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Integrate: 1) [(x-1) x4+2x3-x2+2x+1 ]/x2(x+1) 2) (x2-1)/(x3 2x4-2x2+1 ) ...
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∫x2/√1-x4dx ...
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if a x x x x b x x = f(x)-xf/(x). x x c x x x x d then f(x)=? ...
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*Image* ...
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\hspace{-16}$In how many ways can the selection of $\bf{8}$ letters be done frm $\bf{24}$ letters\\\\ of which $\bf{8}$ are $\bf{'a'}$ and $\bf{8}$ are $\bf{'b'}$ and rest are unlike. ...
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\hspace{-16}$In how many ways can the selection of $\bf{8}$ letters be done frm $\bf{24}$ letters\\\\ of which $\bf{8}$ are $\bf{'a'}$ and $\bf{8}$ are $\bf{'b'}$ and rest are unlike. ...
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\hspace{-16}\bf{(1)\;\; \int\frac{1}{(1+x^4)^{\frac{1}{4}}}dx}$\\\\\\ $\bf{(2)\;\; \int\frac{1}{(1-x^4)^{\frac{1}{4}}}dx}$\\\\\\ $\bf{(3)\;\;\int\frac{1}{(1+x^4)}dx}$\\\\\\ $\bf{(4)\;\;\int\frac{1}{(1+x^6)}dx}$ ...
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\hspace{-18}$Integer values of $\bf{x}$ for which $\bf{x^4+x^3+x^2+x+1}$ is a perfect square. ...
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\hspace{-18}$All positive Integer ordered pairs $\bf{(x,y)}$ for which $\bf{\binom{x}{y} = 120}$ ...
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\hspace{-18}$All positive Integer ordered pairs $\bf{(x,y)}$ for which $\bf{\binom{x}{y} = 120}$ ...
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1+x2y2)dx=ydx+xdy ...
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\hspace{-16}$Determine all pairs $\bf{(a, b)}$ of natural numbers, for which the number\\\\ of $ \bf{a ^ 3 + 6ab + 1} $ and $ \bf{b ^ 3 + 6ab + 1}$ are cubes of natural numbers. ...
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\bf{\int\frac{1}{(x^2-x+1)\sqrt{x^2+x+1}}dx} ...
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It is given that a1=1 and an=n(an-1+1). Define a sequence pn as pn=(1+ 1/a1 )(1+ 1/a2 )...(1+ 1/an ). Find lim(n→∞)pn. ...
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Limn→ ∞ [ n2 . ∫ ( tan-1 nx)/( sin-1 nx) dx Integration upper limit:- ( 1/n ) Integration lower limit:- { (1+ n ) / n } ...
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Given a polynomial of n degree such that f(x)+f(1/x)=f(x)*f(1/x) Find the polynomial ...
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Given a polynomial of n degree such that f(x)+f(1/x)=f(x)*f(1/x) Find the polynomial ...
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\hspace{-16} $ Minimum value of $\bf{\left|z-1-i \right| + \left |z+2-3i \right| + \left |z+3+2i \right|}$\\\\\\ where $\bf{z = x+iy}$ and $\bf{i = \sqrt{-1}}$ ...
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\hspace{-16}$Total Real solution of the equations in Diff. cases\\\\\\ (i) $\bf{2^x = 1+x^2}$\\\\\\ (ii) $\bf{e^x = x^2}$ ...
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\hspace{-16}$Total Real solution of the equations in Diff. cases\\\\\\ (i) $\bf{2^x = 1+x^2}$\\\\\\ (ii) $\bf{e^x = x^2}$ ...
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\hspace{-16}$If $\bf{\mathbb{S} = \sum_{r=4}^{1000000}\frac{1}{r^{\frac{1}{3}}}}.$ Then value of $\bf{\left[\mathbb{S}\right] = }$\\\\\\ where $\bf{[x] = }$ Greatest Integer function ...
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\hspace{-16}$(1): The no. of Integer ordered pairs $\bf{(x,y)}$ in $\bf{x^2+y^2 = 2013}$\\\\\\ $(2):$ The no. of Integer ordered pairs $\bf{(x,y,z)}$ in $\bf{\begin{Vmatrix} \bf{x=yz} \\ \bf{y=zx} \\ \bf{z=xy} \end{Vmatrix}}$ ...
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\hspace{-16}$(1): The no. of Integer ordered pairs $\bf{(x,y)}$ in $\bf{x^2+y^2 = 2013}$\\\\\\ $(2):$ The no. of Integer ordered pairs $\bf{(x,y,z)}$ in $\bf{\begin{Vmatrix} \bf{x=yz} \\ \bf{y=zx} \\ \bf{z=xy} \end{Vmatrix}}$ ...
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\hspace{-16}$If $\bf{f(x)=\frac{x^2}{1+x^2}.}$ Then Determine value of the following expression\\\\\\ $\bf{f\left(\frac{1}{2000}\right)+f\left(\frac{2}{2000}\right)+...+f\left(\frac{1999}{2000}\right)+f\left(\frac{2000}{2000} ...
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\hspace{-16}$In a $\bf{â–³ABC}$ If $\bf{P}$ be a point which is Inside the $\bf{â–³ABC}$ such that\\\\ Area of $\bf{â–³APB=}$ Area of $\bf{â–³BPC=}$ Area of $\bf{â–³CPA}$.\\\\ Then prove that the point $\bf{P}$ is the centroi ...
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\hspace{-16}$Minimum value of $\bf{n\in\mathbb{N},}$ whic has ......\\\\ $\bf{(i)\;\; 16-}$ divisers.\\\\ $\bf{(ii)\;\; 19-}$ divisers.\\\\ $\bf{(iii)\;\; 24-}$ divisers.\\\\ $\bf{(iv)\;\; 25-}$ divisers.\\\\ $\bf{(v)\;\; 26- ...
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\hspace{-16}$Minimum value of $\bf{n\in\mathbb{N},}$ whic has ......\\\\ $\bf{(i)\;\; 16-}$ divisers.\\\\ $\bf{(ii)\;\; 19-}$ divisers.\\\\ $\bf{(iii)\;\; 24-}$ divisers.\\\\ $\bf{(iv)\;\; 25-}$ divisers.\\\\ $\bf{(v)\;\; 26- ...
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\hspace{-16}$Minimum value of $\bf{n\in\mathbb{N},}$ whic has ......\\\\ $\bf{(i)\;\; 16-}$ divisers.\\\\ $\bf{(ii)\;\; 19-}$ divisers.\\\\ $\bf{(iii)\;\; 24-}$ divisers.\\\\ $\bf{(iv)\;\; 25-}$ divisers.\\\\ $\bf{(v)\;\; 26- ...
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\hspace{-16}$Minimum value of $\bf{n\in\mathbb{N},}$ whic has ......\\\\ $\bf{(i)\;\; 16-}$ divisers.\\\\ $\bf{(ii)\;\; 19-}$ divisers.\\\\ $\bf{(iii)\;\; 24-}$ divisers.\\\\ $\bf{(iv)\;\; 25-}$ divisers.\\\\ $\bf{(v)\;\; 26- ...
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\hspace{-16}$If $\bf{a+b=8}$ and $\bf{ab+c+d=23}$ and $\bf{ad+bc=28}$ and $\bf{cd=12}$.\\\\ Then value of \\\\ $\bf{(i)\;\;\;a+b+c+d=}$\\\\ $\bf{(ii)\;\;ab+bc+cd+da=}$\\\\ $\bf{(iii)\; abcd=}$ ...
