Let f- -0- - R be defined as fx-sin x- if x is irrational and x- -0- - tan3x- if x is rational and x- -0- - The number of points in -0- - at which the function f is continuous is -

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Continuity of a Function

Let f: [0, ...

Question

Let $f : [0, π] \to R$ be defined as $f (x) = {\begin{matrix} \begin{matrix} sin x, & if x is irrational and x \in [0, π] {tan}^{3} x, & if x is rational and x \in [0, π] \end{matrix} \end{matrix}$ . The number of points in $[0, π]$ at which the function $f$ is continuous is $?$

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f (x)

[0, π]

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x & , 0 \leq x \leq π / 4 2 x cot x + b & , \frac{π}{4} < x \leq \frac{π}{2} a cos 2 x - b sin x & , \frac{π}{2} < x < π \end{matrix}

f

[0, π]

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(1 + | sin x |)}^{\frac{a}{| sin x |}}, & - \frac{π}{6} < x < 0 b, & x = 0 e^{\frac{tan 2 x}{tan 3 x}}, & 0 < x < \frac{π}{6} \end{matrix}

f

x = 0

x \in (0, π) \cup (π, 2 π),

cosec x \in [1, \frac{2}{\sqrt{3}}] for x \in

f (x) = \begin{matrix} (1 + | s i n x |)^{\frac{a}{| s i n x |}}, & - \frac{π}{6} < x < 0 b, & x = 0 e^{\frac{t a n 2 x}{t a n 3 x}}, & 0 < x < \frac{π}{6} \end{matrix}

k

f

f (x) =

{\begin{matrix} k x^{2}, if x \leq π cos x, if x > π \end{matrix}

x =

π

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