)/2

3.         If two roots of the equation x3 + mx2 + 11x - n = 0 are 2 and 3, then value of m + n is

(A) -1                         (B) -2                        (C) - 3            (D) none  of these Hints and Solutions

1.         D

We know that the roots of x2 + x + 1 = 0 are w, w2. Let a = w, b = w2, then
a+b=w + w2 = -1 and ab = w.w2 = w3 = 1

Now, a19 = w19 = (w3)6 w = w and b7 = (w2)7 = w14 = (w3)4w2 = w2.

Hence the equation whose roots are a19, b7 is x2 + x + 1 = 0

2.         C

3.         A

We have 23 + m(22) + 11(2) - n = 0 and 33 + m(32) + 11(3) - n = 0

4m - n = - 30 and 9m - n = -60

Solving we get m = -6, n = 6 Thus m + n = 0. 