JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 15)
If the numbers appeared on the two throws of a fair six faced die are $$\alpha$$ and $$\beta$$, then the probability that $$x^{2}+\alpha x+\beta>0$$, for all $$x \in \mathbf{R}$$, is :
$$\frac{17}{36}$$
$$
\frac{4}{9}
$$
$$\frac{1}{2}$$
$$\frac{19}{36}$$
Explanation
For $x^{2}+\alpha x+\beta>0 \forall x \in R$ to hold, we should have $\alpha^{2}-4 \beta<0$
If $\alpha=1, \beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices
If $\alpha=2, \beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices
If $\alpha=3, \beta$ can be $3,4,5,6$ i.e., 4 choices
If $\alpha=4, \beta$ can be 5 or 6 i.e., 2 choices
If $\alpha=6$, No possible value for $\beta$ i.e., 0 choices
Hence total favourable outcomes
$$ \begin{aligned} &=6+5+4+2+0+0 \\\\ &=17 \end{aligned} $$
Total possible choices for $\alpha$ and $\beta=6 \times 6=36$
Required probability $=\frac{17}{36}$
If $\alpha=1, \beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices
If $\alpha=2, \beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices
If $\alpha=3, \beta$ can be $3,4,5,6$ i.e., 4 choices
If $\alpha=4, \beta$ can be 5 or 6 i.e., 2 choices
If $\alpha=6$, No possible value for $\beta$ i.e., 0 choices
Hence total favourable outcomes
$$ \begin{aligned} &=6+5+4+2+0+0 \\\\ &=17 \end{aligned} $$
Total possible choices for $\alpha$ and $\beta=6 \times 6=36$
Required probability $=\frac{17}{36}$
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