# JEE Main 2019: 100 Most Important Solved Questions of Mathematics for April Session

In this article, engineering aspirants will find the most important solved questions based on the important topics of Mathematics for JEE Main 2019. Generally, the questions are not repeated in JEE Main but the concepts are frequently repeated.

*Most Important Maths Solved Questions for JEE Main 2019*

Get 100 most important solved questions of Maths for JEE Main April 2019. In JEE Main 2019 January exam, several questions were taken directly from NCERT and NCERT Exemplar books of class 11 and class 12. Few questions were also repeated from JEE Main Previous Years’ question papers. The National Testing Agency has scheduled the second phase of JEE Main 2019 between April 7 and 20 in fully computer based test mode. Students, who were not able to perform well in JEE Main (I) 2019 because of any reason, are given another opportunity to increase their All India Ranks in JEE Main 2019. The all India of ranking of candidates will be released taking into consideration better of the two NTA Scores of all candidates appeared in Jan 2019 and April 2019 examinations. Here, we are going to provide you 100 solved questions of Mathematics which are very important from the viewpoint of JEE Main April 2019. The solutions are elaborate and easy to understand. Questions are collected from NCERT textbooks of class 11 and class 12, previous years’ question papers and JEE Main 2019 January exam papers.

**JEE Main 2019 January Exam: Download Question Papers of all shifts in PDF format**

**JEE Main 2019: Planning for 2nd Attempt in April? Check 3 Takeaways from Exams Conducted in January**

Students will get questions from all important topics like Application Of Derivatives, Vector, Limits, 3-Dimensional Geometry, Definite Integrals, Area Under Curve, Trigonometric Equations, Indefinite Integrals, Probability, Straight Lines, Circle, Complex Numbers, Height & Distance, Statistics, Binomial Theorem, Hyperbola, Permutation & Combination, Determinants and Matrix, Sets & Relations, Parabola, Differentiability, Mathematical Reasoning, Differential Equations, Sets & Relations, Progression & Series.

All the aspirants need to keep a track of their preparedness by practicing more and more questions based on each important topic for JEE Main test. Students can download the complete collection of the questions in the PDF form with the help of the link given at the end of this article.

**Some sample questions are given below:**

**Question:**

The smaller of 99^{100} + 100^{100} and 101^{100}, is

(a) 99^{100} + 100^{100}

(b) Both are equal

(c) 101^{100}

(d) None of these

**Solution: (a)**

**Question:**

If three successive terms of a G.P with common ratio *r* (*r* > 1) form the sides of a triangle ABC and [*r*] denotes greatest integer function, then [*r*] + [−*r*] is

(a) 0

(b) 1

(c) −1

(d) None of these

**Solution: (c)**

Let us suppose the sides of the triangle be *a*, *ar* and *ar*^{2} (being the largest side as *r* > 1).

**Question:**

Eighteen guests have to be seated, half on each side a long table. Four particular guest desires to sit on one particular side and three others on the other side. The number of ways in which the seating arrangement can be made, is

**Question:**

The sum of all four digit numbers that can be med using the digits 1, 2, 3, 4, when repetition of digits is not allowed, is

(a) 36600

(b) 66000

(c) 36000

(d) 66660

**Solution: (d)**

**JEE Main 2019: Attempt Mock Tests at “NTA Student” mobile app, also locate nearest Practice Centre**

**Question:**

The locus of the centre of a circle, which touches externally the circle *x*^{2} + *y ^{2} *- 6

*x*- 6

*y*+ 14 = 0 and also touches the

*y*-axis, is given by the equation

(a) *x*^{2} - 6*x* - 10*y* + 14 = 0

(b) *x*^{2} -10*x* - 6*y* + 14 = 0

(c) *y*^{2 }- 6 *x *- 10*y* + 14 = 0

(d)* y*^{2} - 10*x *- 6*y* + 14 = 0** **

**Solution: (d)**

We have *x*^{2} + *y ^{2} *- 6

*x*- 6

*y*+ 14 = 0

Let C be the centre of the circle. So, the coordinates of the centre of the circle is C (3, 2).

Since CA and CB are perpendicular to PA and PB, CP is the diameter of the circumcircle of triangle PAB.

So, the equation of the circle with the end points of the diameter as (1, 8) and (3, 2) is

**To download the complete file in PDF form, click on the following link:**

**JEE Main 2019: 100 Most Important Solved Questions of Mathematics for April Session**

## Comments