math assignment solutions

ASSIGNMENTS In the second year Pre- University course each student has to submit 5 assignments - during the academic year- of any type mentioned below for deciding 5marks, as prescribed by the Department of PUE. Each assignment has to be valued for 10 marks and finally the total of the 5 assignm..

reference books for assignments

1. A modern Introduction to Ancient Indian Mathematics by Dr. T. S. Bhanu Murthy, Wiley Eastern Ltd., (New Age Intermational), New Delhi, 1992. 2. History of Ancient Indian Mathematics by Dr. C.N. Srinivasa Iyengar, The world Press Pvt Ltd., Calcutta, 1967. 3. Malhematics in Ancient and Medieval India By ..

particular solution

By assigning particular values to $a$ and $b$, say $a=2$ and $b=5$ in (2) we get $f (x,\ y,\ 2,5)=0$. Such a solution is called a "Particular solution'' of the differential equation. ..

solution

Trigonometric Functions and Identities LEVEL-I 1. Here L.H.S. $\sin^{2}12^{\circ}+\sin^{2}21^{\circ}+(\sin^{2}39^{\circ}-\sin^{2}9^{\circ})+(\sin^{2}48^{\circ}-\sin^{2}18^{\circ})$ $\Rightarrow\sin^{2}12^{\circ}+\sin^{2}21^{\circ}+\sin(48^{\circ}).\sin(30^{\circ})+\sin(30^{\circ}).\sin(66^{\circ})$ ..

solutions

LEVEL II 1. We know from Napier's analogy that : $\qquad\qquad{\displaystyle \tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\frac{A}{2}}$ $\qquad\qquad{\displaystyle \tan\left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot\frac{C}{2}}$ $\qquad\qquad{\displaystyle \tan\left(\frac{C-A}{2}\right)=\frac{c-a}{c+a..

Solution

="font54" style="font-style: italic;"> x sin a) dx J oosx J cosx J, . - '' . : - = cos a jdx + sin a Jtan x dx = (cos a) x + sin a log | sec x \ + c 4. Let x-a = t => x = t + a and dx = dt "..

Solutions

The greatest common factor is 8 6Ans : Given equation is xy + 3y 2x 6 = 0 arrange them in an order now. We get xy 2x + 3y 6 = 0 x (y 2) + 3 (y 2) = 0 (x + 3) (y 2) = 0 So here (x + 3) and ( y 2) are the factors 7Ans: Given that. 5(A) +6(O) =$40 Each orange is $3 So we get 5(A) + 6(O) = $40 5(A) + 6(3) = 40 5(A) + 18 ..

Solution

="margin-bottom: 0.0001pt; text-indent: 0in; line-height: 17pt; background: none repeat scroll 0% 0% transparent;"> K '9- Fig. 7.10 (iv) cos Let tan '4 = 0. Th..

solution of triangles

In a triangle there are six elements. $viz$. three sides $a,b,c$ and three angles. $A,B,C$ when any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the..

characteristics of a solution

E is a point on side $AD$ produced of a parallelogram ABCD and $BE$ iniersects $C\mathrm{D}$ at F. Prove that$\triangle A\mathrm{BE}\sim\triangle CFB$. Solution: In $\triangle ABE$ and $\triangle CFB$, we have $\angle AEB=\angle CBF$ [Alternate angles] and $\angle A=\angle C$ [Opposite ang..

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