MCQ 1512 Marks
If $f (x)=\left\{\begin{array}{ll} ax +1 & , x \leq \frac{\pi}{2} \\ \sin x+ b & , x>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then
- A
$a=1, b=0$
- B
$a=b \frac{\pi}{2}+1$
- ✓
$b =\frac{ a \pi}{2}$
- D
$a = b =\frac{\pi}{2}$
AnswerCorrect option: C. $b =\frac{ a \pi}{2}$
(C)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$.
$\therefore \quad \lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f (x)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}(a x+1)=\lim _{x \rightarrow \frac{\pi}{2}}(\sin x+b)$
$\Rightarrow a \cdot \frac{\pi}{2}+1=1+ b \quad \Rightarrow b =\frac{ a \pi}{2}$
View full question & answer→MCQ 1522 Marks
$f (x)=\left\{\begin{array}{ll}\frac{x^2-4}{x-2}+ a , & \text { for } x<2 \\ 8, & \text { for } x=2 \\ x+ b +4, & \text { for } x>2\end{array}\right.$ is continuous at $x=2$, then the values of a and b are respectively
Answer(B)
Since $f (x)$ is continuous at $x=2$.
$\therefore \quad f (2)=\lim _{x \rightarrow 2^{-}} f (x)$
$\Rightarrow f (2)=\lim _{x \rightarrow 2}\left(\frac{x^2-4}{x-2}+ a \right) \Rightarrow 8=4+ a$
$\Rightarrow a =4$
Also, $f (2)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow f (2)=\lim (x+ b +4) \Rightarrow 8=6+ b$
$\Rightarrow b =2$
View full question & answer→MCQ 1532 Marks
If $f (x)=\left\{\begin{array}{r}x+\lambda, x<3 \\ 4, x=3 \\ 3 x-5, x>3\end{array}\right.$ is continuous at $x=3$, then
Answer(D)
Since $f (x)$ is continuous at $x=3$.
$\therefore f (3)=\lim _{x \rightarrow 3^{-}} f (x)$
$\Rightarrow 4=3+\lambda$
$\Rightarrow \lambda=1$
View full question & answer→MCQ 1542 Marks
If $f (x)=\left\{\begin{array}{rc}x^2+ k & ; \quad x \geq 0 \\ -x^2- k & ; \quad x<0\end{array}\right.$ is continuous at $x=0$, then k is equal to
Answer(A)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim (2 x+1)=\lim \left(3-k x^2\right)$
$\Rightarrow 2+1=3-k(1)^2$
$\Rightarrow k =0$
View full question & answer→MCQ 1552 Marks
If $f (x)=\left\{\begin{array}{ll}2 x+1, & x \leq 1 \\ 3- k x^2, & x>1\end{array}\right.$ is continuous at $x=1$, then the value of k is
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(-x^2- k \right)=\lim _{x \rightarrow 0^{+}}\left(x^2+ k \right)$
$\Rightarrow- k = k$
$\Rightarrow k =0$
View full question & answer→MCQ 1562 Marks
$f(x)=\left\{\begin{array}{cc}3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5\end{array}\right.$ is continuous at $x=5$, find k .
- A
$\frac{4}{7}$
- B
$\frac{2}{7}$
- ✓
$\frac{7}{2}$
- D
$\frac{3}{7}$
AnswerCorrect option: C. $\frac{7}{2}$
(C)
$\lim _{x \rightarrow 5^{-}} f (x)=\lim _{x \rightarrow 5}(3 x-8)=7$
$\lim _{x \rightarrow 5^{+}} f (x)=\lim _{x \rightarrow 5} 2 k =2 k$
Since $f (x)$ is continuous at $x=5$.
$\therefore \quad \lim _{x \rightarrow 5^{-}} f (x)=\lim _{x \rightarrow 5^{+}} f (x)$
$\Rightarrow 7=2 k \Rightarrow k =\frac{7}{2}$
View full question & answer→MCQ 1572 Marks
If $f(x)=\left\{\begin{array}{ll}k x^2 & \text { if } x \leq 2 \\ 3 & \text { if } x>2\end{array}\right.$ is continuous at $x=2,$ then the value of $k$ is
- A
- B
- ✓
$\frac{3}{4}$
- D
$\frac{4}{3}$
AnswerCorrect option: C. $\frac{3}{4}$
(C)
$f(2)=k(2)^2=4 k$
$\lim _{x \rightarrow 2^{+}} f (x)=\lim _{x \rightarrow 2^{+}} 3=3$
Since the function is continous at $x=2$,
$\lim _{x \rightarrow 2^{+}} f (x)= f (2)$
$\Rightarrow 4 k =3$
$\Rightarrow k =\frac{3}{4}$
View full question & answer→MCQ 1582 Marks
If $f (x)=\left\{\begin{array}{cl}2 x+5 & ; x>1 \\ k & ; x=1, \\ 8 x-1 & ; x<1\end{array}\right.$ is continuous at $x=1$, then the value of k is
Answer(C)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad \lim _{x \rightarrow 1^{-}} f (x)= f (1)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}(8 x-1)= k$
$\Rightarrow k =7$
View full question & answer→MCQ 1592 Marks
The function ' f ' is defined by $f (x)=2 x-1 \ if \ x>2, f (x)= k$ if $x=2$ and $x^2-1$ if $x<2$ continuous, then the value of $k$ is equal $t_0$
Answer(B)
Since $f (x)$ is continuous at $x=2$.
$\therefore \lim _{x \rightarrow 2^{-}} f (x)= f (2)$
$\Rightarrow \lim _{x \rightarrow 2^{-}}\left(x^2-1\right)= k$
$\Rightarrow k =3$
View full question & answer→MCQ 1602 Marks
If $f(x)=\left\{\begin{array}{cl}x: & 0 \leq x<\frac{1}{2} \\ 1-x: & \frac{1}{2} \leq x<1\end{array}\right.$, then
- ✓
$f ( x )$ is continuous at $x=\frac{1}{2}$
- B
$f ( x )$ is discontinuous at $x=\frac{1}{2}$
- C
$\lim _{x \rightarrow \frac{1}{2}^{-}} f(x)=1$
- D
$\lim _{x \rightarrow \frac{1}{2}^{+}} f(x)=1$
AnswerCorrect option: A. $f ( x )$ is continuous at $x=\frac{1}{2}$
(A)
$f \left(\frac{1}{2}\right)=1-\frac{1}{2}=\frac{1}{2}$
$\lim _{x \rightarrow \frac{1}{2}^{-}} f (x)=\lim _{x \rightarrow \frac{1}{2}^{-}}(x)=\frac{1}{2}$
$\lim _{x \rightarrow \frac{1}{2}^{+}} f(x)=\lim _{x \rightarrow \frac{1^{+}}{2}}(1-x)=1-\frac{1}{2}=\frac{1}{2}$
$\therefore \lim _{x \rightarrow \frac{1}{2}^{-}} f (x)=\lim _{x \rightarrow \frac{1}{2}^{+}} f (x)= f \left(\frac{1}{2}\right)$
$\therefore f (x)$ is continuous at $x=\frac{1}{2}$.
View full question & answer→MCQ 1612 Marks
$f(x)=\left\{\begin{array}{cl}2 x+1, & x<1 \\ 2, & x=1 \ is\\ x^2+1, & x>1\end{array}\right.$
Answer(C)
$\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{-}}(2 x+1)=3 \neq f (1)$
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^2+1\right)=2=f(1)$
View full question & answer→MCQ 1622 Marks
The function $f (x)=\frac{\log (1+ ax )-\log (1- bx )}{x}$ is not defined at $x=0$. The value which should be assigned to f at $x=0$ so that it is continuous at $x=0$, is
- A
$a - b$
- ✓
$a+b$
- C
$\log a+\log b$
- D
$\log a-\log b$
Answer(B)
For $f (x)$ to be continuous at $x=0$,
$f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{\log (1+ ax )-\log (1- bx )}{x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{\frac{ a }{1+ a x}+\frac{ b }{1- b x}}{1}$
$\Rightarrow f (0)= a + b$
View full question & answer→MCQ 1632 Marks
If $f(x)=\left\{\begin{array}{c}\frac{1-\cos x}{x}, x \neq 0 \\ k , x=0\end{array}\right.$ is continuous at $x=0$, then $k =$
- ✓
$0$
- B
$\frac{1}{2}$
- C
$\frac{1}{4}$
- D
$-\frac{1}{2}$
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}=\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{x}{4}$
$\therefore \quad f(0)=2(1)(0)=0$
Alternate method:
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\rightarrow f (0)=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \sin x=0$
View full question & answer→MCQ 1642 Marks
If f is continuous at $x=0$, where $f (x)=\frac{\left( e ^{3 x}-1\right) \sin x}{x^2}, x \neq 0$, then $f (0)=$
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right) \sin x}{x^2}$
$=\lim _{x \rightarrow 0} \frac{ e ^{3 x}-1}{3 x} \times 3 \times \frac{\sin x}{x}=1 \times 3 \times 1$
$\therefore f(0)=3$
View full question & answer→MCQ 1652 Marks
In order that the function $f (x)=(x+1)^{\frac{1}{x}}$ is continuous at $x=0, f (0)$ must be defined as
- A
$f(0)=0$
- ✓
$f(0)=e$
- C
$f(0)=\frac{1}{e}$
- D
$f(0)=1$
AnswerCorrect option: B. $f(0)=e$
(B)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}= e$
View full question & answer→MCQ 1662 Marks
If $f (x)=\left\{\begin{array}{cl}\frac{\log _e x}{x-1}, & x \neq 1 \\ k , & x=1\end{array}\right.$ is continuous at $x=1$, then the value of k is
Answer(B)
Since $f (x)$ is continuous at $x=1$.
$\therefore f (1)=\lim _{x \rightarrow 1} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 1} \frac{\log x}{x-1}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow 1} \frac{\frac{1}{x}}{1}=1$
View full question & answer→MCQ 1672 Marks
Let $f(x)=\left\{\begin{array}{ll}\frac{\sin \pi x}{5 x} ; & x \neq 0 \\ k ; & x=0\end{array}\right.$.If $f(x)$ is continuous at $x=0$, then $k =$
- ✓
$\frac{\pi}{5}$
- B
$\frac{5}{\pi}$
- C
- D
$0$
AnswerCorrect option: A. $\frac{\pi}{5}$
(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\sin \pi x}{5 x}$
$\Rightarrow k =\lim _{x \rightarrow 0}\left(\frac{\sin \pi x}{\pi x}\right) \cdot \frac{\pi}{5}$
$\Rightarrow k =(1) \cdot \frac{\pi}{5}$
$\Rightarrow k =\frac{\pi}{5}$
View full question & answer→MCQ 1682 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{\sin 3 x}{x}, & x \neq 0 \\ \frac{k}{2}, & x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ is
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow \frac{ k }{2}=\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \cdot 3$
$\Rightarrow \frac{k}{2}=3 \Rightarrow k=6$
View full question & answer→MCQ 1692 Marks
If $f (x)=\frac{2 x+\tan x}{x}, x \neq 0$, is continuous at $x=0$, then $f (0)$ equals
Answer(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{2 x+\tan x}{x}$
$=\lim _{x \rightarrow 0}\left(2+\frac{\tan x}{x}\right)=2+1=3$
View full question & answer→MCQ 1702 Marks
If $f (x)=\sin x-\cos x, x \neq 0$, is continuous at $x=0$, then $f (0)$ is equal to
Answer(B)
Since $f (x)$ is continuous $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}(\sin x-\cos x)$
$=\sin 0-\cos 0=-1$
View full question & answer→MCQ 1712 Marks
If the function$f(x)=\left\{\begin{array}{cc}\frac{x^2-(A+2) x+A}{x-2}, & \text { for } x \neq 2 \\2, & \text { for } x=2\end{array}\right. $is continuous at $x=2$, then
- ✓
$A=0$
- B
$A=1$
- C
$A=-1$
- D
$A=2$
Answer(A)
Since $f (x)$ is continuous at $x=2$.
$\therefore f (2)=\lim _{x \rightarrow 2} f (x)$
$\Rightarrow 2=\lim _{x \rightarrow 2} \frac{x^2-( A +2) x+ A }{x-2}$
$\Rightarrow 2=\lim _{x \rightarrow 2} \frac{x(x-2)- A (x-1)}{x-2}$,
which is true if $A =0$
View full question & answer→MCQ 1722 Marks
For the function $f (x)=\left\{\begin{array}{cc}\frac{x^3- a ^3}{x- a }, & x \neq a \\ b , & x= a \end{array}\right.$
If $f (x)$ is continuous at $x= a$, then b is equal to
- A
$a^2$
- B
$2 a ^2$
- ✓
$3 a ^2$
- D
$4 a^2$
AnswerCorrect option: C. $3 a ^2$
(C)
Since $f (x)$ is continuous at $x= a$.
$\therefore \quad f ( a )=\lim _{x \rightarrow a } f (x)$
$\Rightarrow b =\lim _{x \rightarrow a } \frac{x^3- a ^3}{x_{-} a }$
$\Rightarrow b =3 a ^{3-1}=3 a ^2$
View full question & answer→MCQ 1732 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{x^2-9}{x-3} ; & \text { if } x \neq 3 \\ 2 x+k ; & \text { otherwise }\end{array}\right.$, is continuous at $x=3$, then $k =$
Answer(B)
Since $f (x)$ is continuous at $x=3$.
$\therefore f (3)=\lim _{x \rightarrow 3} f (x)$
$\Rightarrow 2(3)+ k =\lim _{x \rightarrow 3} \frac{x^2-9}{x-3}$
$\Rightarrow 6+ k =\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{x-3}$
$\Rightarrow 6+ k =\lim _{x \rightarrow 3}(x+3)$
$\Rightarrow 6+ k =6 \Rightarrow k =0$
View full question & answer→MCQ 1742 Marks
If function $f(x)=\left\{\begin{array}{r}\frac{x^2-1}{x-1}, \text { when } x \neq 1 \\ k, \text { when } x=1\end{array}\right.$ is continuous at $x=1$, then the value of k will be
Answer(B)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad f (1)=\lim _{x \rightarrow 1} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 1} \frac{x^2-1}{x-1}$
$\Rightarrow k =\lim _{x \rightarrow 1}(x+1)$
$\Rightarrow k =2$
View full question & answer→MCQ 1752 Marks
If $f(x)=\frac{x^2-10 x+25}{x^2-7 x+10}$ for $x \neq 5$ and $f$ is continuous at $x=5$, then $f (5)=$
Answer(A)
Since $f (x)$ is continuous at $x=5$.
$\therefore f (5)=\lim _{x \rightarrow 5} f (x)=\lim _{x \rightarrow 5} \frac{x^2-10 x+25}{x^2-7 x+10}$
$=\lim _{x \rightarrow 5} \frac{(x-5)^2}{(x-2)(x-5)}=\frac{5-5}{5-2}=0$
View full question & answer→MCQ 1762 Marks
If $f (x)=\left\{\begin{array}{cl}\left(1+\frac{4 x}{5}\right)^{\frac{1}{x}}, & x \neq 0 \\ e ^{\frac{4}{5}}, & x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0} f(x)=e^{\frac{2}{5}}$
- B
$\lim _{x \rightarrow 0} f(x)$ does not exist
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is discontinuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0}\left(1+\frac{4 x}{5}\right)^{\frac{1}{x}}$
$=\left[\lim _{x \rightarrow 0}\left(1+\frac{4 x}{5}\right)^{\frac{5}{4 x}}\right]^{\frac{4}{5}}= e ^{\frac{4}{5}}= f (0)$
View full question & answer→MCQ 1772 Marks
If $f (x)=\left\{\begin{array}{cc}\frac{\sin x}{x}+\cos x, & x \neq 0 \\ 2, & x=0\end{array}\right.$, then
- A
$f (x)$ is discontinuous at $x=0$
- B
$\lim _{x \rightarrow 0} f (x)=1$
- ✓
$f (x)$ is continuous at $x=0$
- D
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}+\cos x\right)$
$\therefore f (x)$ is continuous at $x=0$
View full question & answer→MCQ 1782 Marks
$\text {If } f(x)=\left\{\begin{array}{ll}\frac{1}{2} \sin x^2, & x \neq 0 \\0, & x=0\end{array}\right. \text {, then }$
- A
$\lim _{x \rightarrow 0} f(x)=\frac{1}{2}$
- B
$f (x)$ is discontinuous at $x=0$
- ✓
$f (x)$ is continuous at $x=0$
- D
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{1}{2} \sin x^2=0= f (0)$
$\therefore \quad f (x)$ is continuous at $x=0$.
View full question & answer→