Class 12th

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

Quantitative Aptitude Prep Tips for MBA

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

Vitamin B₁ is thiamine while vitamin B₆ is pyridoxine

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

Reducing power for group-15 hydrides increases down the group, so BiH₃ is the strongest reducing agent.

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.

Sulphide ion (s^2-) form ores commonly with Pb & Ag as PbS and Ag₂S

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

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

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