Mathematics Test - 2

\(\log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a+\log b)\)
\(\left.\log \left(\frac{a-b}{2}\right)=\frac{1}{2} \log (a \times b) \quad \ldots \log (a \cdot b)=\log a+\log b\right)\)
\(\therefore \frac{a-b}{2}=a b^{\frac{1}{2}} \quad \ldots .\) removing \(\log\)
Squaring both the sides we get
\(\therefore \frac{(a-b)^{2}}{4}=a b\)
\(\therefore a^{2}+b^{2}-2 a b=4 a b\)
\(\therefore a^{2}+b^{2}-6 a b=0\)

\begin{equation}\begin{array}{l}
m=\log 20 \\
n=\log 25 \\
2 \log (x-4)=2 \log 20-\log 25 \\
\log (x-4)^{2}=\log 20^{2}-\log 25 \\
=\log \left(\frac{20^{2}}{25}\right)=\log \frac{4 \times 5 \times 4 \times 5}{5 \times 5} \\
\log (x-4)^{2}=\log 4^{2} \\
x-4=4 \ldots \text { removing } \log \\
x=8
\end{array}\end{equation}

\(\log _{1-x}(x-2) \geq-1\)
Above equation is valid when, \(S_{1}=\{x-2>0\}\)
\(S_{2}=\{1-x>0,1-x \neq 1\}\)
Intersection of \(S_{1}\) and \(S_{2}\) is \(x=\{\emptyset\}\).

\(a^{2}=\log x, b^{3}=\log y\)
\(3 a^{2}-2 b^{3}=6 \log z\)
\(x=10^{a^{2}}, y=10^{b^{3}}, z=10^{\frac{1}{6}\left(3 a^{2}-2 b^{3}\right)}\)....(i)
\(\therefore z=\frac{10^{a^{2} / 2}}{10^{b^{3} / 3}}\)....(ii)
From 1 and 2 we get
\(\therefore z=\frac{x^{\frac{1}{2}}}{y^{\frac{1}{3}}}\) and \(y=\frac{x^{\frac{3}{2}}}{z^{3}}\)

$\text{Let }y=\{x:{\log }_{\frac{1}{3}} {\log }_4(x^2-5)>0\}$
$\Rightarrow y=\{x:0<{\log }_4(x^2-5)<1\}$
$\Rightarrow y=\{x:4^0<x^2-5<4^1\}\dots ..[\because a^{{\log }_{a}b}=b]$
$\Rightarrow y=\{x:1<x^2-5<4\}$
$\Rightarrow y=\{x:6<x^2<9\}$
$\Rightarrow y=(-3,-\sqrt{6})\cup (\sqrt{6},3)$

Hence option C is correct.

\(\frac{n^{2}}{(2 n-1)(2 n+1)}\)
\(=\frac{4 n^{2}}{4(2 n-1)(2 n+1)}\)
\(=\frac{1}{4}\left[\frac{4 n^{2}-1+1}{4 n^{2}-1}\right]\)
\(=\frac{1}{4}\left[1+\frac{1}{4 n^{2}-1}\right]\)
\(=\frac{1}{4}\left[1+\frac{1}{2}\left(\frac{2 n+1-(2 n-1)}{4 n^{2}-1}\right)\right]\)
\(=\frac{1}{4}\left[1+\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)\right]\)
Consider \(\frac{1}{2 n-1}-\frac{1}{2 n+1}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5} \ldots \frac{1}{2 n-1}-\frac{1}{2 n+1}\)
\(=1-\frac{1}{2 n+1}\)
\(=\frac{2 n}{2 n+1}\)
Hence applying summation gives us
\(\frac{n}{4}+\frac{1}{8} \sum\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)\)
\(=\frac{n}{4}+\frac{2 n}{8(2 n+1)}\)
\(=\frac{n}{4}+\frac{n}{4(2 n+1)}\)
\(=\frac{n(2 n+1)+n}{4(2 n+1)}\)
\(=\frac{n(2 n+2)}{4(2 n+1)}\)
\(=\frac{n(n+1)}{2(2 n+1)}\)

\(P(n)=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots \ldots+\frac{1}{\sqrt{n}}\)....(i)
Put \(n=2\) in (1)
\(P(2)=1+\frac{1}{\sqrt{2}}>\sqrt{2}\)
Put \(n=3\) in ( 1\()\)
\(P(3)=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}\)
So, let us assume
\(P(n)=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots \ldots+\frac{1}{\sqrt{n}}>\sqrt{n}\)
So, we will check for \(P(n+1)\)
\(P(n+1)=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots \ldots+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\)
\(P(n+1)>\sqrt{n}+\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n(n+1)}+1}{\sqrt{n+1}} \ldots(2)\)
since, \(\sqrt{n(n+1)}>n\)
\(\Rightarrow \sqrt{n(n+1)}+1>n+1\)
\(\Rightarrow \frac{\sqrt{n(n+1)}+1}{\sqrt{n+1}}>\frac{n+1}{\sqrt{n+1}}\)
\(\Rightarrow \frac{\sqrt{n(n+1)}+1}{\sqrt{n+1}}>\sqrt{n+1}\)
From
(2) and (3), it follows that \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots \ldots+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1}\)
Hence, by mathematical induction method, we can say that \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots \ldots+\frac{1}{\sqrt{n}}>\sqrt{n} \quad\) for \(n \in N, n \geq 2\)

The converse of \(p \rightarrow(q \rightarrow r)\) is, \(\equiv(q \rightarrow r) \rightarrow p \equiv(\sim q \vee r) \rightarrow p \equiv(q \wedge \sim r) \rightarrow p\)

Let's take option \(A\), \((\sim p \vee q) \equiv(p \wedge q)\)
\(L H S=(\sim p \vee q)=p \rightarrow q\)
\(R H S=(p \wedge q)\)
Clearly, \(L H S \neq R H S\)
Hence, option \(A\) is incorrect.
Option \(B ,(p \rightarrow q) \equiv(\sim q \rightarrow \sim p)\)
\(R H S=(\sim q \rightarrow \sim p)\)
\(=\sim(\sim q) \vee \sim p\)
\(=q \vee \sim p\)
\(=\sim p \vee q\)
\(=p \rightarrow q\)
\(=L H S\)
Hence, option \(B\) is correct
Option \(C\) \(\sim(p \rightarrow \sim q) \equiv(p \wedge \sim q)\)
\(L H S=\sim(p \rightarrow \sim q)\)
\(=\sim[(\sim p) \vee(\sim q)]\)
\(=\sim[\sim(p \wedge q)]\)
\(=p \wedge q\)
\(\neq R H S\)
Hence, option \(C\) is incorrect.
Option \(D , \sim(p \leftrightarrow q) \equiv(p \rightarrow q) \vee(q \rightarrow p)\)
\(R H S=(p \rightarrow q) \vee(q \rightarrow p)\)
\(=(\sim p \vee q) \vee(\sim q \vee p)\)
\(L H S=\sim(p \leftrightarrow q)\)
\(=\sim[(p \rightarrow q) \wedge(q \rightarrow p)]\)
\(=\sim[(\sim p \vee q) \wedge(\sim q \vee p)]\)
\(=[\sim(\sim p \vee q)] \vee[\sim(\sim q \vee p)]\)
\(=[p \wedge(\sim q)] \vee[q \wedge(\sim p)]\)
Hence, \(L H S \neq R H S\)
Hence, option \(D\) is incorrect.

When 12 divided by a positive integer \(n,\) the remainder is \((n-3)\) Thus, \(12=n k+(n-3),\) where is a \(k\) positive integer
\(\Rightarrow 12=n k+n-3\)
\(\Rightarrow 15=n(k+1)\)
The divisors of 15 are 1,3,5,15 Thus, the possible values of \((n, k+1)\) are (1,15),(15,1),(3,5),(5,3) Now, (1,15),(15,1),(3,5) does not satisfy our initial equation \(12=n k+(n-3)\)
Therefore \(n=5\) is the possible value.