MCQ
If $x$ is real and $k = \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}},$ then
- ✓$\frac{1}{3} \le k \le 3$
- B$k \ge 5$
- C$k \le 0$
- DNone of these
We have ${x^2}(k - 1) + x(k + 1) + k - 1 = 0$
As given, $x$ is real ==> ${(k + 1)^2} - 4{(k - 1)^2} \ge 0$
==> $3{k^2} - 10k + 3 \ge 0$
Which is possible only when the value of $k$ lies between the roots of the equation $3{k^2} - 10k + 3 = 0$
That is, when $\frac{1}{3} \le k \le 3$ {Since roots are $\frac{1}{3}$ and $3$}
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The maximum value of $k$ for which the sum $\sum_{i=0}^{k}\left(\begin{array}{c}10 \\ i\end{array}\right)\left(\begin{array}{c}15 \\ k-i\end{array}\right)+\sum_{i=0}^{k+1}\left(\begin{array}{c}12 \\ i\end{array}\right)\left(\begin{array}{c}13 \\ k+1-i\end{array}\right)$ exists, is equal to ...... .