Let - denote the greatest integer function and fx-x- e1-x-1e1-x-1 - x0- If fx is continuous at x-0- then the value of a-b is

R.H.L.=x→ 0^+lim[x](e^1/x 1e^1/x+1 )+a =h→ 0lim[h](1 e^ 1/h1+e^ 1/h )+a =0× 1+a =a L.H.L.=x→ 0^ lim[x](e^1/x 1e^1/x+1 ) =h→ 0lim[ h](e^ 1/h 1e^ 1/h+1 ) = 1× ( ...

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Standard XII

Mathematics

Integral as Antiderivative

Let [.] denot...

Question

Let $[.]$ denote the greatest integer function and $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} [x] (\frac{e^{1 / x} - 1}{e^{1 / x} + 1}), & x < 0 b, & x = 0 [x] (\frac{e^{1 / x} - 1}{e^{1 / x} + 1}) + a, & x > 0 \end{matrix}$ . If $f (x)$ is continuous at $x = 0$ , then the value of $a + b$ is

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Solution

The correct option is A $2$
$R . H . L . = lim x \to 0^{+} [x] (\frac{e^{1 / x} - 1}{e^{1 / x} + 1}) + a$
$= lim h \to 0 [h] (\frac{1 - e^{- 1 / h}}{1 + e^{- 1 / h}}) + a$
$= 0 \times 1 + a$
$= a$

$L . H . L . = lim x \to 0^{-} [x] (\frac{e^{1 / x} - 1}{e^{1 / x} + 1})$
$= lim h \to 0 [- h] (\frac{e^{- 1 / h} - 1}{e^{- 1 / h} + 1})$
$= - 1 \times (- 1)$
$= 1$
and $f (0) = b$
Since, $f (x)$ is continuous at $x = 0,$
$∴ L . H . L . = R . H . L . = f (0)$
$\Rightarrow a = b = 1$

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Similar questions

f (x) = \frac{e^{1 / x} - 1}{e^{1 / x} + 1}, x \neq 0, f (0) = - 1

Is function continuous at x=0?

Q. STATEMENT-1 :

lim x \to 0 [x] {\frac{e^{1 / x} - 1}{e^{1 / x} + 1}}

(where [.] represents the greatest integer function) does not exist.
STATEMENT-2 :

lim x \to 0 (\frac{e^{1 / x} - 1}{e^{1 / x} + 1})

does not exists.

Q. If

f (x) = \frac{e^{1 / x} - 1}{e^{1 / x} + 1}, x \neq 0

and

f (0) = 0

, then

f (x)

Q. Let

f (x) = g (x) \frac{e^{1 / x} - e^{- 1 / x}}{e^{1 / x} + e^{- 1 / x}}

and

x \neq 0

where g is a continuous function.

Then

{lim}_{x \to 0} f (x)

exists if?

The function $f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} 0, x = 0 \end{matrix}, x \neq 0$

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Integral as Antiderivative

Standard XII Mathematics

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Let [.] denote the greatest integer function and f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩[x](e1/x−1e1/x+1),x<0b,x=0[x](e1/x−1e1/x+1)+a ,x>0. If f(x) is continuous at x=0, then the value of a+b is

The correct option is A 2R.H.L.=limx→0+[x](e1/x−1e1/x+1)+a =limh→0[h](1−e−1/h1+e−1/h)+a =0×1+a =a L.H.L.=limx→0−[x](e1/x−1e1/x+1) =limh→0[−h](e−1/h−1e−1/h+1) =−1×(−1) =1 and f(0)=b Since, f(x) is continuous at x=0, ∴L.H.L.=R.H.L.=f(0) ⇒a=b=1