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

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(D) The given inequality is $kx^2 - 2x + k \geq 0$.Case $1$: If $k = 0$,the inequality becomes $-2x \geq 0$,which implies $x \leq 0$. This holds for a

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$[-1, \infty)$

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(D) The given inequality is $kx^2 - 2x + k \geq 0$.Case $1$: If $k = 0$,the inequality becomes $-2x \geq 0$,which implies $x \leq 0$. This holds for at least one real $x$,so $k = 0$ is a solution.Case $2$: If $k 
eq 0$,the inequality $kx^2 - 2x + k \geq 0$ holds for at least one real $x$ if the maximum value of the expression $f(x) = kx^2 - 2x + k$ is $\geq 0$ (if $k  0$).Alternatively,we can write $k(x^2 + 1) \geq 2x$,which implies $k \geq \frac{2x}{x^2 + 1}$.Let $f(x) = \frac{2x}{x^2 + 1}$. The range of $f(x)$ is $[-1, 1]$.For the inequality $k \geq f(x)$ to hold for at least one $x$,$k$ must be greater than or equal to the minimum value of $f(x)$.Thus,$k \geq -1$ and $k 
eq 0$ (from Case $1$,$k=0$ is included).Therefore,the set of values is $k \in [-1, \infty)$.

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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