Maths

HCF = $\frac{HCFof(81,27,9)}{LCMof(16,20,40)}=\frac{9}{80}$  

Quantitative Aptitude Prep Tips for MBA

$\begin{array}{l}=\sqrt[]{(l^2+b^2+h^2)}\\ =\sqrt[]{100+100+225}\\ =\sqrt[]{425}=5\sqrt[]{17} m\end{array}$

$\frac{417}{4}$ gives a remainder of 1.

For all numbers which are in the form  x^1/x and all x is greater than 3, the one with the greatest base is the smallest number.

f (t) = t³ – 6t² + 9t + 3

$\Rightarrow f\text{'}\left(t\right)=3t^2-12t+9=3\left(t-1\right)\left(t-3\right)$

$\therefore f\text{'}\left(t\right)=0\Rightarrow t=1,3$

$\Rightarrow f\left(1\right)=1,f\left(3\right)=-3$

$g\left(x\right)=\{\begin{array}{l}f\left(x\right),0\le x<1\\ 1,1\le x\le 3g\left(x\right)iscontinuous\\ 4-x,3<x\le 4\end{array}$

Hence g (x) is not differentiable at only x = 3.

$\frac{1}{\alpha \left(\alpha +1\right)\left(\alpha +2\right)..........\left(\alpha +20\right)}=\frac{A_0}{\alpha }+\frac{A_1}{\alpha +1}+\frac{A_2}{\alpha +2}+......+\frac{A_{20}}{\alpha +20}$

Solving by partial fraction, we get

$A_{13}=\frac{-1}{14!7!},A_{14}=\frac{-1}{14!6!}and{A}_{15}=\frac{-1}{14!5!}$

$\therefore \frac{A_{14}+A_{15}}{A_{13}}=\frac{3}{10}\Rightarrow {\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2={\left(\frac{3}{10}\right)}^2\Rightarrow 100{\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2=9$

$\frac{log\left(\left(2x+5\right)\left(x+1\right)\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-4=0\Rightarrow \frac{log\left(2x+5\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-3=0$

$\Rightarrow x\in \left(-1,0\right)\cup \left(0,\infty \right)$

$\frac{log\left(2x+5\right)}{log\left(x+1\right)}=1\Rightarrow x=-4$ (not possible) and $\frac{log\left(2x+5\right)}{log\left(x+1\right)}=2\Rightarrow x=2$

Hence only one solution is possible.

${\displaystyle \int dy}={\displaystyle \int \frac{cos\left(\frac{1}{2}co{s}^{-1}\left(e^{-x}\right)\right)}{e^x\sqrt[]{1-{\left(e^{-x}\right)}^2}}}dxputcos2\theta =e^{-x}\Rightarrow -2sin2\theta d\theta =-e^{-x}dx$

$\therefore {\displaystyle \int dy}={\displaystyle \int \frac{2cos\theta sin2\theta d\theta }{\sqrt[]{1-co{s}^{2}2\theta }}\Rightarrow y=2sin\theta +c\Rightarrow y=-1,\theta =0,c=0}$

$\therefore y=2\sqrt[]{\frac{1-e^{-x}}{2}}-1at\left(\alpha ,0\right),e^{\alpha }=2$

$?\left|{\overrightarrow{v}}_1\right|=\left|{\overrightarrow{v}}_2\right|\Rightarrow p^2-p-2=0\Rightarrow p-1,2$ we take only p = 2 (p > 0)

$\therefore cos\theta =\frac{\overrightarrow{v_2}.\overrightarrow{v_2}}{\left|\overrightarrow{v_1}\right|\left|\overrightarrow{v_2}\right|}=\frac{4\sqrt[]{3}+3}{13}\Rightarrow tan\theta =\frac{6\sqrt[]{3}-2}{4\sqrt[]{3}+3}=\frac{\alpha \sqrt[]{3}-2}{4\sqrt[]{3}+3}\therefore \alpha =6$

$A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\Rightarrow \left|A\right|=4\Rightarrow \left|3adj\left(2A^{-1}\right)\right|=\left|3.2^{2}adj\left(A^{-1}\right)\right|$

$\Rightarrow 12^3\left|adj\left(A^{-1}\right)\right|=12^3{\left|A^{-1}\right|}^2=\frac{12^3}{{\left|A\right|}^2}=\frac{12*12*12}{16}=108$