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Q.
Given that a function $ f(x) = [x]^2 + x^2 $ , where $ [x] $ is the greatest integer less than or equal to $ x $ . If $ f(x) > 25 $ , then the value of $ x $ is
J & K CETJ & K CET 2018
Solution:
We have, $f(x) = [x]^2 + x^2$
Given, $f(x) > 25 $
$\therefore [x]^2 + x^2 > 25$
$\Rightarrow (x - \{x\})^2 + x^2 > 25$
$[\because x = [x] + \{x\}$ where $ 0 \le \{x\} <1 ]$
$\Rightarrow x^2 + \{x\}^2 +x^2 - 2x\{x\} > 25$
$\Rightarrow 2x^2 - 2x\{x\} + \{x\}^2 >25$
$\Rightarrow 2x(x - \{x\}) + \{x\}^2 >25$
$\Rightarrow 2x[x] +\{x\}^2 > 25$
$\Rightarrow 2x[x] \ge 24\,\,\,[\because 0 \le \{x\}^2 < 1]$
$\Rightarrow x[x] \ge 12$, which is possible for $x \ge 4$ or $x \le -4$
$\therefore $ Value of $x$ is the member of the set
$\{x | x \le -4$ or $ x \ge 4\}$