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\displaystyle \int \frac{x^3-2}{\sqrt{(x^3+1)^2}}dx

If \displaystyle I_{m,n} = \int\cos (mx).\sin(nx)dx, Then \displaystyle 7I_{4,3}-4I_{3,2}=

If \vec{a}\;\;,\vec{b}\;\;,\vec{c} are Three Unit Vector satisfying \mid \vec{a}-\vec{b}\mid^2+\mid \vec{b}-\vec{c}\mid^2+\mid \vec{c}-\vec{a}\mid^2 = 9 then \mid 2\vec{a}+5\vec{b}+5\vec{c}\mid =

I have Tried for lower bond like \displaystyle \int_{0}^{1}e^{-x^2}dx\geq \int_{0}^{1}e^{-x}dx=(1-\frac{1}{e}) So I am getting (ii) option How can I calculate for Upper Bond,Help required... Thanks

If \mathbb{S} be The Area of The Region enclosed by y=e^{-x^2}\;,x=0\;,y=0\;,x=1. Then Which one is Right options: \displaystyle (a)\;\;\mathbb{S}\geq \frac{1}{e} \displaystyle (b)\;\;\mathbb{S}\geq 1-\frac{1}{e} \displaystyle (c)\;\;\mathbb{S}\leq...

Thanks Carrotsticks , after seeing Your Solution, I get This Method...

Thanks Friends got it

Let p(x) be a real polynomials of Least Degree Which has Local Maximum at x=1 and Local Minimum at x=3.If p(1)=6 and p(3)=2. Then p^{'}(0)=

\hspace{-16}%20$Here%20$\bf{\mid%20z%20\mid%20=%202}$%20and%20Here%20We%20have%20to%20find%20the%20value%20of%20$\bf{(a+bz)}$\\\\\\%20$\bf{\mid%20a+bz\mid%20\leq%20\mid%20a%20\mid+\mid%20b%20\mid%20.\mid%20z%20\mid\Leftrightarrow%20\mid%20a+bz\mid%20\leq%20a+2b\;\;.\;%20a,b\in%20\mathbb{R}}$\\\\\...

Thanks friends but answer given is abc = 1

if \mid z \mid= 2 , then find the locus of complex number (a + bz) where a\;,b\in \mathbb{R}

\hspace{-16}$Let%20$a,b,c$%20be%20distinct%20Complex%20no.%20such%20that%20$\mid%20a\mid%20=%20\mid%20b%20\mid%20=%20\mid%20c\mid%3E0$\\\\%20If%20$a+bc\;,b+ca\;,c+ab%20\in%20\mathbb{R}$.%20Then%20$abc=$

\hspace{-16}$Find%20Complex%20no.%20$\mathbf{z}$%20that%20Satisfy%20the%20equation\\\\%20$\mathbf{\left|%20z%20\right|%20+%2020\left|%20{\dfrac{{z%20-%201-%20i}}{{z%20+%202+%20i}}}%20\right|%20=%203\sqrt{5}}$

\hspace{-16}$If $\mathbf{z_{1}}$ and $\mathbf{z_{2}}$ are two Complex no. such that $\mathbf{\mid z_{1}\mid = 3}$\\\\ and $\mathbf{\mid z_{2}\mid = 4}$ and $\mathbf{\mid z_{1}-z_{2}\mid = \sqrt{37}}$. Then $\mathbf{\frac{z_{1}}{z_{2}}=}$

Thanks Carrotsticks

Area of Region bounded by the locus of z that satisfying \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4}

i have tried like in that way \mid z_{k} \mid = 1\Leftrightarrow \mid z_{k} \mid ^2=1\Leftrightarrow z_{k}\bar{z_{k}} = 1\forall k=1,2,3 Now \displaystyle \frac{z_{1}z_{2}}{(z_{1}-z_{2})^2}+\frac{z_{2}z_{3}}{(z_{2}-z_{3})^2}+\frac{z_{3}z_{1}}{(z_{3}-z_{1})^2} = -1 OR \displaystyle...

\hspace{-16}$Here%20Equ.%20of%20Quarter%20Circle%20$\mathbf{OA}$%20is\\\\%20$\mathbf{(x-2)^2+y^2=4\Leftrightarrow%20y=\pm%20\sqrt{4-(x-2)^2}}$\\\\%20So%20equation%20of%20$\mathbf{OA}$%20is%20$\mathbf{y=\sqrt{4-(x-2)^2}}$\\\\%20Similarly%20equation%20of%20$\mathbf{AB}$%20is%20$\mathbf{y=2}$\\\\%20...

if z_{1},z_{2},z_{3} are distinct Complex no. and \mid z_{1}\mid = \mid z_{2}\mid = \mid z_{3}\mid = 1 and satisfy the relation \displaystyle \frac{z_{1}z_{2}}{(z_{1}-z_{2})^2}+\frac{z_{2}z_{3}}{(z_{2}-z_{3})^2}+\frac{z_{3}z_{1}}{(z_{3}-z_{1})^2} = -1. then prove that z_{1},z_{2},z_{3} are...