Match List-I with List-II and select the correct answer using the code given below the list. List-I List-II (I)If limn→∞(n12(1+1n)(11223344…nn)1n2)=e−(pq)then the value of p/q is greater than(P)5(II)If (7!)! is divisible by (7!)k!⋅(6!) then the maximum value of k is(Q)6(III)The value of limx→0+ x(xn−1) is less than or equal to(R)15(IV)If α,β,γ are the roots of the equation x3−2x2−1=0 and Tn=αn+βn+γn then the value of T11−T8T10 is greater than(S)16 (T)14The correct option is:

(P) limn→∞(n−12(1+1n)(11⋅22⋅33⋅44…nn)1n2)=epqL.H.S. limn→∞(11⋅22⋅33⋅44…nnn+(n22+n))1n2 =elimx→∞1n2ln1n22n235n5⋯n2n2=elimx→n21n r=−1nlnrxnx =elimx→xn1n∑ (xn)ln(xn)=e∫01x⋅lnx⋅dx=e(x22lnxx24)|01 =e−14(Q) 7! objects are divided in 6! groups consisting of 7 objects each.Number of groups =(7!)!(7!)6!(6!)!∈NMaximum value of k=6 (R) limx→0+x(xx−1)=elimx→0+(xx−1)/nx=elimx→0+(xx−1)Apply L. H. Rule=elimx→0+−((nx x)−2⋅(1x)=elimx→0+−xx[x(lnx)2+x(lnx)3]=e0=1As limx→0+x⋅(lnx)2=limx→0−(lnx)21/x=limx→0+2lnx⋅1x(−1x2) =limx→0+−2x(lnx)=limx→0+−2lnx(1x)=limx→0+−2(1x)−(1x2)=limx→0+2x=0Similarly, limx→0+x(lnx)3=0α3−2α2−1=0 ⇒αn+3−2αn+2−αn=0⇒∑αn+3−αn=2∑αn+2

Match List-I with List-II and select the correct answer using the code given below the list. List-I List-II (I)If limn→∞(n12(1+1n)(11223344…nn)1n2)=e−(pq)then the value of p/q is greater than(P)5(II)If (7!)! is divisible by (7!)k!⋅(6!) then the maximum value of k is(Q)6(III)The value of limx→0+ x(xn−1) is less than or equal to(R)15(IV)If α,β,γ are the roots of the equation x3−2x2−1=0 and Tn=αn+βn+γn then the value of T11−T8T10 is greater than(S)16 (T)14The correct option is:

(P) limn→∞(n−12(1+1n)(11⋅22⋅33⋅44…nn)1n2)=epqL.H.S. limn→∞(11⋅22⋅33⋅44…nnn+(n22+n))1n2 =elimx→∞1n2ln1n22n235n5⋯n2n2=elimx→n21n r=−1nlnrxnx =elimx→xn1n∑ (xn)ln(xn)=e∫01x⋅lnx⋅dx=e(x22lnxx24)|01 =e−14(Q) 7! objects are divided in 6! groups consisting of 7 objects each.Number of groups =(7!)!(7!)6!(6!)!∈NMaximum value of k=6 (R) limx→0+x(xx−1)=elimx→0+(xx−1)/nx=elimx→0+(xx−1)Apply L. H. Rule=elimx→0+−((nx x)−2⋅(1x)=elimx→0+−xx[x(lnx)2+x(lnx)3]=e0=1As limx→0+x⋅(lnx)2=limx→0−(lnx)21/x=limx→0+2lnx⋅1x(−1x2) =limx→0+−2x(lnx)=limx→0+−2lnx(1x)=limx→0+−2(1x)−(1x2)=limx→0+2x=0Similarly, limx→0+x(lnx)3=0α3−2α2−1=0 ⇒αn+3−2αn+2−αn=0⇒∑αn+3−αn=2∑αn+2

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Match List-I with List-II and select the correct answer using the code given below the list.
List-I List-II
(I) If $\lim_{n \to \infty} (n^{\frac{1}{2} (1 + \frac{1}{n})} {(1^{1} 2^{2} 3^{3} 4^{4} \dots n^{n})}^{\frac{1}{n^{2}}}) = e^{- (\frac{p}{q})}$ then the value of $p / q$ is greater than (P) 5
(II) If $(7!)!$ is divisible by ${(7!)}^{k!} \cdot (6!)$ then the maximum value of k is (Q) 6
(III) The value of $\lim_{x \to 0} + x^{(x^{n} - 1)}$ is less than or equal to (R) $\frac{1}{5}$
(IV) If $α, β, γ$ are the roots of the equation $x^{3} - 2 x^{2} - 1 = 0$ and $T_{n} = α^{n} + β^{n} + γ^{n}$ then the value of $\frac{T_{11} - T_{8}}{T_{10}}$ is greater than (S) $\frac{1}{6}$
(T) $\frac{1}{4}$
The correct option is:

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(I)-(R, S) (II)-(Q, S) (III)-(P) (IV)-(S, T)

(I)-(P, S) (II)-(Q) (III)-(P, Q) (IV)-(R, S, T)

(I)-(R) (II)-(Q, S) (III)-(S) (IV)-(P, T)

(I)-(R, S) (II)-(Q) (III)-(P, Q) (IV)-(R, S, T)

answer is D.

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Detailed Solution

(P) $\lim_{n \to \infty} (n^{- \frac{1}{2} (1 + \frac{1}{n})} {(1^{1} \cdot 2^{2} \cdot 3^{3} \cdot 4^{4} \dots n^{n})}^{\frac{1}{n^{2}}}) = e^{\frac{p}{q}}$
L.H.S. $\lim_{n \to \infty} {(\frac{1^{1} \cdot 2^{2} \cdot 3^{3} \cdot 4^{4} \dots n^{n}}{n^{+ (\frac{n^{2}}{2} + n})})}^{\frac{1}{n^{2}}}$
$= e^{lim_{x \to \infty} \frac{1}{n^{2}} [ln (\frac{1}{n} \frac{2^{2}}{n^{2}} \frac{3^{5}}{n^{5}} \dots \frac{n^{2}}{n^{2}})]} = e^{\underset{x \to n^{2}}{l i m} \frac{1}{n}_{r = - 1}^{n} l n (\frac{r^{x}}{n^{x}})}$
$= e^{\underset{x \to x_{n}}{l i m} \frac{1}{n} \sum (\frac{x}{n}) l n (\frac{x}{n})} = e^{\int_{0}^{1} x \cdot l n x \cdot d x} = {e^{(\frac{x^{2}}{2} l n x \frac{x^{2}}{4})} |}_{0}^{1} = e^{- \frac{1}{4}}$
(Q) 7! objects are divided in 6! groups consisting of 7 objects each.
Number of groups $= \frac{(7!)!}{{(7!)}^{6!} (6!)!} \in N$
Maximum value of $k = 6$
Question Image
(R) ${lim}_{x \to 0^{+}} x^{(x^{x} - 1)} = e^{{lim}_{x \to 0^{+}} (x^{x} - 1) / n x} = e^{{lim}_{x \to 0^{+}} (x^{x} - 1)}$
Apply L. H. Rule
$= e^{lim_{x \to 0^{+}} - ({(nx x)}^{- 2} \cdot (\frac{1}{x})} = e^{lim_{x \to 0^{+}} - x^{x} [x {(l n x)}^{2} + x {(l n x)}^{3}]} = e^{0} = 1$

As ${lim}_{x \to 0^{+}} x \cdot {(l n x)}^{2} = {lim}_{x \to 0^{-}} \frac{{(l n x)}^{2}}{1 / x} = {lim}_{x \to 0^{+}} \frac{2 l n x \cdot \frac{1}{x}}{(\frac{- 1}{x^{2}})} = lim_{x \to 0^{+}} - 2 x (ln x) = lim_{x \to 0^{+}} \frac{- 2 ln x}{(\frac{1}{x})} = lim_{x \to 0^{+}} \frac{- 2 (\frac{1}{x})}{- (\frac{1}{x^{2}})} = lim_{x \to 0^{+}} 2 x = 0$

Similarly, ${lim}_{x \to 0^{+}} x {(l n x)}^{3} = 0$

$α^{3} - 2 α^{2} - 1 = 0 \Rightarrow α^{n + 3} - 2 α^{n + 2} - α^{n} = 0 \Rightarrow \sum α^{n + 3} - α^{n} = 2 \sum α^{n + 2}$

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