JEE Mains · Maths · STD 12 - 1. relation and function
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 \(5\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(\operatorname{af}(x)+\alpha f\left(\frac{1}{x}\right)=b x+\frac{\beta}{x}.....(1)\) replace \(x\) by \(\frac{1}{x}\) af \(\left(\frac{1}{x}\right)+\alpha f(x)=\frac{b}{x}+\beta x.....(2)\) \((1)+(2)\)…
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