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Chain Rule of Differentiation

A point of di...

Question

$A$ point of discontinuity of $f (x) = t a n x$ is

x = 0

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x = \frac{π}{4}

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x = \frac{π}{3}

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x = \frac{π}{2}

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Solution

The correct option is C $x = \frac{π}{2}$
The function defined as $f (x) = t a n x$ is everywhere continuous except at the points where it is undefined.
So, $tan x$ is not defined at every odd multiple of $\frac{π}{2}$
So a point of discontinuity of $f (x)$ is $\frac{π}{2}$

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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 - tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{4}]

f (x)

[0, \frac{π}{2}]

f (\frac{π}{4})

f (x) = [4 + 3 cos x],

x \in

(- \frac{π}{2}, \frac{π}{2}),

[x] =

x

f (x)

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

f (x) = tan x, 0 \leq x \leq \frac{π}{4}

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