Find the possible values $k$ can take for the given inequality.
$\frac{2 (4 k + 1)}{3} \geq \frac{k (6 + 5) - 3}{2}$

k \leq - 1

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k \geq - 1

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k \leq \frac{13}{17}

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k \geq \frac{13}{17}

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Solution

The correct option is C $k \leq \frac{13}{17}$
Given, $\frac{2 (4 k + 1)}{3} \geq \frac{k (6 + 5) - 3}{2}$
$\Rightarrow \frac{8 k + 2}{3} \geq \frac{11 k - 3}{2}$
$\Rightarrow \frac{8 k}{3} + \frac{2}{3} \geq \frac{11 k}{2} - \frac{3}{2}$
Add $- \frac{11 k}{2}$ and $- \frac{2}{3}$ both sides, we get
$\frac{8 k}{3} + \frac{2}{3} - \frac{11 k}{2} - \frac{2}{3} \geq \frac{11 k}{2} - \frac{3}{2} - \frac{11 k}{2} - \frac{2}{3}$
$\Rightarrow \frac{8 k}{3} - \frac{11 k}{2} \geq - \frac{3}{2} - \frac{2}{3}$
$\Rightarrow \frac{16 k - 33 k}{6} \geq \frac{- 9 - 4}{6}$
$\Rightarrow - \frac{17 k}{6} \geq - \frac{13}{6}$
$\Rightarrow k \leq \frac{13}{17}$

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k

k_{1}

k_{2}

\to V

\to V = k_{1} \to A + k_{2} \to B

\to V = 2 i - j

\to A = 3 i - 2 j

\to B = 3 i + 3 j .

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