Maths Binomial Theorem

Coeff of x? in (x²+1/bx)¹¹: T? = ¹¹C? (x²)¹¹? (1/bx)? = ¹¹C? x²²? ³? b?
22-3r=7 ⇒ 3r=15 ⇒ r=5. Coeff is ¹¹C? /b?
Coeff of x? in (x-1/bx²)¹¹: T? = ¹¹C? (x)¹¹? (-1/bx²)? = ¹¹C? x¹¹? ³? (-1)? b?
11-3r=-7 ⇒ 3r=18 ⇒ r=6. Coeff is ¹¹C? (-1)? /b? = ¹¹C? /b?
¹¹C? /b? = ¹¹C? /b? ⇒ b = ¹¹C? /¹¹C? = (11-5+1)/5 = 7/5. This differs from the solution.
Let's check the exponents again.
x? : 22-2r-r=7 => 22-3r=7 => 3r=15 => r=5. Correct.
x? : 11-r-2r=-7 => 11-3r=-7 => 3r=18 => r=6. Correct.
The given solution has b=1.

$T_{r+1}{=}^{10}{C}_r{\left(t{x}^{\frac{1}{5}}\right)}^{10-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$

According to question, 10 – 2r = 0 ⇒ r = 5

$\therefore T_6{=}^{10}{C}_5*{\left(1-x\right)}^{\frac{1}{2}}$

T₆ is maximum, when f(x) = $x{\left(1-x\right)}^{\frac{1}{2}}$ is maximum.

$f\text{'}\left(x\right)={\left(1-x\right)}^{\frac{1}{2}}-\frac{x}{2\sqrt[]{1-x}}=\frac{2\left(1-x\right)-x}{2\sqrt[]{1-x}}$

For maximum, $f\text{'}\left(x\right)=0\Rightarrow x=\frac{2}{3}$

$\therefore T_6{=}^{10}{C}_5\frac{2}{3}.{\left(\frac{1}{3}\right)}^{\frac{1}{2}}=\frac{2\left(10!\right)}{3\sqrt[]{3}{\left(5!\right)}^2}$

= A + B (say)

(1 + x)¹⁵ =

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

put  = x = -1

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

$B{=}^{14}{C}_{13}=2^{13}$    

$\therefore A+B=2^{13}-14$     

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

= 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

 $\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{n}{{\left(n+r\right)}^2}}$

$\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{1}{n}.\frac{1}{{\left(1+\frac{r}{n}\right)}^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{{\left(1+x\right)}^2}={\left[-\frac{1}{1+x}\right]}_0^1}=-\frac{1}{2}-\left(-1\right)=\frac{1}{2}}$

${\left(2x^r+\frac{1}{x^2}\right)}^{10}$

Let the constant term is (k + 1)^th term.

${\therefore }^{10}{C}_k{\left(2x^r\right)}^{10-k}.x^{-2k}$              

${=}^{10}{C}_k{2}^{10-k}.x^{10r-rk-2k}$              

              For constant term 10r – rk – 2k = 0.

$\therefore k=\frac{10r}{r+2}k\in I$              

$\therefore r=3,k=6andr=8,k=8$               

For k = 6

For k = 8  

$\therefore r=8$

$f\left(x\right)=\frac{5x+3}{6x-\alpha }=y.....................(i)$

$\Rightarrow x=\frac{\alpha y+3}{6y-5}\Rightarrow f^{-1}\left(x\right)=\frac{\alpha x+3}{6x-5}..............(ii)$

According to question, $f\left(x\right)=f^{-1}\left(x\right)\therefore$  from (i) and (ii) we get $\propto =5$

$\frac{x+1}{a}=\frac{y}{1}=\frac{z-1}{a}...........\left(i\right)$

$\frac{x+2}{a}=\frac{y}{1}=\frac{z}{\frac{3}{b}}..........\left(ii\right)$

Let A (-1, 0, 1) and B (-2, 0, 0) $$ $\therefore$ direction ratios of AB = -1, 0, -1

$$ $\therefore$  lines are coplanar

$\therefore \left|\begin{array}{ccc}\hfill a\hfill & \hfill 1\hfill & \hfill a\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill \frac{3}{b}\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right|=0\Rightarrow b=1anda\in R-\left\{0\right\}$