JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 15)
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
2
Explanation
$$af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x}$$ ........(i)
Replace $$x $$ with $$ {1 \over x}$$
$$af\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x$$ ..... (ii)
(i) + (ii)
$$(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x + {1 \over x}} \right)(b + \beta )$$
$${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} = {{\beta + b} \over {a + \alpha }} = {2 \over 1} = 2$$
Replace $$x $$ with $$ {1 \over x}$$
$$af\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x$$ ..... (ii)
(i) + (ii)
$$(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x + {1 \over x}} \right)(b + \beta )$$
$${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} = {{\beta + b} \over {a + \alpha }} = {2 \over 1} = 2$$
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