Question

Match the following. The codes for the lists have choices $\left(A\right),\left(B\right),\left(C\right)and\left(D\right)$ out of which only ONE is correct.
Codes :
P Q R S
(A)$4321$
(B)$2143$
(C)$4312$
(D)$2431$

Answer 1

Code $\left(C\right):\left(P\right)-4,\left(Q\right)-3,\left(R\right)-1,\left(S\right)-2$

$\left(P\right)$ Limit is easily reducible to $cose{c}^{-1}\left(\begin{array}{c}\frac{k^2}{2}\end{array}\right)$ by L Hospital rule which exist if $\frac{k^2}{2}\ge 1$.

$\left(Q\right)k{x}^2+(3-2k)x-6=(kx+3)(kx-2)$

$\Rightarrow 4\le \frac{3}{k}\le 5\Rightarrow -\frac{3}{4}\le k\le -\frac{3}{5}$.

$\left(R\right)(2k+1,k-1)$ is an interior point

$\Rightarrow (2k+1{)}^2+(k-1{)}^2-2(2k+1)-4(k-1)-4<0$

$\Rightarrow 0<k<\frac{6}{5}.....\left(1\right)$

$\Rightarrow$Centre $(1,2)$ and point $(2k+1,k-1)$ must lie on opposite side of chord $x+y-z=0$

$\Rightarrow k<\frac{2}{3}......\left(2\right)$

$\Rightarrow 0<k<\frac{2}{3}$

$\left(S\right)x>-\frac{5}{2},-4<x<4\Rightarrow x\in \left(\begin{array}{c}-\frac{5}{2},4\end{array}\right)$

$\Rightarrow {\log }_5\left(\begin{array}{c}\frac{16-x^2}{2x+5}\end{array}\right)\le 1$

$\Rightarrow \frac{16-x^2}{2x+5}\le 5^1$

$\Rightarrow x\in (-\infty ,-9)\cup [-1,\infty ]$

$\therefore x\in [-1,4]$