Download App

STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
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

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

Similar questions

The area of the region  

$A=\left\{(x, y):|\cos x-\sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\right\}$

→
If $1, \log _{10}\left(4^{x}-2\right)$ and $\log _{10}\left(4^{x}+\frac{18}{5}\right)$ are in
arithmetic progression for a real number $x$ then the value of the determinant $\left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{array}\right|$ is equal to ...... .
→
$xy$ - plane divides the line joining the points $(2, 4, 5)$ and $(-4, 3, -2)$ in the ratio
→
If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$ $(i = \sqrt { - 1} ),$then ${A^{ - 1}}$=
→
$\int_{}^{} {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)} \,dx = $
→
Let $S=\left\{\left(\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right) ; a, b \in\{1,2,3, \ldots 100\}\right\}$ and let $T_{n}=\left\{A \in S: A^{n(n+1)}=I\right\}$. Then the number of elements in $\bigcap \limits_{n=1}^{100} T_{n}$ is
→
${d \over {dx}}[(1 + {x^2}){\tan ^{ - 1}}x] = $
→
The equation of the circle which touches $x$ - axis and whose centre is $(1, 2)$ is
→
If $x = {\log _b}a,\,\,y = {\log _c}b,\,\,\,z = {\log _a}c$, then $xyz$ is
→
Let $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},.....,$  $({x_i} \ne \,0\,for\,\,i\, = 1,2,....,n)$  be in $A.P.$  such that  $x_1 = 4$ and $x_{21} = 20.$ If $n$  is the least positive integer for which $x_n > 50,$  then $\sum\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}}} \right)} $  is equal to.
→