Match each entry in List – I to the correct entries in List – II.
Column – I Column – II
P) Real number(s) $x$ satisfying the equation $2^{x} + 3^{x} - 4^{x} + 6^{x} - 9^{x} = 1.$ Then $x =$ 1) 6
Q) Real number(s) $x$ for which $\frac{8^{x} + 27^{x}}{12^{x} + 18^{x}} = \frac{7}{6} .$ Then $x =$ 2) 1
R) $\sqrt[3]{45 + 29 \sqrt{2}} + \sqrt[3]{45 - 29 \sqrt{2}} =$ 3) 0
S) If $x^{5} - 5 q x + 4 r$ is divisible by ${(x - 2)}^{2} .$ Then the value of $q$ 4) 16
5) An irrational number
The correct option is

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$P - 3, Q - 2, R - 1, S - 4$

$P - 3, Q - 2, R - 5, S - 4$

$P - 2, Q - 1, R - 1, S - 3$

$P - 2, Q - 1, R - 4, S - 5$

answer is A.

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Detailed Solution

If $x^{5} + 5 q x + 4 r$ is divisible by ${(x - 2)}^{2} .$ Find the value of $q$ and $r .$
$p (x) = x^{5} - 5 q x + 4 r$ is divisible by $(x - 2)$
$p (2) = 0 \Rightarrow 32 - 10 q + 4 r = 0$
$5 q = 16 + 2 r$
$p (x) = x^{5} - 16 x - 2 r x + 4 r$
$= x (x^{2} + 4) (x + 2) (x - 2) - 2 r (x - 2)$
$= (x - 2) [x (x + 2) (x^{2} + 4) - 2 r]$
Let $Q (x) = x (x + 2) (x^{2} + 4) - 2 r$
Again $Q (x)$ is divisible by $x - 2$ and by factor theorem
$Q (2) = 0$
$2.4.8 - 2 r = 0 \Rightarrow r = 32 a n d q = 16$

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	Column – I		Column – II
P)	Real number(s) $x$ satisfying the equation $2^{x} + 3^{x} - 4^{x} + 6^{x} - 9^{x} = 1.$ Then $x =$	1)	6
Q)	Real number(s) $x$ for which $\frac{8^{x} + 27^{x}}{12^{x} + 18^{x}} = \frac{7}{6} .$ Then $x =$	2)	1
R)	$\sqrt[3]{45 + 29 \sqrt{2}} + \sqrt[3]{45 - 29 \sqrt{2}} =$	3)	0
S)	If $x^{5} - 5 q x + 4 r$ is divisible by ${(x - 2)}^{2} .$ Then the value of $q$	4)	16
		5)	An irrational number