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1.Relation and Function
medium
જો $f:[1,\; + \infty ) \to [2,\; + \infty )$ માટે વિધેય $f(x) = x + \frac{1}{x}$ આપેલ હોય તો ${f^{ - 1}}$ મેળવો.
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) Here $y = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$
Now, ${x^2} – yx + 1 = 0$ or $x = \frac{{y \pm \sqrt {{y^2} – 4} }}{2}$
$\therefore$ ${f^{ – 1}}(x) = \frac{{x \pm \sqrt {{x^2} – 4} }}{2} \ge 1$
Since ${f^{ – 1}}[2,\,\infty ) \to [1,\,\infty )$ take only positive sign
$\therefore$ ${f^{ – 1}}(x) = \frac{{x + \sqrt {{x^2} – 4} }}{2}$.
Standard 12
Mathematics
