Let -x- denote the greatest integer less than or equal to x- Then the value of - for which the function fx - sin -x2-x2- x- 0 - x - 0- is continuous at x - 0 is

The correct option is A α = sin (1) lim_x→ 0sin [ x^2][ x^2] = sin (1) if f(x) is continuous at x = 0 , then lim_x→ 0 f(x) = f(0) = α ...

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Standard XII

Mathematics

Domain

Let [x] den...

Question

Let $[x]$ denote the greatest integer less than or equal to $x$ . Then the value of $α$ for which the function $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin [- x^{2}]}{[- x^{2}]}, & x \neq 0 α, & x = 0 \end{matrix}$ is continuous at $x = 0$ is

α = 0

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α = sin (- 1)

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α = sin (1)

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α = 1

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Solution

The correct option is A $α = sin (1)$
$lim x \to 0 \frac{sin [- x^{2}]}{[- x^{2}]} = sin (1)$
if $f (x)$ is continuous at $x = 0$ , then $lim x \to 0 f (x) = f (0) = α = sin (1)$

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Similar questions

Q. Let

[x]

denote the greatest integer less than or equal to

x

. Then the value of

α

for which the function

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin [- x^{2}]}{[- x^{2}]}, & x \neq 0 α, & x = 0 \end{matrix}

is continuous at

x = 0

Q. Let

[x]

denotes the greatest integer less than or equal to

x

. Then, the value of

α

for which the function

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin [- x^{2}]}{[- x^{2}]} & x \neq 0 α & x = 0 \end{matrix}

is continuous at

x = 0

, is

Q. Let

α \in R

be such that the function

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{{cos}^{- 1} (1 - {x}^{2}) {sin}^{- 1} (1 - {x})}{{x} - {x}^{3}}, & x \neq 0 α, & x = 0 \end{matrix}

is continuous at

x = 0,

where

{x} = x - [x],

[x]

is the greatest integer less than or equal to

x .

Then :

Q. Let

f : R \to R

be defined as

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{3}}{(1 - cos 2 x)^{2}} \cdot {log}_{e} (\frac{1 + 2 x e^{- 2 x}}{(1 - x e^{- x})^{2}}) & , & x \neq 0 α & , & x = 0 \end{matrix}

f

is continuous at

x = 0

, then

α

is equal to:

Q. Let

f (x) = \frac{\sqrt{s g n (α x^{2} + α x + 1)}}{{cot}^{- 1} (x^{2} - α)}

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α

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[Note : sgn k denotes signum function of k.]

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Let [x] denote the greatest integer less than or equal to x. Then the value of α for which the function f(x)=⎧⎪⎨⎪⎩sin[−x2][−x2],x≠0α,x=0 is continuous at x=0 is

The correct option is A α=sin(1)limx→0sin[−x2][−x2]=sin(1)if f(x) is continuous at x=0, then limx→0f(x)=f(0)=α=sin(1)

Let $[x]$ denote the greatest integer less than or equal to $x$ . Then the value of $α$ for which the function $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin [- x^{2}]}{[- x^{2}]}, & x \neq 0 α, & x = 0 \end{matrix}$ is continuous at $x = 0$ is

The correct option is A $α = sin (1)$
$lim x \to 0 \frac{sin [- x^{2}]}{[- x^{2}]} = sin (1)$
if $f (x)$ is continuous at $x = 0$ , then $lim x \to 0 f (x) = f (0) = α = sin (1)$