Using $L-$ Hospital’s rule,
$y = \mathop {\lim }\limits_{x \to 0} \frac{{{4^x}\log 4 - {9^x}\log 9}}{{({4^x} + {9^x}) + x({4^x}\log 4 + {9^x}\log 9)}}$
==> $y = \frac{{\log 4 - \log 9}}{2}$
==> $y = \frac{{\log {{\left( {\frac{2}{3}} \right)}^2}}}{2} = \log \frac{2}{3}$.
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$(i)$ Let $\alpha_1$ be the total number of ways in which the committee can be formed such that the committee has $5$ members, having exactly $3$ boys and $2$ girls.
$(ii)$ Let $\alpha_2$ be the total number of ways in which the committee can be formed such that the committee has at least $2$ members, and having an equal number of boys and girls.
$(iii)$ Let $\alpha_3$ be the total number of ways in which the committee can be formed such that the committee has $5$ members, at least $2$ of them being girls.
$(iv)$ Let $\alpha_4$ be the total number of ways in which the committee can be formed such that the committee has $4$ members, having at least $2$ girls and such that both $M _1$ and $G _1$ are $NOT$ in the committee together.
| $LIST I $ | $LIST I $ |
| $P$ The value of $\alpha_1$ is | $1$ $136$ |
| $Q$ The value of $\alpha_2$ is | $2$ $189$ |
| $R$ The value of $\alpha_3$ is | $3$ $192$ |
| $S$ The value of $\alpha_4$ is | $4$ $200$ |
| $5$ $381$ | |
| $6$ $461$ |
The correct option is: