Let f- R- R be defined byfx-sin-x-2 if x-0 2 if x-0- -sin x-x-x3 - if x-0where - y - denotes the integral part of y- If f is continuous at x-0- then - - -

X gt;0⇒[x ]=0 x lt;0 ⇒[sin x x/x^3 ]= 1 ∴=2=β 1 ∴β =1 ...

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Byju's Answer

Standard XII

Mathematics

Addition Rule of Differentiation

Let f: R→ R b...

Question

Let f: $R \to R$ be defined by $f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} α + \frac{s i n [x]}{2} & i f x > 0 2 & i f x = 0 β + [\frac{s i n x - x}{x^{3}}] & i f x < 0 \end{matrix}$ where $[y]$ denotes the integral part of y. If f is continuous at $x = 0, t h e n β - α =$

-1

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Solution

The correct option is B
1

$x > 0 \Rightarrow [x] = 0$
$x < 0 \Rightarrow [\frac{s i n x - x}{x^{3}}] = - 1$
$∴\propto= 2 = β - 1$
$∴ β - \propto= 1$

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} α + \frac{sin [x]}{x} & i f x > 0 2 & i f x = 0 β + [\frac{sin x - x}{x^{3}}] & i f x < 0 \end{matrix}

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Let f: $R \to R$ be defined by $f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} α + \frac{s i n [x]}{2} & i f x > 0 2 & i f x = 0 β + [\frac{s i n x - x}{x^{3}}] & i f x < 0 \end{matrix}$ where $[y]$ denotes the integral part of y. If f is continuous at $x = 0, t h e n β - α =$