All the values of k such that the quadratic e | vedclass

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Question

(A) The given quadratic expression is $f(x) = 2kx^2 - (4k+1)x + 2$.
We can factorize this as $f(x) = (2x - 1)(kx - 2)$.
The roots of $f(x) = 0$ are

Vedclass · Accepted Answer

$[-\frac{2}{3}, -\frac{1}{2}]$

Vedclass · Answer

(A) The given quadratic expression is $f(x) = 2kx^2 - (4k+1)x + 2$. We can factorize this as $f(x) = (2x - 1)(kx - 2)$. The roots of $f(x) = 0$ are $x = \frac{1}{2}$ and $x = \frac{2}{k}$. For the expression to be negative,$x$ must lie between the roots. Case $1$: If $k > 0$,then $\frac{1}{2} < x < \frac{2}{k}$. The integers $x$ satisfying this are $1, 2, 3$. For exactly three integers to exist,we need $3 < \frac{2}{k} \le 4$. Solving $3 < \frac{2}{k}$ gives $k < \frac{2}{3}$. Solving $\frac{2}{k} \le 4$ gives $k \ge \frac{1}{2}$. So,$k \in [\frac{1}{2}, \frac{2}{3})$. Case $2$: If $k < 0$,then $\frac{2}{k} < x < \frac{1}{2}$. The integers $x$ are $-1, -2, -3$. For exactly three integers,we need $-4 \le \frac{2}{k} < -3$. Solving $\frac{2}{k} < -3$ gives $k > -\frac{2}{3}$. Solving $-4 \le \frac{2}{k}$ gives $k \le -\frac{1}{2}$. So,$k \in [-\frac{2}{3}, -\frac{1}{2}]$. Since the provided options do not match these intervals,the question or options may be flawed.

All the values of $k$ such that the quadratic expression $2kx^2 - (4k+1)x + 2$ is negative for exactly three integral values of $x$,lie in the interval

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