The number of values of k for which the equat | vedclass

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

Question

(A) Let $f(x) = x^2 - 3x + k$. For the roots $\alpha$ and $\beta$ to lie in the interval $(0, 1)$ with $\alpha 
eq \beta$,the following conditions mu

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $f(x) = x^2 - 3x + k$. For the roots $\alpha$ and $\beta$ to lie in the interval $(0, 1)$ with $\alpha 
eq \beta$,the following conditions must be satisfied:$1$. Discriminant $\Delta > 0$: $\Delta = (-3)^2 - 4(1)(k) = 9 - 4k > 0 \implies k $2$. $f(0) > 0$: $0^2 - 3(0) + k > 0 \implies k > 0$.$3$. $f(1) > 0$: $1^2 - 3(1) + k > 0 \implies 1 - 3 + k > 0 \implies k > 2$.$4$. The vertex of the parabola $x = -\frac{b}{2a}$ must lie in $(0, 1)$: $x = -\frac{-3}{2(1)} = \frac{3}{2} = 1.5$.Since the vertex $1.5$ does not lie in the interval $(0, 1)$,it is impossible for both roots to lie in $(0, 1)$.Thus,there are no such values of $k$.

The number of values of $k$ for which the equation $x^2 - 3x + k = 0$ has two real and distinct roots lying in the interval $(0, 1)$ is

Similar Questions

The number of integral values of $k$,for which one root of the equation $2x^2-8x+k=0$ lies in the interval $(1,2)$ and its other root lies in the interval $(2,3)$,is:

If one root of $x^2+px-q^2=0$, where $p$ and $q$ are real, is less than $2$ and the other is greater than $2$, then:

If the graph of $y = ax^2 + bx + c$ $(a, b, c \in R)$ is as shown in the figure,where $D = b^2 - 4ac$,which of the following is incorrect?

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$. If $b$ can take any real value,what is the range of $m(b)$?

If $\alpha, \beta$ are the roots of $x^2 - 3x + a = 0, a \in R$ and $\alpha < 1 < \beta$,then :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module