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Showing posts from May, 2021

### Mathematics Assignments | PDF | 8 to 12

PDF Files of Mathematics Assignments From VIII Standard to XII Standard PDF of mathematics Assignments for the students from VIII standard to XII standard.These assignments are strictly according to the CBSE and DAV Board Final question Papers

### Math Assignment Class XI Ch -1 Set Theory

Mathematics Assignment Math Assignment  Class XI  Chapter 1  Set Theory, Extra and important questions on Set Theory useful for Mat and Applied Math students.  Class - XI | Subject Mathematics | Chapter - 1 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   Ï†  e   {0,    åœŸ 1,    åœŸ 2,    åœŸ 3}

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

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 Basic points of  Chapter 3 Pair of Linear Equations in two variable . ASSIGNMENT BASED ON CH-3 CLASS 10

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

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 =