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### $y = \frac{{{x^2} - \alpha \beta }}{{2x - \alpha - \beta }}$$x \in \mathbb{R} - \left\{ {\frac{{\alpha + \beta }}{2}} \right\}$

Assuming $\beta > \alpha$ which of the following option(s) is/are correct?

(A) $y \le \alpha$
(B) $y \ge \beta$
(C) $\alpha \le y \le \beta$
(D) $y \in \mathbb{R}$

Solution

${x^2} - 2xy + [(\alpha + \beta )y - \alpha \beta ] = 0$

$\because x \in \mathbb{R},\Delta \ge 0$

$\therefore 4{y^2} - 4[(\alpha + \beta )y - \alpha \beta ] \ge 0$

$\Rightarrow {y^2} - \alpha y - \beta y + \alpha \beta \ge 0$

$\Rightarrow (y - \alpha )(y - \beta ) \ge 0$

The wave-curve is shown below for the expression $(y - \alpha )(y - \beta )$ assuming $\beta > \alpha$.

Hence, (A) & (B).

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$