Class 12th

${\left[e^x\right]}^2+\left[e^x\right]-2=0$              

Let $\left[e^x\right]=t$   

->t² + t – 2 = 0 Þ (t + 2) (t – 1) = 0

t = -2 $\left[e^{-x}\right]$ -> = -2 not possible

$\therefore t=\left[e^x\right]=1\Rightarrow 1\le e^x<2$

$\Rightarrow 0\le x<ln2$              

Five years ago the ratio of their ages was 5:3 with the total age being 64 years. Their individual ages will be 40 and 24 years.

Today their ages will be 45 years and 29 years and after 9 years their ages will be 54 and 38 years.

Required ratio will be 27 : 19.

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt[]{2km}}=\frac{h}{2\left(ev\right)m}$

$\lambda \alpha \frac{1}{\sqrt[]{m}}\Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\sqrt[]{\frac{m_p}{m_e}}$

${\displaystyle \underset{}{\overset{}{\int }}\frac{si{n}^{2}xdx}{e^{\left\{\frac{x}{\pi }\right\}}}=l}$

$f\left(x\right)=\frac{si{n}^{2}x}{e^{\left\{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.\right\}}}$ is periodic with period =

$l=100{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{si{n}^{2}x}{e^{x/\pi }}dx.}$

$=50{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1-cos2x}{e^{x/\pi }}}dx=50{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }dx-50}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }}cos2xdx$

$=50\pi \left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)\left[1-\frac{1}{1+4{\pi }^2}\right]=\frac{50\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)4{\pi }^3}{1+4{\pi }^2}$

$\therefore \alpha =200\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)$

612 = 212 * 312. For perfect cubes, power must be multiple of 3.

Therefore, power of 2 equal 20, 23, 26, 29, 212 and power of 3 equals 30, 33, 36, 39, 312 So, number of divisors equal 5* 5 = 25.

New rent = 2730 per month.  Rs. 2720 * 12 = Rs. 32640 per year

Net for X = 70 * 4 + 50 * 3 + 66 * 5 = 760

For Y = 50 * 4 + 70 * 8 = 760

For Z = 40 * 7 + 24 * 5 = 400

The shares will 12920, 12920 and 6800.

When T2 is at 45 points, T3 is at 40 points, that is, T2 can beat T3 by 5 points in a 45 point game. In a 765 point game, T2 can beat T3 by $\frac{765}{45}*5$ $\frac{}{}$ = 85 points.

Selling price = Rs. 16.56

Cost price for 38/% profit = 1.38 C.P. = 16.56

C.P. = $\frac{15}{1.25}$ $\frac{}{}$ = Rs. 12

$\begin{array}{l}\frac{Chemical one }{Chemical second}=\frac{\left(13.5-12\right)}{\left(12-9\right)}\\ =\frac{1.5}{3}=\frac{1}{2}\end{array}$

Let the price of P2 per kg be Rs. X. Then the price of P1 per kg = Rs. 3X.

1 kg of P3 P3contains

$\left(\frac{2}{7}\right)*1,i.e.,\frac{2}{7}kgofAand\frac{5}{7}kgofB$

Price of 1 kg of P3 = $\left(\frac{2}{7}\right)*3X+\left(\frac{5}{7}\right)X=\left(\frac{11}{7}\right)X$ $\frac{}{}\frac{}{}\frac{}{}$

P2y the given condition,

$\begin{array}{l}\frac{11X}{7}=10.4-1.6=Rs.8.8\\ \Rightarrow X=8.8*\left(\frac{7}{11}\right)=5.6\end{array}$

Hence, the price of P1 per kg = 3X = 3 * 5.6 = Rs. 16.8

Price of P2 per KG = Rs5.6

Let X = working time

$\frac{74}{100}*5574300$

Solve to get X = 6 hours.