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1.Relation and Function
hard
If $a+\alpha=1, b+\beta=2$ and $\operatorname{af}(x)+\alpha f\left(\frac{1}{x}\right)=b x+\frac{\beta}{x}, x \neq 0,$ then the value of expression $\frac{ f ( x )+ f \left(\frac{1}{ x }\right)}{ x +\frac{1}{ x }}$ is ..... .
A
$2$
B
$1$
C
$4$
D
$5$
(JEE MAIN-2021)
Solution
$\operatorname{af}(x)+\alpha f\left(\frac{1}{x}\right)=b x+\frac{\beta}{x}…..(1)$
replace $x$ by $\frac{1}{x}$
af $\left(\frac{1}{x}\right)+\alpha f(x)=\frac{b}{x}+\beta x…..(2)$
$(1)+(2)$
$(a+\alpha) f(x)+(a+\alpha) f\left(\frac{1}{x}\right)=x(b+\beta)+(b+\beta) \frac{1}{x}$
$\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{b+\beta}{a+\alpha}=\frac{2}{1}=2$
Standard 12
Mathematics
