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1.Relation and Function
medium
यदि $f:[1,\; + \infty ) \to [2,\; + \infty )$, $f(x) = x + \frac{1}{x}$, तब ${f^{ - 1}}$=
A
$\frac{{x + \sqrt {{x^2} - 4} }}{2}$
B
$\frac{x}{{1 + {x^2}}}$
C
$\frac{{x - \sqrt {{x^2} - 4} }}{2}$
D
$1 + \sqrt {{x^2} - 4} $
(IIT-2001)
Solution
(a) यहाँ$y = x + \frac{1}{x}$ जबकि $x \ge 1$ तथा $y \ge 2$
अब ${x^2} – yx + 1 = 0$ या$x = \frac{{y \pm \sqrt {{y^2} – 4} }}{2}$
$\therefore$ ${f^{ – 1}}(x) = \frac{{x \pm \sqrt {{x^2} – 4} }}{2} \ge 1$ जब $x \ge 2$
लेकिन $x \ge 2$ के लिए $\frac{{x – \sqrt {{x^2} – 4} }}{2} \ge 1$
$\therefore$ ${f^{ – 1}}(x) = \frac{{x + \sqrt {{x^2} – 4} }}{2}$.
Standard 12
Mathematics
