\(f(x) = (x -3)^{n_1}(x - 5)^{n_2}\)
\(f'(x) = n_2 (x - 3)^{n_1} (x- 5)^{n_2 - 1} + n_1(x - 3)^{n_1}(x - 5)^{n_2}\)
\(= (x - 3)^{n_1 - 1} (x- 5)^{n_2 - 1} (n_2 (x - 3) + n_1 (x - 5))\)
\(= (x - 3)^{n_1 - 1} (x - 5)^{n_2 - 1 }((n_1 + n_2) x - (3n_2+ 5n_1))\)
(A) If n1 = 3, n2 = 4 then
\(f'(x) = (x - 3)^2 (x - 5)^3 (7x - 27)\)
f'(x) change sign from +Ve to -Ve through \(\frac{27}7\).
\(\therefore \alpha = \frac{27}{7} \in (3, 5)\) is point of local maxima for f(x).
(B) If n1 = 4, n2 = 3 then
\(f'(x) = (x - 3)^3 (x - 5)^2 (7x - 29)\)
f'(x) changes sign from positive to negative through \(\frac{29}7\).
\(\therefore\alpha = \frac{29}{7} \in (3, 5)\) is point of local minima for f(x).
(C) If n1 = 3, n2 = 5 then
\(f'(x) = (x - 3)^2 (x - 5)^4 (8x - 30)\)
f'(x) change sign from -Ve to +Ve through \(\frac{30}8\).
\(\therefore\alpha = \frac{30}{8} \in (3, 5)\) is point of local minima for f(x).
Hence, option (C) is not true.
(D) If n1 = 4, n2 = 6 then
\(f'(x) = (x - 3)^3 (x - 5)^5 (10x - 38)\)
f'(x) change sign from +Ve to -Ve through \(\frac{38}{10}\).
\(\therefore\alpha = \frac{38}{10} \in (3, 5)\) is point of local maxima for f(x).
