Maths

$f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=x^5+64$

$\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}\left(\frac{f\left(1\right)}{0}\right)f\left(1\right)=0$

Let f(x) = x⁵ + ax⁴ + bx³ + cx² + dx + e

  f(1) = 0 Þ 1 + a + b + c + d + e = 0

$\Rightarrow x^5+\left(a+5\right)x^4+\left(b+4a+20\right)x^3+\left(c+3b+12a\right)x^2+\left(d+2c+6b\right)x+e+d+2c=x^5+64$  

 Þ e + d + 2c = 64

b + 4a + 20 = 0                 c – 1 – a – b = 64

c + 3b + 12a = 0

d + 2c + 6b = 0                 a + b – c = -65

1 + a + b + c + d + e = 0   13a + 4b = -65

b + 4c = -20 * 4

-3a = 15 a = -5

$\begin{array}{l}f\left(x\right)=x^5-5x^4+60x^2-120x+64\\ f\text{'}\left(x\right)=5x^4-20x^3+12x-120\\ f\text{'}\left(1\right)=5-20+120-120=-15\end{array}$

$t_n=\frac{1}{2^n.3^{11-n}}={\left(\frac{1}{3}\right)}^{11}.{\left(\frac{3}{2}\right)}^n$

$S_{10}={\left(\frac{1}{3}\right)}^{11}.\frac{\left(\frac{3}{2}\right).\left({\left(\frac{3}{2}\right)}^{10}-1\right)}{\frac{3}{2}-1}$

$=\frac{1}{3^{10}}.\frac{3^{10}-2^{10}}{2^{10}}$

$=1{+}^{10}{C}_1.2^1{+}^{10}{C}_2.2^2+....{+}^{10}{C}_9.2^9$

k = 21 + 6m ® R = 3

= A + B (say)

(1 + x)¹⁵ =

differentiating 15 (1 + x)¹⁴ =   $\sum r{.}^{15}{C}_r{x}^{r-1}$

put  = x = -1

$\Rightarrow 0={-}^{15}{C}_1+2{.}^{15}{C}_2-.......=A$      

  $B{=}^{14}{C}_{13}=2^{13}$            

$\therefore A+B=2^{13}-14$           

$f\left(g\left(x\right)\right)=x\forall x\in R$

$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)$

For y = g (x)

x = y³ + y – 5

$\frac{dx}{dy}=3y^2+1\Rightarrow g\text{'}\left(63\right)=\frac{1}{3\left(16\right)+1}=\frac{1}{49}$

$g\text{'}\left(63\right)=\left(\frac{dy}{dx}\right)$

$\left(\begin{array}{l}x=63\\ y=4\end{array}\right)$

Let direction ratio of the normal to the required plane are l, m, n

$\begin{array}{l}\therefore 3l+m-2n=0\\ \therefore 2l-5m-n=0\\ \therefore \frac{l}{-11}=\frac{m}{-1}=\frac{n}{-17}\end{array}$             

Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

$\Rightarrow 11x+y+17z+38=0$           

$e^{\left(co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+.....\infty \right)}lo{g}_{e}2$  

$=e^{\frac{co{s}^{2}x}{1-co{s}^{2}x}lo{g}_{e}2}=e^{co{t}^{2}xlo{g}_{e}2}=2^{co{t}^{2}x}$          

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t=2^{co{t}^{2}x}=8=2^3$

$\Rightarrow co{t}^{2}x=3=co{t}^2\frac{\pi }{6}$

$\frac{2sinx}{sinx+\sqrt[]{3}cosx}=\frac{2*\frac{1}{2}}{\frac{1}{2}+\sqrt[]{3}*\frac{\sqrt[]{3}}{2}}=\frac{1}{2}$            

$A\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)$

$A\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}-1\\ 0\\ 1\end{array}\right)$

$A\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)$ $\Rightarrow A\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

$AB=C\Rightarrow A=C{B}^{-1}$

$\left|A-2l\right|=\left(-4\right)\left(-1\right)+3\left(-1\right)+1\left(-1\right)$

= 4 – 3 – 1 = 0

$\mathcal{l}\left(-1,0,1\right)+m\left(-1,1,0\right)=\left(-4,3,1\right)\Rightarrow -\mathcal{l}-m=-4$

$m=3,\mathcal{l}=1$

Any point on line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$       

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$   

  $=3\pi -\frac{3\sqrt[]{3}}{2}$           

$\therefore$ Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$