Relations and Functions

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                          9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

$30{}^{30}{C}_0+29{}^{30}{C}_1+.....+2{}^{30}{C}_{28}+1{.}^{30}{C}_{29}=n.2^m$

$\therefore n=15,m=0$

$\therefore n+m=15+30=45$

$\sqrt[]{3}co{s}^{2}x=\left(\sqrt[]{3}-1\right)cosx+1$

$\left(\sqrt[]{3}cosx+1\right)\left(cosx-1\right)=0$

$\therefore cosx=-\frac{1}{\sqrt[]{3}}\left(rejected\right)$

Hence, cos x = 1 x = 0 one solution

R = { (P, Q) |P and Q are at the same distance from the origin}.

at $\left(1,-1\right),x^2+y^2={\left(1\right)}^2+{\left(-1\right)}^2=1+1=2$

 $\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a+2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a+3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a+4\hfill & \hfill 1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a\hfill & \hfill 1\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|,$ [using properties]

$=0+\left(a+1\right)\left(a+2\right)\left(-1\right)+2\left(a+3\right)\left(a+4\right)0\left(a+2\right)\left(a+3\right)+4\left(\left(a+2\right)\left(a+3\right)-3\left(a+3\right)\left(a+4\right)\right)$

$=2\left(a+2\right)-2\left(a+3\right)=2\left(a+2-a-3\right)=-2$

$\frac{si{n}^{-1}x}{a}=\frac{co{s}^{-1}x}{b}=\frac{ta{n}^{-1}y}{c}=\lambda$

$\therefore \frac{si{n}^{-1}x}{a}=\lambda$

$\therefore si{n}^{-1}x=a\lambda$

$x=sina\lambda .........\left(i\right)$

$x=cosb\lambda ..........\left(ii\right)$

$y=tanc\lambda ..........\left(iii\right)$

From (i) & (ii), we get, $sina\lambda =cosb\lambda =sin\left(\frac{\pi }{2}-b\lambda \right)$

$\therefore a\lambda =\frac{\pi }{2}-b\lambda \Rightarrow \left(a+b\right)\lambda =\frac{\pi }{2}\Rightarrow \lambda =\frac{\pi }{2\left(a+b\right)}...........\left(iv\right)$

From (iii) y = tan c $\lambda$

$\Rightarrow cos\left(\frac{2c\pi }{2\left(a+b\right)}\right)=\frac{1-y^2}{1+y^2}\Rightarrow cos\left(\frac{\pi c}{a+b}\right)=\frac{1-y^2}{1+y^2}$

$\underset{x\to 7}{lim}\frac{18-\left[1-x\right]}{\left[x-3a\right]}$

exist &   $a\in I.$

$=\underset{x\to 7}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}$

exist

RHL =   $\underset{x\to 7^{+}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$

$LHL=\underset{x\to 7^{-}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{24}{6-3a}\left[a\ne 2\right]$

LHL = RHL

$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$

$\therefore a=-6$

$\underset{x\to 0}{lim}\frac{\alpha x\left(1+x+\frac{x^2}{2}+....\right)-\beta \left(x-\frac{x^2}{2}+\frac{x^3}{3}+....\right)+\gamma x^2\left(1-x\right)}{x^3}=10$

For limit to exist a - b = 0 . (i), $\alpha +\frac{\beta }{2}+\gamma =0.........(ii)$

and $\frac{\alpha }{2}-\frac{\beta }{2}-\gamma =10$ . (iii)

Solving (i), (ii) and (iii) we get α = 6, β = 6 and $\lambda =-9\Rightarrow \alpha +\beta +\lambda =3$

  $f\left(x\right)=\left[2x^2+1\right]$

  $g\left(x\right)=\{\begin{array}{l}2x-3,x<0\\ 2x+3,x\ge 0\end{array}$

  $f\left(g\left(x\right)\right)=\left[2{\left(g\left(x\right)\right)}^2\right]+1$              

 for 0 < x < 1

-3 < 2x 3 < -1

2 < 2 (2x 3)² < 18

31 + 15 = 46

Case 1 : If f (3) = 3 then f (1) and f (2) take 1 or 2

No. of ways = 2.6 = 12

Case 2 : If f (3) = 5 then f (1) and f (2) take 2 or 3

OR 1 and 4

No. of ways = 2.6.2 = 24

Similarly for all other cases

Total no. of ways 90.