Let f- be defined as fx-x31-cos 2x2-log e 1-2xe-2x1-xe-x2 - x- 0- - x-0-If f is continuous at x - 0- then - is equal to-

Given f(x) is continuous at x = 0 ∴α =lim _x→ 0x^3(1 cos 2x)^2log _e ( 1+2xe^ 2x(1 xe^ x)^2 ) α =lim _x→ 0x^44sin ^4x.1xlog _e ( e^2x+2xx^2 2xe^x+e^2x ) =14lim_ ...

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Byju's Answer

Standard XII

Mathematics

Right Hand Derivative

Let f:ℝ→ℝ be ...

Question

Let $f : R \to R$ be defined as $f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{3}}{(1 - cos 2 x)^{2}} \cdot {log}_{e} (\frac{1 + 2 x e^{- 2 x}}{(1 - x e^{- x})^{2}}) & , & x \neq 0 α & , & x = 0 \end{matrix}$
If $f$ is continuous at $x = 0$ , then $α$ is equal to:

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Solution

The correct option is B $1$
Given $f (x)$ is continuous at $x = 0$
$∴ α = lim x \to 0 \frac{x^{3}}{(1 - cos 2 x)^{2}} {log}_{e} (\frac{1 + 2 x e^{- 2 x}}{(1 - x e^{- x})^{2}})$
$α = lim x \to 0 \frac{x^{4}}{4 {sin}^{4} x} . \frac{1}{x} {log}_{e} (\frac{e^{2 x} + 2 x}{x^{2} - 2 x e^{x} + e^{2 x}})$
$= \frac{1}{4} lim x \to 0 {\frac{ln (e^{2 x} + 2 x)}{x} - \frac{ln (x^{2} - 2 x e^{x} + e^{2 x})}{x}}$
On applying L'hospital rule
$= \frac{1}{4} lim x \to 0 {\frac{2 e^{2 x} + 2}{e^{2 x} + 2 x} - \frac{2 x - 2 e^{x} (x + 1) + 2 e^{2 x}}{x^{2} - 2 x e^{x} + e^{2 x}}}$
$= \frac{1}{4} (4 - 0) = 1$

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Let f:R→R be defined as f(x)=⎧⎪ ⎪⎨⎪ ⎪⎩x3(1−cos2x)2⋅loge(1+2xe−2x(1−xe−x)2),x≠0α,x=0 If f is continuous at x=0, then α is equal to:

Let $f : R \to R$ be defined as $f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{3}}{(1 - cos 2 x)^{2}} \cdot {log}_{e} (\frac{1 + 2 x e^{- 2 x}}{(1 - x e^{- x})^{2}}) & , & x \neq 0 α & , & x = 0 \end{matrix}$
If $f$ is continuous at $x = 0$ , then $α$ is equal to: