If a + $\alpha$ = 1, b + $\beta$ = 2 and $af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x},x \ne 0$, then the value of the expression ${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}}$ is __________.
Answer (integer)
2
Solution
$af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x}$ ........(i)<br><br>Replace $x$ with ${1 \over x}$<br><br>$af\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x$ ..... (ii)<br><br>(i) + (ii)<br><br>$$(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x + {1 \over x}} \right)(b + \beta )$$<br><br>$${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} = {{\beta + b} \over {a + \alpha }} = {2 \over 1} = 2$$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
This question is part of PrepWiser's free JEE Main question bank. 195 more solved questions on Sets, Relations and Functions are available — start with the harder ones if your accuracy is >70%.