Let f (x) = x^ 2 x+1 , x -1 . The value of α for which f (a ) = a, (a ≠ 0) is
Mathematics · JEE Main · NTA Exams — Sets, Relations and Functions
Let f (x) = \(\frac{\alpha x^{2}}{x+1}, x \neq-1\). The value of α for which
f (a ) = a, (a ≠ 0) is
f (a ) = a, (a ≠ 0) is
- \(1-\frac{1}{a}\)
- \(\frac{1}{\mathrm{a}}\)
- \(1+\frac{1}{\mathrm{a}}\)
- \(\frac{1}{\mathrm{a}}-1\)
Answer
(C) 1+ 1 a
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