List of Chapters (Std. XII)

Paper 1

  1. Mathematical Logic (Misc. Ex. 1)
  2. Matrices (Misc. Ex. 2A  Misc Ex. 2B)
  3. Trigonometric Functions (Misc. Ex. 3A  Misc. Ex. 3B  Misc. Ex. 3C)
  4. Pair of Straight Lines (Misc. Ex. 4)
  5. Vectors (Misc. Ex. 5A  Misc Ex. 5B)
  6. Three Dimensional Geometry  (Misc. Ex. 6)
  7. Line (Misc. Ex. 7)
  8. Plane (Misc. Ex. 8)
  9. Linear Programming (Misc. Ex. 9)

Paper 2

  1. Continuity (Misc. Ex. 1)
  2. Differentiation
  3. Applications of Derivatives
  4. Integration
  5. Definite Integral
  6. Applications of Definite Integral
  7. Differential Equations
  8. Probability Distribution
  9. Binomial Distribution
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Points of Concurrency in a Triangle

Medians of a Triangle are Concurrent. The Point of Concurrency is the Centroid.

A median of a triangle is the line joining the midpoint of a side and the opposite vertex.

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Angle Bisectors of a Triangle are Concurrent. The Point of Concurrency is the Incenter.

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Altitudes of a Triangle are Concurrent. The Point of Concurrency is the Orthocenter.

An altitude is the perpendicular line drawn from a vertex on the opposite side of the triangle.

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Perpendicular Bisectors of a Triangle are Concurrent. The Point of Concurrency is the Circumcenter.

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Volume of Parallelopiped and Tetrahedron

A parallelopiped is a 3-D object, each of whose faces is a parallelogram.

A rectangular parallelopiped is the one, whose faces are rectangular.

A cube is a rectangular parallelopiped, whose edges are of equal length.

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A tetrahedron is a 3-D object, whose all faces are triangular. It can be shown that a parallelopiped can be decomposed into 6 tetrahedra.

Image Source : https://www.dune-project.org/

For a visual proof :  1 Parallelopipded  = 6 Tetrahedra


 

Let there be 2 such objects, whose coterminus edges are identical. Then, the volume of the tetrahedron is {\dfrac 1 6}th of that of the parallelopiped. The expression can also be obtained using the following theorem :

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Vector Algebra : Theorems (I)

In this blog, we will prove fundamental theorems in vector algebra. These theorems are in accordance with the +2 curriculum.

Theorem 1 : Collinear Vectors

vectors_theorems01-00.jpgTheorem 2 : Non-collinear and Coplanar Vectors

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Theorem 3 : Coplanar Vectors

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Theorem 4 : Non-coplanar Vectors

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Trigonometric Functions

Preliminaries

  • We were introduced to the trigonometric functions in class {XI} with their general definitions for any angle {\theta}. The specialty of these functions is that their values get repeated after an interval. This is because after every interval of {2 \pi^c}, the angles have same initial and terminal arms.
  • Basics : If we fix the initial arm of the angle {\theta} as positive {X} axis, with vertex at origin. Let {P (x,y)} be ANY point on the terminal arm of the angle. Let {r} be {\sqrt {x^2 + y^2}}. Then,

{sin (\theta) = \frac {y}{r}, \ cos (\theta) = \frac {x}{r}, \ tan (\theta) = \frac {y}{x}, \ cosec (\theta) = \frac {r}{y}, \ sec (\theta) = \frac {r}{x}, \ cot (\theta) = \frac {x}{y}}

  • Periodicity : The characteristic of trigonometric functions, which repeat their values after a fixed interval, is known as periodicity. The smallest non-negative interval is known as the fundamental period.

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  • Trigonometric Ratios of Standard Angles

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Main Content

  • Equations and Solutions

The value which satisfies an equation is known as a solution. For example, {2x+4=0} is an equation. On simplifying,

{2x= -4,}

{So, \ x = \frac {-4}{2} = -2}

So, the solution is {x= -2}.

  • Trigonometric Equations

The equations involving trigonometric functions are known as trigonometric equations. The unknown values, which are to be found, represent the measure of an angle. For example, {sin (\theta)= \frac 1 {\sqrt 2}} is satisfied by {\theta = \frac {\pi}{4}} or {45^o}.

  • Principal Solutions

The solution, which lies between {0} and {2 \pi}, is known as the principal solution.

{0 \le \theta < 2 \pi}

  • General Solutions

We saw that the measure of full circle is {2 \pi} radians or {360^o}. Hence, after adding {2 \pi} to above angle, we will still get a solution. So,

{sin (2 \pi + \frac {\pi}{4}) = \frac {1}{\sqrt 2}}

Similarly,

{sin (4 \pi + \frac {\pi}{4}) = \frac {1}{\sqrt 2}}

Thus, there are infinitely many solutions to above equations, apart from {\theta = \frac \pi 4}. These solutions are known as general solutions. Consider another example:

{sin \ \theta = 0 \Rightarrow \theta = n \pi, n \in \mathbb Z}

Note: We will have to use allied angles formulas and the formula sheet to solve problems of this kind, where we are asked to find the general solutions.


  • Polar Coordinates

The Cartesian coordinates {(x,y)} and the polar coordinates {(r, \theta)} are inter-convertible. See the figure below:

polar

 

{x= r \ cos \ \theta, y = r \ sin \ \theta }

{r^2 = x^2 + y^2, tan \ \theta = \frac y x, \theta = tan^{-1} \frac y x}


  • Solving a Triangle

A triangle has 3 sides, say, {a,b,c} and 3 angles, {\angle A, \angle B, \angle C}. By solving a triangle, we mean, to find values of all lengths of sides and all angles of a triangle. See the figure below.

 

triangle

The side opposite to {\angle A} is denoted by {a} and so on.

There are certain rules to be used to solve these problems. These are:

  • Sine Rule

{\dfrac {a}{sin \ A} = \dfrac {b}{sin \ B}= \dfrac {c}{sin \ C}=2 R}

{R} is the radius of the circumcircle.

  • Cosine Rule

{a^2 = b^2 + c^2 -2bc cos \ A}

{b^2 = a^2 + c^2 -2ac cos \ B}

{c^2 = a^2 + b^2 - 2ab cos \ A}

  • Projection Rule

{a = b cos \ C + c cos \ B}

{b = a cos \ C + c cos \ A}

{c = a cos \ B + b cos \ A}

  • Half Angle Formulas

{a+b+c = 2s, \ s} is the semi-perimeter.

{sin \ \dfrac A 2 = \sqrt {\dfrac {(s-b)(s-c)}{bc}}}

{cos \ \dfrac A 2 = \sqrt {\dfrac {s(s-a)}{bc}}}

{tan \ \dfrac A 2 = \sqrt { \dfrac {(s-b)(s-c)}{s(s-a)}}}

  • Area of a Triangle

The area of triangle is given by,

{area = \dfrac 1 2 \ bc \ sin \ A = \dfrac 1 2 \ ac \ sin \ B = \dfrac 1 2 \ ab \ sin \ C}

It is also given by

{area = \sqrt {s(s-a)(s-b)(s-c)}}. This is known as Heron’s formula.

  • Napier’s Analogies

{tan \left ( \dfrac {B-C}{2} \right ) = \dfrac {b-c}{b+c}cot \ \dfrac A 2}


  • Inverse Trigonometric Functions

If a function {f} is defined from set A to set B (which is one-one and onto), we can define an inverse function, {f^{-1}} from set B to set A. So, if {x \in A, y \in B}, then {y = f(x)} and {x = f^{-1} y}.

If {sin \ \dfrac {\pi}{6} = \dfrac 1 2}, then {sin^{-1} \frac 1 2 = \dfrac \pi 6}. If {cos \ \dfrac {\pi}{4} = \dfrac 1 {\sqrt 2}}, then {sin^{-1} \dfrac 1 {\sqrt 2} = \dfrac \pi 4}.

Note that {sin^{-1} y} is different from {[sin (y)]^{-1}}. {[sin (y)]^{-1}} is {\frac 1 {sin (y)}}, which is {cosec (y)}.

On the other hand, {sin^{-1} y} is the angle, whose sine is {y}.

Corresponding to each of six trigonometric functions, we have 6 inverse trigonometric functions, i.e. {sin^{-1} y, cos^{-1}y, tan^{-1}y, sec^{-1}y, cosec^{-1}y, cot^{-1}y}.

Recall : Principal value is the value of angle, which lies between {0} and {2 \pi}.


  • Properties of Inverse Trigonometric Functions

Set I

{sin^{-1} \dfrac 1 x = cosec^{-1}x, \ for \ x \ge 1, x \le -1}

{cos^{-1} \dfrac 1 x = sec^{-1}x, \ for \ x \ge 1, x \le -1}

{tan^{-1} \dfrac 1 x = cot^{-1} x, for \ x > 0}

Set II

{sin^{-1} (-x) = - sin^{-1}x, -1 \le x \le 1}

{tan^{-1} (-x) = - tan^{-1}x, x \in \mathbb R}

{cosec^{-1} (-x) = - cosec^{-1} x, for \ x \ge 1, x \le -1}

Set III

{cos^{-1} (-x) = \pi - cos^{-1}x, -1 \le x \le 1}

{cot^{-1} (-x) = \pi - cot^{-1}x, x \in \mathbb R}

{sec^{-1} (-x) =\pi - sec^{-1} x, for \ x \ge 1, x \le -1}

Set IV

{sin^{-1}(x)+ cos^{-1}(x) = \dfrac {\pi}{2}, -1 \le x \le 1}

{tan^{-1}(x)+ cot^{-1}(x) = \dfrac {\pi}{2}, x \in \mathbb R}

{cosec^{-1}(x)+ sec^{-1}(x) = \dfrac {\pi}{2}, -1 \le x \le 1}

Set V

{tan^{-1}x + tan^{-1}y = tan^{-1} \dfrac {x+y}{1-xy}, when \ x,y > 0 \ and \ xy<1}

{tan^{-1}x + tan^{-1}y = \pi + tan^{-1} \dfrac {x+y}{1-xy}, when \ x,y > 0 \ and \ xy > 1}

{tan^{-1}x - tan^{-1}y = tan^{-1} \dfrac {x-y}{1+xy}, when \ x,y > 0 }

Set VI

{2 tan^{-1}x = sin^{-1} \dfrac {2x}{1+x^2}, for \ -1 \le x \le 1}

{2 tan^{-1}x = cos^{-1} \dfrac {1-x^2}{1+x^2}, for \ x \ge 0}

{2 tan^{-1}x = tan^{-1} \dfrac {2x}{1-x^2}, for \ -1 < x < 1}

Indeterminate Forms

  • Standard Limits

{\lim \limits_{x \to 0} \frac {sin(x)}{x} = 1, \lim \limits_{x \to 0} \frac {tan(x)}{x} = 1, \lim \limits_{x \to 0} \frac {1 - cos(x)}{x^2} = \frac 1 2}

{\lim \limits_{x \to a } \frac {x^n - a^n}{x-a} = n \cdot x^{n-1}, \lim \limits_{x \to 0} \frac {a^x -1 }{x} = ln(a), \lim \limits_{x \to 0} (1+x)^{\frac 1 x} = e}

{\lim \limits_{x \to 0+} ln(x) = -\infty , \lim \limits_{x \to \infty} ln(x) = \infty}

  • Introduction

Consider 2 functions, {f(x)} and {g(x)}. Let {x=a} be a value at which these functions are defined.

I) If {f(x) =sin (x), g(x) =x} and {a=0}, then the limit {\lim \limits_{x \to 0} \frac {sin(x)}{x}} takes the form {\frac 0 0}.

II) If {f(x)= \frac 1 x, g(x) = tan \Big( \frac \pi 2 - x\Big), a= 0}, then {\lim \limits_{x \to 0} \frac {\frac 1 x}{ tan \Big( \frac \pi 2 - x\Big)}} takes the form {\frac {\infty}{\infty}}

III) If {f(x) = x, g(x) = ln(x)} and {a=0}, then the limit {\lim \limits_{x \to 0} x \cdot ln(x)} takes the form {0 \times \infty}

IV) If {f(x)= sin(x), g(x)=x}, and {a=0}, then the limit {\lim \limits_{x \to 0} [{sin(x)}]^x} takes the form {0^0}.

V) If {f(x)= cos(x), g(x)=tan (\frac {\pi}{2} - x)}, then the limit {\lim \limits_{x \to 0} {[cos(x)]}^{tan (\frac \pi 2 -x)}} takes the form {1^\infty}.

VI) If {f(x)= \frac 1 x} and {g(x)= sin(x)} and {a=0}, then the limit {\lim \limits_{x \to 0} \Big(\frac {1}{x} \Big)^{sin(x)}} takes the form {\infty^0}

VII) If {f(x)= tan (\frac \pi 2 -x)} and {g(x)= \frac 1 x} and {x=0}, then the limit {\lim \limits_{x \to a} tan (\frac \pi 2 -x) - \frac 1 x} takes the form {\infty - \infty}.

These are known as the indeterminate forms. The limits are evaluated either by L’Hosptial’s rule or by substituting an equivalent infinitesimal.

  • L’Hosptial’s Rule (French : {Lopital})

The rule can be proved using Taylor’s theorem. It says, if {f(x)} and {g(x)} are {0} at {x = a} or {\lim \limits_{x \to a} f(x)=0} and {\lim \limits_{x \to a} g(x)=0}, then

{\lim \limits_{x \to a } \frac {f(x)}{g(x)} = \lim \limits_{x \to a} \frac {f'(x)}{g'(x)}}

This rule is sometimes applied on {n}th derivatives, if all derivatives of lesser orders are {0}.

  • Equivalent Infinitesimal

This is used for evaluation of {\frac 0 0} form. One of the functions can be replaced by another, if they both converge to {0} at a point and the limit of their ratio at that point is {1}. For example,

{x \sim sin(x), x \sim tan^{-1}x, \sim ln(1+x)}

  • Hint for MCQs

One can try substituting a value of {x} closer to (but not equal to) the actual limit. Evaluate the function using the calculator. The answer will be closer to the actual limit. We’ve actually used the concept of limit here.

Explanation: Consider the limit

{\lim \limits_{x \to 0} [1+ sin(x)]^{cot(x)}}

This is of the form {1^{\infty}}. Let’s put {x=0.1} in the function {[1+ sin(x)]^{cot(x)}}.

We get {[1+sin(0.1)]^{cot(0.1)}} as {2.58160}.

On substituting {x = 0.01}, a value closer to {0}, we get {2.7046}.

On substituting {x = 0.001}, a value closer to {0} and {< 0.01}, we get {2.7169}. Clearly, the limit is {e}.