QJEEM22Ma2bcc: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2bcc Solution: If the numbers appeared on the two throws of a fair six faced…

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

$$\begin{array}{rl}& =6+5+4+2+0+0\\ \\ & =17\end{array}$$

Total possible choices for $\alpha$ and $\beta=6 \times 6=36$

Required probability $=\frac{17}{36}$