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Let $f(\mathrm{a})=\frac{\mathrm{a}-1}{\mathrm{a}+1} . \quad[2017-\mathrm{I}]$
Consider the following :
1- $ f(2 \mathrm{a})=f(\mathrm{a})+1$
2- $f\left(\frac{1}{\mathrm{a}}\right)=-f(\mathrm{a})$
Which of the above is/are correct?
Options:
Consider the following :
1- $ f(2 \mathrm{a})=f(\mathrm{a})+1$
2- $f\left(\frac{1}{\mathrm{a}}\right)=-f(\mathrm{a})$
Which of the above is/are correct?
Solution:
2997 Upvotes
Verified Answer
The correct answer is:
2 only
$\mathrm{f}(\mathrm{a})=\frac{\mathrm{a}-1}{\mathrm{a}+1}$
$\mathrm{f}(2 \mathrm{a})=\frac{2 \mathrm{a}-1}{2 \mathrm{a}+1}$
$\mathrm{f}(\mathrm{a})+1=\frac{\mathrm{a}-1}{\mathrm{a}+1}+1=\frac{2 \mathrm{a}}{\mathrm{a}+1}$
So, $\mathrm{f}(2 \mathrm{a}) \neq \mathrm{f}(\mathrm{a})+1$
Now, $f\left(\frac{1}{a}\right)=\frac{\frac{1}{a}-1}{\frac{l}{a}+1}=\frac{1-a}{1+a}=-\left(\frac{a-1}{a+1}\right)=-f(a)$
$\mathrm{f}(2 \mathrm{a})=\frac{2 \mathrm{a}-1}{2 \mathrm{a}+1}$
$\mathrm{f}(\mathrm{a})+1=\frac{\mathrm{a}-1}{\mathrm{a}+1}+1=\frac{2 \mathrm{a}}{\mathrm{a}+1}$
So, $\mathrm{f}(2 \mathrm{a}) \neq \mathrm{f}(\mathrm{a})+1$
Now, $f\left(\frac{1}{a}\right)=\frac{\frac{1}{a}-1}{\frac{l}{a}+1}=\frac{1-a}{1+a}=-\left(\frac{a-1}{a+1}\right)=-f(a)$
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