Given quadratic equation is 

\(x^2 + 2(k - 1)x + (k + 5) = 0\)

By comparing with ax² + bx + c = 0, we obtain

\(a = 1>0, b=2(k - 1)\) & \(c = k + 5\)

Let α and \(\beta \) are roots of quadratic equation .

Let \(f(x) = x^2 + 2(k - 1)x + k + 5\).

(a) Given that \(\alpha < 2 < \beta\)

Required condition are 

(i) \(D < 0\)

(ii) \(a f(2) < 0\)

(i) \(D = b^2 - 4ac > 0\)

⇒ \(4(k - 1)^2 - 4(k + 5) > 0\)

⇒ \(4k^2 - 8k + 4 -4k - 20 > 0\)

⇒ \(4k^2- 12k - 16> 0\)

⇒ \(k^2 - 3k - 4 > 0\)

⇒ \((k - 4)(k + 1) > 0\)

⇒ \(k < -1 \;or\; k> 4\)   .....(1)

(ii) \(a f(2) < 0\)

⇒ \(1.(4 + 4k - 4 + k + 5) < 0\)

⇒ \(5k + 1 < 0\)

⇒ \(k < \frac{-1}5\)   ......(2)

\(\therefore\) From (1) & (2), we get

\(k< - 1\)

⇒ \(k \in (-\infty, -1)\)

(b) Given that \(\alpha < 2 < \beta\)

same as question part (a).

\(\therefore\) \(k \in (-\infty, -1)\).

(c) Quadratic equation has roots of opposite sign.

\(\therefore\) Product of roots = negative < 0.

⇒ \(\frac ca < 0\)

⇒ \(\frac{k + 5}1 < 0\)

⇒ \(k< - 5\)

⇒ \(k \in(-\infty, - 5)\)

(d) Case I: \(\alpha, \beta < 5\)

Required conditions are 

(i) \(D \ge 0\)

(ii) \(a f(5) > 0\)

(iii) \(5 > \frac{-b}{2a}\)

(i) \(D \ge 0\)

⇒ \(4k^2 - 12k - 16 \ge 0\)

⇒ \(k^2 - 3k - 4 \ge 0\)

⇒ \((k - 4) (k + 1) \ge 0\)

⇒ \(k \le -1 \; or\; k \ge 4\)  ....(1)

(ii) \(a f(5) > 0\) 

⇒ \(1(25 + 10k - 10 + 10) > 0\)

⇒ \(25 + 10k > 0\)

⇒ \(k > \frac{-25}{10}\)

⇒ \(k > \frac{-5}2\)    .....(2)

(iii) \(5 > \frac{-b}{2a}\) 

⇒ \(5 > \frac{-2(k - 1)}{2(1)}\)

⇒ \(5 > 1 - k\)

⇒ \(k > - 4\)    ....(3)

From (1), (2) & (3), we obtain

\(k\in(\frac{-5}2 , -1] \cup [4,\infty)\)    .....(4)

Case II: \(\alpha, \beta < 5\)

Required conditions are 

(i) \(D \ge 0\)

(ii) \(a f(5) > 0\)

(iii) \(5 > \frac{-b}{2a}\) 

From (i), we get 

\(k \le -1\;or\; k\ge 4 \)   .....(1b)

From (ii), we get 

\(k > \frac{-5}2\)   .....(2b)

From (iii), we get 

\(k < -4\)   ......(3b)

From(1b), (2b), (3b), we obtain 

\(x\in \phi\)  ....(5)

\(\therefore\) From (4) & (5), we get 

\(x\in (\frac{-5}2, -1] \cup [4, \infty)\).

(e) has both roots greater than 5 is solution of part (d) case II.

\(\therefore\) This case is not possible because solution set is empty set.

\(\therefore\) Both roots of given quadratic equation never be greater than 5.

(f) Quadratic equation has exactly one root in (1, 3).

\(\therefore\) \(f(1) f(3) < 0\)

⇒ \((1 + 2k - 2 + 5 + k) (9 + 6k - 6 + k + 5) < 0\)

⇒ \((3k + 4) (7k + 8) < 0\)

⇒ \(\frac{-4}3 < k < \frac{-8}7\)

⇒ \(k \in (\frac{-4}3, \frac{-8}7)\).