If f x -2cos x 1 0 0 2cos x 1 0 1 2cos x then df-dx at x-2 is

The correct option is D 2 f( x ) = 2cos x amp; 1 amp; 0 0 amp; 2cos x amp; 1 0 amp; 1 amp; 2cos x df d x = ...

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

Cofactor

If f x =2co...

Question

If $f (x) = ∣ ∣ ∣ ∣ \begin{matrix} 2 c o s x & 1 & 0 0 & 2 c o s x & 1 0 & 1 & 2 c o s x \end{matrix} ∣ ∣ ∣ ∣$ then $\frac{d f}{d x}$ at $x = \frac{π}{2}$ is

0

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

2

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

\frac{π}{2}

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

8

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 cos x & 1 & 0 x - \frac{π}{2} & 2 cos x & 1 0 & 1 & 2 cos x \end{matrix} ∣ ∣ ∣ ∣ ∣

\frac{d f}{d x}

x = \frac{π}{2}

| x |

\frac{d f}{d x}

- \frac{π}{4}

f (x) = \frac{1 - tan x}{4 x - π}

x \neq π / 4

x \in [0, \frac{π}{2}]

f (x)

[0, \frac{π}{2}]

f (\frac{π}{4})

f (x) = {(sin x)}^{\frac{1}{π - 2 x}}, x \neq \frac{π}{2}

f (x)

x = \frac{π}{2}

f (\frac{π}{2})

g (x) = f (tan x) + f (cot x)

x \in (\frac{π}{2}, π)

f^{''} (x) < 0 \forall x \in (\frac{π}{2}, π),

Join BYJU'S Learning Program