Question

Let f(x) =Then at x = np +$\frac { 1 } { 2 }$p, f(x) has –

• A. Minima when n = 0, 2, 4,…,
• B. Minima when n = 1, 3, 5,….,
• C. Maxima when n = 1, 3, 5,….,
• D. ) None of these

Answer 1

Answer: (B)

Minima when n = 1, 3, 5,….,

Answer 2

f(x) = $\frac { \cos x } { x } = 0$Ž cos x = 0

Ž x = ± p/2, 3p/2

f ''(x) = $\frac { x ( - \sin x ) - \cos x } { x ^ { 2 } }$

f ''(x)x = p/2 = – ve –® maximum

f(x) x = $\frac { 3 \pi } { 2 }$= +ve –® minimum