IITJEE

Assignment 1 Relations and functions
class 12 CBSE
IITJEE

Assignment -1 (Relations and Functions)

Assignment class 12 Relations of Unit – I, as per Latest Syllabus For CBSE board. Due to COVID 19 there is a change in testing of class 12th board is proposed by CBSE.

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Assignment on set theory for Practice
class 11th
IITJEE

Assignment Set theory – 2

Set theory Assignment – 2 Set theory Assignment is the method to make the set theory perfect. The best way to develop the required skills is that practice sufficient assignment on the topic Set theory.  Question 1. 1. If A and B be two sets such that n(A)=3 and n(B)=6. Find (i)  Minimum number of […]

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Homogeneous equation in two variable
IITJEE

Homogeneous Equation

Homogeneous equation in two variable. $$ $$ Homogeneous equation in two variable i.e $ax^2 + 2hxy +by^2 = 0$  always represent pair of line. Since $g=f=c=0\;\Rightarrow bg^2+ch^2+af^2-2hfg-abc=0\,\forall\,x \in R$ Pair of lines represented by homogeneous equation in two variables are $L_1:y – m_1x = 0$ and $L_2 : y – m_2x = 0$, where $m_1,m_2$ […]

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IITJEE Mock test
IITJEE

IITJEE mains maths mock test

IITJEE mains is scheduled in the coming few days. There is no better way to prepare for IITJEE mains now, do practice for the number of mock tests. We will provide a series of iitjee mains mock test. What is expected from students? Attempt the iitjee mains mock test paper seating continuous forĀ  01 Hours. […]

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Pair of lines
class 11th
IITJEE

Pair of straight line.

Equation of pair of straight line.                     What does concept of pair of straight line represent.                     Let $L_1\,:\,a_1x+b_1y+c_1=0$ and  $L_2\,:\,a_2x+b_2y+c_2=0$ be two lines, then pair of straight line (also called as combined equation of given lines) […]

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Parallelogram area
class 11th
IITJEE

Area of a parallelogram.

  Area of a parallelogram. The area of a parallelogram ABCD can be found by calculating area of two triangles ABD and triangle BCD.  In $\Delta$ADB ,  $Sin\theta = \cfrac{P_1}{AD}\;\Rightarrow AD=BC=\cfrac{P_1}{Sin\theta}$   $$ $$ In $\Delta$BCD , $Sin\theta =\cfrac{P_2}{DC}\;\Rightarrow DC=AB=\cfrac{P_2}{Sin\theta}$                    $\therefore$ Area  ABCD = $2.(\cfrac{1}{2}.AB.AD.Sin\theta)$     […]

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