117372
For what natural numbers \(n \in N\), the inequality \(2^n>n+1\) is valid ?
1 \(\forall \mathrm{n} \in \mathrm{N}\)
2 \(\forall \mathrm{n} \geq 2\)
3 \(\forall 1 \leq \mathrm{n} \leq 3\)
4 \(\forall \mathrm{n} \in \mathrm{N}-\{2,3\}\)
Explanation:
B Given, \(\mathrm{n} \in \mathrm{N}\) \(2^{\mathrm{n}}>\mathrm{n}+1\) \(\mathrm{n}=1\) \(2^1>1+1\) \(2>2 \quad \text { It is not possible }\) \(\mathrm{n}=2\) \(2^2>2+1\) \(4>3 \quad \text { It is possible }\)Hence, \(\mathrm{n} \geq 2\).
Shift-II
Sets, Relation and Function
117374
If \(f(x)=x\left(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right), x>1\). Then,
117412
The domain of the function \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) is
1 \((-8,1) \cup\left(\frac{3}{5}, 2\right)\)
2 \(\left(1, \frac{5}{3}\right)\)
3 \([-8,1] \cup\left[\frac{5}{3}, 2\right]\)
4 \((-8,1] \cup\left[\frac{5}{3}, 2\right)\)
Explanation:
D Given that , \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) Case - (I) : \(\sec ^{-1}(3 x-4)\), it is defined as - \(3 x-4 \leq-1 \text { or } 3 x-4 \geq 1\) \(3 x \leq 3 \text { or } 3 x \geq 5\) \(x \leq 1 \text { or } x \geq \frac{5}{3}\) Case - (II) : \(\tanh ^{-1}\left(\frac{x+3}{5}\right)\), it is defined as - \(-1\lt \frac{x+3}{5}\lt 5\) \(-5\lt x+3\lt 5\) \(-8\lt x\lt 2\) From case - (I) and (II), we get - Domain of \(f(x), x \in(-8,1] \cup\left[\frac{5}{3}, 2\right)\).
117372
For what natural numbers \(n \in N\), the inequality \(2^n>n+1\) is valid ?
1 \(\forall \mathrm{n} \in \mathrm{N}\)
2 \(\forall \mathrm{n} \geq 2\)
3 \(\forall 1 \leq \mathrm{n} \leq 3\)
4 \(\forall \mathrm{n} \in \mathrm{N}-\{2,3\}\)
Explanation:
B Given, \(\mathrm{n} \in \mathrm{N}\) \(2^{\mathrm{n}}>\mathrm{n}+1\) \(\mathrm{n}=1\) \(2^1>1+1\) \(2>2 \quad \text { It is not possible }\) \(\mathrm{n}=2\) \(2^2>2+1\) \(4>3 \quad \text { It is possible }\)Hence, \(\mathrm{n} \geq 2\).
Shift-II
Sets, Relation and Function
117374
If \(f(x)=x\left(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right), x>1\). Then,
117412
The domain of the function \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) is
1 \((-8,1) \cup\left(\frac{3}{5}, 2\right)\)
2 \(\left(1, \frac{5}{3}\right)\)
3 \([-8,1] \cup\left[\frac{5}{3}, 2\right]\)
4 \((-8,1] \cup\left[\frac{5}{3}, 2\right)\)
Explanation:
D Given that , \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) Case - (I) : \(\sec ^{-1}(3 x-4)\), it is defined as - \(3 x-4 \leq-1 \text { or } 3 x-4 \geq 1\) \(3 x \leq 3 \text { or } 3 x \geq 5\) \(x \leq 1 \text { or } x \geq \frac{5}{3}\) Case - (II) : \(\tanh ^{-1}\left(\frac{x+3}{5}\right)\), it is defined as - \(-1\lt \frac{x+3}{5}\lt 5\) \(-5\lt x+3\lt 5\) \(-8\lt x\lt 2\) From case - (I) and (II), we get - Domain of \(f(x), x \in(-8,1] \cup\left[\frac{5}{3}, 2\right)\).
117372
For what natural numbers \(n \in N\), the inequality \(2^n>n+1\) is valid ?
1 \(\forall \mathrm{n} \in \mathrm{N}\)
2 \(\forall \mathrm{n} \geq 2\)
3 \(\forall 1 \leq \mathrm{n} \leq 3\)
4 \(\forall \mathrm{n} \in \mathrm{N}-\{2,3\}\)
Explanation:
B Given, \(\mathrm{n} \in \mathrm{N}\) \(2^{\mathrm{n}}>\mathrm{n}+1\) \(\mathrm{n}=1\) \(2^1>1+1\) \(2>2 \quad \text { It is not possible }\) \(\mathrm{n}=2\) \(2^2>2+1\) \(4>3 \quad \text { It is possible }\)Hence, \(\mathrm{n} \geq 2\).
Shift-II
Sets, Relation and Function
117374
If \(f(x)=x\left(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right), x>1\). Then,
117412
The domain of the function \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) is
1 \((-8,1) \cup\left(\frac{3}{5}, 2\right)\)
2 \(\left(1, \frac{5}{3}\right)\)
3 \([-8,1] \cup\left[\frac{5}{3}, 2\right]\)
4 \((-8,1] \cup\left[\frac{5}{3}, 2\right)\)
Explanation:
D Given that , \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) Case - (I) : \(\sec ^{-1}(3 x-4)\), it is defined as - \(3 x-4 \leq-1 \text { or } 3 x-4 \geq 1\) \(3 x \leq 3 \text { or } 3 x \geq 5\) \(x \leq 1 \text { or } x \geq \frac{5}{3}\) Case - (II) : \(\tanh ^{-1}\left(\frac{x+3}{5}\right)\), it is defined as - \(-1\lt \frac{x+3}{5}\lt 5\) \(-5\lt x+3\lt 5\) \(-8\lt x\lt 2\) From case - (I) and (II), we get - Domain of \(f(x), x \in(-8,1] \cup\left[\frac{5}{3}, 2\right)\).
117372
For what natural numbers \(n \in N\), the inequality \(2^n>n+1\) is valid ?
1 \(\forall \mathrm{n} \in \mathrm{N}\)
2 \(\forall \mathrm{n} \geq 2\)
3 \(\forall 1 \leq \mathrm{n} \leq 3\)
4 \(\forall \mathrm{n} \in \mathrm{N}-\{2,3\}\)
Explanation:
B Given, \(\mathrm{n} \in \mathrm{N}\) \(2^{\mathrm{n}}>\mathrm{n}+1\) \(\mathrm{n}=1\) \(2^1>1+1\) \(2>2 \quad \text { It is not possible }\) \(\mathrm{n}=2\) \(2^2>2+1\) \(4>3 \quad \text { It is possible }\)Hence, \(\mathrm{n} \geq 2\).
Shift-II
Sets, Relation and Function
117374
If \(f(x)=x\left(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right), x>1\). Then,
117412
The domain of the function \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) is
1 \((-8,1) \cup\left(\frac{3}{5}, 2\right)\)
2 \(\left(1, \frac{5}{3}\right)\)
3 \([-8,1] \cup\left[\frac{5}{3}, 2\right]\)
4 \((-8,1] \cup\left[\frac{5}{3}, 2\right)\)
Explanation:
D Given that , \(f(x)=\sec ^{-1}(3 x-4)+\tanh ^{-1}\left(\frac{x+3}{5}\right)\) Case - (I) : \(\sec ^{-1}(3 x-4)\), it is defined as - \(3 x-4 \leq-1 \text { or } 3 x-4 \geq 1\) \(3 x \leq 3 \text { or } 3 x \geq 5\) \(x \leq 1 \text { or } x \geq \frac{5}{3}\) Case - (II) : \(\tanh ^{-1}\left(\frac{x+3}{5}\right)\), it is defined as - \(-1\lt \frac{x+3}{5}\lt 5\) \(-5\lt x+3\lt 5\) \(-8\lt x\lt 2\) From case - (I) and (II), we get - Domain of \(f(x), x \in(-8,1] \cup\left[\frac{5}{3}, 2\right)\).
