##### JEE Main Solved Mathematics Sample Paper Set-XII

Find the pdf version of the JEE Main 2017 Solved Mathematics Sample Paper Set-XII. This sample exam will be very helpful for the IIT JEE Main examinations scheduled for April 2, 8, and 9, 2017. This exam consists of thirty mathematical questions. Subject Experts at Jagranjosh have designed each question of this exam with great attention because of its significance from an examination standpoint.

IIT JEE Examination is administered to grant admission to numerous top engineering colleges, including IITs, NITs, and other prestigious engineering institutes.

JEE Main and JEE Advanced are the two levels of this examination.

About Paper

This article’s JEE Main Solved Mathematics Sample Paper Set-XII has thirty questions from Mathematics. The questions in this exam were drawn from the whole course outline. Subject Experts at Jagranjosh have designed each question of this exam with great attention because of its significance from an examination standpoint.

**Relevance of Example Paper**

Sample questions and questions from past years are useful for evaluating your preparation and time management. It will also assist in improving your examination performance.

**Questions**

**1.** If a and b are roots of the equation x^{2} + x + 1 = 0. The equation whose roots are a^{19}, b^{7} is

**(A)** x^{2} – x -1 = 0 **(B)** x^{2} – x + 1 = 0 **(C)** x^{2} + x -1 = 0 **(D)** x^{2} + 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 x^{3} + mx^{2} + 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 x^{2} + x + 1 = 0 are w, w^{2}. Let a = w, b = w^{2}, then

a+b=w + w^{2} = -1 and ab = w.w^{2} = w^{3} = 1

** **Now, a^{19} = w^{19} = (w^{3})^{6} w = w and b^{7} = (w^{2})^{7} = w^{14} = (w^{3})^{4}w^{2} = w^{2}.

** **Hence the equation whose roots are a^{19}, b^{7} is x^{2} + x + 1 = 0

**2. C**

**3. A**

We have 2^{3} + m(2^{2}) + 11(2) – n = 0 and 3^{3} + m(3^{2}) + 11(3) – n = 0

4m – n = – 30 and 9m – n = -60

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