Match the columns:
$\begin{matrix} Column I & Column II (a) & X e F_{6} & (p) & Xe(excess) + F_{2} (b) & X e F_{4} & (q) & X e + F_{2} (1 : 4) (c) & X e F_{2} & (r) & X e + F_{2} (1 : 5) (s) & X e + F_{2} (1 : 20) \end{matrix}$

a - s

b - r

c - p

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a - r

b - s

c - p

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a - p

b - r

c - q

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a - s

b - r

c - q

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Solution

The correct option is A $a - s$
$b - r$
$c - p$
$X e (E x c e s s) + F_{2} \to X e F_{2}$

$X e 1 + : 2 F_{2} 5 \to X e F_{4}$

$X e 1 + : 3 F_{2} 20 \to X e F_{6}$

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Similar questions

\begin{matrix} Column I & Column II (a) & X e F_{6} & (p) & Xe(excess) + F_{2} (b) & X e F_{4} & (q) & X e + F_{2} (1 : 4) (c) & X e F_{2} & (r) & X e + F_{2} (1 : 5) (s) & X e + F_{2} (1 : 20) \end{matrix}

\begin{matrix} Column I & Column II (a) & X e F_{6} & (p) & Xe(excess) + F_{2} (b) & X e F_{4} & (q) & X e + F_{2} (1 : 4) (c) & X e F_{2} & (r) & X e + F_{2} (1 : 5) (s) & X e + F_{2} (1 : 20) \end{matrix}

f (n) = α^{n} + β^{n}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣ ∣ ∣ ∣ = k (1 - α)^{2} (1 - β)^{2} (α - β)^{2}

k

= α^{n} + β^{n} a n d ∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 + f (1) & 1 + f (2) 1 + f (1) & 1 + f (2) & 1 + f (3) 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} ∣ ∣ ∣ ∣ ∣ = k (1 - α)^{2} (1 - β)^{2} (α - β)^{2},

f (x) = \int_{0}^{x} \frac{d t}{2 + t^{4}}

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