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August 17, 2021
Assignment -1 (Relations and Functions)

July 9, 2021
Assignment Set theory – 2

April 8, 2020
Homogeneous Equation

April 8, 2020
IITJEE mains maths mock test

April 4, 2020
Pair of straight line.

March 23, 2020
Area of a parallelogram.

February 24, 2020
Angle Bisector

February 20, 2020
Distance of a point from line.

February 10, 2020
Principle of Inclusion and Exclusion .

January 16, 2020
Concept of rotation

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.

Read MoreSet 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 […]

Read MoreHomogeneous 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 $m_1,m_2$ […]

Read MoreIITJEE 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. […]

Read MoreEquation 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) […]

Read MoreArea 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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