Posts

Showing posts from May, 2021

Featured Posts

Math Assignment Class XI Ch -1 Set Theory

Image
Mathematics Assignment Class - XI | Subject Mathematics | Chapter - 1 Math Assignment  Class XI  Chapter 1  Set Theory, Extra and important questions on Set Theory useful for Mat and Applied Math students.  MATHEMATICS ASSIGNMENT  SET THEORY For Non-Medical /Applied Math Students ******************************************************* Question 1 Write the following sets in tabular form a){ x : x = 2n + 1 ; n ∈ N}     b) {x : x – 1 = 0}        c) {x : x = prime number and    x ≤ 7 ; x ∈ N     d) {x : x ∈ I,   3x - 2 = 3}     e) {x : x 2  ≤ 10 ,   x  ∈ Z }      Answers of Q 1  a  {3, 5, 7, 9, ........}  b  {1}  c  { 2, 3, 5, 7}  d   φ  e   {0,    土 1,    土 2,    土 3}

Math Assignment Class X Ch-3 | Linear Equations in Two Variables

Image
  CHAPTER 3    CLASS   X PAIR OF LINEAR EQUATIONS IN TWO VARIABLE Extra questions of chapter 3 class 10 : Pair of Linear Equations in Two Variable with answer and  hints . Useful math assignment for the students of class 10 For better results Students should learn all the  basic points of  Chapter 3 Pair of Linear Equations in two variable . Student should revise N C E R T book thoroughly with examples. Now revise this assignment. This assignment integrate the knowledge of the students. ASSIGNMENT BASED ON CH-3 CLASS 10

NCERT Sol Maths Class XII Ch-6 | Application of Derivatives

Image
  Solution of Important Question of NCERT Book Class XII Chapter 6 Matrices Get Free NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivative.  Solution of important questions  of NCERT book  solved by Expert Teachers as per NCERT (CBSE) Book guidelines.   Exercise 6. 3 Q 18    For the curve    y = 4x 3  – 2x 5 , find all the points at which the tangent passes through the origin. Solution: Given equation of the curve is    y = 4x 3  – 2x 5 ………….(1) Let (h, k) be any point on this curve. Let tangent to the curve at    point    (h, k) passes through the origin (0, 0) Equation of the curve at point (h, k) becomes y = 4x 3  – 2x 5 ,    ⇒     k = 4h 3  – 2h 5    ………..(2) Differentiating eqn. (1)    w.r.t. x we get \[\frac{dy}{dx}=4\times 3x^{2}-2\times 5x^{4}=12x^{2}-10x^{4}\] \[\left (\frac{dy}{dx} \right )_{(h,\;k)}=12h^{2}-10h^{4}\] Equation of the tangent at (h, k) is given by y – k = (4h 3  – 2h 5 )( x – h ) This tangent passes through the origion (0, 0) 0 – k =

Breaking News

Popular Post on this Blog

Lesson Plan Maths Class 10 | For Mathematics Teacher

Lesson Plan Math Class 10 (Ch-1) | Real Numbers

Lesson Plan Maths Class XII | For Maths Teacher

SUBSCRIBE FOR NEW POSTS

Followers