Let \(f: R \rightarrow R\) be defined as \(f(x)=\left\{\begin{array}{cl}\frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{e}\left(\frac{1+2 x e^{-2 x}}{\left(1-x e^{-x}\right)^{2}}\right) & , x \neq 0 \\ \alpha & , x=0\end{array}\right.\). If \(f\) is continuous at \(x=0\), then \(\alpha\) is equal to :

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Continuity and Differentiability

Let

f : R \to R

be defined as

f (x) = {\frac{x ^{3}}{( 1 - c o s 2 x ) ^{2}} lo g_{e} (\frac{1 + 2 x e ^{- 2 x}}{( 1 - x e ^{- x} ) ^{2}}) α, x \neq = 0, x = 0

. If

f

is continuous at

x = 0

, then

α

is equal to :

Answer: (B)

Since

1 - cos 2 x = 2 sin^{2} x

, the denominator is

4 sin^{4} x

. For small

x

ln (1 + 2 x e^{- 2 x}) = 2 x e^{- 2 x} + O (x^{2})

and

- 2 ln (1 - x e^{- x}) = 2 x e^{- x} + O (x^{2})

, so the logarithm as a whole behaves like

4 x

x \to 0 lim \frac{1}{x} ln (\frac{1 + 2 x e ^{- 2 x}}{( 1 - x e ^{- x} ) ^{2}}) = 4

Writing the function so that both of these appear,

x \to 0 lim f (x) = x \to 0 lim \frac{x ^{4}}{4 sin ^{4} x} \cdot \frac{1}{x} ln (\frac{1 + 2 x e ^{- 2 x}}{( 1 - x e ^{- x} ) ^{2}}) = \frac{1}{4} \times 4 = 1

Continuity at

x = 0

requires

α

to equal this limit, so

α = 1

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option