MCQ 1012 Marks
The value of $f(0)$ so that the function $f(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}, x \neq 0$, is continuous at $x=0$, is
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{\frac{1}{3}(27-2 x)^{\frac{-2}{3}}(-2)}{-\frac{3}{5}(243+5 x)^{\frac{-4}{5}}(5)}=2$
View full question & answer→MCQ 1022 Marks
If $f (x)=\frac{2-\sqrt{x+4}}{\sin 2 x} ;(x \neq 0)$, is continuous function at $x=0$, then $f(0)$ equals
- A
$\frac{1}{4}$
- B
$-\frac{1}{4}$
- C
$\frac{1}{8}$
- ✓
$-\frac{1}{8}$
AnswerCorrect option: D. $-\frac{1}{8}$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{\left(-\frac{1}{2 \sqrt{x+1}}\right)}{2 \cos 2 x}=-\frac{1}{8}$
View full question & answer→MCQ 1032 Marks
If $f (x)=\frac{( a +x)^2 \sin ( a +x)- a ^2 \sin a }{x}, x \neq 0$, then the value of $f (0)$ so that f is continuous at $x=0$ is
AnswerCorrect option: B. $a^2 \cos a+2 a \sin a$
(B)
For $f (x)$ to be continuous at $x=0$,
$f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{( a +x)^2 \sin ( a +x)- a ^2 \sin a }{x}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{2(a+x) \sin (a+x)+(a+x)^2 \cos (a+x)}{1}$
$\Rightarrow f(0)=2 a \sin a+a^2 \cos a$
View full question & answer→MCQ 1042 Marks
If $f (x)=\frac{1-\sin x}{(\pi-2 x)^2}$, when $x \neq \frac{\pi}{2}$ and $f \left(\frac{\pi}{2}\right)=\lambda$, then $f (x)$ will be continuous function at $x=\frac{\pi}{2}$, when $\lambda=$
- ✓
$\frac{1}{8}$
- B
$\frac{1}{4}$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: A. $\frac{1}{8}$
(A)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$.
$\therefore \quad f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)$
$\Rightarrow \lambda=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{(\pi-2 x)^2}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-4(\pi-2 x)}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x}{-4(-2)}$
$\Rightarrow \lambda=\frac{1}{8}$
View full question & answer→MCQ 1052 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}, & x \neq \frac{\pi}{6} \\ a, & x=\frac{\pi}{6}\end{array}\right.$ is continuous at $x=\frac{\pi}{6}$, then $a=$
- A
$\sqrt{3}$
- ✓
$\frac{1}{\sqrt{3}}$
- C
$-\sqrt{3}$
- D
$-\frac{1}{\sqrt{3}}$
AnswerCorrect option: B. $\frac{1}{\sqrt{3}}$
(B)
Since $f(x)$ is continuous at $x=\frac{\pi}{6}$,
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{6}} f(x)=f\left(\frac{\pi}{6}\right)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}= a$
Applying L'Hospital rule to L.H.S, we get
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \cos x+\sqrt{3} \sin x}{6}=a$
$\Rightarrow \frac{3\left(\frac{\sqrt{3}}{2}\right)+\sqrt{3}\left(\frac{1}{2}\right)}{6}= a$
$\Rightarrow \frac{4 \sqrt{3}}{12}=a \Rightarrow a=\frac{1}{\sqrt{3}}$
View full question & answer→MCQ 1062 Marks
If $f(x)=\frac{1-\tan x}{1-\sqrt{2} \sin x}, x \neq \frac{\pi}{4}$, is continuous $x=\frac{\pi}{4}$ and $f\left(\frac{\pi}{4}\right)=k$, then $k=$
- ✓
- B
$2 \sqrt{2}$
- C
- D
$4 \sqrt{2}$
Answer(A)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{1-\sqrt{2} \sin x}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^2 x}{-\sqrt{2} \cos x}=\frac{-2}{-1}=2$
View full question & answer→MCQ 1072 Marks
Let $f(x)=\left\{\begin{array}{cl}\frac{\tan x-\cot x}{x-\frac{\pi}{4}}, & x \neq \frac{\pi}{4} \\ \text { a, } & x=\frac{\pi}{4}\end{array}\right.$
The value of a so that $f(x)$ is continuous , $x=\frac{\pi}{4}$, is
Answer(B)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)$
$\Rightarrow a =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan x-\cot x}{x-\frac{\pi}{4}}$
Applying L'Hospital rule on R.H.S., we get
$a =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sec ^2 x+\operatorname{cosec}^2 x}{1}$
$\Rightarrow a =(\sqrt{2})^2+(\sqrt{2})^2=4$
View full question & answer→MCQ 1082 Marks
Let $f: R \rightarrow R$ be defined by
$f(x)=\left\{\begin{array}{cc}\frac{\cos 3 x-\cos x}{x^2}, & x \neq 0 \\\lambda, & x=0\end{array} .\right.$
If f is continuous at $x=0$, then $\lambda$ is equal to
Answer(B)
Since $f (x)$ is continuous $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow \lambda=\lim _{x \rightarrow 0} \frac{\cos 3 x-\cos x}{x^2}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow 0} \frac{-3 \sin 3 x+\sin x}{2 x}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow 0} \frac{-9 \cos 3 x+\cos x}{2} \Rightarrow \lambda=\frac{-9+1}{2}=-4$
View full question & answer→MCQ 1092 Marks
If a function of defined by
$f(x)=\left\{\begin{array}{cc}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & \text { if } x \neq \frac{\pi}{4} \\k, & \text { if } x=\frac{\pi}{4} \end{array}\right.$
is continuous at $x=\frac{\pi}{4}$, then $k =$
- ✓
$\frac{1}{4}$
- B
- C
$-\frac{1}{4}$
- D
AnswerCorrect option: A. $\frac{1}{4}$
(A)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$,
$\therefore \quad f\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f(x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4} \quad \Rightarrow k =\frac{1}{4}$
View full question & answer→MCQ 1102 Marks
If the function $f(x)=\left\{\begin{array}{cc}\frac{k \cos x}{\pi-2 x}, & \text { when } x \neq \frac{\pi}{2} \\ 3, & \text { when } x=\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then $k =$
Answer(B)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$.
$\therefore \quad f \left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f (x)$
$\Rightarrow 3=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{ k \cos x}{\pi-2 x}\right)$
Applying L'Hospital rule on R.H.S., we get
$3=\lim _{x \rightarrow \frac{\pi}{2}} \frac{ k (-\sin x)}{-2}$
$\Rightarrow 3-\frac{ k }{2} \Rightarrow k -6$
View full question & answer→MCQ 1112 Marks
$\begin{array}{lc}\text {Let } f (x)=\left\{\begin{array}{rc}(1+|\sin x|)^{\frac{1}{|\sin x|}}, & -\frac{\pi}{6} < x<0 \\ b, & x=0 \\ e ^{\frac{\tan 2 x}{\tan 3 x}}, & 0 < x<\frac{\pi}{6}\end{array}\right. \end{array}$
Then the values of $a$ and $b$ if $f$ is continuous at $x=0$, are respectively
- A
$\frac{2}{3}, \frac{3}{2}$
- ✓
$\frac{2}{3}, e ^{\frac{2}{3}}$
- C
$\frac{3}{2}, e ^{\frac{3}{2}}$
- D
AnswerCorrect option: B. $\frac{2}{3}, e ^{\frac{2}{3}}$
(B)
For $f (x)$ to be continuous at $x=0$, we must have
$\lim _{x \rightarrow 0^{-}} f (x)= f (0)=\lim _{x \rightarrow 0^{+}} f (x)$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0^{+}} e ^{\tan 2 x / \tan 3 x}$
$=\lim _{x \rightarrow 0^{+}} e ^{\left(\frac{\tan 2 x}{2 x} \times 2 x\right) /\left(\frac{\tan 3 x}{3 x} \times 3 x\right)}$
$= e ^{\frac{2}{3}}$
$f (0)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow b = e ^{\frac{2}{3}}$
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{ a /|\sin x|}$
$= e ^{\lim _{x \rightarrow 0}\left(|\sin x| \times \frac{ a }{|\sin x|}\right)}= e ^{ a }$
$f (0)=\lim _{x \rightarrow 0^{-}} f (x)$
$\Rightarrow b = e ^{ a } \Rightarrow e ^{\frac{2}{3}}= e ^{ a }$
$\Rightarrow a =\frac{2}{3}$
View full question & answer→MCQ 1122 Marks
If $f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, & x \neq 0 \\ 2 k, & x=0\end{array}\right.$ is continuous at $x=0$, then the value of k is equal to
- ✓
$\frac{3 \pi}{10}$
- B
$\frac{3 \pi}{5}$
- C
$\frac{\pi}{10}$
- D
$\frac{3 \pi}{2}$
AnswerCorrect option: A. $\frac{3 \pi}{10}$
(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 2 k =\lim _{x \rightarrow 0} \frac{3 \sin \pi x}{5 x}$
$=\lim _{x \rightarrow 0} \frac{3 \sin \pi x}{5(\pi x)} \times \pi=\frac{3 \pi}{5}$
$\therefore \quad k =\frac{3 \pi}{10}$
View full question & answer→MCQ 1132 Marks
$\begin{array}{rlr}\text {If } f(x) & =\frac{\cos x-\sin x}{\cos 2 x}, x \neq \frac{\pi}{4} \\& =k, \quad x=\frac{\pi}{4}\end{array}$
is continuous at $x=\frac{\pi}{4}$, then the value of $k$ is
- A
$\sqrt{2}$
- ✓
$\frac{1}{\sqrt{2}}$
- C
$2 \sqrt{2}$
- D
$\frac{1}{2 \sqrt{2}}$
AnswerCorrect option: B. $\frac{1}{\sqrt{2}}$
(B)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos 2 x}$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos ^2 x-\sin ^2 x}$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{(\cos x-\sin x)(\cos x+\sin x)}$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1}{\cos x+\sin x}=\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 1142 Marks
If $f(x)=\left\{\begin{array}{c}\frac{\sin [x]}{[x]+1}, \text { for } x>0 \\ \frac{\cos \frac{\pi}{2}[x]}{[x]}, \text { for } x<0 ; \\ k, \text { for } x=0\end{array}\right.$
where $[x]$ denotes the greatest integer less than or equal to $x$, then in order that f be continuous at $x=0$, the value of k is
AnswerCorrect option: A. Equal to $0$
(A)
For $f (x)$ to be continuous at $x=0$,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0^{-}} f (x)$
$\Rightarrow k =\lim _{ h \rightarrow 0} f (0- h )=\lim _{ h \rightarrow 0} \frac{\cos \frac{\pi}{2}[0- h ]}{[0- h ]}$
$\Rightarrow k =\lim _{ h \rightarrow 0} \frac{\cos _2^\pi[ h ]}{[- h ]}$
$\Rightarrow k =\lim _{ h \rightarrow 0} \frac{\cos \left(-\frac{\pi}{2}\right)}{-1}$
$\Rightarrow k =0$
View full question & answer→MCQ 1152 Marks
If $f (x)=\left\{\begin{array}{c} ax ^2- b , \text { when } 0 \leq x<1 \\ 2, \text { when } x=1 \\ x+1, \text { when } l < x \leq 2\end{array}\right.$ is continuous at
$x=1$, then the most suitable values of $a , b$ are
- A
$a=2, b=0$
- B
$a=1, b=-1$
- C
$a=4, b=2$
- ✓
Answer(D)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad f (1)=\lim _{x \rightarrow 1^{-}} f (x)$
$\Rightarrow 2=\lim _{x \rightarrow 1}\left(a x^2-b\right)$
$\Rightarrow 2=a-b$
The values of $a$ and $b$ in options (A), (B) and (C) satisfies this relation.
$\therefore $ option (D) is the correct answer.
View full question & answer→MCQ 1162 Marks
The function $f (x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$ is not defined at $x=\pi$. The value of $f (\pi)$, so that $f (x)$ is continuous at $x=\pi$, is
- A
$-\frac{1}{2}$
- B
$\frac{1}{2}$
- ✓
- D
Answer(C) For $f (x)$ to be continuous at $x=\pi$,
$f(\pi)=\lim _{x \rightarrow \pi}
f(x)=\lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x}$
Applying L'Hospital rule on R.H.S., we get
$f(\pi)=\lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x}$
$\Rightarrow f (\pi)=-1$
View full question & answer→MCQ 1172 Marks
The value of k which makes $f(x)=\left\{\begin{array}{c}\sin \frac{1}{x}, x \neq 0 \\ k , x=0\end{array}\right.$ continuous at $x=0$ is
Answer(D) If $x \rightarrow 0$, then the value of $\sin \frac{1}{x}$ passes through $[-1,1]$ infinitely many ways, therefore, limit of the function docs not exist at $x=0$. Hence, there is no value of k for which the function is continuous at $x=0$.
View full question & answer→MCQ 1182 Marks
$\begin{aligned}f(x) & =\frac{1-\cos 3 x}{x \tan x}, \text { for } x \neq 0 \\& =k, \text { for } x=0\end{aligned}$
If $f (x)$ is continuous at $x=0$, the value of k is
- A
$\frac{3}{2}$
- B
$\frac{5}{2}$
- C
$\frac{7}{2}$
- ✓
$\frac{9}{2}$
AnswerCorrect option: D. $\frac{9}{2}$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{1-\cos 3 x}{x \tan x}$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{1-\cos 3 x}{x^2} \times \frac{1}{\frac{\tan x}{x}}$
$\Rightarrow k =\frac{3^2}{2} \times l \quad \ldots .\left[\because \lim _{x \rightarrow 0}\left(\frac{1-\cos k x}{x^2}\right)=\frac{ k ^2}{2}\right]$
$\Rightarrow k =\frac{9}{2}$
View full question & answer→MCQ 1192 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & ; \text { when } x<0 \\ a & ; \text { when } x=0 \\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4} & ; \text { when } x>0\end{array}\right.$, is continuous at $x=0$, then the value of ' $a$ ' will be
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0^{-}} f (x)$
$\Rightarrow a =\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^2}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2 2 x}{x^2}$
$=2 \lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{(2 x)^2} \times 4=2 \times 4=8$
View full question & answer→MCQ 1202 Marks
Function $f (x)=\frac{1-\cos 4 x}{8 x^2}$, where $x \neq 0$ and $f (x)= k$, where $x=0$ is a continuous function at $x=0$, then the value of k will be
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{8 x^2}=\lim _{x \rightarrow 0} \frac{2 \sin ^2 2 x}{8 x^2}$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{4 x^2}=1$
View full question & answer→MCQ 1212 Marks
If $f (x)=\frac{\tan \left(x^2-x\right)}{x}, x \neq 0$, is continuous at $x=0$, then $f(0)$ is
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{\tan \left(x^2-x\right)}{x}$
$=\lim _{x \rightarrow 0} \frac{\tan [x(x-1)]}{x(x-1)} \times(x-1)$
$=1 \times(-1)=-1$
View full question & answer→MCQ 1222 Marks
$\begin{aligned}\text { If } f(x) & =\frac{x^4-64 x}{\sqrt{x^2+9}-5}, x \neq 4 \\& =k \quad, x=4\end{aligned}$
is continuous at $x=4$, then $k =$
Answer(D)
Since $f (x)$ is continuous at $x=4$.
$\therefore \quad f (4)=\lim _{x \rightarrow 4} f (x)$
$=\lim _{x \rightarrow 4} \frac{x^4-64 x}{\sqrt{x^2+9}-5}$
$=\lim _{x \rightarrow 4} \frac{x\left(x^3-64\right)\left(\sqrt{x^2+9}+5\right)}{\left(x^2+9\right)-25}$
$=\left(\lim _{x \rightarrow 4} \frac{x^3-4^3}{x^2-4^2}\right)\left[\lim _{x \rightarrow 4} x\left(\sqrt{x^2+9}+5\right)\right]$
$=\frac{3}{2}(4)[4(\sqrt{16+9}+5)]$
$=240$
View full question & answer→MCQ 1232 Marks
If $f(x)=\left\{\begin{array}{lll}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & \text { if } & -1 \leq x<0 \\ \frac{2 x+1}{x-1} & \text { if } & 0 \leq x \leq 1\end{array}\right.$ is continuous at $x=0$, then the value of k is
- A
$k=1$
- ✓
$k=-1$
- C
$k =0$
- D
$k=2$
AnswerCorrect option: B. $k=-1$
(B)
Since $f (x)$ is continuous at $x=0$,
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+ k x}-\sqrt{1- k x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x+1}$
Applying L'Hospital rule on L.H.S, we get
$\lim _{x \rightarrow 0} \frac{\frac{ k }{2 \sqrt{1+ k x}}-\frac{(- k )}{2 \sqrt{1- k x}}}{1}=-1$
$\Rightarrow \frac{ k }{2}+\frac{ k }{2}=-1 \quad \Rightarrow k =-1$
View full question & answer→MCQ 1242 Marks
If $f (x)=\frac{\sqrt{x+3}-2}{x^3-1}, x \neq 1$, is continuous at $x=1$, then $f (1)$ is
- A
- B
$\frac{1}{8}$
- ✓
$\frac{1}{12}$
- D
AnswerCorrect option: C. $\frac{1}{12}$
(C)
Since $f (x)$ is continuous at $x=1$.
$\therefore f (1)=\lim _{x \rightarrow 1} f (x)=\lim _{x \rightarrow 1} \frac{\sqrt{x+3-2}}{x^3-1}$
$=\lim _{x \rightarrow 1} \frac{\sqrt{x+3}-2}{x^3-1^3} \times \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}$
$=\lim _{x \rightarrow 1} \frac{x-1}{(x-1)\left(x^2+x+1\right)(\sqrt{x+3}+2)}$
$=\frac{1}{3(4)}=\frac{1}{12}$
View full question & answer→MCQ 1252 Marks
If $f (x)=\frac{x- a }{\sqrt{x}-\sqrt{ a }}, x \neq a$, is continuous at $x= a$, then $f(a)$ is equal to
- A
$\sqrt{ a }$
- ✓
$2 \sqrt{ a }$
- C
- D
AnswerCorrect option: B. $2 \sqrt{ a }$
(B)
Since $f (x)$ is continuous at $x= a$.
$\therefore \quad f(a)=\lim _{x \rightarrow a} f(x)$
$=\lim _{x \rightarrow a } \frac{x- a }{\sqrt{x}-\sqrt{ a }} \times \frac{\sqrt{x}+\sqrt{ a }}{\sqrt{x}+\sqrt{ a }}$
$=\lim _{x \rightarrow a}(\sqrt{x}+\sqrt{a})=\sqrt{a}+\sqrt{a}=2 \sqrt{a}$
View full question & answer→MCQ 1262 Marks
If $f (x)=\left\{\begin{array}{cc}\frac{x^6-\frac{1}{64}}{x^3-\frac{1}{8}}, & x \neq \frac{1}{2} \\ k , & x=\frac{1}{2}\end{array}\right.$ is continuous at $x=\frac{1}{2}$, then the value of k is
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{5}$
AnswerCorrect option: C. $\frac{1}{4}$
(C)
Since $f (x)$ is continuous at $x=\frac{1}{2}$.
$\therefore \quad f \left(\frac{1}{2}\right)=\lim _{x \rightarrow \frac{1}{2}} f (x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{1}{2}} \frac{x^6-\frac{1}{64}}{x^3-\frac{1}{8}}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow \frac{1}{2}} \frac{6 x^5}{3 x^2}=\lim _{x \rightarrow \frac{1}{2}} 2 x^3=2\left(\frac{1}{2}\right)^3=\frac{1}{4}$
View full question & answer→MCQ 1272 Marks
Let $f (x)=\left\{\begin{array}{lc}5^{\frac{1}{x}} ; & x<0 \\ \lambda[x] ; & x \geq 0, \lambda \in R \end{array}\right.$, then at $x=0$
AnswerCorrect option: A. f is continuous whatever $\lambda$ may be
(A)
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}} 5^{\frac{1}{x}}=\lim _{ h \rightarrow 0} 5^{-\frac{1}{h}}=0$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0^{+}} \lambda[x]=0$, for all $\lambda \in R$
$f(0)=\lambda(0)=0$
$\therefore f$ is continuous at $x=0$, whatever $\lambda$ may be.
View full question & answer→MCQ 1282 Marks
Which of the following functions is continuous at $x=0$ ?
- A
$f(x)=\left\{\begin{array}{cl}\frac{\sin 2 x}{x} ; & x \neq 0 \\ 1 ; & x=0\end{array}\right.$
- B
$f(x)=\left\{\begin{array}{cl}(1+x)^{\frac{1}{x}} ; & x \neq 0 \\ 1 ; & x=0\end{array}\right.$
- C
$f (x)=\left\{\begin{array}{cc} e ^{\frac{-1}{x}} ; & x \neq 0 \\ 1 ; & x=0\end{array}\right.$
- ✓
$f(x)=\left\{\begin{array}{cc}\frac{3 x+4 \tan x}{x} ; & x \neq 0 \\ 7 ; & x=0\end{array}\right.$
AnswerCorrect option: D. $f(x)=\left\{\begin{array}{cc}\frac{3 x+4 \tan x}{x} ; & x \neq 0 \\ 7 ; & x=0\end{array}\right.$
(D)
$\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=2 \neq f (0)$
$\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e \neq f(0)$
$\lim _{x \rightarrow 0} e ^{\frac{-1}{x}}=\lim _{x \rightarrow 0} \frac{1}{ e ^{\frac{1}{x}}}=\frac{1}{ e ^{\infty}}=\frac{1}{\infty}=0 \neq f (0)$
$\lim _{x \rightarrow 0}\left(\frac{3 x}{x}+\frac{4 \tan x}{x}\right)=3+4=7= f (0)$
$\therefore f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1292 Marks
If $f(x)=\left\{\begin{array}{c}\frac{x}{e^{\frac{1}{x}}+1}, \text { when } x \neq 0 \\ 0, \text { when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x)=1$
- B
$\lim _{x \rightarrow 0^{-}} f(x)=1$
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is not continuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)$
$=\lim _{h \rightarrow 0} \frac{-h}{e^{\frac{-1}{h}}+1}=\lim _{h \rightarrow 0} \frac{-h}{1+\frac{1}{e^{\frac{1}{h}}}}=0$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{h}{e^{\frac{1}{h}}+1}=0$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)= f (0)$
$\therefore \quad f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1302 Marks
If $f (x)=\left\{\begin{array}{cc}x^* \sin \frac{1}{x} ; & x \neq 0 \\ 0 ; & x=0\end{array}\right.$ is continuous an $x=0$, then
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0} f (x)= f (0)$
$\therefore \quad \lim _{x \rightarrow 0} x^{ a } \sin \frac{1}{x}=0$, if $a >0$
View full question & answer→MCQ 1312 Marks
If $f(x)=\left\{\begin{array}{c}x^2 \sin \frac{1}{x} ; \text { when } x \neq 0 \\ 0 ; \text { when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x)=1$
- B
$\lim _{x \rightarrow 0^{-}} f(x)=-1$
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is discontinuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f(x)-\lim _{x \rightarrow 0} x^2 \sin \frac{1}{x}$, but $-1 \leq \sin \frac{1}{x} \leq 1$ and $x \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow 0^{+}} f (x)=0=\lim _{x \rightarrow 0^{-}} f (x)= f (0)$
$\therefore f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1322 Marks
If $f (x)=\left\{\begin{array}{r}\sin ^{-1}|x| \text {; when } x \neq 0 \\ 0 \text {; when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x) \neq 0$
- B
$\lim _{x \rightarrow 0^{-}} f(x) \neq 0$
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is not continuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f(x)=\sin ^{-1}(0)=0=f(0)$
$\therefore \quad f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1332 Marks
If $f(x)=\left\{\begin{array}{c}1 ; \text { when } 0< x \leq \frac{3 \pi}{4} \\ 2 \sin \frac{2}{9} x ; \text { when } \frac{3 \pi}{4}< x<\pi\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=0$
- B
$f (x)$ is continuous at $x=\pi$
- ✓
$f(x)$ is continuous at $x=\frac{3 \pi}{4}$
- D
$f (x)$ is discontinuous at $x=\frac{3 \pi}{4}$
AnswerCorrect option: C. $f(x)$ is continuous at $x=\frac{3 \pi}{4}$
(C)
Here, $f \left(\frac{3 \pi}{4}\right)=1$ and $\lim _{x \rightarrow \frac{3 \pi^{-}}{4}} f (x)=1$
$\lim _{x \rightarrow \frac{3 \pi^{+}}{4}} f (x)=\lim _{ h \rightarrow 0} 2 \sin \frac{2}{9}\left(\frac{3 \pi}{4}+ h \right)$
$=2 \sin \frac{\pi}{6}=1$
$\therefore \quad f (x)$ is continuous at $x-\frac{3 \pi}{4}$.
View full question & answer→MCQ 1342 Marks
If $f (x)=|x-2|$, then
- A
$\lim _{x \rightarrow z^{+}} f(x) \neq 0$
- B
$\lim _{x \rightarrow 2^{-}} f (x) \neq 0$
- C
$\lim _{x \rightarrow 2^{+}} f(x) \neq \lim _{x \rightarrow 2^{-}} f(x)$
- ✓
$f (x)$ is continuous at $x=2$
AnswerCorrect option: D. $f (x)$ is continuous at $x=2$
(D)
Here, f(2) = 0
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{ h \rightarrow 0} f (2- h )=\lim _{ h \rightarrow 0}|2- h -2|=0$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}|2+h-2|=0$
$\therefore \quad f (x)$ is continuous at $x=2$.
View full question & answer→MCQ 1352 Marks
If $f(x)$ is continuous on $[-4,2]$, where $f(x)=\left\{\begin{array}{ll}6 b-3 a x, & \text { for }-4 \leq x<-2 \\ 4 x+1, & \text { for }-2 \leq x \leq 2\end{array}\right.$, then $a+b=$
- A
$\frac{1}{6}$
- B
$-\frac{1}{6}$
- C
$\frac{7}{6}$
- ✓
$-\frac{7}{6}$
AnswerCorrect option: D. $-\frac{7}{6}$
(D)
Since $f(x)$ is continuous on $[-4,2]$.
$\therefore \quad$ it is continuous at $x=-2$.
$\therefore \quad \lim _{x \rightarrow-2^{-}} f (x)=\lim _{x \rightarrow-2^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow-2^{-}}(6 b-3 a x)=\lim _{x \rightarrow-2^{+}}(4 x+1)$
$\Rightarrow 6 b-3 a (-2)=4(-2)+1$
$\Rightarrow 6 b+6 a=-7$
$\Rightarrow a + b =-\frac{7}{6}$
View full question & answer→MCQ 1362 Marks
If $f(x)$ is continuous in $[-2,2]$. where$f(x)=\left\{\begin{array}{ll}x+a, & x<0 \\x, & 0 \leq x<1, \text { then } \\b-x, & x \geq 1\end{array}\right.$
Answer(A)
Since $f(x)$ is continuous in $[-2,2]$.
$\therefore $ it is continuous at $x=0$ and $x=1$.
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}(x+a)=\lim _{x \rightarrow 0^{+}} x$
$\Rightarrow a=0$
Also, $\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}} x=\lim _{x \rightarrow 1^{+}}( b -x)$
$\Rightarrow l = b -1 \Rightarrow b=2$
View full question & answer→MCQ 1372 Marks
If the function $f(x)=\left\{\begin{array}{cc}5 x-4, & \text { if } 0< x \leq 1 \\ 4 x^2+3 b x, & \text { if } 1< x< 2\end{array}\right.$ is continuous at every point of its domain, they value of $b$ is
Answer(A)
Since $f (x)$ is continuous at every point of its domain.
∴ it is continuous at $x=1$.
$\therefore \quad \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1}(5 x-4)=\lim _{x \rightarrow 1}\left(4 x^2+3 b x\right)$
$\Rightarrow 1=4+3 b \Rightarrow b=-1$
View full question & answer→MCQ 1382 Marks
If f(x) = $\frac{x+1}{(x-2)(x-5)}$, then in $[0,1], f(x)$ is
- ✓
- B
- C
continuous except at x = 0
- D
discontinuous except at x = 0
Answer(A)
$f (x)$ being a rational function, is continuous in $[0,1]$ except at those points where the denominator $(x-2)(x-5)=0$
i.e. when $x=2$ or $x=5$
Since $2,5 \notin[0,1]$
$\therefore f (x)$ is continuous in $[0,1]$.
View full question & answer→MCQ 1392 Marks
If $f (x)=\sqrt{x-2}, 2< x< 4$, then
- ✓
f(x) is continuous in (2, 4)
- B
f(x) is discontinuous in (2, 4)
- C
f(x) is continuous in (2, 4) except at x = 3
- D
f(x) is discontinuous in (2, 4) except at x = 3
AnswerCorrect option: A. f(x) is continuous in (2, 4)
(A)
$\lim _{x \rightarrow 3} f (x)=\lim _{x \rightarrow 3} \sqrt{x-2}=1$
$f(3)=\sqrt{3-2}=1$
$\therefore \quad \lim _{x \rightarrow 3} f (x)= f (3)$
$\therefore f (x)$ is continuous at $x=3$.
Since $3 \in(2,4)$
$\therefore f (x)$ is continuous in $(2,4)$.
View full question & answer→MCQ 1402 Marks
Function $f (x)=\left\{\begin{array}{cc}x-1, & x<2 \\ 2 x-3, & x \geq 2\end{array}\right.$ is continuous
Answer(A)
For $x<2, f (x)=x-1$
Since f is a polynomial function, it is continuous for all $x<2$.
For $x>2, f (x)=2 x-3$
Since f is a polynomial function, it is continuous for all $x>2$.
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2}(x-1)=1$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(2 x-3)=1$
$f(2)=1$
$\therefore f (x)$ is continuous for all real values of $x$.
View full question & answer→MCQ 1412 Marks
If $f (x)=\left\{\begin{array}{l}x, x \geq 0 \\ x^2, x<0\end{array}\right.$, then $f (x)$ is
- ✓
- B
- C
continuous on R except at x = 0
- D
discontinuous on R except at x = 0
Answer(A)
For $x>0, f (x)=x$
Since f is a polynomial function, it is continuous for all $x>0$.
For $x<0, f (x)=x^2$
Since f is a polynomial function, it is continuous for all $x<0$.
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0} x^2=0$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0^{+}} x=0$
$f(0)=0$
$\therefore f (x)$ is continuous at $x=0$.
$\therefore f (x)$ is continuous on R .
View full question & answer→MCQ 1422 Marks
If f and g are both continuous at $x= a$, then $f - g$ is
View full question & answer→MCQ 1432 Marks
The points at which the function $f (x)=\frac{x+1}{x^2+x-12}$ is discontinuous are
- A
$-3, 4$
- ✓
$3, -4$
- C
- D
$-3, -4$
AnswerCorrect option: B. $3, -4$
(B)
$f (x)=\frac{x+1}{(x-3)(x+4)}$
$\therefore f (x)$ is discontinuous at $x=3,-4$.
View full question & answer→MCQ 1442 Marks
$f (x)=\frac{x^2+x-2}{x^2-3 x+2}$ is discontinuous at $x=$
Answer(B)
$f (x)$ is discontinuous, when $x^2-3 x+2=0$
i.e., $(x-1)(x-2)=0 \Rightarrow x=1, x=2$
View full question & answer→MCQ 1452 Marks
If $f (x)=\left\{\begin{array}{l}x \sin x ; 0 < x \leq \frac{\pi}{2} \\ \frac{\pi}{2} \sin (\pi+x) ; \quad \frac{\pi}{2}< x< \pi\end{array}\right.$, then
- ✓
$f (x)$ is discontinuous at $x=\frac{\pi}{2}$
- B
$f (x)$ is continuous at $x=\frac{\pi}{2}$
- C
$f (x)$ is continuous at $x=0$
- D
AnswerCorrect option: A. $f (x)$ is discontinuous at $x=\frac{\pi}{2}$
(A)
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x)=\lim _{x \rightarrow \frac{\pi}{2}} x \sin x=\frac{\pi}{2}$
$\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\pi}{2} \sin (\pi+x)=\frac{-\pi}{2}$
$\therefore f (x)$ is discontinuous at $x=\frac{\pi}{2}$.
View full question & answer→MCQ 1462 Marks
If $f (x)=\left\{\begin{array}{c}\frac{x^2-4 x+3}{x^2-1}, \text { for } x \neq 1 \\ 2, \text { for } x=1\end{array}\right.$, then
- A
$\lim _{x \rightarrow 1^{+}} f(x)=2$
- B
$\lim _{x \rightarrow 1^{-}} f(x)=3$
- ✓
$f (x)$ is discontinuous at $x=1$
- D
$f (x)$ is continuous at $x=1$
AnswerCorrect option: C. $f (x)$ is discontinuous at $x=1$
(C)
$\lim _{x \rightarrow 1} f (x)=\lim _{x \rightarrow 1} \frac{x^2-4 x+3}{x^2-1}$
$=\lim _{x \rightarrow 1} \frac{(x-3)}{(x+1)}=-1$
$f(1)=2$
$\therefore \quad \lim _{x \rightarrow 1} f (x) \neq f (1)$
$\therefore f (x)$ is discontinuous at $x=1$.
View full question & answer→MCQ 1472 Marks
$\begin{aligned}\text {If} \ f (y) & =y^2-y-1, & & \text { for } 0 \leq y<2 \\ & =4 y+1, & & \text { for } 2 \leq y \leq 4, \text { then }\end{aligned}$
- A
$f (y)$ is continuous at $y=2$
- ✓
$f (y)$ is discontinuous at $y=2$
- C
$\lim _{y \rightarrow 2^{-}} f (y)=9$
- D
$\lim _{y \rightarrow 2^{+}} f (y)=1$
AnswerCorrect option: B. $f (y)$ is discontinuous at $y=2$
(B)
$\lim _{y \rightarrow 2^{-}} f (y)=\lim _{y \rightarrow 2^{-}}\left(y^2-y-1\right)$
$=4-2-1=1$
$\lim _{y \rightarrow 2^{+}} f (y)=\lim _{y \rightarrow 2^{+}}(4 y+1)$
$=8+1=9$
$\therefore \quad \lim _{y \rightarrow 2^{-}} f (y) \neq \lim _{y \rightarrow 2^{+}} f (y)$
$\therefore f (y)$ is discontinuous at $y=2$.
View full question & answer→MCQ 1482 Marks
If $f(x)=\left\{\begin{array}{l}1+x^2, \text { when } 0 \leq x \leq 1 \\ 1-x, \text { when } x>1\end{array}\right.$, then
- A
$\lim _{x \rightarrow 1^{+}} f(x) \neq 0$
- B
$\lim _{x \rightarrow 1^{-}} f(x) \neq 2$
- ✓
$f (x)$ is discontinuous at $x=1$
- D
$f (x)$ is continuous at $x=1$
AnswerCorrect option: C. $f (x)$ is discontinuous at $x=1$
(C)
$\lim _{x \rightarrow 1^{+}} f (x)=\lim _{x \rightarrow 1}(1-x)=0$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(1+x^2\right)$
$=1+1^2$
$=2$
$\therefore \quad \lim _{x \rightarrow 1^{+}} f (x) \neq \lim _{x \rightarrow 1^{-}} f (x)$
$\therefore f (x)$ is discontinuous at $x=1$.
View full question & answer→MCQ 1492 Marks
If $f(x)=\left\{\begin{array}{l}\frac{5}{2}-x, \text { when } x<2 \\ 1, \text { when } x=2 \\ x-\frac{3}{2}, \text { when } x>2\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=2$
- ✓
$f (x)$ is discontinuous at $x=2$
- C
$\lim _{x \rightarrow 2} f (x)=1$
- D
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=2$
(B)
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{5}{2}-x\right)=\frac{1}{2}$
$\lim _{x \rightarrow 2^{+}} f (x)=\lim _{x \rightarrow 2}\left(x-\frac{3}{2}\right)=\frac{1}{2}$ and $f (2)=1$
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{+}} f (x) \neq f (2)$
$\therefore f (x)$ is discontinuous at $x=2$.
View full question & answer→MCQ 1502 Marks
If $f (x)=\left\{\begin{array}{c}x^2 ; \text { when } x \leq 1 \\ x+5 ; \text { when } x>1\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=1$
- ✓
$f (x)$ is discontinuous at $x=1$
- C
$\lim _{x \rightarrow 1^{+}} f(x)=1$
- D
$\lim _{x \rightarrow 1^{-}} f(x)=6$
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=1$
(B)
$\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1} x^2=1$
$\lim _{x \rightarrow 1^{+}} f (x)=\lim _{x \rightarrow 1}(x+5)=6$
$\therefore \quad f (x)$ is discontinuous at $x=1$.
View full question & answer→