The post JEE Conic Sections: Ellipse appeared first on Magoosh JEE Blog.
]]>So, let’s learn about ellipses in this blog post!
Hey there! Let’s start this chapter with a small exercise. No, not the JEE problems, but a simple, fun exercise. Pick up any coin from your wallet. Found it? Now switch off the lights and shine a torch or flashlight onto the coin pressed against a wall. Slowly pull the coin away from the wall. You’d see a shadow of the circle. Now, rotate the coin slightly in any direction. What shaped shadow did you observe? It looks like a squashed circle, doesn’t it? Well, we have a word for it–ellipse!
Ellipse is a part of conic sections, that is, it can be obtained as a cross-section of a cone. If you cut a cone parallel to its base, you get a circle. But if you make a cut on the cone at an angle through its curved face, you get an ellipse, as shown in the figure below. More precisely, an ellipse is any plane section of a cone, not containing the cone’s apex, and with a slope less than the slope of the lines on the cone.
In fact, you could call a circle a special type of ellipse!
Now that we have a brief idea of what an ellipse is and what it looks like, let’s describe the ellipse mathematically.
Ellipse can be described as a curve in which the ratio of the distance between any point P on the curve and a fixed point F, and the distance between that point (P) and a fixed line L is a constant real number greater than/equal to zero and less than one.
Too much jargon? Don’t worry. Let’s break it down step-by-step.
Take a fixed point F, say (0, 0).
Now take a fixed line L, say x = 1, or, x — 1 = 0.
Now, take a general point P(x, y). Drop a perpendicular from point P onto line L. Let the perpendicular from P and line L intersect at P’, that is, the (perpendicular) distance between P and L is PP’. The distance between P and F is PF.
We have,
PF/PP’ = constant ∊ (0, 1). Let’s set this constant to be 0.5.
⇒PF/PP’ = 0.5
⇒√((x — 0)^{2} + (y — 0)^{2})/(|x — 1|/√(1^{2})) = 0.5
⇒2√(x^{2} + y^{2}) = |x — 1|
⇒4(x^{2} + y^{2}) = (x — 1)^{2}
⇒4x^{2} + 4y^{2} = x^{2}–2x + 1
⇒3x^{2} + 4y^{2} + 2x — 1 = 0 is the equation of the ellipse.
If you sketch the curve, you’d get something like this:
The curve in green is the desired ellipse. Every point on this curve is twice as far from the line x — 1 = 0 as it is from the point (0, 0). You can check and verify it for any point on this ellipse.
The fixed point F is known as ‘focus’ of the ellipse, and the fixed line L is known as the ‘directrix’ of the ellipse.
The ratio, (Distance of any point from Focus)/(Distance of the same point from Directrix) is known as the eccentricity of the ellipse, denoted by ‘e’, where 0 < = e < 1 Eccentricity is a measure of the degree of resemblance of an ellipse to a circle. For a circle, eccentricity will be zero. That is why we consider a circle as a special type of ellipse.
Every ellipse has 2 foci and 2 corresponding directrices, due to symmetry. Given a focus and a directrix, we can uniquely define an ellipse. Likewise, an ellipse can be uniquely defined by giving the focus and its eccentricity.
Ellipse can also be described in another, simpler way. Ellipse can be described as a curve in which the sum of distances between any point P on the curve and 2 fixed points is a constant number, greater than the distance between the fixed points.
Got confused? No worries, let’s simplify this as well.
Take 2 points, say F1(–4, 0) and F2(4, 0). Now, take a general point P(x, y). The distance of the point P from F1 is given by PF1 = √((x– (–4))^{2} +(y — 0)^{2}). Similarly, the distance of the point P from F2 is given by PF2 = √((x — 4)^{2} + (y — 0)^{2}). The distance between F1 and F2 is √((4– (–4))^{2} + (0 — 0)^{2}) = 8 units.
We have, PF1 + PF2 = constant, where constant > F1F2
⇒√((x– (–4))^{2} + (y-0)^{2}) + √((x — 4)^{2} + (y — 0)^{2}) = constant, constant > 8
Let’s take constant = 10
⇒√(x^{2} + 8x + 16 + y^{2}) + √(x^{2} –8x + 16 + y^{2}) = 10
⇒√(x^{2} + 8x + 16 + y^{2}) = 10 — √(x^{2} — 8x + 16 + y^{2})
⇒x^{2} + 8x + 16 + y^{2} = 100 + x^{2} — 8x + 16 + y^{2} — 20√(x^{2} –8x + 16 + y^{2})
⇒25 — 4x = 5√(x^{2} –8x + 16 + y^{2})
⇒625 + 16x^{2} — 200x = 25x^{2} — 200x + 400 + 25y^{2}
⇒9x^{2} + 25y^{2} — 225 = 0 is the desired ellipse
If you take any point P(x, y) located on this ellipse, the sum of its distances from the points F1(–4, 0) and F2(4, 0) will be 10. Try to check and confirm it!
In general, for any ellipse, the fixed points F1 and F2 are known as the ‘foci’ of the ellipse. Yes, these are the same foci that we discussed in the previous description of foci and directrices. Isn’t it interesting? Hold your excitement, there is a lot more interesting stuff coming up.
Also, the fixed sum of distances is known as the length of ‘major axis’ of the ellipse, denoted by ‘2a’. The major axis is the line joining the two farthest points in a given ellipse. For example, in the ellipse described above, the x-axis is the major axis as it is the line joining (–5, 0) and (5, 0), which are clearly the farthest points in that ellipse. Length of the major axis is the distance between the farthest points, given by √((5–(–5))^{2} + (0 — 0)^{2}) = 10 units.
Also, the ratio (F1F2)/(2a) is nothing but the eccentricity (e) of the given ellipse, that is,
e = (F1F2)/(2a) = 8/10 = 0.8
Note that this is the same eccentricity about which we learned in the previous section. It comes full circle.
Let’s quickly summarize our discussion on ellipse,
In this post, we learned some important theoretical concepts of an Ellipse. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>The post JEE Conic Sections: Hyperbola appeared first on Magoosh JEE Blog.
]]>The mathematical concept of Hyperbola is essential for engineering and architectural studies. It’s quite obvious that Hyperbola is very important for JEE. All its properties must be at your fingertips in order to solve the questions asked in JEE.
You can think of a hyperbola to be two curves that resemble infinite bows.
A hyperbola is the locus of all points in a plane, the difference of whose distances from two fixed points (focus) is constant.
Now let me introduce you to the various terms related to hyperbola:
Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0
The expression B^{2} — 4AC is called the discriminant, which is used to determine the type of conic section.
For a hyperbola, B^{2} — 4AC > 0.
A hyperbola with Center as Origin (0, 0) and both the foci on the x-axis is given by the equation:
The properties of the above hyperbola are:
Now you might be wondering about ‘e’ used in various formulae. ‘e’ is known as the Eccentricity of a conic section. It is the ratio between the focus and the directrix which remains constant for a given conic section. For a hyperbola, e > 1.
As the name suggests, it is the distance of any arbitrary point on the hyperbola from the focus.
Let P be any point present on the hyperbola,
|S_{1}P — S_{2}P| = 2a
Clearly, the focal distance of any point on the hyperbola is constant and is equal to the length of the Transverse Axis.
The directions of the axes should be parallel to the co-ordinate axis.
The parametric equations of the general hyperbola being discussed are:
x = a sec θ , y = b tan θ
Or
x = a cosh θ , y = b sinh θ
1) The equation of a chord joining two points P( a sec θ₁ , b tan θ₁) and Q( a sec θ₂ , b tan θ₂) on the hyperbola
2) The equation of the chord of the hyperbola bisected at point (x₁, y₁) is given by,
T = S₁
3) Equations of the chord of contact of tangents drawn from a point (x₁, y₁) to the hyperbola is given by,
1) In point form the equation of the tangent to the hyperbola is,
2) In parametric form the equation of the tangent to the hyperbola is,
3) In Slope form the equations of the tangent to the hyperbola is,
y = mx ± √(a²m² — b²)
Here, m is the slope of the normal the tangent to the hyperbola.
4) The line y = mx + c touches the hyperbola if c² = a²m² — b².
1) Point Form: In point form the equation of normal to the hyperbola is,
2) Parametric Form: The equation of normal to the hyperbola at a point P(a secθ, b tanθ) is,
ax cosθ + by cotθ = a² + b²
3) Slope Form: The equation of normal to the hyperbola is,
Here, m is the slope of the normal to the hyperbola being discussed.
4) A maximum of 4 normals can be drawn to a hyperbola from a point P(x₁, y₁)
1) Diameter: Diameter is the locus of mid-points of all the parallel chords of the hyperbola. The equation of the diameter, bisecting a system of parallel chords with each of slope m to the hyperbola is given by,
2) Conjugate Diameter: The diameters of a hyperbola are said to be conjugate if each diameter bisects the chords parallel to the other.
The diameters y = m₁x and y = m₂x are conjugate if,
m₁m₂ = b²/a²
3) In a pair of conjugate diameters of a hyperbola, only one of them meets the hyperbola in real points while the other meets the hyperbola at imaginary points.
The locus of points of intersection of the tangents to the hyperbola, which are perpendicular to each other.
The equation of Director Circle is,
x² + y² = a² — b²
A hyperbola for which the asymptotes are perpendicular is known as a rectangular hyperbola.
In other words, it is a hyperbola with a = b.
Therefore, the eccentricity of a rectangular hyperbola is √2 .
The equation for a rectangular hyperbola can be reduced to,
xy = a²
1) Asymptotes are the co-ordinate axes, x = 0 and y = 0.
2) e = √2.
3) Center is O(0, 0).
4) Foci are S(√2c,√2c) and S₁(–√2c, –√2c).
5) Vertices are A(c, c) and A₁(–c, –c).
6) Length of Latus Rectum = 2√2c.
7) In parametric form, the equation is x = ct and y = c/t.
2) The combined equation of pair of tangents drawn from an external point P(x₁,y₁)
is SS₁–T².
3) The equation of a chord of the hyperbola whose mid-point is (x₁,y₁) is given by
T = S₁.
4) Eccentricity of a Rectangular Hyperbola is √2 and the angle between asymptotes is 90°.
5) If a triangle is inscribed in a hyperbola, then it’s orthocenter lies on the hyperbola.
In this blog post, you learnt about some important concepts about Hyperbola. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>The post JEE Conic Sections: Parabola appeared first on Magoosh JEE Blog.
]]>Let’s start by picturing two lines intersecting each other. Now make one of the lines (called the Generating Line) revolve around the other. We get what is called a double cone. Let another plane now intersect it (obviously, not where the cone’s vertex lies–in which case you get straight lines!!).
What you get on the plane is called a conic section. The following are five conic sections and they are characterized by what is known as eccentricity (often denoted by ‘e’).
So let’s get started!
We are given a straight line (directrix) and a point (focus). Then on a plane, the parabola is the set of all points such that they are equidistant from the directrix and the focus. The vertex is the point closest to the directrix. A line parallel to the directrix and passing through the focus cuts a conic section in two points, right? This particular line segment is called the latus rectum.
Taking the directrix as x + a = 0 and the focus as (a, 0), the equation of a right open horizontal parabola is: y^{2} = 4 a x
Thus, the parametric coordinates are: (at^{2}, 2at)
For latus rectum, we set x = a and get: l = 4 a
It can be easily shown that a point (x_{1}, y_{1}) is an interior point if S_{1} > 0, lies on the parabola if S_{1} = 0 and exterior to it if S _{1} < 0.
Focal Chord:
Any chord of the focus of a parabola is called a focal chord. It can be easily proved that if t_{1} and t_{2} are the parameters for its endpoints, then t_{1} t_{2} = –1.
(Complete description using Desmos Graphing Calculator)
The point of intersection of two tangents (drawn at t_{1} and t_{2}) is (at_{1}t_{2}, a(t_{1}+t_{2})).
Now, I am about to tell you something interesting! Just try to think over it. The tangents that are drawn at the end of a focal chord intersect at right angles on the directrix (Hint: t_{1}t_{2} = –1 and hence m_{1}m_{2} = –1). This also means that a circle drawn with a focal chord as the diameter touches the directrix! Don’t worry, see a complete proof here.
For any exterior point (x_{1}, y_{1}), the equation T^{2} = SS_{1} gives the tangents to the parabola from the given point. When S_{1} = 0, the point lies on the parabola and T = 0 which is actually true. Also notice that if S_{1} < 0, then T^{2} >0, and hence no tangent exists from the interior of a parabola.
When the tangents from an exterior point (x_{1}, y_{1}) are drawn, the points where they touch can be joined to form ‘The Chord of Contact’: T_{1} = 0
When an interior point (x_{1},y_{1}) is given, then the chord through the point such it is the midpoint is given by:
T = S_{1}
Don’t worry. The following examples will clear all your doubts.
Question 1
For the parabola y^{2} = 16 x, determine the angle between the tangents drawn from (–8,4).
Answer: c) 72°
Solution:
For the parabola y^{2} = 4 (4) x, a = 4
Putting (x_{1}, y_{1}) = (–8, 4) in the equation T^{2} = SS_{1} gives:
(4y — 2(4)(x — 8))^{2} = (y^{2} — 16x)(4^{2} — 16(–8))
=> x^{2} — xy –2y^{2} + 20x + 8y + 64 = 0
=> (x + y + 4)(x — 2y + 16) = 0
Hence, the slopes are m_{1} = ½ and m_{2} = –1
The angle between the tangents is tanθ = mod((m_{2} — m_{1})/(1 — m_{1}m_{2})) = 3
Since tanθ > 1 => θ > 45o
Also, θ < 90^{o}
Checking the options, the option (c) is the most appropriate.
You should try more and more problems on this topic. After all, practice makes a man perfect!
Moving on, let us study the normal to a parabola!
(Visualizing normals using Desmos Graphing Calculator)
Thus, if the slope of a normal is given, the equation of the normal can be rewritten as:
y = m x — (a m^{3} + 2 a m)
Moreover, you can even find the point to which the normal corresponds if its slope is given (using the fact that t = –m).
(at^{2}, 2at) = (am^{2}, –2am)
If you are given a point (x_{1}, y_{1}), then you can also find the possible slopes for a normal at that point.
y = m x — (am^{3} + 2am) ⇒ am^{3} + m (2a — x) + y = 0
Interesting! Through a single point on a parabola, as many as three normal could pass!
Reflection property of a parabola: A ray of light parallel to the axis of parabola reflects and passes through the focus. This reflection property is exploited to make spherical mirrors used in physics to remove any spherical aberrations.
In this blog post, you learnt about some important concepts about Parabola. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>Coordinate geometry carries huge weightage in the JEE mathematics syllabus, and conic sections is an important topic in JEE coordinate geometry. You can expect around two to five questions on this topic in the JEE paper. So, it is very important for you to have good command over this topic if you want to score high. Read on to know some important things about the JEE conic sections!
Technically, you would define a conic section as a curve formed by the intersection of a plane with the surface of a double-napped right circular cone. And if you don’t know what a double-napped right circular cone is—let’s make it simple.
Take a right circular cone and stack an inverted right circular cone on its vertex. What you get is a double-napped right circular cone. When you take into account all possible orientations of the intersecting plane with the double-napped cone, you get four curves, namely: a circle, an ellipse, a parabola, and a hyperbola.
There is another definition that you will find more useful for the JEE. A conic section is a locus of all points P such that its distance from a fixed point F (called the focus of the conic), is a constant multiple e, the eccentricity, of the distance from P to a fixed line L, called the directrix of the conic.
This gives us the working mathematical form for a conic section,
Now, depending on the value of e, the curves exhibit varying properties. If 0<e<1, we get a closed curve called ellipse. When e=1, we get a parabola and for e >1, we have a hyperbola.
Question 1
Here is a question that will make you more comfortable with the concept.
Given the equation Identify if it is a
A) Parabola
B) Hyperbola
C) Circle
D) Ellipse
The correct answer is A). It is parabola. With what I told you above, you may find the question mind-boggling! If not, you are really good in algebraic manipulation. Anyway, let me show you how to do it. The aim is to convert the equation 16x^{2} + 9y^{2} –92x –256y –24xy + 76 = 0 into generic conic section formula . We proceed as follows.
You can compare it with the general form. What you get is focus F ≡(1,2) and the equation of directrix 3x + 4y + 7. Also, the eccentricity is 1. So, you can safely say that it is a parabola.
At times, the necessary factorization may not be trivial. In fact, you may not find one at all. In such a case, we employ a different method. The generic equation for a conic section is a general second-degree equation Ax^{2} + 2Bxy + Cy^{2} + 2Dx + 2Ey + F = 0.
In the light of the above conditions, you can easily determine what conic section the equation
16x^{2} + 9y^{2} –92x –256y –24xy + 76 = 0 represents.
B^{2} — 4AC = 24^{2} — 4 x 16 x 9 = 576 — 576 = 0. Hence, a parabola.
Now, looking at the general second-order equation, does it remind you of something else you may have studied earlier? To make things more interesting, it is also the equation for a pair of straight lines. You may be wondering how a single equation could represent both the pair of straight lines and conic sections. Well, give the statement some thought.
A pair of straight lines is itself a conic section. Confused? Don’t be. Suppose you have a double-napped right circular cone. Take a plane such that it passes through the vertex of the double-napped cone and also contains its axis. What curve will you get? A pair of straight lines!
For the time being and also for the rest of the article, let us assume that a pair of straight lines and conic sections are two independent entities. Now that you know all this, you want to distinguish a conic section second-order equation from a pair of straight lines second-order equation.
Given an equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, it will represent a pair of straight lines if abc + 2fgh — af^{2} — ch^{2} — bg^{2} = 0, otherwise any of the four conic sections.
Question 2
Here is another example for you to practice.
What does the given equation 5x^{2} — y^{2} –4xy + 8x — 2y +3 = 0 represent?
A) A pair of straight lines
B) Circle
C) Ellipse
D) Parabola
The correct answer is A) a pair of straight lines. You can use the above formulae to identify the nature of the equation. Also, if you factorize it, you will get (5x + y + 3)(x — y + 1) = 0, which are the equations of the two lines the equation represents.
So far, we have only talked about the classification of conic sections. Sometimes, we are also interested in the coordinates of the center of the conic sections. You would ask why? Well, most of the time the properties we are interested in are unaffected by the transformation of coordinates. In such cases, you can shift the center of the conic to a more desirable coordinate to make calculations simple and easy to perform.
Now that you know how you can exploit it, you need to know how to determine the center of a conic. The easiest way to do so is to partially differentiate the conic equation with respect to x and y, and simultaneously solve for x and y from the two equations obtained.
Consider the general equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = f(x,y). Therefore, . When you solve for x and y, you get the coordinate of the center of the conic.
When working with conic sections, there are a few terms that you should know.
Principal Axis: In the case of hyperbola and ellipse, there exist two foci—in contrast to the single focus in a parabola. We define the Principal Axis as a straight line perpendicular to the directrix, which passes through the focus (foci).
Focal Parameter: It is the distance between the focus and the directrix.
Focal Length: For a parabola, the focal length is the distance between the focus and the vertex. For ellipse and hyperbola, it is defined as the distance from the center to any of the two foci.
Latus Rectum: The latus rectum of a conic section is the chord (line segment) that passes through the focus, is perpendicular to the major axis, and has both endpoints on the curve. The latus rectum satisfies some properties which are given below:
You may find it difficult to wrap your head around the generalized form. Studying their properties will give you a hard time if you are allergic to mathematics. In such cases, the standard form of conics come to the rescue.
In all the above equations, the center is (0,0) and the major axis is along x-axis while the minor axis is along y-axis.
In this blog post, we learned about the conic sections, which are present in the JEE maths syllabus. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning!
For more information about JEE Conic Sections, check out the following Magoosh resources:
Additional resources:
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]]>The post Important Formulae for JEE Mathematics (Free PDF Download) appeared first on Magoosh JEE Blog.
]]>As a bonus, we’ve created a free, downloadable list of all the important JEE math formulae you need to know.
A formula is a mathematical expression that is used to describe a scientific phenomenon mathematically. They are derived from previously known facts after rigorous research, experimentation and brainstorming. It is simply amazing how a single formula works wonders and decodes the science of nature into concepts that we can understand and apply.
In an exam, formulae make problem-solving so much easier. Many complex questions can be solved in one go using the correct JEE math formula. Yes, it is good to work through the basics, but when the questions get complex, involve multiple concepts (which they usually do in an exam like JEE) and you are expected to solve them within a time limit, formulae are your saviour.
However, it is also a bad practice to be unfamiliar with the basics and the derivations. JEE is an exam that tests your ability to apply formulae accurately. In some questions, you are required to apply a modified version of the formula or a step of the derivation is used. So, it is also important to know where the formula is coming from.
We understand that memorizing maths formulae for IIT JEE can be a cumbersome task. But as explained above, it is an inevitable part of your curriculum. Here are a few tips that can help you sail through smoothly.
It is a good practice to write down all the maths formulae for IIT JEE chapter wise and topic wise in one place or in a separate notebook. This saves you the hassle of going through the book repeatedly to search for a particular formula. For revision, there could be nothing better than this. When you have all the formulae in one place, it becomes easier for your brain to access the information when required. Also, this helps you avoid confusion which you might have in two or more identical formulae.
We have included a downloadable resource for you below that has a compendium of all the important formulae for JEE Mathematics!
Practising questions related to a formula embed it deep into your mind. When you are first given a formula, don’t just try to mug up. First, solve some questions related to it by referring to it. With each question, you’ll notice it becomes easier to remember and apply the formula. This is important as you cannot blindly apply formulae in the exam. You have to be thorough with the concepts and have to know what to apply, where, and how. Practising questions develop that skill of application, while at the same time making it easier for you to remember.
What we are trying to say is that maths formulae for IIT JEE are not that difficult to remember. On the contrary, they make our lives so much easier. You need not be nervous about them but just work through them smartly in order to make them work your way.
All the best!
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]]>Let’s first talk about how important trigonometry is for the JEE (both Main and Advanced). If you check previous years’ papers, you will find that hardly 1 – 2 questions in trigonometry were asked in the JEE Main. The same goes for JEE Advanced! Yet many professors and mentors recommend studying this chapter thoroughly. But why?
Although the chapter initially seems a little boring, trust me, it is
quite the opposite. It is really useful in solving the majority of problems that show up on the JEE. So, how should you prepare for trigonometry in order to use it for the majority of questions found in the JEE? I will list some reference books which I used to refer for studying or practising questions during my JEE Preparation.You may also go for R. D. Sharma (even if it is not at par with level of the JEE Advanced, almost every CBSE student has it for school purposes). Besides this, previous years’ papers and some practice papers should be more than enough.
Plane Trigonometry by S. L. Loney is a must-have for the JEE. It covers almost everything in trigonometry, and teaches concepts in a well defined and easy manner. Besides this, lots of new formulae are given which can be memorised (as they are not too complex) and widely used in various other parts. One of them is given below.
tan(a + b + c + …) = (S1 – S3 + S5…) / (1 – S2 + S4…) where Sn is the sum of all possible products of separate tangents (products in n) taken at the time. For example, S1 is tan(a) + tan(b) +… and so on for Sn.
Using this, we can easily calculate values of tan(2x) or any tan(nx) just by taking a = b = c = …
This probably would take a lot of time to do manually–but it can be done in a few seconds using this formula. There are lots of other surprises in this book. This was just a small example.
Doing these two things will make your concepts strong and you will excel in the subject.
So how do you complete these two steps in a systematic manner? For that, you need to break this chapter into various parts according to your wish. For instance, you can break the chapter into sections where each part will consist of 4 – 5 formulae and their applications. Study these parts accordingly.
Another method includes writing formulae again and again on a regular basis without actually proving them. This method can be mainly used in case we are not getting to the proof of any formula; for example sin(c+d) = sin(c)cos(d) + cos(c)sin(d). Its proof in S. L. Loney is actually a figure based proof which in turn is derived for a general triangle. Also, it can be cumbersome to always prove a sin(c) + sin(d) kind of formulae. So writing these again and again and solving questions which mainly include basic substitution of formula can to be helpful.
So far we’ve addressed memorizing formulae–but what about the questions in which you have to apply these formulae? For that you have to understand which formula you should use for a particular problem in order to save precious time. This makes your solution less crowded and mostly error-free.
The only solution to this problem is to practice more and more questions. I used to use M. L. Khanna as it contains a lot of questions to solve. This book contains questions of all levels ranging from a very basic level to a bunch of advanced level problems. Solutions are also provided.
You can also opt for some other book which provides you with questions. If you do not go to any coaching center or you don’t have a teacher who is on par with the JEE Advanced level, then a book with questions and their elaborate solutions will be very useful.
Finally, I would like to say that trigonometry is not a tough chapter and can be understood easily with some practice and revision. Since it is easier than other Mathematics chapters, it is more scoring too. Hence, practice the questions well to score good in JEE Maths. Good Luck!
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]]>Maths is not just a subject for the JEE, rather it is a tool that runs deep within all the subjects one prepares for. Be it Physics or Chemistry, Maths is always required in its conceptual level for all. JEE maths as a subject sounds tough to some people, but for an honest aspirant, it is a journey full of fun and learning. JEE maths deepened my interest in Maths as a subject and made me see its implications and use as a tool for various aspects. My advice to you will be to focus on Maths during your JEE preparation because it is the most fun if you put effort into learning the concepts. The following tips will help you enjoy and crack JEE Maths with ease.
While preparing for JEE Maths, being regular is really important. Without regular and diligent study it would get tough to finish the course on time, and you will eventually lose confidence in yourself.
Tips for being regular would be to stay motivated and fresh always. Have a fixed target and be motivated to achieve it. Do not procrastinate on tasks, and try to set achievable but challenging targets for yourself. In Maths, it’s important to realize that almost all of the concepts are equally important, and the questions will test you on all of them. Inadequate knowledge of any concept can ruin your chances of cracking the JEE. So be regular in reading the concepts and solving problems.
By this, I mean that you should not try to solve all the books you hear of and end up completing none. I would suggest you pick a good book and complete that book honestly and fully. Completing a book gives you immense confidence and a boost to your JEE preparation. I always preferred solving a single book completely rather than buying different books and partially solving each of them. Solving a single book for a particular chapter helps you cover the entire course and all the concepts with great clarity and helps you get a taste of all the types of problems you may face in the JEE. Other than a book you should use previous year archives of JEE problems for solving questions and solve sample papers from various institutes. So do not run around with a lot books. Take one and complete it diligently.
The key aspect of preparing for JEE Maths is an extensive practice of comprehensive problems which test the conceptual grip of the person rather than testing just the applications of a formula. Solve numerous problems from previous years’ papers and your reference book so as to get yourself clear of all the problems you may face in the examination. I would list some sources for practicing the problems:
Diligent practice from these sources guarantees clarity in your concepts and ease in cracking JEE Maths.
While solving maths problems in the JEE, one needs to maintain a clean rough work. It helps to ensure easy backtracking in case of some error in the solving process. Always try to maintain a clean rough work and try to write all of the important steps. From my personal experience, I found that clean rough work helps to minimize errors in solving questions and improves the speed of solving. It also helped me to develop a good intuition which proved to be useful to skip some steps while solving a problem as well as to keep my mind active during the exam. Some ways to maintain a clean rough work are:
In JEE Maths, you will face many problems which test you on the boundary value of various problems, or some unique cases which won’t be so obvious. So try to be cautious with these cases. Always keep your eyes open for the cases like:
These cases can be crucial for the exams, as such things make a person regret not thinking of it earlier. It is crucial that you do not commit blunders in these areas, and then losing precious marks.
The availability of multiple choice questions has degraded the skills of pure maths that JEE used to work upon. Children use an elimination strategy to effectively find out answers in the exam. But I would suggest you go subjective in your preparations.
Always try to solve the questions with a conceptual clarity. Don’t be simply answer oriented. It helps to improve your thinking ability and reduces the chances of losing marks due to blunders. Solve subjective problems which will help to clear your concepts. Try solving the comprehension type problems of Calculus. They used to be my personal favorite considering the conceptual copiousness they had. Your preparation should be such that any question that provokes thinking should excite you.
But of course, your aim is to get marks and that does not come by just thinking in the exams. Try to be subjective in your preparations but make sure that you are reaching correct answers and not just arriving at more questions.
Some important chapters for JEE Maths are:
Hope these tips help you in your studies. Good luck!
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]]>The syllabus for differential calculus in JEE Maths is vast. A fair number of questions in JEE Mains, as well as JEE Advanced, are taken from differential calculus. Being a highly method-based and technique-based topic, it is one of the most scoring chapters in IIT JEE Maths.
The chapters that constitute JEE Maths differential calculus are as follows.
Functions form the skeleton of differential calculus. They provide the framework for calculus to develop. A good understanding of the fundamentals will help you get a clear picture of calculus. You should have an idea about the mathematical definition of functions, and the different kinds of mapping such as injection, bijection, and surjection.
A typical question deals with the calculation of domain and range for a given function. So be sure to practice a lot of questions on domain and range calculation. It will develop your intuition for differential calculus.
Inverse functions is another hot topic that demands your attention. Problems based on it generally require a deeper understanding of functions and their properties. An effective approach is the graphical analysis of functions. You will be able to solve questions in comparatively less time with graphs. A good book for graphs in JEE Maths is “Play with Graphs” by Amit M. Agarwal.
It is essential that you thoroughly understand the concept of limits. It is a fundamental idea which is used extensively in differential calculus. You should be able to calculate limits for most of the functions which are given to you.
Try out all the problems with different methods. It will help you comprehend the idea of limits in a much better way. L’Hopital Rule is the most important technique for evaluation of limits, so become competent enough to apply L’Hopital Rule.
The other method that comes in handy is the series expansion for limit evaluation. You should remember the exponential series, the logarithmic series, and the trigonometric series for cosine and sine functions.
The idea of limits is a prerequisite to continuity. Here, the concepts that you learn in functions become important. You will be able to solve most of the problems if you are quick at drawing accurate graphs. It is a good idea to keep the properties of continuous functions and the theorems associated with them in the back of your mind.
This is a relatively easy chapter. You can crack it if you remember the differentiation formulas and practice a sufficient number of questions on them. You are not expected to derive formulas during exam hours–so do not forget what you learn.
The only thing that demands your attention is the definition of the derivative. I highly recommend that you remember it. Some problems from other chapters can be solved quickly and easily if you know what a derivative is.
This is the most important chapter. It sums up everything you have learned so far. Some of the important tools that you must know are maxima and minima, inflection points, Rolle’s theorem, Lagrange’s mean value theorem, increasing and decreasing function, concavity of a function, convexity of a function, and approximating the value of a function.
The above shows the order in which differential calculus should be taught. An important thing to note is each chapter derives itself from the concepts learned in chapters prior to it. So, it is advised you do not have any weak links in these chapters.
There are a lot of books available on differential calculus for JEE. The books by RD Sharma and HC Verma alone may not be enough for the JEE Main. Arihant series (“Differential Calculus” by Amit M. Agarwal) is one of the best books for calculus for IIT JEE. It covers all the concepts in detail and abounds with objective as well as subjective questions. There are past year questions as well, which gives you an idea of how the JEE tests its candidates. Each chapter includes a large number of solved examples and illustrations which are sufficient for your JEE preparation. If you like challenges, it also offers very long questions which are meant for Olympiads.
Another best book for calculus for IIT JEE is Arihant’s “Play with Graphs” by Amit M. Agarwal. This book gives insight into the graphical nature of problems. It is relatively short and will provide you with a set of tools that will help you solve some of the questions efficiently.
Some other books that you may read are “Cengage Calculus” by G.Tewani and “Mathematics for IIT JEE” by R.D Sharma. Another book that qualifies as a best book for calculus for IIT JEE, and which you may find interesting is “Calculus in One Variable” by I.A. Maron. It presents a large number of theorems and formulas on differential calculus. Before you try this book, you should be aware of the fact that “Calculus in One Variable” is not JEE specific. Therefore, there may be some concepts which are not a part of the JEE syllabus.
With all these books, you will have more than enough on your plate. I suggest that you stick with one book that you find the most interesting and easy to read.
There are a number of standard revision packages available from different coaching institutes. You can try FIITJEE Grand Master Package or Resonance Rank Booster. It is pretty good for revision during the final months of JEE preparation.
Some of the test series offered by coaching institutes contain thought-provoking problems on differential calculus. It is a good idea to try them out. Don’t forget to solve the previous years’ questions on differential calculus.
In some topics of differential calculus, students are prone to make silly mistakes. Limit calculation and domain/range finding are a few of them. Be mindful of such errors. Practice as many problems as you can and always verify your answers.
Differential calculus in JEE Maths needs a methodical thought process. There are a lot of standard methods and techniques that you need to remember. Deriving them during exams will be a waste of time as well as effort. You will do well if you memorise each and every one of them. Although you do not need to know the derivations of the theorems in differential calculus, a bit of knowledge about some of them will do you no harm.
Some of the techniques learned in differential calculus can also be applied in algebra as well as coordinate geometry. At times, such an approach will drastically reduce the time it takes to solve problems.
Apart from your ability to solve the problems, you should also focus on improving your time. You should do it once you have completed all the chapters and developed enough confidence in your accuracy. Do not sacrifice your accuracy for time. Differential calculus constitutes a big part of the JEE Maths syllabus. So, it is advised not to ignore this chapter.
Lastly, while you prepare for your JEE, keep a positive attitude and try to stay happy. All the best!
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]]>One of the main themes I want to highlight is the idea of culmination topics–which will be clear as the article progresses. The concept is simple. In each class of topics, for example, Calculus, Algebra, Combinatorics and the like, the initial chapters will build ideas up. Each chapter will flow logically from the previous one. The final chapters will involve all the concepts that you have assimilated — and those are the ones that carry the maximum weight.
Although not so significant by weight, Trigonometric identities is a prerequisite for several other chapters. You’ll also need to learn how to solve Trigonometric Equations with ease. Inverse Trigonometric Functions as a topic is fairly popular in the JEE and tends to be involved in a few quality crossovers with Calculus. Don’t be surprised if you see a Calculus question that requires deep knowledge of Inverse Trigonometry.
Once Quadratic Equations sets the stage, Sequences and Series will feature quite a bit in your preparation. Quite a few questions are asked in the JEE from it. The best part? It is easy to score marks in such topics since the problems are relatively easier.
Determinants and Matrices may appear cumbersome at this level, but this is a fairly straightforward and scoring topic for the JEE. Also, it is a powerful coding tool, so you will be seeing more of it later.
This class carries the most weight. Once the notion of a Function is introduced, discussions about Limits and their application to determine Continuity start making sense. Soon you go one step further to Differentiability and Methods of Differentiation. But the topic that actually tends to fetch a bulk of the marks in the JEE is Application of Derivatives – which is an apt conclusion.
Indefinite Integrals are challenging, but the JEE has not asked as many questions from here as from the topics that follow. Definite Integrals is definitely scoring – it’s not very tough and you often find very elegant solutions to problems! However, the examiner’s favourite is the ‘integrating act’ – Differential Equations and Area Under The Curve. Differential Equations, at the JEE level, have standard templates. Master that, and your score will shoot up!
This category is particularly enjoyable if you’re inclined towards CSE. And it helps that it includes a bunch of scoring topics for JEE. Once you acquaint yourself with the intricacies of Binomial Theorem and PnC, you are ready for the Main Event – Probability!
The beauty is that these chapters offer a lot of scope for creativity – you can have fun discovering multiple elegant methods and boost your JEE score while you’re at it.
All three topics can be considered scoring but Probability is the examiners’ favourite.
Here is another one of the JEE Mains maths important topics. This starts off with Straight Lines and then moves on to discussions on Conic Sections. Parabola deserves a special mention here because of its simplicity – you don’t want to miss out on marks from here. Circle has significant weight as a topic too – you’ll notice that the questions are thematic and standard, so it’s easy to score as well.
The topic of Complex Numbers also falls into this category. It is more powerful than the regular 2D coordinate system and problems utilize the properties of all the curves you have studied. This makes it a quintessential Culmination Topic. Often the Geometrical Approach (cis θ) is much more elegant than the brute force x+ iy.
You won’t find a bulk of marks coming from Complex Numbers, but in a quality paper, mastery over this topic makes a huge difference. This is what separates toppers from dreamers.
Moving further on, Vectors and 3D Geometry are a more general application of the ideas we have built up so far. The problems are thematic and quite frequent, so these are scoring topics for JEE. In fact, quite a few JEE level problems from this topic are from school level books like RD Sharma or ML Agarwal.
Now we have identified a bunch of Culmination Topics which are going to build your score. The next question is, how do we go about consolidating?
An important observation is that Culmination Topics will be your strengths if and only if you’ve grasped what preceded it reasonably well. The key isn’t the mindless rote learning of results and formulae – rather, it is the conceptual clarity that matters.
The associated ideas should be intuitive, just like second nature! To get there, well-planned practice is a must.
Next, observe problems from the Culmination Topic in question. More often than not, you’ll identify themes. The problems won’t be exactly the same. There won’t be a set formula. If there were, writing this article would be pointless. What I mean is, the general plan of approach will more or less be the same.
The algorithm will be quite similar for most problems of a particular theme. A striking example is Linear Differential Equations. The plan is simple – get it into the format, find the integrating factor, multiply, and solve. Your ingenuity lies in the execution. Maybe you’ll have to make a clever substitution or multiply both sides by a function to even get to the format, or whatever.
So it’s crucial to identify the themes. Once you’ve done that, you’ll gain confidence as the general algorithm runs in your head. Problem after problem, you’ll tell yourself, “I got this.” That’s how scores are built!
It’s easy to score in JEE Mains maths important topics once you know exactly what you are doing, and why you are doing it.
Culmination topics are scoring topics because they solve real-life problems as well! This is engineering — you systematically apply the knowledge you have!
This systematization is the key to making the most of scoring topics for JEE. Take some time, you’ll find a way yourself. Good luck!
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]]>Basically, integration means simply adding up pretty naively. However this adding up is done on infinitesimally (tends to 0) small intervals. It is also defined as the reverse process of differentiation. This kind of integration is done without any upper and lower limits, hence the name indefinite.
Where, C represents the integration constant which would not be present in case of definite integration.
Now, basically the formulas for integration can be generally proved as the antiderivative.
So, by definition,
Similarly the others follow, and you can find them easily in any CBSE book–the best being NCERT mathematics parts one and two. Here are some of the special integrals which you should keep in your mind while solving the problems:
Now, let’s get to the content of JEE integration. The questions are usually based on the following three categories.
To solve this type of problem, the first thing you should do is take the highest power of x (in this case 4) outside. The problem is designed in such a way that the differential of the term inside the bracket will be 1 + (1/x^{4}), which is exactly equal to the term outside the bracket 1/x^{5} (with maybe some considerations of constant terms).
In this case, you should try to eliminate by substituting x with asin^{2/3} θ, so that a gets cancelled out from both numerator and denominator. Then it becomes a trivial problem. In case you see there is a term of x^{4} along with some other terms. Try to substitute x with tan θ. If there is only 1 + x^{4} in the denominator then consider dividing numerator and denominator by x^{2} terms and proceed by method 1.
Get a strong hold on this matter from now on, as this is the most frequently asked topic from indefinite integration in the JEE. What does this mean? Recursion, huh?
One probably hardly remembers what is and in the recursive approach you are asked to find what is .
Sounds interesting, let’s check out:
Or,
Therefore, we have successfully got our recurrence relation. If we know I_{0}, we can find I_{2} and then I_{4} and hence all the values that n can take.
So, after discussing some important problems on JEE integration, it’s time to learn about the materials required to strengthen your JEE integration skills. Here are the best books in the order of precedence that you should definitely follow:
The book comes fully solved, so it won’t be a problem if you are preparing without coaching. Always remember to attempt the solved exercises before the unsolved ones. There’s a plethora of them there, so manage your time as you have to attempt other chapters as well.
It starts by discussing all the methods of integration following examples that make understanding very easy. Recursion in this book is explained very well.
I may be repeating myself, but it is important for you to know that you should do recursion very well if you want to ace JEE integration. Along with these qualities, it comes with solved solutions for almost all of the problems. There are two sections, one MCQ (multiple choice questions) and the other section is for subjective questions. Few problems in the subjective part are out of the current JEE standards, and also they do not have solutions.
Apart from the general books, you can also refer to the coaching material given by different institutes. Though the theory given in these materials isn’t really great, the problems given there are generally quite good. They are sometimes really thought-provoking.
So if you use these materials, read the theory from the books mentioned above, and solve the exercises from these materials. One disadvantage is that they do not always come fully solved as they are meant to be discussed in the classes. However you should definitely attempt them if you feel you have enough time to solve the other chapters.
So, apart from what we discussed above, indefinite integration is all about memorizing the formulae the correct way and assembling them the right way. Keep those formulae fresh in your mind as well as the integration of
tan x = log(|sec x|), cosec x = log (|tan x/2|), cot x = log(|sin x|) and similarly some others.
Another method which should have been mentioned much earlier is the Integration by parts. You should master it very soon. Don’t try to remember it by the formulae given in the book, you’ll find it extremely difficult. Instead, try to remember by the means of an example:
I know, it’s a bit difficult at first, but later on, you will be able to do it mentally. Always remember to integrate one term at first, then differentiate the other term, thus integrating the whole.
I hope you have got some ideas about indefinite integration and how to master it for the JEE. So, what are you waiting for? Go on, start your beautiful journey!
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