If fx= (e^(2x+2x)-1)/(2x+2x) : x≠0 1 : x=0 , then (where, [.] represents the greatest integer function)

Correct answer: (D) fx is discontinuous at x=0. limx→0^(+)fx=limx→0^(+)f0+h =limh→0^(+)(e^(2h+2h)-1)/(2h+2h) =limh→0^(+)(e^2h-1)/2h=1 and limx→0^(-)fx=limh→0^(+)f0-h =limh→0(e^(-2h-2h)-1)/(-2h-2h) =limh→0(e^(-1-2h)-1)/(-1-2h)=(e^(-1)-1)/(-1) =1-1/e Since, limx→0^(+)fx≠limx→0^(-)fx, therefore, fx is not continuous at x=0.

Q6 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If $f (x) = \{\begin{cases} \frac{e^{[2 x] + 2 x} - 1}{[2 x] + 2 x} & : & x \neq 0 \\ 1 & : & x = 0 \end{cases}$ , then (where, [.] represents the greatest integer function)

A
$\underset{x \to 0^{+}}{l i m} f (x) = - 1$
B
$\underset{x \to 0^{-}}{l i m} f (x) = \frac{1}{e} - 1$
C
$f (x)$ is differentiable at $x = 0$
Correct answer
D
$f (x)$ is discontinuous at $x = 0$

Solution

\underset{x \to 0^{+}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} f (0 + h)

= \underset{h \to 0^{+}}{l i m} \frac{e^{[2 h] + 2 h} - 1}{[2 h] + 2 h}

= \underset{h \to 0^{+}}{l i m} \frac{e^{2 h} - 1}{2 h} = 1

and

\underset{x \to 0^{-}}{l i m} f (x) = \underset{h \to 0^{+}}{l i m} f (0 - h)

= \underset{h \to 0}{l i m} \frac{e^{[- 2 h] - 2 h} - 1}{[- 2 h] - 2 h}

= \underset{h \to 0}{l i m} \frac{e^{- 1 - 2 h} - 1}{(- 1 - 2 h)} = \frac{e^{- 1} - 1}{- 1}

= 1 - \frac{1}{e}

Since,

\underset{x \to 0^{+}}{l i m} f (x) \neq \underset{x \to 0^{-}}{l i m} f (x),

therefore,

f (x)

is not continuous at

x = 0 .

Hence, the correct answer is option: 'D'.

If fx=e2x+2x-12x+2x:x≠01:x=0, then (where, [.] represents the greatest integer function)

Solution

More Continuity and Differentiability Questions

If $f (x) = \{\begin{cases} \frac{e^{[2 x] + 2 x} - 1}{[2 x] + 2 x} & : & x \neq 0 \\ 1 & : & x = 0 \end{cases}$ , then (where, [.] represents the greatest integer function)