If fx-1-sin x-2x2- x -2- x-2- is continuous at x-2- then the value of - is

For f(x) to be continuous at x=π2, we must have lim_x→π2 f(x)=f ( π2 )=λ ⇒λ= lim_x→π2 f(x)=lim_x→π2(1 sin x)(π 2x)^2 nbsp; nbsp; Using L Hospital's Rule =l ...

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Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

If fx=1-sin x...

Question

If $f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - sin x)}{(π - 2 x)^{2}}, & x \neq \frac{π}{2} λ, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , then the value of $λ$ is

\frac{1}{8}

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\frac{1}{2}

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\frac{1}{4}

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1

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Solution

The correct option is A $\frac{1}{8}$
For $f (x)$ to be continuous at $x = \frac{π}{2}$ , we must have
$lim x \to \frac{π}{2} f (x) = f (\frac{π}{2}) = λ$
$\Rightarrow λ = lim x \to \frac{π}{2} f (x) = lim x \to \frac{π}{2} \frac{(1 - sin x)}{(π - 2 x)^{2}}$
Using L Hospital's Rule
$= lim x \to \frac{π}{2} \frac{- cos x}{- 4 (π - 2 x)}$
$= \frac{1}{4} \cdot lim x \to \frac{π}{2} \frac{- sin x}{- 2} = \frac{1}{8}$

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Similar questions

Q. If

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{2} 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}

then

f (x)

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x = \frac{π}{2}

, if

Q. If

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}

is continuous at

x = \frac{π}{2}

, then the value of

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

Q. If

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

is continuous at

x = \frac{π}{2},

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f (\frac{π}{2})

Q. If

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

so that

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, then

Q. If

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - sin x}{(π - 2 x)^{2}} \cdot \frac{log sin x}{log (1 + π^{2} - 4 π x + x^{2})}, & x \neq \frac{π}{2} k, & x = \frac{π}{2} \end{matrix}

is continuous at

x = \frac{π}{2}

, then

k

is equal to.

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Substitution Method to Remove Indeterminate Form

Standard XII Mathematics

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If f(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩(1−sinx)(π−2x)2,x≠π2λ,x=π2 is continuous at x=π2, then the value of λ is

The correct option is A 18For f(x) to be continuous at x=π2, we must have limx→π2f(x)=f(π2)=λ ⇒λ=limx→π2f(x)=limx→π2(1−sinx)(π−2x)2 Using L Hospital's Rule =limx→π2−cosx−4(π−2x) =14⋅limx→π2−sinx−2=18

If $f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{(1 - sin x)}{(π - 2 x)^{2}}, & x \neq \frac{π}{2} λ, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , then the value of $λ$ is