Question

The function f(x) = (4sin² x – 1)ⁿ (x² – x + 1), n Ī N, has a local minimum at x = $\frac { \pi } { 6 }$ , then –

• A. n can be any even number
• B. n can be any odd number
• C. n can be odd prime number
• D. ) n can be any natural number

Answer 1

Answer: (A)

n can be any even number

Answer 2

f(x) = (4sin² x – 1)ⁿ (x² – x + 1)

since x² – x + 1 > 0 " x Ī R

f(p/6) = 0

f(p/6)⁺ = $\lim _ { x \rightarrow \frac { \pi ^ { + } } { 6 } }$ (4sin²x – 1)ⁿ (x² – x + 1) = ® 0⁺

f(p⁺/6) = $\lim _ { x \rightarrow \frac { \pi ^ { - } } { 6 } }$ (4sin² x – 1)ⁿ (x² – x + 1)

= ® (0)^–n (a positive value)

f(p^–/6) > 0 if n is an even no.