Question

Let f(x) = (x − 1)⁴ (x − 2)ⁿ , n ∈ N. Then f(x) has 

(A) Local minimum at x = 2 if n is even 

(B) Local minimum at x = 1 if n is odd 

(C) Local maximum at x = 1 if n is odd 

(D) Local minimum at x = 1 if n is even

Answer 1

Answer is (A), (C), (D)

f(x) = (x − 1)⁴ (x − 2)ⁿ , n ∈ N (1)

Therefore, f′(x) = 4 (x − 1)³ (x − 2)ⁿ + (x − 1)⁴ n(x − 2)^{n − 1}

= (x − 1)³ (x − 2)^{n − 1} (4x − 8 + nx − n)

= (x − 1)³ (x − 2)^{n − 1} [(n + 4) x − (n + 8)]

If n is odd, then f′(x) > 0 if x < 1 and sufficiently close to 1 and f′(x) < 0 if x > 1 and sufficiently close to 1. Therefore, x = 1 is point of local maximum.

Similarly, if n is even, then x = 1 is a point of local minimum. 

Further if n is even, then f′(x) < 0 for x < 2 and sufficiently close to 2 and f′(x) > 0 for x > 2 and sufficiently close to 2. Therefore, x = 2 is a point of local minimum.