C Given, \(f(x)=\frac{x}{x-1}\) \(x . f(x)-f(x)=x\) \(x f(x)-x=f(x)\) \(x[f(x)-1]=f(x)\) \(x=\frac{f(x)}{f(x)-1}\) Then, \(f(3 x)=\frac{3 x}{3 x-1}\) Put the value of \(x\) by equation (i) in equation (ii), we get \(f(3 x)=\frac{3 \times \frac{f(x)}{f(x)-1}}{3 \times \frac{f(x)}{f(x)-1}-1}\) \(f(3 x)=\frac{\frac{3 f(x)}{f(x)-1}}{\frac{3 f(x)-\{f(x)-1\}}{f(x)-1}}\) \(f(3 x)=\frac{3 f(x)}{3 f(x)-f(x)+1}\) \(f(3 x)=\frac{3 f(x)}{2 f(x)+1}\)
SRMJEEE-2015
Sets, Relation and Function
116852
for \(f(x)=[x]\), where \([x]\) is the greatest integer function, which of the following is true, for every \(x \in R\).
1 \([\mathrm{x}]+1=\mathrm{x}\)
2 \([x]+1>x\)
3 \([x]+1 \leq x\)
4 \([\mathrm{x}]+1\lt \mathrm{x}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=[\mathrm{x}]\) We know that, the greatest integer function is also known as the step function. Greatest integer function is a function that gives the greatest integer less than or equal to a given number. It means, \([\mathrm{x}]=\mathrm{n}\), where, \(\mathrm{n} \leq \mathrm{x}\lt \mathrm{n}+1\) and ' \(\mathrm{n}\) ' is an integer. Ex. [5.2] \(=5\) as, \(5 \leq 5.2\lt 6\) and \([-5.3]=-6\), as \(-6 \leq-5.3\lt -5\) Since, \(x=[x]+\{x\} \Rightarrow\{x\}=x-[x]\) Where \([\mathrm{x}]=\) Greatest integer function \(\{\mathrm{x}\}=\) Fractional part Then, \(0 \leq\{\mathrm{x}\}\lt 1\) So, \(\quad 0 \leq \mathrm{x}-[\mathrm{x}]\lt 1\) \(0+[\mathrm{x}] \leq \mathrm{x}\lt 1+[\mathrm{x}]\) \({[\mathrm{x}] \leq \mathrm{x}\lt [\mathrm{x}]+1}\) Hence, \([\mathrm{x}]+1>\mathrm{x}\)
C Given, \(f(x)=\frac{x}{x-1}\) \(x . f(x)-f(x)=x\) \(x f(x)-x=f(x)\) \(x[f(x)-1]=f(x)\) \(x=\frac{f(x)}{f(x)-1}\) Then, \(f(3 x)=\frac{3 x}{3 x-1}\) Put the value of \(x\) by equation (i) in equation (ii), we get \(f(3 x)=\frac{3 \times \frac{f(x)}{f(x)-1}}{3 \times \frac{f(x)}{f(x)-1}-1}\) \(f(3 x)=\frac{\frac{3 f(x)}{f(x)-1}}{\frac{3 f(x)-\{f(x)-1\}}{f(x)-1}}\) \(f(3 x)=\frac{3 f(x)}{3 f(x)-f(x)+1}\) \(f(3 x)=\frac{3 f(x)}{2 f(x)+1}\)
SRMJEEE-2015
Sets, Relation and Function
116852
for \(f(x)=[x]\), where \([x]\) is the greatest integer function, which of the following is true, for every \(x \in R\).
1 \([\mathrm{x}]+1=\mathrm{x}\)
2 \([x]+1>x\)
3 \([x]+1 \leq x\)
4 \([\mathrm{x}]+1\lt \mathrm{x}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=[\mathrm{x}]\) We know that, the greatest integer function is also known as the step function. Greatest integer function is a function that gives the greatest integer less than or equal to a given number. It means, \([\mathrm{x}]=\mathrm{n}\), where, \(\mathrm{n} \leq \mathrm{x}\lt \mathrm{n}+1\) and ' \(\mathrm{n}\) ' is an integer. Ex. [5.2] \(=5\) as, \(5 \leq 5.2\lt 6\) and \([-5.3]=-6\), as \(-6 \leq-5.3\lt -5\) Since, \(x=[x]+\{x\} \Rightarrow\{x\}=x-[x]\) Where \([\mathrm{x}]=\) Greatest integer function \(\{\mathrm{x}\}=\) Fractional part Then, \(0 \leq\{\mathrm{x}\}\lt 1\) So, \(\quad 0 \leq \mathrm{x}-[\mathrm{x}]\lt 1\) \(0+[\mathrm{x}] \leq \mathrm{x}\lt 1+[\mathrm{x}]\) \({[\mathrm{x}] \leq \mathrm{x}\lt [\mathrm{x}]+1}\) Hence, \([\mathrm{x}]+1>\mathrm{x}\)
C Given, \(f(x)=\frac{x}{x-1}\) \(x . f(x)-f(x)=x\) \(x f(x)-x=f(x)\) \(x[f(x)-1]=f(x)\) \(x=\frac{f(x)}{f(x)-1}\) Then, \(f(3 x)=\frac{3 x}{3 x-1}\) Put the value of \(x\) by equation (i) in equation (ii), we get \(f(3 x)=\frac{3 \times \frac{f(x)}{f(x)-1}}{3 \times \frac{f(x)}{f(x)-1}-1}\) \(f(3 x)=\frac{\frac{3 f(x)}{f(x)-1}}{\frac{3 f(x)-\{f(x)-1\}}{f(x)-1}}\) \(f(3 x)=\frac{3 f(x)}{3 f(x)-f(x)+1}\) \(f(3 x)=\frac{3 f(x)}{2 f(x)+1}\)
SRMJEEE-2015
Sets, Relation and Function
116852
for \(f(x)=[x]\), where \([x]\) is the greatest integer function, which of the following is true, for every \(x \in R\).
1 \([\mathrm{x}]+1=\mathrm{x}\)
2 \([x]+1>x\)
3 \([x]+1 \leq x\)
4 \([\mathrm{x}]+1\lt \mathrm{x}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=[\mathrm{x}]\) We know that, the greatest integer function is also known as the step function. Greatest integer function is a function that gives the greatest integer less than or equal to a given number. It means, \([\mathrm{x}]=\mathrm{n}\), where, \(\mathrm{n} \leq \mathrm{x}\lt \mathrm{n}+1\) and ' \(\mathrm{n}\) ' is an integer. Ex. [5.2] \(=5\) as, \(5 \leq 5.2\lt 6\) and \([-5.3]=-6\), as \(-6 \leq-5.3\lt -5\) Since, \(x=[x]+\{x\} \Rightarrow\{x\}=x-[x]\) Where \([\mathrm{x}]=\) Greatest integer function \(\{\mathrm{x}\}=\) Fractional part Then, \(0 \leq\{\mathrm{x}\}\lt 1\) So, \(\quad 0 \leq \mathrm{x}-[\mathrm{x}]\lt 1\) \(0+[\mathrm{x}] \leq \mathrm{x}\lt 1+[\mathrm{x}]\) \({[\mathrm{x}] \leq \mathrm{x}\lt [\mathrm{x}]+1}\) Hence, \([\mathrm{x}]+1>\mathrm{x}\)
C Given, \(f(x)=\frac{x}{x-1}\) \(x . f(x)-f(x)=x\) \(x f(x)-x=f(x)\) \(x[f(x)-1]=f(x)\) \(x=\frac{f(x)}{f(x)-1}\) Then, \(f(3 x)=\frac{3 x}{3 x-1}\) Put the value of \(x\) by equation (i) in equation (ii), we get \(f(3 x)=\frac{3 \times \frac{f(x)}{f(x)-1}}{3 \times \frac{f(x)}{f(x)-1}-1}\) \(f(3 x)=\frac{\frac{3 f(x)}{f(x)-1}}{\frac{3 f(x)-\{f(x)-1\}}{f(x)-1}}\) \(f(3 x)=\frac{3 f(x)}{3 f(x)-f(x)+1}\) \(f(3 x)=\frac{3 f(x)}{2 f(x)+1}\)
SRMJEEE-2015
Sets, Relation and Function
116852
for \(f(x)=[x]\), where \([x]\) is the greatest integer function, which of the following is true, for every \(x \in R\).
1 \([\mathrm{x}]+1=\mathrm{x}\)
2 \([x]+1>x\)
3 \([x]+1 \leq x\)
4 \([\mathrm{x}]+1\lt \mathrm{x}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=[\mathrm{x}]\) We know that, the greatest integer function is also known as the step function. Greatest integer function is a function that gives the greatest integer less than or equal to a given number. It means, \([\mathrm{x}]=\mathrm{n}\), where, \(\mathrm{n} \leq \mathrm{x}\lt \mathrm{n}+1\) and ' \(\mathrm{n}\) ' is an integer. Ex. [5.2] \(=5\) as, \(5 \leq 5.2\lt 6\) and \([-5.3]=-6\), as \(-6 \leq-5.3\lt -5\) Since, \(x=[x]+\{x\} \Rightarrow\{x\}=x-[x]\) Where \([\mathrm{x}]=\) Greatest integer function \(\{\mathrm{x}\}=\) Fractional part Then, \(0 \leq\{\mathrm{x}\}\lt 1\) So, \(\quad 0 \leq \mathrm{x}-[\mathrm{x}]\lt 1\) \(0+[\mathrm{x}] \leq \mathrm{x}\lt 1+[\mathrm{x}]\) \({[\mathrm{x}] \leq \mathrm{x}\lt [\mathrm{x}]+1}\) Hence, \([\mathrm{x}]+1>\mathrm{x}\)
