Maths

${\displaystyle \sum _{i=0}^k\left(\begin{array}{l}10\\ i\end{array}\right)}\left(\begin{array}{l}15\\ K-i\end{array}\right)+{\displaystyle \sum _{i=0}^{k+1}\left(\begin{array}{l}12\\ i\end{array}\right)\left(\begin{array}{l}13\\ k+1-i\end{array}\right)}$  

= Equating the coefficient of $x^{k}in{\left(1+x\right)}^{25}{=}^{25}{C}_k$ .(i)

= Equating the coefficient of   $x^{k+1}in{\left(1+x\right)}^{12}{\left(x+1\right)}^{13}$

From (i) and (ii)

Hence   $26\ge k+1\Rightarrow k\le 25$

But maximum value of k is not defined bonus

$L_1=\frac{x-\lambda }{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z}{-\frac{1}{2}}$           

$\therefore SD=\frac{\left|-2\lambda +3\left(2\lambda +\frac{1}{2}\right)+\lambda \right|}{\sqrt[]{14}}=\frac{\left|5\lambda +\frac{3}{2}\right|}{\sqrt[]{14}}$           

$5\lambda +\frac{3}{2}=-\frac{7}{2}\Rightarrow 5\lambda =-5\Rightarrow \lambda =-1$           

$\therefore \left|\lambda \right|=1$

$x\frac{dy}{dx}+y=b{x}^4$       

$\frac{dy}{dx}+\frac{y}{x}=b{x}^3$           

As per question

$\frac{32b}{25}+\left(10-b\right)ln2-\frac{b}{32}=\frac{62}{5}$           

$\left(10-b\right)ln2=\frac{62}{5}-\frac{31}{25}b=\frac{31}{5}\left(2-\frac{b}{5}\right)$           
$\therefore b=10$

S = 432 km/hr

tan 60° =   $\frac{h}{y}$

$h=\sqrt[]{3}$ y

$tan30°=\frac{h}{x+y}$           

$?x=432*\frac{1000}{3600}*20$           

x = 2400 m

y = 1200 m

$\therefore h=1200\sqrt[]{3}m$          

y = ax² + bx + c

a + b + c = 2

$\therefore \frac{dy}{dx}=2ax+b\Rightarrow {\frac{dy}{dx}|}_{\left(0,0\right)}=b=1$    

It passes through (0, 0)

0 = 0 + 0 + c -> c = 0

a = 1

$\therefore a=1,b=1,c=0$           

(A)  

tautology

           (B)            

                             tautology

$tan\left(\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}\right)$

$Put\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}=\theta$

$sin4\theta =\frac{\sqrt[]{63}}{8}\therefore cos4\theta =\frac{1}{8}$

$co{s}^2\theta =\frac{7}{8}\Rightarrow se{c}^2\theta =\frac{8}{7}$

$ta{n}^2\theta =\frac{8}{7}-1=\frac{1}{7}$

$tan\theta =\frac{1}{\sqrt[]{7}}$

y = x² + 4

x² = y – 4

y = 4x – 1

$PQ=\frac{\left|4*\frac{1}{2}t-\frac{1}{4}{t}^2\right|-4-1}{\sqrt[]{17}}$           

$p=\frac{\left|t^2-8t-2\right|}{4\sqrt[]{17}}$          

$\frac{dp}{dt}=\frac{1}{4\sqrt[]{17}}$           (2t – 8) for maximum / minimum, $\frac{dp}{dt}=0\Rightarrow t=4$  

$also\frac{d^{2}p}{d{t}^2}>0att=4$           

Hence the closest point becomes at t = 4 is (2, 8)

  $\left|\begin{array}{cc}\hfill f\left(X\right)\hfill & \hfill f\text{'}\left(x\right)\hfill \\ \hfill f\text{'}\left(x\right)\hfill & \hfill f\text{'}\text{'}\left(x\right)\hfill \end{array}\right|=0$         

$f\left(x\right)f\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)f\text{'}\left(x\right)$

$\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\left(x\right)}=\frac{f\text{'}\left(x\right)}{f\left(x\right)},$ Integrating on both sides

$\frac{f\text{'}\left(x\right)}{f\left(x\right)}=2,$ again integrating on both side

ln f(x) = 2^X + k

f(x) = e^{2x + k}

f(0) = e^k = e^k = 1-> k = 0

$\therefore f\left(x\right)=e^{2x}\left[\therefore e=2.718\right]$           

$\therefore e^2\in \left(6,9\right)$           

$\sim \left(\sim p\wedge \left(p\vee q\right)\right)=pV\left(\sim p\wedge \sim q\right)$

$\begin{array}{l}=\left(pv\sim p\right)\wedge \left(pv\sim q\right)\\ =pv\sim q\end{array}$