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Question

Match the following. The codes for the lists have choices $$(A),\,(B),\,(C)\;and\;(D)$$ out of which only ONE is correct.
Codes :
$$\;\;\;\;\;\;\;$$ P Q R S
(A)$$\;\;\;\;\;4\;3\;2\;1$$
(B)$$\;\;\;\;\;2\;1\;4\;3$$
(C)$$\;\;\;\;\;4\;3\;1\;2$$
(D)$$\;\;\;\;\;2\;4\;3\;1$$


Solution

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

$$(P)$$ Limit is easily reducible to $$cosec^{-1}\begin{pmatrix}\displaystyle\frac{k^2}{2}\end{pmatrix}$$ by L Hospital rule which exist if $$\displaystyle\frac{k^2}{2}\,\ge\,1$$.


$$(Q)\;kx^2+(3-2k)x-6=(kx+3)(kx-2)$$

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


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

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

$$\Rightarrow 0\,<\,k\,<\,\displaystyle\frac{6}{5}\;\;\;\;\;\;\;\;\;\;\;\;\;\;.....(1)$$

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

$$\Rightarrow k\,<\,\displaystyle\frac{2}{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;......(2)$$

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


$$(S)\;x\,>\,-\displaystyle\frac{5}{2},\,-4\,<\,x\,<\,4\,\Rightarrow\,x\,\in\,\begin{pmatrix}-\displaystyle\frac{5}{2},\,4\end{pmatrix}$$

$$\Rightarrow \log_5\begin{pmatrix}\displaystyle\frac{16-x^2}{2x+5}\end{pmatrix}\,\le\,1$$

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

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

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

Mathematics

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