Discuss the continuity of fx - sin2x-1- cos2x for 0 - x -2 cos- - 2x for -2 -x - - at x - -2

For function to be continuous RHL=LHL at x=π/2 . LHL=lim_x→ (π/2)^ sin2x√(1 cos2x)=0√(1 ( 1))=0 RHL=lim_x→(π/2)^+cosxπ 2x=lim_x→ (π/2)^+ sin ...

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

Continuity of a Function

Discuss the c...

Question

Discuss the continuity of
$f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{s i n 2 x}{\sqrt{1 - c o s 2 x}} f o r 0 < x \leq π / 2 \frac{c o s}{π - 2 x} f o r \frac{π}{2} < x < π \end{matrix}$ at $x = π / 2$

Open in App

Solution

Suggest Corrections

Similar questions

x = \frac{π}{2}

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

x \neq \frac{π}{2}

= 3

x = \frac{π}{2}

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

x = \frac{π}{2},

f (\frac{π}{2})

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - {sin}^{3} x)}{3 {cos}^{2} x}, & x < \frac{π}{2} a, & x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}}, & x > \frac{π}{2} \end{matrix}

x = \frac{π}{2}

{(\frac{b}{a})}^{5 / 3}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} | x - 3 |; 0 \leq x < 1 sin x; 1 \leq x \leq \frac{π}{2} {log}_{\frac{π}{2}} x; \frac{π}{2} < x < 3 \end{matrix}

[0, 3)

Join BYJU'S Learning Program