Warning: preg_match(): Compilation failed: quantifier does not follow a repeatable item at offset 1 in /home/customer/www/mathsdiscussion.com/public_html/wp-includes/class-wp.php on line 223
Algebra - Best Maths Practice Material


Algebra for IITJEE

Complex Numbers.

(1)   How to find Argument of a complex number.   ( Video Link Only )     

(3)   Conjugate of Complex Number. ( Video Link Only )

Complex Number.-Discussion Forum

$(Cos\theta +iSin\theta)^{2n+1}$ $$=\sum_{r=0}^{2n+1} {{2n+1}\choose r} Cos^{2n+1-r}\theta (i Sin\theta)^r$$ $\Rightarrow Cos(2n+1)\theta $  $$= \sum_{r=0}^{n} {{2n+1}\choose {2r} }Cos^{2n+1-2r}\theta (i Sin\theta)^{2r}$$

Permutation & Combination

From $^{(m-n)}P_r\,=\,30$ $$ $$ $(m-n)!=5.6.(m-n-r)!$ $\therefore m-n-r=4$ $$ $$ and $\, m-n=6\, ----(1)$ $$\therefore r=2$$ And $^{(m+n)}P_r\,=\,90$

Principle of Inclusion and Exclusion


In the equation LHS $\ge 1$ where as RHS $\le 1$ $$ $$ Hence LHS=RHS=1 is only possible solution. $\therefore x=2nπ\,,\forall n\in I$ for RHS = 1 and for same x LHS =1 if $a(2nπ)=mπ \,,m\in I$

Leave a Reply

Your email address will not be published. Required fields are marked *