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The smallest value of $k$, for which both the roots of the equation $x^2-8 k x+16\left(k^2-k+1\right)=0$ are real, distinct and have values atleast 4 , is
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The correct answer is:
2
(i) Given, $x^2-8 k x+16\left(k^2-k+1\right)=0$
Now, $\quad D=64\left\{k^2-\left(k^2-k+1\right)\right\}$
$$
\begin{aligned}
& =64(k-1)>0 \\
& \therefore \quad k>1 \\
& 16-32 k+16\left(k^2-k+1\right) \geq 0 \\
& \Rightarrow \quad k^2-3 k+2 \geq 0 \\
& \Rightarrow(k-2)(k-1) \geq 0 \Rightarrow k \leq 1 \text { or } k \geq 2 \\
&
\end{aligned}
$$
Hence, $k=2$
Now, $\quad D=64\left\{k^2-\left(k^2-k+1\right)\right\}$
$$
\begin{aligned}
& =64(k-1)>0 \\
& \therefore \quad k>1 \\
& 16-32 k+16\left(k^2-k+1\right) \geq 0 \\
& \Rightarrow \quad k^2-3 k+2 \geq 0 \\
& \Rightarrow(k-2)(k-1) \geq 0 \Rightarrow k \leq 1 \text { or } k \geq 2 \\
&
\end{aligned}
$$
Hence, $k=2$
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