Question

Set of real values of k if the equation x2 - (k-1)x + K2 = 0 has atleast one root in (1,2) is
(A) (2, 4)
(B) (-1, 1/3]
(C) (3)
(D) o​

Answer 1

$\textbf{Given:}$

$x^2-(k-1)x+k^2=0\;\text{has atleast one root in (1,2)}$

$\textbf{To find:}$

$\text{The real values of k}$

$\textbf{Solution:}$

$\text{Since the roots of x^2-(k-1)x+k^2=0 are real, we have}$

$b^2-4ac\;{\geq}\;0$

$(k-1)^2-4(1)k^2\;{\geq}\;0$

$k^2+1-2k-4k^2\;{\geq}\;0$

$1-2k-3k^2\;{\geq}\;0$

$3k^2+2k-1\;{\leq}\;0$

$(k+1)(3k-1)\;{\leq}\;0$

$\implies\,k\,\in\,[-1,\frac{1}{3}]$

$\textbf{Answer:}$

$\textbf{Option (B) is correct}$