MCQ
The coefficient of $x^4$ in ${\left[ {\frac{x}{2}\,\, - \,\,\frac{3}{{{x^2}}}} \right]^{10}}$ is :
- ✓$\frac{{405}}{{256}}$
- B$\frac{{504}}{{259}}$
- C$\frac{{450}}{{263}}$
- D$\frac{{405}}{{512}}$
Applying to the above question we get $T_{r+1}=(-1)^{r+10} C_{r} x^{10-r} 2^{r-10} 3^{r} x^{-2 r}$
$=(-1)^{r \cdot 10} C_{r} x^{10-3 r} 2^{r-10} 3^{r} \ldots$
For the coefficient of $x^{4}$
$10-3 r=4$
$6=3 r$
$r=2$
Substituting in (i) we get $T_{3}={ }^{10} C_{2} x^{4} 2^{-8} 3^{2}$
$=\frac{10.9 .3^{2}}{2 ! 2^{8}}$
$=\frac{405}{256}$
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$A=\left\{(x, y) \in R \times R \mid 2 x^{2}+2 y^{2}-2 x-2 y=1\right\}$
$B=\left\{(x, y) \in R \times R \mid 4 x^{2}+4 y^{2}-16 y+7=0\right\} \text { and }$
$C=\left\{(x, y) \in R \times R \mid x^{2}+y^{2}-4 x-2 y+5 \leq r^{2}\right\}$
Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is equal to:
$x \operatorname{cosec} \alpha-y \sec \alpha=\operatorname{kcot} 2 \alpha$ and $x \sin \alpha+y \cos \alpha=k \sin 2 \alpha$
respectively, then $\mathrm{k}^{2}$ is equal to :