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Question
If the function f(x) = x3 − 9kx2 + 27x + 30 is increasing on R, then
Options
−1 ≤ k < 1
k < −1 or k > 1
0 < k < 1
−1 < k < 0
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Solution
\[f\left( x \right) = x^3 - 9k x^2 + 27x + 30\]
\[f'\left( x \right) = 3 x^2 - 18kx + 27\]
\[ = 3 \left( x^2 - 6kx + 9 \right)\]
\[\text { Given: f(x) is increasing on R } . \]
\[ \Rightarrow f'\left( x \right) > 0 \text { for all } x \in R\]
\[ \Rightarrow 3 \left( x^2 - 6kx + 9 \right) > 0 \text { for all } x \in R\]
\[ \Rightarrow x^2 - 6kx + 9 > 0 \text { for all } x \in R\]
\[ \Rightarrow \left( - 6k \right)^2 - 4\left( 1 \right)\left( 9 \right) < 0 \left[ \because a x^2 + bx + c > \text { 0 for all }x \in R \Rightarrow a > \text{0 and Disc}< 0 \right]\]
\[ \Rightarrow 36 k^2 - 36 < 0\]
\[ \Rightarrow k^2 - 1 < 0\]
\[ \Rightarrow \left( k + 1 \right)\left( k - 1 \right) < 0\]
\[\text { It can be possible when } \left( k + 1 \right) < 0 \text { and } \left( k - 1 \right) > 0 . \]
\[ \Rightarrow k < - 1 \text { and } k > 1 (\text { Not possible })\]
\[or \left( k + 1 \right) > 0 \text { and } \left( k - 1 \right) < 0\]
\[ \Rightarrow k > - 1 \text { and } k < 1\]
\[ \Rightarrow - 1 < k < 1\]
\[\text { Disclaimer: (a) part should be } - 1 < k < 1 \text { instead of }-1 \leq k < 1 .\]
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