Fx-1-2sin x-2cos 2 x- 0- x-2 is maximum at-

The correct option is B x=π6 f(x)=1+2sin x+2cos ^ 2 x ⇒f'(x)=2cos x+2(2cos x )( sin x) ⇒f'(x)=2cos x(1 sin x) For max/min, f'(x)=0⇒co ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Local Maxima

fx=1+2sin x+2...

Question

$f (x) = 1 + 2 sin x + 2 {cos}^{2} x$ , $0 \leq x \leq \frac{π}{2}$ is maximum at-

x = \frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = \frac{π}{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = \frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, & - π \leq x \leq \frac{- π}{2} a cos (x) + b, & \frac{- π}{2} < x < 0 - 2 + cos x, & 0 \leq x \leq \frac{π}{2} c sin x, & \frac{π}{2} < x \leq π \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

[- π, π],

f (x)

α

β

f (x)

f (x) = - 2 sin x,

- π \leq x \leq - \frac{π}{2}

= α sin x + β,

- \frac{π}{2} < x < \frac{π}{2}

= cos x,

\frac{π}{2} \leq x \leq π

[- π, π]

K

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{4}{5})}^{\frac{tan 4 x}{tan 5 x}}, & 0 < x < \frac{π}{2} k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2},

f (x) = sin x (1 + cos x), x \in (0, \frac{π}{2})

f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] .

[0, \frac{π}{2}]

f (\frac{π}{4}) =

Join BYJU'S Learning Program