If fx-1-sin xsin 2x- x -2 is continuous at x-2- then the value of f -2 is

For continuity of f(x) at x=π2, we must have f ( π2 )=lim_x→π21 sin xsin 2x (00 form) nbsp; nbsp; Using L Hospital's Rule =lim_x→π2 cos x2 cos 2x= cosπ22 c ...

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Standard XII

Mathematics

Properties of Determinants

If fx=1-sin x...

Question

If $f (x) = \frac{1 - sin x}{sin 2 x}, x \neq \frac{π}{2}$ is continuous at $x = \frac{π}{2},$ then the value of $f (\frac{π}{2})$ is

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Solution

The correct option is A $0$
For continuity of $f (x)$ at $x = \frac{π}{2}$ , we must have
$f (\frac{π}{2}) = lim x \to \frac{π}{2} \frac{1 - sin x}{sin 2 x} (\frac{0}{0} form)$
Using L Hospital's Rule
$= lim x \to \frac{π}{2} \frac{- cos x}{2 cos 2 x} = \frac{- cos \frac{π}{2}}{2 cos π} = 0$

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Properties of Determinants

Standard XII Mathematics

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If f(x)=1−sinxsin2x,x≠π2 is continuous at x=π2, then the value of f(π2) is

The correct option is A 0For continuity of f(x) at x=π2, we must have f(π2)=limx→π21−sinxsin2x (00 form) Using L Hospital's Rule =limx→π2−cosx2cos2x=−cosπ22cosπ=0

If $f (x) = \frac{1 - sin x}{sin 2 x}, x \neq \frac{π}{2}$ is continuous at $x = \frac{π}{2},$ then the value of $f (\frac{π}{2})$ is

The correct option is A $0$
For continuity of $f (x)$ at $x = \frac{π}{2}$ , we must have
$f (\frac{π}{2}) = lim x \to \frac{π}{2} \frac{1 - sin x}{sin 2 x} (\frac{0}{0} form)$
Using L Hospital's Rule
$= lim x \to \frac{π}{2} \frac{- cos x}{2 cos 2 x} = \frac{- cos \frac{π}{2}}{2 cos π} = 0$