Let fx-1-tan x-4x-x-4-x-0-4- lf fx is continuous in -0-2- then f-4 is-

The correct option is D 1/2 Substitute x=π4+t lim t→ o1 tan ( π4+t )4t =lim_t→ 01 ( 1+tan t )1 tan t4t =lim_t→ 0 2tan t ( 1 ta ...

Let fx=1-ta...

Question

Let $f (x) = \frac{1 - tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{4}]$ . lf $f (x)$ is continuous in $[0, \frac{π}{2}]$ then $f (\frac{π}{4})$ is:

- \frac{1}{2}

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\frac{1}{2}

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Similar questions

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) = \frac{1 - t a n x}{1 - \sqrt{2} s i n x}

x \neq \frac{π}{4}

x = \frac{π}{4}

f (\frac{π}{4})

f (x) = \frac{1 - tan x}{4 x - π}, x \neq \frac{π}{4}

lim x \to \frac{π}{4} f (x) =

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

f (x)

x = \frac{π}{2}

f (\frac{π}{2})

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} {(1 - | tan x |)}^{a / | tan x |}, & - \frac{π}{4} < x < 0 \end{matrix} \begin{matrix} b, & x = 0 \end{matrix} \begin{matrix} e^{\frac{sin 3 x}{sin 2 x}}, & 0 < x < \frac{π}{4} \end{matrix} \end{matrix}

x = 0