Inverse Trigonometric Functions

$\left|\frac{x^2+4x+2}{x^2+3}\right|\le 1$

${\left(x^2-4x+2\right)}^2\le {\left(x^2+3\right)}^2$

$\Rightarrow \left(2x^2-4x+5\right)\left(-4x-1\right)\le 0$

$\Rightarrow -4x-1\le 0\to x\ge -\frac{1}{4}$

f : {1, 3, 5, 7, ….,99} {2, 4, 6, 8, …….100}

$f\left(3\right)\ge f\left(9\right)\ge f\left(15\right)......\ge f\left(99\right)$

$3+\left(n-1\right)6=99\Rightarrow n=17$

cases f (3) > f (9) > f (15) ……. > f (99)

$\therefore$ from the set {2, 4, 6, …., 100}

17 distinct numbers can be selected in 50C17 ways again remaining {1, 5, 7, 11, ….} can map in 33! ways

$\therefore$ total number of such required functions

${=}^{50}{C}_{17}*33!$

$=\frac{50!}{33!17!}*33!{=}^{50}{P}_{33}$

use $si{n}^{-1}x=co{s}^{-1}\sqrt[]{1-x^2}$

$ta{n}^{-1}\sqrt[]{1-x^2}=co{t}^{-1}\frac{1}{\sqrt[]{1-x^2}}$

$si{n}^{-1}\frac{x}{\sqrt[]{1-x^2}}=co{s}^{-1}\frac{\sqrt[]{1-2x^2}}{\sqrt[]{1-x^2}}$

Sum of roots b = 1 + 2 $\frac{\left(k^2-1\right)}{k^2-2}$

Product of roots 5 = 2 $\left(\frac{k^2-1}{k^2-2}\right)$

b = 4, k² = $\frac{1}{3}$

$\frac{3x^2-9x+17}{x^2+3x+10}=\frac{5x^2-7x+19}{3x^2+5x+12}$

$1+\frac{2x^2-12x+7}{x^2+3x+10}=1+\frac{2x^2-12x+7}{3x^2+5x+12}$

$\left(2x^2-12x+7\right)\left(\frac{1}{x^2+3x+10}-\frac{1}{3x^2+5x+12}\right)=0$

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

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

For y = g (x)

x = y³ + y – 5

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

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

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

$f\left(-2\right)+f\left(3\right)=0.$ One root of f (x) = 0 is (-1)

Let's assume other root to be a

$\therefore f\left(x\right)=a\left(x+1\right)\left(x-\alpha \right)$

Given that f (-2) + f (3) = 0

a (-2 + 1) (-2 -a) + a (3 + 1) (3 - a) = 0

Þ 14 -3a = 0 Þ a =    $\frac{14}{3}$

$\therefore$ sum of roots =   $\frac{14}{3}-1=\frac{11}{3}$

 $\overline{x}=6=\frac{a+b+8+5+10}{5}\Rightarrow a+b=7$ …… (i)

and

${\sigma }^2=\frac{a^2+b^2+8^2+5^2+10^2}{5}-6^2=6.8$

a² + b² = 25 ……. (ii)

From (i) and (ii) (a, b) = (3, 4) or (4, 3)

Now mean deviation about mean

$M=\frac{1}{5}\left(3+2+2+1+4\right)=\frac{12}{5}$

25M = 60

Given system of equations

αx + y + z = 5

x + 2y + 3z = 4, has infinite solution

x + 3y + 5z =β

$\begin{array}{c}\hfill \hfill \end{array}$ $\therefore \Delta =\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right|$  = 0

α (1) – 1 (2) + 1 (1) = 0

α= 1 and

${\Delta }_3=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \beta \hfill \end{array}\right|=0$

β= 3

$\therefore \left(\alpha ,\beta \right)=\left(1,3\right)$