MCQ 512 Marks
The points of discontinuity of $\tan x$ are
AnswerCorrect option: C. $(2 n+1) \frac{\pi}{2}, n \in I$
(C)
Let $f (x)=\tan x$
The point of discontinuity of $f (x)$ are those points where $\tan x$ is infinite.
i.e., $\tan x=\infty$
$\Rightarrow x=(2 n +1) \frac{\pi}{2}, n \in I$
View full question & answer→MCQ 522 Marks
The number of discontinuities of the greatest integer function $f (x)=[x], x \in\left(-\frac{7}{2}, 100\right)$ is equal to
Answer(D)
Given, $f (x)=[x], x \in(-3.5,100)$
As we know greatest integer function is discontinuous on integer values.
In given interval, the integer values are $(-3,-2,-1,0, \ldots, 99)$.
$\therefore \quad$ the total number of integers are 103 .
View full question & answer→MCQ 532 Marks
For the function $f(x)=\left\{\begin{array}{r}\frac{\sin ^2 a x}{x^2}, \text { when } x \neq 0 \\ 1, \text { when } x=0\end{array}\right.$ which one is a true statement?
- A
$f (x)$ is continuous at $x=0$, when $a \neq \pm 1$
- ✓
$f (x)$ is discontinuous at $x=0$, when $a \neq \pm 1$
- C
$\lim _{x \rightarrow 0} f (x)= a$
- D
$\lim _{x \rightarrow 0} f(x)=a^3$
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=0$, when $a \neq \pm 1$
(B)
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin ^2 a x}{(a x)^2} a^2=a^2$ and $f(0)=1$.
$\therefore f (x)$ is discontinuous at $x=0$, when $a \neq \pm 1$
View full question & answer→MCQ 542 Marks
If $f (y)=\frac{\left( e ^{2 y}-1\right) \cdot \sin y}{y^2} $, for $y \neq 0$
$=4$ , for $y=0$, then
- ✓
$f (y)$ is discontinuous at $y=0$
- B
$f (y)$ is continuous at $y=0$
- C
$\lim _{y \rightarrow 0} f (y)$ does not exist
- D
AnswerCorrect option: A. $f (y)$ is discontinuous at $y=0$
(A)
$\lim _{y \rightarrow 0} f (y)=\lim _{y \rightarrow 0} \frac{\left( e ^{2 y}-1\right)}{2 y} \times 2 \times \frac{\sin y}{y}$
$=\log e \times 2 \times 1=2$
and $f(0)=4$
$\therefore f (y)$ is discontinuous at $y=0$.
View full question & answer→MCQ 552 Marks
If $f: R \rightarrow R$ is defined by
$f(x)=\left\{\begin{array}{ll}x-1, & \text { for } x \leq 1 \\2-x^2, & \text { for } 1 < x \leq 3 \\x-10, & \text { for } 3 < x< 5 \\2 x, & \text { for } x \geq 5\end{array}\right.$
then the set of points of discontinuity of $f$ is
- A
$R -\{1,3,5\}$
- B
$\{1,3,5\}$
- C
$R-\{1,5\}$
- ✓
$\{1,5\}$
AnswerCorrect option: D. $\{1,5\}$
(D)
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2-x^2=1$
$f(1)=1-1=0$
$\lim _{x \rightarrow 1} f(x) \neq f(1)$
$\therefore \quad$ The function is discontinuous at $x=1$
$\lim _{x \rightarrow 5} f (x)=\lim _{x \rightarrow 5} x-10=-5$
$f(5)=2(5)=10$
$\lim _{x \rightarrow 5} f(x)+f(5)$
$\therefore \quad$ The function is discontinuous at $x=5$
$\lim _{x \rightarrow 3^{+}} f (x)=x-10=-7$
$f(3)=2-3^2=-7$
$\lim _{x \rightarrow 3} f(x)=f(3)$
$\therefore \quad$ The function is continuous at $x=3$
View full question & answer→MCQ 562 Marks
If $f(x)=\left\{\begin{array}{ll}4-3 x ; & 0 < x \leq 2 \\ 2 x-6 ; & 2 < x \leq 3 \\ x+5 ; & 3 < x \leq 6\end{array}\right.$, then $f(x)$ is
- ✓
continuous at $x=2$ and discontinuous at $x=3$
- B
continuous at $x=3$ and discontinuous at $x=2$
- C
continuous at $x=2$ and $x=3$
- D
discontinuous at $x=2$ and $x=3$
AnswerCorrect option: A. continuous at $x=2$ and discontinuous at $x=3$
(A)
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{-}}(4-3 x)=4-6=-2$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x-6)=4-6=-2$
$f(2)=4-3(2)=-2$
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(2 x-6)=6-6=0$
$\lim _{x \rightarrow 3^{+}} f (x)=\lim _{x \rightarrow 3^{+}}(x+5)=3+5=8$
$\therefore \quad \lim _{x \rightarrow 3^{-}} f (x) \neq \lim _{x \rightarrow 3^{+}} f (x)$
$\therefore f (x)$ is continuous at $x=2$ and discontinuous at $x=3$.
View full question & answer→MCQ 572 Marks
The value(s) of r for which the function
$f(x)=\left\{\begin{array}{cc}1-x , & x<1 \$1-x)(2-x) , & 1 \leq x \leq 2 \\3-x , & x>2\end{array}\right.$
fails to be continuous is (are)
Answer(B)
$\begin{array}{l}\lim _{x \rightarrow 1^{-}} f(x)=0, \lim _{x \rightarrow 1^{+}} f(x)=0 \text { and } f(1)=0\end{array}$
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
$\therefore f (x)$ is continuous at $x=1$.
$\lim _{x \rightarrow 2^{-}} f (x)=0$ and $\lim _{x \rightarrow 2^{+}} f (x)=1$
$\therefore \lim _{x \rightarrow 2^{-}} f (x) \neq \lim _{x \rightarrow 2^{+}} f (x) =1$
$\therefore f (x)$ is not continuous at $x=2$.
View full question & answer→MCQ 582 Marks
The number of discontinuities in R for the function $f(x)=\frac{x-1}{x^3+6 x^2+11 x+6}$ is
Answer(A)
$f (x)=\frac{x-1}{x^3+6 x^2+11 x+6}$
$=\frac{x-1}{(x+1)(x+2)(x+3)}$
∴ the points of discontinuity are $x=-1, x=-2$ and $x=-3$.
$\therefore \quad$ The number of discontinuities is 3 .
View full question & answer→MCQ 592 Marks
The function $f (x)=\frac{2 x^2+7}{x^3+3 x^2-x-3}$ is discontinuous for
AnswerCorrect option: C. $x=1, x=-1, x=-3$ only
(C)
$f (x)=\frac{2 x^2+7}{x^2(x+3)-1(x+3)}=\frac{2 x^2+7}{\left(x^2-1\right)(x+3)}$
$=\frac{2 x^2+7}{(x-1)(x+1)(x+3)}$
∴ the points of discontinuity are
$x=1, x=-1$ and $x=-3$ only.
View full question & answer→MCQ 602 Marks
The function $f(x)=\frac{4-x^2}{4 x-x^3}$ is
- A
discontinuous at only one point
- B
discontinuous exactly at two points
- ✓
discontinuous exactly at three points
- D
AnswerCorrect option: C. discontinuous exactly at three points
(C)
$f (x)=\frac{4-x^2}{x\left(4-x^2\right)}=\frac{4-x^2}{x(2+x)(2-x)}$
Since $f (x)$ does not exist at $x=0,2,-2$.
∴ there are three points of discontinuity.
View full question & answer→MCQ 612 Marks
The number of points at which the function $f ( x )=\frac{1}{\log |x|}$ is discontinuous are
Answer(C)
Since $f (x)$ is not defined at $x=0,1,-1$ and at all other points $f (x)$ is continuous.
∴ the given function is discontinuous at 3 points.
View full question & answer→MCQ 622 Marks
For the function $f (x)=\left\{\begin{array}{ll}\frac{ e ^{1 / x}-1}{ e ^{1 / x}+1}, & , x \neq 0 \\ 1 & , x=0\end{array}\right.$, which of the following is correct?
- A
$\lim _{x \rightarrow 0^{+}} f(x)=-1$
- B
$f (x)$ is continuous at $x=0$
- C
$\lim _{x \rightarrow 0^{-}} f(x)=1$
- ✓
$f (x)$ is not continuous at $x=0$
AnswerCorrect option: D. $f (x)$ is not continuous at $x=0$
(D)
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \frac{e^{\frac{-1}{h}}-1}{e^{\frac{-1}{h}}+1}=\lim _{h \rightarrow 0} \frac{e^{\frac{1}{{ }^{\frac{1}{h}}}-1}}{\frac{1}{e^{\frac{1}{h}}}+1}=\frac{0-1}{0+1}=-1$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{e^{\frac{1}{h}}-1}{e^{\frac{1}{h}}+1}=\lim _{h \rightarrow 0} \frac{1-\frac{1}{e^{\frac{1}{h}}}}{1+\frac{1}{e^{\frac{1}{h}}}}=\frac{1-0}{1+0}=1$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x) \neq \lim _{x \rightarrow 0^{+}} f (x)$
$\therefore f (x)$ is not continuous at $x=0$.
View full question & answer→MCQ 632 Marks
The function $f(x)=|x|+\frac{|x|}{x}$ is
- A
- B
Discontinuous at the origin because $|x|$ is discontinuous there
- ✓
Discontinuous at the origin because $\frac{|x|}{x}$ is discontinuous there
- D
Discontinuous at the origin because both $|x|$ and $\frac{|x|}{x}$ are discontinuous there
AnswerCorrect option: C. Discontinuous at the origin because $\frac{|x|}{x}$ is discontinuous there
(C)
$|x|$ is continuous at $x=0$ and $\frac{|x|}{x}$ is discontinuous at $x=0$.
$\therefore f (x)=|x|+\frac{|x|}{x}$ is discontinuous at $x=0$.
View full question & answer→MCQ 642 Marks
The function $f(x)=\frac{|x|}{x^2+2 x}, x \neq 0$ and $f(0)=0$ is not continuous at $x=0$ because
- A
$\lim _{x \rightarrow 0} f(x) \neq f(0)$
- B
$\lim _{x \rightarrow 0^{+}} f(x)$ does not exist
- C
$\lim _{x \rightarrow 0^{-}} f(x)$ does not exist
- ✓
$\lim _{x \rightarrow 0} f (x)$ does not exist
AnswerCorrect option: D. $\lim _{x \rightarrow 0} f (x)$ does not exist
(D)
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}} \frac{|x|}{x^2+2 x}$
$=\lim _{x \rightarrow 0} \frac{-x}{x^2+2 x}=-\frac{1}{2}$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x^2+2 x}$
$=\lim _{x \rightarrow 0} \frac{x}{x^2+2 x}=\frac{1}{2}$
$\therefore \quad \lim _{x \rightarrow 0} f (x)$ does not exist.
View full question & answer→MCQ 652 Marks
If $f (x)=\left\{\begin{array}{r}\frac{x-|x|}{x} ; \text { when } x \neq 0 \\ 2 ; \text { when } x=0\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=0$
- ✓
$f (x)$ is discontinuous at $x=0$
- C
$\lim _{x \rightarrow 0} f(x)=2$
- D
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=0$
(B)
$\lim _{x \rightarrow 0^{-}} f (x)=1+1=2$
$\lim _{x \rightarrow 0^{+}} f(x)=0$
$\therefore \quad f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 662 Marks
The function $\frac{\sin x}{|x|}$
AnswerCorrect option: B. is discontinuous at $x=0$
(B)
$\lim _{x \rightarrow 0^{+}} \frac{\sin x}{|x|}=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
and $\lim _{x \rightarrow 0^{-}} \frac{\sin x}{|x|}=\lim _{x \rightarrow 0} \frac{\sin x}{-x}=-1$
∴ the given function is discontinuous at $x=0$.
View full question & answer→MCQ 672 Marks
The function $f (x)=\frac{|3 x-4|}{3 x-4}$ is discontinuous at
- A
$x=4$
- B
$x=\frac{3}{4}$
- ✓
$x=\frac{4}{3}$
- D
$x=\frac{2}{3}$
AnswerCorrect option: C. $x=\frac{4}{3}$
(C)
As $\frac{|x|}{x}$ is discontinuous at $x=0$.
$\therefore \quad \frac{|3 x-4|}{3 x-4}$ is discontinuous at $3 x-4=0$.
$\therefore \quad x=\frac{4}{3}$
View full question & answer→MCQ 682 Marks
$\begin{aligned}\text {If} f (x) & =\frac{|x|}{x}, \text { for } x \neq 0 \\ & =1, \text { for } x=0,\end{aligned}$
then $f (x)$ is
AnswerCorrect option: B. discontinuous at $x=0$
(B)
When $x<0,|x|=-x$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0} \frac{-x}{x}=\lim _{x \rightarrow 0}(-1)=-1$
When $x>0,|x|=x$
$\therefore \quad \lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0} \frac{x}{x}=\lim _{x \rightarrow 0}(1)=1$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x) \neq \lim _{x \rightarrow 0^{+}} f (x)$
$\therefore f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 692 Marks
If $f(x)=\left\{\begin{array}{r}\frac{x^4-16}{x-2}, \text { when } x \neq 2 \\ 16, \text { when } x=2\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=2$
- ✓
$f (x)$ is discontinuous at $x=2$
- C
$\lim _{x \rightarrow 2} f(x)=16$
- D
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=2$
(B)
$\lim _{x \rightarrow 2} f (x)=\lim _{x \rightarrow 2} \frac{x^4-16}{x-2}$
$=\lim _{x \rightarrow 2} \frac{(x-2)(x+2)\left(x^2+4\right)}{x-2}$
$=\lim _{x \rightarrow 2}(x+2)\left(x^2+4\right)=32$ and $f(2)=16$
$\therefore \quad \lim _{x \rightarrow 2} f (x) \neq f (2)$
$\therefore f (x)$ is discontinuous at $x=2$.
View full question & answer→MCQ 702 Marks
If $f(x)=\left\{\begin{array}{ll}e^{\frac{1}{x}}, & \text { when } x \neq 0 \\ 1, & \text { when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x)=e$
- B
$\lim _{x \rightarrow 0^{+}} f(x)=0$
- ✓
$f (x)$ is discontinuous at $x=0$
- D
$f (x)$ is continuous at $x=0$
AnswerCorrect option: C. $f (x)$ is discontinuous at $x=0$
(C)
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} e^{-1 / h}=0$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} e^{1 / h}=\infty$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x) \neq \lim _{x \rightarrow 0^{+}} f (x)$
$\therefore f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 712 Marks
$\text { If } f(x)=\left\{\begin{array}{ll}\frac{\sin (a+1) x+\sin x}{x}, & x<0 \\c, & x=0 \\\frac{\sqrt{x+b x^2}-\sqrt{x}}{b \sqrt{x}}, & x>0\end{array}\right.$
is continuous at $x=0$, then
- A
$a=-2, b=0, c \neq 0$
- ✓
$a=-2, b \neq 0, c=0$
- C
$a=2, b=0, c \neq 0$
- D
$a=2, b \neq 0, c=0$
AnswerCorrect option: B. $a=-2, b \neq 0, c=0$
(B)
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}} \frac{\sin ( a +1) x+\sin x}{x}$
$=\lim _{x \rightarrow 0^{-}}\left[\frac{\sin (a+1) x}{(a+1) x} \times(a+1)+\frac{\sin x}{x}\right]$
$=a+1+1$
$=a+2$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+b x^2}-\sqrt{x}}{b \sqrt{x}}$
$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}\left(\begin{array}{ll}\sqrt{1+b x}-1\end{array}\right)}{b \sqrt{x}}$
$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1+ b x}-1}{b}=\frac{0}{b}=0$, if $b \neq 0$
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow a +2=0= c$
$\Rightarrow a =-2, c =0$
$\therefore \quad a=-2, b \neq 0$ and $c=0$
View full question & answer→MCQ 722 Marks
If $f (x)=|x|+|x-1|$, then
- A
$f (x)$ is continuous at $x=0$ only
- B
$f (x)$ is continuous at $x=1$ only
- ✓
$f (x)$ is continuous at both $x=0$ and $x=1$
- D
$f (x)$ is discontinuous
AnswerCorrect option: C. $f (x)$ is continuous at both $x=0$ and $x=1$
(C)
Given, $f (x)=|x|+|x-1|$
$\therefore f (x)=\left\{\begin{array}{ll}-x-(x-1), & \text { if } x<0 \\ x-(x-1), & \text { if } 0 \leq x<1 \\ x+(x-1), & \text { if } x \geq 1\end{array}\right.$
$\therefore f (x)=\left\{\begin{array}{ll}-2 x+1, & \text { if } x<0 \\ 1, & \text { if } 0 \leq x<1 \\ 2 x-1, & \text { if } x \geq 1\end{array}\right.$
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0}(-2 x+1)=1$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0} 1=1$
$f(0)=1$
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)= f (0)$
$\therefore f (x)$ is continuous at $x=0$.
$\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1} 1=1$
$\lim _{x \rightarrow 1^{+}} f (x)=\lim _{x \rightarrow 1}(2 x-1)=1$
$f(1)=2(1)-1=1$
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
$\therefore f (x)$ is continuous at $x=1$.
View full question & answer→MCQ 732 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{x-2}{|x-2|}+a, & x<2 \\ a+b, & x=2 \\ \frac{x-2}{|x-2|}+b, & x>2\end{array}\right.$ is continuous at $x=2$, then $a + b =$
Answer(C)
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{ h \rightarrow 0} f (2- h )$
$=\lim _{h \rightarrow 0} \frac{2-h-2}{|2-h-2|}+a$
$=\lim _{h \rightarrow 0}\left(-\frac{h}{h}+a\right)=a-1$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)$
$=\lim _{h \rightarrow 0} \frac{2+h-2}{|2+h-2|}+b=b+1$ and $f(2)=a+b$
Since $f (x)$ is continuous at $x=2$,
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)= f (2)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow a -1= a + b = b +1$
$\Rightarrow b =-1$ and $a =1$
View full question & answer→MCQ 742 Marks
If $f ( x )$ is continuous at $x=0$, where$f(x)=\left\{\begin{array}{ll}x^2+a, & x \geq 0 \\2 \sqrt{x^2+1}+b, & x<0\end{array}\right.$and $f\left(\frac{1}{2}\right)=2$, then the values of $a$ and $b$ are respectively
- A
$\frac{7}{4}, \frac{1}{4}$
- ✓
$\frac{7}{4},-\frac{1}{4}$
- C
$\frac{-1}{4}, \frac{7}{4}$
- D
$-\frac{7}{4},-\frac{1}{4}$
AnswerCorrect option: B. $\frac{7}{4},-\frac{1}{4}$
(B)
$f \left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+ a$
$\Rightarrow 2=\frac{1}{4}+a \Rightarrow a=\frac{7}{4}\quad\ldots(i)$
Since $f (x)$ is continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(2 \sqrt{x^2+1}+ b \right)=\lim _{x \rightarrow 0^{+}}\left(x^2+ a \right)$
$\Rightarrow 2 \sqrt{0+1}+b=0+a$
$\Rightarrow 2+b=\frac{7}{4} \quad \ldots[From (i)]$
$\Rightarrow b =-\frac{1}{4}$
View full question & answer→MCQ 752 Marks
$f(x)=\left\{\begin{array}{ll}\frac{x^2-9}{x-3}+a , &x>3 \\ 5 , &x=3 \\ 2 x^2+3 x+b, &x<3\end{array}\right.$
is continuous at $x=3$, then
- A
$a=1, b=-22$
- B
$a=1, b=22$
- C
$a=-1, b=22$
- ✓
$a=-1, b=-22$
AnswerCorrect option: D. $a=-1, b=-22$
(D)
Since $f (x)$ is continuous at $x=3$.
$\therefore \quad \lim _{x \rightarrow 3^{-}} f (x)=\lim _{x \rightarrow 3^{+}} f (x)= f (3)$
$\lim _{x \rightarrow 3^{-}} f (x)= f (3)$
$\Rightarrow \lim _{x \rightarrow 3^{-}}\left(2 x^2+3 x+ b \right)=5$
$\Rightarrow 2(3)^2+3(3)+b=5$
$\Rightarrow b =-22$
Also, $\lim _{x \rightarrow 3^{+}} f (x)= f (3)$
$\Rightarrow \lim _{x \rightarrow 3^{+}}\left(\frac{x^2-9}{x-3}+ a \right)=5$
$\Rightarrow(3+3+a)=5$
$\Rightarrow a =-1$
View full question & answer→MCQ 762 Marks
The value of $p$ for which the function
$f(x)=\left\{\begin{array}{ll}\frac{\left(4^{\prime}-1\right)^3}{\sin \left(\frac{x}{p}\right) \log \left(1+\frac{x^2}{3}\right)} & , x \neq 0 \$12)(\log 4)^3 & , x=0\end{array}\right. $
may be continuous at $x=0$, is
Answer(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 12(\log 4)^3=\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)^3}{\sin \frac{x}{p} \log \left(1+\frac{x^2}{3}\right)}$
$\Rightarrow 12(\log 4)^3$ $=\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)^3 \times \frac{\left(\frac{x}{ p }\right)}{\left(\sin \frac{x}{ p }\right)} \times \frac{ p }{\frac{\log \left(1+\frac{1}{3} x^2\right)}{\frac{x^2}{3} \times 3}}$
$\Rightarrow 12(\log 4)^3=(\log 4)^3(1)\left(\frac{3 p }{1}\right)$
$\Rightarrow p =4$
View full question & answer→MCQ 772 Marks
The value of $f(0)$ so that the function $f (x)=\frac{\log \left(\sec ^2 x\right)}{x \sin x}, x \neq 0$, is continuous at $x=0$ is
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{\log \left(\sec ^2 x\right)}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x} \times \frac{\tan ^2 x}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x} \times \frac{\frac{\tan ^2 x}{x^2}}{\frac{\sin x}{x}}$
$=1 \times \frac{1^2}{1}=1$
View full question & answer→MCQ 782 Marks
If $f (x)=\frac{\log _{ e }\left(1+x^2 \tan x\right)}{\sin x^3}, x \neq 0$ is continuous at $x=0$, then $f (0)$ must be defined as
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{\log _e\left(1+x^2 \tan x\right)}{\sin x^3}$
$=\lim _{x \rightarrow 0}\left(\frac{\log _{ e }\left(1+x^2 \tan x\right)}{x^2 \tan x} \cdot \frac{x^2 \tan x}{\sin x^3}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{\log _{ e }\left(1+x^2 \tan x\right)}{x^2 \tan x} \cdot \frac{x^3}{\sin x^3} \cdot \frac{\tan x}{x}\right)$
$\therefore \quad f(0)=1$
View full question & answer→MCQ 792 Marks
$\begin{aligned}f(x) & =\frac{\left(3^{\sin x}-1\right)^2}{x \log (1+x)}, x \neq 0 \\& =k \quad, x=0\end{aligned}$
If f is continuous at $x=0$, then $k =$
- A
$\frac{1}{2} \log 3$
- B
$\log 3$
- C
$2 \log 3$
- ✓
$(\log 3)^2$
AnswerCorrect option: D. $(\log 3)^2$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\left(3^{\sin x}-1\right)^2}{x \log (1+x)}$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\left(\frac{3^{\sin x}-1}{\sin x}\right)^2 \cdot\left(\frac{\sin x}{x}\right)^2}{\frac{\log (1+x)}{x}}$
$\Rightarrow k =\frac{(\log 3)^2 \times(1)^2}{1}=(\log 3)^2$
View full question & answer→MCQ 802 Marks
$\text { If } f(x)=\left\{\begin{array}{cl}\frac{\log (1+2 a x)-\log (1-b x)}{x}, & x \neq 0 \\k, & x=0 \end{array}\right. \text { is }$
continuous at $x=0$, then k is equal to
- ✓
$2 a+b$
- B
$2 a-b$
- C
$b -2 a$
- D
$a+b$
AnswerCorrect option: A. $2 a+b$
(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\log (1+2 a x)-\log (1- b x)}{x}$
$\Rightarrow k =\lim _{x \rightarrow 0}\left[\frac{\log (1+2 a x)}{2 a x} \times 2 a +\frac{\log (1- b x)}{- b x} \times b \right]$
$\Rightarrow k =2 a + b$
View full question & answer→MCQ 812 Marks
$\begin{array}{rlrl}\text {If } f(x) & =\frac{\log x-\log 7}{x-7}, & x \neq 7 \\& =k ,& x=7\end{array}$
is continuous at $x=7$, then the value of $k$ is
- A
$\frac{1}{3}$
- B
$\frac{1}{5}$
- ✓
$\frac{1}{7}$
- D
$\frac{1}{9}$
AnswerCorrect option: C. $\frac{1}{7}$
(C)
Since $f (x)$ is continuous at $x=7$.
$\therefore \quad f(7)=\lim _{x \rightarrow 7} f(x)$
$\Rightarrow k =\lim _{x \rightarrow 7} \frac{\log x-\log 7}{x-7}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow 7} \frac{\frac{1}{x}}{1}=\frac{1}{7}$
View full question & answer→MCQ 822 Marks
$\begin{array}{rlrl}f(x) & =\frac{\log (1+k x)}{\sin x}, & x \neq 0 \\& =5 , & x=0\end{array}$
If f is continuous at $x=0$, then $k =$
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 5=\lim _{x \rightarrow 0} \frac{\log (1+ k x)}{\sin x}$
$\Rightarrow 5=\lim _{x \rightarrow 0} \frac{\frac{\log (1+ k x)}{ k x} \times k }{\frac{\sin x}{2}}$
$\Rightarrow 5=\frac{1 \times k }{1} \Rightarrow k =5$
View full question & answer→MCQ 832 Marks
Function $f(x)=\left\{\begin{array}{ll}\left(\log _2 2 x\right)^{\log _x 8} ; & x \neq 1 \\ (k-1)^3 ; & x=1\end{array}\right.$ is continuous at $x=1$, then $k =$ __________
- ✓
$e+1$
- B
$e^{1 / 3}$
- C
$e ^3$
- D
$e-1$
Answer(A)
$\lim _{x \rightarrow 1}\left(\log _2 2 x\right)^{\log _x 8}$
$=\lim _{x \rightarrow 1}\left[\log _2 2+\log _2 x\right]^{\log _x 2^3}$
$=\lim _{x \rightarrow 1}\left[1+\log _2 x\right]^{3 \log _x 2}=\lim _{x \rightarrow 1}\left[1+\log _2 x\right]^{\frac{3}{\log _2 x}}$
$= e ^{\lim _{x \rightarrow 1} \log _2 x \times \frac{3}{\log _2 x}}= e ^3$
Since the function is continuous at $x=1$,
$\therefore \quad \lim _{x \rightarrow 1} f(x)=f(1)$
$\Rightarrow e ^3=( k -1)^3$
$\Rightarrow e = k -1$
$\Rightarrow k = e + l$
View full question & answer→MCQ 842 Marks
If $f (x)$ is continuous at $x=\frac{\pi}{2}$, where $f(x)=(\sin x)^{\frac{1}{\pi-2x}}$, for $x \neq \frac{\pi}{2}$, then $f\left(\frac{\pi}{2}\right)=$
Answer(C)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$,
$\therefore f \left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f (x)$
$=\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{1}{\pi-2 x}}$
$=\lim _{x \rightarrow \frac{\pi}{2}}[1+(\sin x-1)]^{\frac{1}{\pi-2 x}}$
$= e ^{\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\sin x-1}{\pi-2 x}\right)}$
$= e ^{\left(-\frac{1}{2} \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)}\right)}$
$= e ^0$
$=1$
View full question & answer→MCQ 852 Marks
$\begin{aligned}\text { If }f(x) & =\left(\sec ^2 x\right)^{\cot ^2 x}, & x \neq 0 \\& =k , &x=0 \end{aligned}$
is continuous at $x=0$, then k is equal to
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim_{x \rightarrow 0} \left(\sec ^2 x\right)^{\cot ^2 x}$
$\Rightarrow k =\lim _{x \rightarrow 0}\left(1+\tan ^2 x\right)^{\frac{1}{\tan ^2 x}}$
$\Rightarrow k = e$
View full question & answer→MCQ 862 Marks
If $f (x)=\left\{\begin{array}{ll}\frac{\sin 3 x}{ e ^{2 x}-1} & ; x \neq 0 \\ k -2 & ; x=0\end{array}\right.$ is continuous at $x=0$, then $k =$
- A
$\frac{9}{5}$
- B
$\frac{2}{3}$
- ✓
$\frac{7}{2}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{7}{2}$
(C)
f is continuous at $x=0$
$\Rightarrow \lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x}{ e ^{2 x}-1}= k -2$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{x}}{\frac{ e ^{2 x}-1}{x}}= k -2$
$\Rightarrow \frac{3 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}}{2 \lim _{x \rightarrow 0} \frac{ e ^{2 x}-1}{2 x}}= k -2$
$\Rightarrow \frac{3}{2}= k -2$
$\Rightarrow k=\frac{7}{2}$
View full question & answer→MCQ 872 Marks
$\begin{aligned}\text { If } f(x) & =\left(\frac{4 x+1}{1-4 x}\right)^{\frac{1}{x}}, x \neq 0 \\& =k, \quad x=0\end{aligned}$
is continuous at $x=0$, then $k =$
- A
$e ^2$
- B
$e ^4$
- C
$e^6$
- ✓
$e^8$
Answer(D)
Since $f (x)$ is continuous at $x=0$.)
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}\left(\frac{4 x+1}{1-4 x}\right)^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0} \frac{\left[(1+4 x)^{\frac{1}{4 x}}\right]^4}{\left[(1-4 x)^{-\frac{1}{4 x}}\right]^{-4}}$
$=\frac{e^4}{e^{-4}}=e^8$
View full question & answer→MCQ 882 Marks
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
$f(0)=e^2$
AnswerCorrect option: C. $f(0)=e$
(C)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}(x+1)^{\cot x}$
$=\lim _{x \rightarrow 0}\left\{(1+x)^{\frac{1}{x}}\right\}^{x \cot x}$
$=\lim _{x \rightarrow 0}\left\{(1+x)^{\frac{1}{x}}\right\}^{\lim _{x \rightarrow 0}\left(\frac{x}{\tan x}\right)}$
$= e ^1= e$
View full question & answer→MCQ 892 Marks
The function $f : R -\{0\} \rightarrow R$ given by $f (x)=\frac{1}{x}-\frac{2}{ e ^{2 x}-1}$ can be made continuous at $x=0$ by defining $f(0)$ as
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{2}{ e ^{2 x}-1}\right)$
$=\lim _{x \rightarrow 0} \frac{ e ^{2 x}-1-2 x}{x\left( e ^{2 x}-1\right)}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{2 e^{2 x}-2}{x\left(2 e^{2 x}\right)+1\left(e^{2 x}-1\right)}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{4 e ^{2 x}}{2 x\left(2 e ^{2 x}\right)+ e ^{2 x}(2)+2 e ^{2 x}}$
$\Rightarrow f (0)=\frac{4}{2+2}=1$
View full question & answer→MCQ 902 Marks
If $f (x)=\frac{ e ^x- e ^{ e }}{2(x-\sin x)}, x \neq 0$ is continuous at $x=0$, then $f (0)=$
AnswerCorrect option: C. $\frac{1}{2}$
(C)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{ e ^x e ^{\sin x}}{2(x-\sin x)}$
$=\frac{1}{2} \lim _{x \rightarrow 0} e ^{\sin x}\left(\frac{ e ^{x-\sin x}-1}{x-\sin x}\right)$
$=\frac{1}{2} \times e ^0 \times 1 \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{ e ^x-1}{x}=1\right]$
$=\frac{1}{2}$
View full question & answer→MCQ 912 Marks
$\begin{aligned}\text { If } f(x) & =\frac{e^{5 x}-e^{2 x}}{\sin 3 x}, x \neq 0 \\& =\frac{k}{2}, \quad x=0 \end{aligned}$
is continuous at $x=0$, the value of k is
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow \frac{ k }{2}=\lim _{x \rightarrow 0} \frac{ e ^{5 x}- e ^{2 x}}{\sin 3 x}$
Applying L'Hospital rule on R.H.S., we get
$\frac{ k }{2}=\lim _{x \rightarrow 0} \frac{5 e ^{5 x}-2 e ^{2 x}}{3 \cos 3 x}$
$\Rightarrow \frac{ k }{2}=\frac{5 e ^0-2 e ^0}{3 \cos 0}=\frac{5-2}{3}=1$
$\Rightarrow k =2$
View full question & answer→MCQ 922 Marks
If f is continuous at $x=0$, where $f (x)=\frac{\left( e ^{3 x}-1\right) \sin x^{\circ}}{x^2}, x \neq 0$, then $f (0)$ is
- A
$\frac{\pi}{30}$
- B
$\frac{\pi}{45}$
- ✓
$\frac{\pi}{60}$
- D
$\frac{\pi}{90}$
AnswerCorrect option: C. $\frac{\pi}{60}$
(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{\left( e ^{3 x}-1\right) \sin x^{\circ}}{x^2}$
$=\lim _{x \rightarrow 0} \frac{ e ^{3 x}-1}{3 x} \times 3 \times \frac{\sin \frac{\pi x}{180}}{\frac{\pi x}{180}} \times \frac{\pi}{180}$
$=1 \times 3 \times 1 \times \frac{\pi}{180}=\frac{\pi}{60}$
View full question & answer→MCQ 932 Marks
If $f (x)$ is continuous at $x=0$, where $f(x)=\frac{8^x-2^x}{k^x-1}$, for $x \neq 0 =2$, for $x=0$, then k is equal to
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 2=\lim _{x \rightarrow 0} \frac{8^x-2^x}{k^x-1}$
$\Rightarrow 2=\lim _{x \rightarrow 0} \frac{2^x\left(\frac{4^x-1}{x}\right)}{\frac{k^x-1}{x}}$
$\Rightarrow 2=\frac{2^0 \log 4}{\log k }$
$\Rightarrow 2 \log k =\log 4$
$\Rightarrow 2 \log k =2 \log 2$
$\Rightarrow k =2$
View full question & answer→MCQ 942 Marks
If $f (x)=\frac{3^x+3^{-x}-2}{x^2}, x \neq 0$, is continuous at $x=0$, then $f(0)$ equals
AnswerCorrect option: B. $(\log 3)^2$
(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{3^x+3^{-x}-2}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(3^x-1\right)^2}{x^2}}{3^x}=\frac{(\log 3)^2}{1}=(\log 3)^2$
View full question & answer→MCQ 952 Marks
The value of f at $x=0$ so that the function $f (x)=\frac{2^x-2^{-x}}{x}, x \neq 0$ is continuous at $x=0$, is
- A
$\log 2$
- B
- C
$e^4$
- ✓
$\log 4$
AnswerCorrect option: D. $\log 4$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{2^x-2^{-x}}{x}\right)$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0}\left[\frac{\left(2^x+2^{-x}\right) \log _e 2}{1}\right]$
$=\left(2^0+2^0\right) \log _{ e } 2$
$\therefore \quad f(0)=2 \log _e 2=\log _e 4$
View full question & answer→MCQ 962 Marks
The function defined by
$f (x)=\left\{\begin{array}{cc}\left(x^2+ e ^{\frac{1}{2-x}}\right)^{-1} & , x \neq 2 \\ k & , x=2\end{array}\right.$, is continuous from right at the point $x=2$, then k is equal to
- A
$0$
- ✓
$\frac{1}{4}$
- C
$-\frac{1}{4}$
- D
AnswerCorrect option: B. $\frac{1}{4}$
(B)
If $f (x)$ is continuous from right at $x=2$, then
$f (2)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow k =\lim _{ h \rightarrow 0} f (2+ h )$
$\Rightarrow k =\lim _{ h \rightarrow 0}\left[(2+ h )^2+ e ^{\frac{1}{2-(2+ h )}}\right]^{-1}$
$\Rightarrow k =\lim _{ h \rightarrow 0}\left[4+ h ^2+4 h+ e ^{\frac{-1}{h}}\right]^{-1}$
$\Rightarrow k =\left(4+0+0+ e ^{-\infty}\right)^{-1}$
$\Rightarrow k =\frac{1}{4}$
View full question & answer→MCQ 972 Marks
If the function $f (x)=\frac{\cos ^2 x-\sin ^2 x-1}{\sqrt{x^2+1}-1}, x \neq 0$, is continuous at $x=0$, then $f (0)$ is equal to
Answer(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{\cos ^2 x-\sin ^2 x-1}{\sqrt{x^2+1}-1}$
$=\lim _{x \rightarrow 0} \frac{\left(\cos ^2 x-1\right)-\sin ^2 x}{\sqrt{x^2+1}-1} \times \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$
$=\lim _{x \rightarrow 0} \frac{\left(-\sin ^2 x-\sin ^2 x\right)\left(\sqrt{x^2+1}+1\right)}{x^2+1-1}$
$=\lim _{x \rightarrow 0} \frac{-2 \sin ^2 x}{x^2} \times\left(\sqrt{x^2+1}+1\right)$
$=-2(1)^2\left(\sqrt{0^2+1}+1\right)=-4$
View full question & answer→MCQ 982 Marks
If $f (x)=\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}, x \neq 0$, is continuous at $x=0$, then $f (0)$ is
Answer(A)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{\frac{\cos x}{2 \sqrt{1+\sin x}}+\frac{\cos x}{2 \sqrt{1-\sin x}}}{1}$
$=\frac{1}{2}(1+1)=1$
View full question & answer→MCQ 992 Marks
$f (x)=\frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos ^2 x}, x \neq \frac{\pi}{2}$. The value of $f \left(\frac{\pi}{2}\right)$ so that f is continuous at $x=\frac{\pi}{2}$ is
- A
$\frac{1}{\sqrt{2}}$
- B
$\frac{1}{2 \sqrt{2}}$
- C
$\frac{1}{3 \sqrt{2}}$
- ✓
$\frac{1}{4 \sqrt{2}}$
AnswerCorrect option: D. $\frac{1}{4 \sqrt{2}}$
(D)
For $f (x)$ to be continuous at $x=\frac{\pi}{2}$,
$f \left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f (x)$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos ^2 x}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2-(1+\sin x)}{\left(1-\sin ^2 x\right)(\sqrt{2}+\sqrt{1+\sin x})}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{(1-\sin x)(1+\sin x)(\sqrt{2}+\sqrt{1+\sin x})}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1}{(1+\sin x)(\sqrt{2}+\sqrt{1+\sin x})}$
$=\frac{1}{(1+1)(\sqrt{2}+\sqrt{1+1})}=\frac{1}{4 \sqrt{2}}$
View full question & answer→MCQ 1002 Marks
If the function $f(x)=\left\{\begin{aligned}(\cos x)^{\frac{1}{x}} & ; x \neq 0 \\ k & ; x=0\end{aligned}\right.$ is continuous at $x=0$, then the value of k is
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow k =\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}$
$\Rightarrow \log k =\lim _{x \rightarrow 0} \frac{1}{x} \log (\cos x)$
Applying L'Hospital rule on R.H.S., we get
$\log k =\lim _{x \rightarrow 0} \frac{-\frac{\sin x}{\cos x}}{1}$
$\Rightarrow \log k =0$
$\Rightarrow k = e ^0=1$
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