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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Column-$I$	Column-$II$
$A$. Imaginary roots	$P$. $a \in (-1, 4)$
$B$. One root less than $3$ and other greater than $3$	$Q$. $a \in (-\infty, -1)$
$C$. One root less than $1$ and other greater than $3$	$R$. $a \in (-\infty, -4/3)$

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

Similar Questions

If the quadratic equation $ax^2+bx+c=0$ $(a>0)$ has two roots $\alpha$ and $\beta$ such that $\alpha < -2$ and $\beta > 2$, then which of the following is true?

Let $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$,where $x \in R$. If $b$ and $c$ are nonzero real numbers such that $\min f(x) > \max g(x)$,then $\left|\frac{c}{b}\right|$ lies in the interval:

Let the set of all values of $p \in R$,for which both the roots of the equation $x^2-(p+2)x+(2p+9)=0$ are negative real numbers,be the interval $(\alpha, \beta]$. Then $\beta-2\alpha$ is equal to

If the roots of the quadratic equation $x^2 - 2kx + k^2 + k - 5 = 0$ are less than $5$,then in which interval does $k$ lie?

Vedclass Test Series

Exam Paper Generator

Online Exam Module