In order that the function f(x) = (x + 1)^ ,x is continuous at x = 0 , f(0) must be defined as

In order that the function f(x) = (x + 1)^ ,x is continuous at x …

Download App

MCQ

In order that the function $f(x) = {(x + 1)^{\cot \,x}}$ is continuous at $x = 0 , f(0)$ must be defined as

A
$f(0) = \frac{1}{e}$
B
$f(0) = 0$
✓
$f(0) = e$
D
None of these

✓

Answer

Correct option: C.

$f(0) = e$

c
(c) For continuity at $x = 0$, we must have $f(0) = \mathop {{\rm{lim}}}\limits_{x \to 0} f(x)$

$\mathop {\lim }\limits_{x \to \alpha } \,\frac{{1 - \cos \,(a{x^2} + bx + c)}}{{{{(x - \alpha )}^2}}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} {\left\{ {{{(1 + x)}^{\frac{1}{x}}}} \right\}^{x\cot x}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} {\left\{ {{{(1 + x)}^{\frac{1}{x}}}} \right\}^{\mathop {{\rm{lim}}}\limits_{x \to 0} \,\left( {\frac{x}{{\tan x}}} \right)}}$ $ = {e^1} = e$.

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

More "M.C.Q (1 Marks)" from 5. continuity and differentiation→5. continuity and differentiation — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The solution of the differential equation $({x^2} + {y^2})dx = 2xydy$ is

→

vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are inclined at angle $\theta=120^\circ.$ if $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=2,$ then $\big[\big(\vec{\text{a}}+3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)\big]^2$ is equal to:

→

A computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20 \%$ and plant $T_2$ produces $80 \%$ of the total computers produced. $7 \%$ of computers produced in the factory turn out to be defective. It is known that

$P$ (computer turns out to be defective given that it is produced in plant $T_1$ )

$=10 P\left(\right.$ computer turns out to be defective given that it is produced in plant $\left.T_2\right)$,

where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is

→

If $\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&x&{{x^2}}\\{{b^2}}&{ab}&{{a^2}} \end{array}} \right|$ $= 0$ , then :

→

If the four points, whose position vectors are $3 \hat{ i }-4 \hat{ j }+2 \hat{ k }, \hat{ i }+2 \hat{ j }-\hat{ k },-2 \hat{ i }-\hat{ j }+3 \hat{ k } \quad$ and $5 \hat{ i }-2 \alpha \hat{ j }+4 \hat{ k }$ are coplanar, then $\alpha$ is equal to

→

Let $A = \{ {x_1},\,{x_2},\,............,{x_7}\} $ and $B = \{ {y_1},\,{y_2},\,{y_3}\} $ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f : A \to B$ that are onto, if there exist exactly three elements $x$ in $A$ such that $f(x)\, = y_2$, is equal to

→

The function $f(x)\, = \left\{ \begin{array}{l}x + 2\,\,\,\,,\,\,\,1 \le x \le 2\\4\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x = 2\\3x - 2\,\,,\,\,\,x > 2\end{array} \right.$ is continuous at

→

The corner points of the feasible region determined by the following system of linear inequalities : $2 x+y \leq 10, x+3 y \leq 15$, $x, y \geq 0$ are $(0,0),(5,0),(3,4)$ and $(0,5)$. Let $Z =p x+q y$ where $p, q>0$, condition on $p$ and $q$ so that the maximum of Z occurs at both $(3,4)$ and $(0,5)$ is

→

The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&0\leq\text{x}<1\\\text{a},&1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous for $0\leq\text{x}<\infty,$ then the most suitable values of a and b are: