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

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(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)  <

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$(-3, \infty)$

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(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:

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