Function f(x)= ( 1 + 1 x )^x then Domain of f (x) is - Vidyadip

MCQ

Function $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ then Domain of $f (x)$ is

A
$(-1, 0) \cup (0, \infty )$
B
$R - { 0 }$
✓
$(-\infty , -1) \cup (0, \infty )$
D
$(0, \infty )$

✓

Answer

Correct option: C.

$(-\infty , -1) \cup (0, \infty )$

c

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $f\,\, \& \,\, g$ are differentiable functions such that $g' (a) = 2 \,\, \& \,\, g(a) = b$ and if $fog$ is an identity function then $f ' (b)$ has the value equal to :

The sum to $50$ terms of series $\frac{3}{{{1^2}}} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + ..... = $

If $\alpha $ and $\beta $ are roots of $a{x^2} + 2bx + c = 0$, then $\sqrt {\frac{\alpha }{\beta }} + \sqrt {\frac{\beta }{\alpha }} $is equal to

Rank of matrix $\left[ {\begin{array}{*{20}{c}}4&1&0&0\\3&0&1&0\\6&0&2&0\end{array}} \right]$ is

If $x = {{2\,t} \over {1 + {t^2}}},\,\,y = {{1 - {t^2}} \over {1 + {t^2}}},$ then ${{d\,y} \over {d\,x}}$ equals

Let $R\,= \{(x,y) : x,y \in N\, and\, x^2 -4xy +3y^2\, =0\}$, where $N$ is the set of all natural numbers. Then the relation $R$ is

If $x$ be real, then the minimum value of ${x^2} - 8x + 17$ is

The locus of a point which divides the join of $A( - 1,\;1)$ and a variable point $P$ on the circle ${x^2} + {y^2} = 4$ in the ratio $3 : 2$, is

Let $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$. Then $18 \int \limits_1^2 f(x) d x$ is equal to:

If $\int_0^{2a} {f(x)\,dx = 2\int_0^a {f(x)\,dx,} } $ then