If the inequality kx^2 - 2x + k geq 0 holds g | vedclass

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(D) The given inequality is $kx^2 - 2x + k \geq 0$.If $k = 0$,the inequality becomes $-2x \geq 0$,which implies $x \leq 0$. Since this holds for at le

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$(-1, \infty)$

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(D) The given inequality is $kx^2 - 2x + k \geq 0$.If $k = 0$,the inequality becomes $-2x \geq 0$,which implies $x \leq 0$. Since this holds for at least one real $x$,$k = 0$ is a solution.If $k 
eq 0$,we can rewrite the inequality as $k(x^2 + 1) \geq 2x$,which gives $k \geq \frac{2x}{x^2 + 1}$.Let $f(x) = \frac{2x}{x^2 + 1}$. To find the range of $f(x)$,we analyze its derivative $f'(x) = \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}$.Setting $f'(x) = 0$ gives $x^2 = 1$,so $x = 1$ or $x = -1$.The maximum value is $f(1) = \frac{2}{2} = 1$ and the minimum value is $f(-1) = \frac{-2}{2} = -1$.Thus,the range of $f(x)$ is $[-1, 1]$.For $k \geq f(x)$ to hold for at least one $x$,$k$ must be greater than or equal to the minimum value of $f(x)$.Therefore,$k \geq -1$. Combining this with the case $k=0$ and the condition for the quadratic to exist,the set of values is $k \in [-1, \infty) \setminus \{0\}$ is incorrect; rather,checking the boundary,the condition $k \geq -1$ covers all valid $k$ values.

If the inequality $kx^2 - 2x + k \geq 0$ holds good for at least one real $x$,then the complete set of values of $k$ is

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