The set of values of a for which the inequali | vedclass

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

Question

(B) Let $f(x) = x^2 - (a + 2)x - (a + 3)$.For $f(x)  <  0$ to have at least one positive root $x$, we consider the following cases:Case-$I$: $f(0)  <

Vedclass · Accepted Answer

$(-3, \infty)$

Vedclass · Answer

(B) Let $f(x) = x^2 - (a + 2)x - (a + 3)$.For $f(x)  Case-$I$: $f(0)  $-(a + 3)   0 \Rightarrow a > -3$.Case-$II$: $f(0) \geq 0$ and the vertex is at $x > 0$ and $D > 0$.$1$) $D = (a + 2)^2 + 4(a + 3) = a^2 + 4a + 4 + 4a + 12 = a^2 + 8a + 16 = (a + 4)^2 > 0$, which is true for all $a 
eq -4$.$2$) $f(0) = -(a + 3) \geq 0 \Rightarrow a \leq -3$.$3$) Vertex $x_v = -\frac{b}{2a} = \frac{a + 2}{2} > 0 \Rightarrow a > -2$.Intersection of $a \leq -3$ and $a > -2$ is empty $(\phi)$.Combining Case-$I$ and Case-$II$, we get $a > -3$.Thus, the set of values is $(-3, \infty)$.

The set of values of $a$ for which the inequality $x^2 - (a + 2)x - (a + 3) < 0$ is satisfied by at least one positive real $x$ is:

Similar Questions

If $f:[1, 2] \rightarrow R$ defined by $f(x) = x^2 + 2kx + k$ is always negative for all $x \in [1, 2]$,then the interval in which $k$ lies is:

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

Number of integral values of $a$ for which the smaller root of the quadratic equation $x^2 - 2ax + a^2 - 4 = 0$ is smaller than $1$ and the bigger root is greater than $6$ is:

For the quadratic equation $x^2 - (K + 1)x + (K^2 + K - 8) = 0$,if one root is greater than $2$ and the other root is less than $2$,then $K$ lies in which interval?

If $y = f(x) = ax^2 + 2bx + c = 0$ has imaginary roots and $4a + 4b + c < 0$,then :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module