### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

### NCERT Sol Maths Class XII Ch-6 | A O D

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 6.3Q 18  For the curve  y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.Solution:Given equation of the curve is  y = 4x3 – 2x5………….(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) becomesy = 4x3 – 2x5,   ⇒    k = 4h3 – 2h5  ………..(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 byy – k = (4h3 – 2h5)( x – h )This tangent passes through the origion (0, 0)0 – k = (12h2 – 10h4)( 0 – h )k = (12h2 – 10h4)h   ⇒    k = 12h3 – 10h5Putting the value of k from equation (2) we get4h3 – 2h5 = 12h3 – 10h5              ⇒  8 h5 - 8 h3 = 0  ⇒   h5 -  h3 = 0                             ⇒  h3(h2 – 1) = 0    ⇒   h3(h + 1)(h – 1) = 0                ⇒  h = 0,  1, -1Now Putting the values of h in eqn.(2)If h = 0, k = 0, If h = 1, then k = 4 (1) 3 - 2 (1) 5 = 4 - 2 = 2If h = -1 then k = 4 (- 1) 3 – 2 (- 1) 5 = - 4 + 2 = -2Required points on the curve at which the tangent passes through the origin are(0, 0), (1, 2), (- 1, - 2)

*************************************************

 Exercise 6.3 Q 23 Prove that the curves  x = y2 and xy = k cut at right angles if  8k2 = 1Solution.   The given equations are  x = y2      ……….  (1)xy = k  …………..  (2)First of all we find the point on the curvePutting eqn(1) in eqn(2) we gety2 x y = k        ⇒     y3 = k     ⇒     y = k1/3Putting the value of y in eqn. (1) we getx = (k1/3)2  =  k2/3Point on the curve is  (k2/3, k1/3 )Now differentiating eqn.(1) w. r. t. x we get$1=2y\frac{dy}{dx}\; \; \Rightarrow\; \; \frac{dy}{dx}=\frac{1}{2y}=m_{1}$Now differentiating eqn.(2) w. r. t. x we get$1.y+x.\frac{dy}{dx}= 0\; \; \Rightarrow \; \; \frac{dy}{dx}=\frac{-y}{x}=m_{2}$Eqn(1) and eqn(2) intersect  Each other at right angles iff$m_{1}\times m_{2}=-1$$\Rightarrow \frac{1}{2y}\times \frac{-y}{x}=-1\; \; \Rightarrow \; \; 2x=1$Putting x = k2/3  we get $2(k^{2/3})=1$Cubing on both sides we get $\left [2(k^{2/3}) \right ]^{3}=(1)^{3}\; \; \Rightarrow\; 8k^{2}=1$

*************************************************

 Exercise 6.4Q 1 (iv)  Using Differential find the approximate value of  (0.009)1/3  up to three decimal places.Solution (0.009)1/3  = (0.008 + 0.001)1/3   = (x + Δx)1/3   (x + Δx)1/3  = (0.008 + 0.001)1/3  x = 0.008     and    Δx = 0.001Let y = x1/3Differentiating it with respect to x we get$\frac{dy}{dx}=\frac{1}{3}x^{1-\frac{1}{3}}=\frac{1}{3}x^{-2/3}=\frac{1}{3}(0.008)^{-2/3}$$\frac{dy}{dx}=\frac{1}{3}\times (0.2)^{-2}=\frac{1}{3\times 0.04}=\frac{1}{0.12}$$(x+\Delta x)^{1/3}=x^{1/3}+\frac{dy}{dx}\times \Delta x$$\left (0.008+0.001 \right )^{1/3}=(0.008)^{1/3}+\frac{1}{0.12}\times 0.001$  $\left (0.009 \right )^{1/3}=0.2+\frac{0.001}{0.120}$$\left (0.009 \right )^{1/3}=0.2+\frac{1}{120}=0.2+0.008$$\left (0.009 \right )^{1/3}=0.208\: \; \; Ans$

*************************************************