MCQ
${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ if and only if:
- ✓$2 \sqrt{2}<\mathrm{k} \leq 3$
- B$2 \sqrt{3}<\mathrm{k} \leq 3 \sqrt{2}$
- C$2 \sqrt{3}<\mathrm{k}<3 \sqrt{3}$
- D$2 \sqrt{2}<\mathrm{k}<2 \sqrt{3}$
$\underbrace{r+1 \geq 0, \quad r \geq 0}_{r \geq 0}$
$\frac{{ }^{n-1} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}+1}}=\mathrm{k}^2-8$
$\frac{\mathrm{r}+1}{\mathrm{n}}=\mathrm{k}^2-8$
$\Rightarrow \mathrm{k}^2-8>0$
$(\mathrm{k}-2 \sqrt{2})(\mathrm{k}+2 \sqrt{2})>0$
$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)$ $......(I)$
$\therefore \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1$
$ \Rightarrow \mathrm{k}^2-8 \leq 1 $
$\mathrm{k}^2-9 \leq 0$
$ -3 \leq \mathrm{k} \leq 3 $ $......(II)$
From equation $(I)$ and $(II)$ we get
$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$
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$y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots . .$ upto $\infty$ terms and $\frac{2+4+6+\ldots+2 \mathrm{y}}{3+6+9+\ldots+3 \mathrm{y}}=\frac{4}{\log _{10} \mathrm{x}}$, then the ordered pair $(x, y)$ is equal to :
| $x_i$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
| $f_i$ | $4$ | $4$ | $\alpha$ | $15$ | $8$ | $\beta$ | $4$ | $5$ |
are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.