Practice and conceptual knowledge are simple ways to command Mathematics for the JEE Mains exam. JEE Mains exam comprises Physics, Chemistry, and Mathematics with equal weightage. Although the syllabus is vast and time is tight, hard work and smart learning techniques always let you stand ahead.

This article will help you understand the difficulty level of Mathematics questions in JEE Mains.

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**Chapter-wise Important Questions for JEE Mains**

**Chapter 1: Sets, Relations, and Functions**

1. If log2x+logx2 = 52 = log2y + logy2 and xy, then find the value of x+y-2.

log2x+logx2 = 52 = log2y + logy2

t + 1t = 52, s + 1s = 52; where t = log2x, s = log2y

t = 2, 12 and s = 2, 12 (as ts, xy)

t = 2, s = 12 and t = 12 , s = 2

log2x = 2 and log2y = 12

x = 4 and y = 2

Hence, x+y-2 = 4 + 2 –2 = 4.

2. What is the range of sin-1(x²+1x²+2)?

Here, x²+1x²+2 = 1-1x²+2

Now, 2 x2 + 2 < for all x R

12 1x²+2>0

–12 –1x²+2<0

12 1-1x²+2<1

6sin-1(1-1x²+2)<2

Hence, Range is [6,2).

**Chapter 2: Complex Numbers and Quadratic Equations**

1. The addition of all positive integral values of ‘a’, a [1, 500] for which the equation [x]³ + x – a = 0 has a solution. Find it.

a is integer then x must be an integer, i.e., [x] = x.

[x]³ + x – a = 0

[x]³ + x = a

1a500

1x7, x 1

ai=x=17(x3+x) = 812

2. Find the least integral value of ‘a’ if (a – 3)x2 + 12x + (a + 6) >0 x R.

ax2+bx+c>0 x R

a > 0, D < 0

(a – 3) > 0 and (a +9)(a – 6), a > 6

Hence, least integral value of a = 7.

**Chapter 3: Matrices and Determinants**

1.Find the value of |a|, if the system of equations x + ay = 0, az + y = 0, ax + z = 0 has an infinite solution.

x + ay = 0 x = – ay

ax + z = 0 -a2y+z=0

az + y = 0 a3y+y=0

a3 = -1 a = -1

Hence, |a| = 1.

2. A square matrix A is said to be nilpotent of index m. If Am=0, now, if for this A

(1 – A)n=1+A+A2+…+Am-1, then find the value of n.

Let B = 1+A+A2+…+Am-1

B (1 – A)n= (1 – A)n(1+A+A2+…+Am-1)

B (1 – A)n = 1 – Am

B (1 – A)n = 1, (as Am=0)

B = (1 – A)-n

n = -1

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**Chapter 4: Permutations and Combinations**

1. If one test (on a screening paper basis) was conducted on batch A, the maximum number of marks obtained is (90 x 3), 270. Four students get marks lower than 80. The coaching institute decided to inform their guardians. Hence, they sent their result to their home. Find the number of ways to put the letters in the wrong envelopes.

The number of ways in which put the letters in the wrong envelopes is

4! x (1 – 11!+12!–13!+14!)

= 4! (12–16+124)

= 12 – 4 + 1 = 9

Hence, they are put in 9 wrong ways.

2. Find the number of different necklaces that can be made from 17 identical pearls and two different diamonds.

The 19 precious stones (17 identicals pearls and two different diamonds) can be arranged in a circle in

18!17! = 18 ways

Now, the mutual arrangement of 2 different diamonds does not give a different necklace (merely gives a minor image).

Hence, the required number of different necklaces = 182 = 9.

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**Chapter 5: Mathematical Induction**

1. If m and n are two positive odd integers with n < m, then find the largest positive integers which divide all the numbers of the type m2-n2.

Let m = (2k + 1) and n = (2k – 1); (as n < m)

m2–n2 = (2k + 1)2–(2k – 1)2

= 4k2 + 1 + 4k – 4k2 – 1 + 4k

= 8k

Now, P(k) = 8k

Hence, P(k) is divisible by 8.

2. Find the greatest positive integer, which divides (n+2)(n+3)(n+4)(n+5)(n+6) for all n N.

Product of r successive integers is divisible by r!.

Therefore, the given expression is divisible by 5!, i.e., 120.

**Chapter 6: Binomial Theorem and its Simple Applications**

1. Find the coefficient of a10b7c3 in the expansion of (bc+ca+ab)10.

The general term in the expansion of (bc+ca+ab)10 is

10!r!s!t! (bc)r (ca)s (ab)t = 10!r!s!t! a(t+s)b(r+t)c(r+s), where r + s + t = 10

For coefficient of a10b7c3, we get

(t+s) = 10, (r+t) = 7, (r+s) = 3

As (r + s + t) = 10, we get r = 0, s = 3, t = 7.

Hence, coefficient of of a10b7c3 in the expansion of (bc+ca+ab)10 = 10!0!3!7! = 120.

2. Find the coefficient of x2009 in (1 + x + x2+x3+x4)¹⁰⁰¹(1 – x)¹⁰⁰².

(1 + x + x2+x3+x4)¹⁰⁰¹(1 – x)¹⁰⁰² = (1 – x)(1 – x5)

Therefore, all the power of x will be (5m) or (5m+1), where m R.

Hence, coefficient of x2009 is 0.

**Chapter 7: Sequence and Series**

1. Let a₁, a₂, a₃, …, a₁₀ be in A.P. and h₁, h₂, h₃, …, h₀ be in H.P. If a₁ = h₁ = 2 and a₁₀ = h₁₀ = 3, then find a₄h₇.

a₁ = h₁ = 2, a₁₀ = h₁₀ = 3

3 = a₁₀ = 2 + 9d

d = 19

a₄ = 2 + 3d = 73

Now, 3 = h₁₀ 13 = 1h₁₀ = 12 + 9D

D = – 154

1h₇ = 12 + 6D = 718

Hence, a₄h₇ = 73 x 187 = 6.

2. If the sum of infinity of the series [3 + (3 + d)14 + (3 + 2d)142+ …] is 449, find the value of d.

S = a1-r + dr(1-r)2

Here, r = 14 and S = 449

449 = a1-r + dr(1-r)2

Hence, d = 2.

**Chapter 8: Limit, Continuity and Differentiability**

1. Let f(x) = x tan-1(x2) + x4. If fk(x) denotes kth derivative of f(x), where x, k N. If f2m(0) 0, m N, then find the value of m.

Let g(x) = x tan-1(x2). It is an odd function.

g2m(x) = 0

Let h(x) = x4

Hence, f(x) = g(x) + h(x)

f2m(0) = g2m(0) + h2m(0) = h2m(0) 0

It happens when 2m = 4 and m = 4.

2. Let f(x) = [x] + |1-x|, -1 x 3 and [x] is the largest integer not exceeding x. Find the number of points where f is not continuous in [-1, 3].

f(x) = [x] + |1-x|, -1 x 3

f(x) = -x, -1 x < 0

f(x) = 1-x, 0 x < 1

f(x) = x, 1 x < 2

f(x) = x + 1, 2 x < 3

f(x) = 5, x = 3

Therefore, f is continuous at x 4 and 5.

**Chapter 9: Integral Calculus**

1. If x = a(t + sint), y = a(1-cost), then find dydx.

dxdt=a(1 +cost), dydt=a sint

dydx = dydt x dtdx

dydx = a sinta(1 +cost) = tant2

2. If x = 0ydu1+9u2, then find the value of k for d2ydx2=ky.

dxdy = 11+9y2

dydx = 1+9y2

d2ydx2 = 1 x 18y2 x 1+9y2 dydx

d2ydx2 = 9y = ky

Hence, k = 9.

**Chapter 10: Differential Equations**

1. Find the order of the differential equation of all tangent lines to the parabola y = x2.

The line x = my + 14m is a tangent to the given parabola for all m. This line represents the one-parameter family of lines. Hence, the order of this differential equation is 1.

2. Find the order of differential equation whose general solution is given by

y = (C₁+C₂) cos(X + C₃) – C₄ex+C₅, where C₁, C₂, C₃, C₄, C₅ are arbitrary constants.

Hence,

Y = Acos(X + C₃) – Bex, where A = (C₁+C₂) and B = C₄eC₅

Hence, there are three independent variables (A, B, C₃).

Therefore, the order of this differential equation is 3.

**Chapter 11: Coordinate Geometry**

1. If the point P (4, -2) is the one end of the focal chord PQ of the parabola y2=x, find the tangent slope at Q.

The equation of the tangent at (4, -2) to y2=x is

-2y = 12 (x + 4) or x + 4y + 4 = 0

Its slope is – 14.

Therefore, the slope of this tangent is 4.

2. A straight line with a negative slope passing through the point (1, 4) meets the coordinate axis at A and B. Find the least value of OA + OB.

xa + yb = 1 (a > 0, b > 0)

1a + 4b = 1

Let S = a + b

S = 11 – 4b + b

S 9

Hence, the least value of OA + OB = 9.

**Chapter 12: Three-Dimensional Geometry**

1. A plane is passed through the middle point of the segments A (-2, 5, 1) and B (6, 1, 5) and is perpendicular to this line. Find its equation.

Plane passes through the midpoint of AB and the direction ratio of normal to plane is proportional to the direction ratio of AB. Therefore, the equation of plane is

(6 + 2)(x – 2) + (1 – 5)(y – 3) + (5 – 1)(z – 3) = 0

2x – y + z = 4

2. Find the shortest distance between the two straight lines x-432 = y+653 = z-324 and 3x-45 = 5y+68 = 2z-39.

Since two lines are intersecting each other, the shortest distance between them will be 0.

**Chapter 13: Vector Algebra**

1. The vectors (2i+3j), (5i+6j), and (8i+j) have their initial points at (1, 1). Find the value of so that the vectors terminate on one straight line.

Since initial point of (2i+3j), (5i+6j), and (8i+j) is (i+j), their terminal points will be (3i+4j), (6i+7j), and (9i+(+1)j).

As all the vectors are terminating at one point, therefore,

3i+3j = ₁ (3i+(+1-7)j)

₁ = 1 and = 9

2. Find the number of vectors of unit length perpendicular to vectors a (1, 1, 0) and b (0, 1, 1).

The vector of unit length perpendicular to the given vectors = (a x b|a||b|). Therefore, there are two such vectors.

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**Chapter 14: Statistics and Probability**

1. There are eight men, seven women, and five children in a family. Their mean age separately is 24, 20, and 9, respectively. What is the mean age of the family?

Here, A₁ = 24, n₁ = 8, A₂ = 20, n₂ = 7, and A₃ = 6, n₃ = 5.

Combined mean = n₁A₁+n₂A₂+ n₃A₃n₁+n₂+n₃ = 36220 = 18.1

Hence, the mean age of the family = 18.1 years.

2. A speaks truth in 60% of cases, and B speaks truth in 70%. What will be the probability that they will say the same thing while describing a single event?

They will say the same thing in two ways, i.e., either truth or lie.

Hence, probability = (0.6 x 0.7) + (1 – 0.6)(1 – 0.7) = 0.54

**Chapter 15: Trigonometry**

1. Find the solution of the equation tan.tan2 = 1.

As tan.tan2 = 1 2tan21-tan2 = 1

3tan2 = 1 tan = 13

Hence, = n6.

2. Find the value of 3 cosec 20° – sec 20°.

3 cosec 20° – sec 20° = 3sin 20° – 1cos 20° = 4 sin 40°sin 40° = 4

**Chapter 16: Mathematical Reasoning**

1. ‘Violets are blue.’ Write the negative of this statement.

Violets are not blue.

2. ‘All prime integers are either even or odd.’ Write the negative of this statement.

All prime integers are not even, and all prime integers are not odd.

Those mentioned above are some important questions from the JEE Mains Mathematics section. Hopefully, these questions will let you understand the important concepts and topics better. And if you find it hard, you can always look for us.