MCQ 11 Mark
If $f ( x )=\lfloor x\rfloor$ for $x \in(-1,2)$, then $f$ is discontinuous at
- A
$x =-1,0,1,2$
- B
$x=-1,0,1$
- ✓
$x=0,1$
- D
$x=2$
AnswerCorrect option: C. $x=0,1$
(C) $x=0,1$
Hint:
$f(x)=\lfloor x\rfloor, x \in(-1,2)$
This function is discontinuous at all integer values of $x$ between -1 and 2 . $\therefore f ( x )$ is discontinuous at $x =0$ and $x =1$.
View full question & answer→MCQ 21 Mark
If $f(x)=\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}$, for $x \neq 0$ and $f(0)=k$, is continuous at $x=0$, then $k$ is
- A
$e ^7$
- B
$e^3$
- ✓
$e^{12}$
- D
$e^{\frac{3}{4}}$
AnswerCorrect option: C. $e^{12}$
(C) $e^{12}$
Hint:
$f (x)$ is continuous at $x=0$
$f(0)=\lim _{x \rightarrow 0} f(x)$
$=\lim _{x \rightarrow 0}\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}$
$=\lim _{x \rightarrow 0}\left[\frac{4\left(1+\frac{5 x}{4}\right)}{4\left(1-\frac{7 x}{4}\right)}\right]^{\frac{4}{x}}$
$=\frac{\lim _{x \rightarrow 0}\left[\left(1+\frac{5 x}{4}\right)^{\frac{4}{5 x}}\right]^5}{\lim _{x \rightarrow 0}\left[\left(1-\frac{7 x}{4}\right)^{\frac{-4}{7 x}}\right]^{-7}}$
$=\frac{ e ^5}{ e ^{-7}} \cdots\left[\because x \rightarrow 0, \frac{5 x}{4} \rightarrow 0, \frac{-7 x}{4} \rightarrow 0, \text { and } \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}= e \right]$
$= e ^{12}$
View full question & answer→MCQ 31 Mark
If $f(x)=\frac{12^x-4^{x^2-3^x+1}}{1-\cos 2 x}$, for $x \neq 0$ is continuous at $x=0$ then the value of $f(0)$ is
AnswerCorrect option: B. $\log 2 \cdot \log 3$
(B) $\log 2 \cdot \log 3$
Hint:
$f (x)$ is continuous at $x=0$.
(given)
$f(0)=\lim _{x \rightarrow 0} f(x)$
$=\lim _{x \rightarrow 0} \frac{12^x-4^x-3^x+1}{1-\cos 2 x}$
$=\frac{1}{2} \lim _{x \rightarrow 0} \frac{4^x\left(3^x-1\right)\left(3^x-1\right)}{\sin ^2 x}$
$=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\left(3^x-1\right)\left(4^x-1\right)}{\sin ^2 x} $
$=\frac{1}{2} \frac{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)}{\left(\lim \frac{\sin x}{x}\right)^2}$
$=\frac{1}{2} \times \frac{(\log 3) \times(\log 4)}{(1)^2} $
$=\frac{1}{2} \times \log 3 \times \log (2)^2$
$=\log 3 \cdot \log 2$
View full question & answer→MCQ 41 Mark
$f(x)=\frac{32^x-8^x-4^x+1}{4^x-2^{x+1}+1}, \text { for } x \neq 0$
$=k, \text { for } x=0$
is continuous at $x=0$, then value of ' $k$ ' is
- ✓
$6$
- B
$4$
- C
$(\log 2)(\log 4)$
- D
$3 \log 4$
Answer$(A) 6$
Hint:
$f (x)$ is continuous at $x=0$.
...(given)
$f(0)=\lim _{x \rightarrow 0} f(x)$
$k =\lim _{x \rightarrow 0} \frac{32^x-8^x-4^x+1}{4^x-2^{x+1}+1}$
$=\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)\left(8^x-1\right)}{\left(2^x-1\right)^2}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)\left(\frac{8^x-1}{x}\right)}{\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)^2}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{8^x-1}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{2^x-1}{x}\right)^2}$
$=\frac{\log 4 \times \log 8}{(\log 2)^2}$
$=\frac{2 \log 2 \times 3 \log 2}{(\log 2)^2}=6$
View full question & answer→MCQ 51 Mark
$f(x)=\frac{\left(16^x-1\right)\left(9^x-1\right)}{\left(27^x-1\right)\left(32^x-1\right)}$, for $x \neq 0$ $=k_{\text {, }}$ for $x=0$
is continuous at $x=0$, then ' $k$ ' =
- A
$\frac{8}{3}$
- ✓
$\frac{8}{15}$
- C
$-\frac{8}{15}$
- D
$\frac{20}{3}$
AnswerCorrect option: B. $\frac{8}{15}$
(B) $\frac{8}{15}$
Hint:
$f (x) \text { is continuous at } x=0$
$f(0)=\lim _{x \rightarrow-0} f(x) $
$k =\lim _{x \rightarrow-0} \frac{\left(16^x-1\right)\left(9^x-1\right)}{\left(27^x-1\right)\left(32^x-1\right)} $
$=\frac{\lim _{x \rightarrow-0}\left(\frac{16^x-1}{x}\right) \times \lim _{x \rightarrow-0}\left(\frac{9^x-1}{x}\right)}{\lim _{x \rightarrow-0}\left(\frac{27^x-1}{x}\right) \times \lim _{x \rightarrow-0}\left(\frac{32^x-1}{x}\right)} $
$=\frac{\log 16 \times \log 9}{\log 27 \times \log 32} \quad \cdots\left[\because \lim _{x \rightarrow 0} \frac{ a ^x-1}{x}=\log a \right] $
$=\frac{4 \log 2 \times 2 \log 3}{3 \log 3 \times 5 \log 2}$
$=\frac{8}{15} $
View full question & answer→MCQ 61 Mark
If $f(x)=a x^2+b x+1$, for $|x-1| \geq 3$ and $=4 x+5$, for $-2is continuous everywhere then,
- ✓
$a =\frac{1}{2}, b =3$
- B
$a=-\frac{1}{2}, b=-3$
- C
$a=-\frac{1}{2}, b=3$
- D
$a=\frac{1}{2}, b=-3$
AnswerCorrect option: A. $a =\frac{1}{2}, b =3$
(A) $a=\frac{1}{2}, b=3$
Hint:
$f(x)=a x^2+b x+1,|x-1| \geq 3$
$=4 x+5 ;-2$
The first interval is
$|x-1| \geq 3 $
$\therefore x-1 \geq 3 \text { or } x-1 \leq-3 $
$\therefore x \geq 4 \text { or } x \leq-2$
$\therefore f ( x )$ is same for $x \leq-2$ as well as $x \geq 4$.
$\therefore f ( x )$ is defined as:
$f(x)=a x^2+b x+1 ; x \leq-2 $
$=4 x+5 ;-2=a x^2+b x+1 ; x \geq 4$
$f(x)$ is continuous everywhere.
$\therefore f ( x )$ is continuous at $x =-2$ and $x =4$.
As $f(x)$ is continuous at $x=-2$,
$\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x) $
$\therefore \lim _{x \rightarrow-2}\left(a x^2+b x+1\right)=\lim _{x \rightarrow-2}(4 x+5) $
$\therefore a(-2)^2+ b (-2)+1=4(-2)+5 $
$\therefore 4 a-2 b+1=-3$
$\therefore 4 a-2 b=-4$
$\therefore 2 a-b=-2 \ldots . .( i )$
$\because f ( x )$ is continuous at $x =4$,
$\lim _{x \rightarrow 4^{-}} f (x)=\lim\limits_{x \rightarrow 4^{+}} f (x) $
$\therefore \lim _{x \rightarrow 4}(4 x+5)=\lim _{x \rightarrow 4}\left(a x^2+b x+1\right) $
$4(4)+5=a(4)^2+ b (4)+1 $
$16 a +4 b +1=21$
$16 a +4 b =20$
$4 a + b =5 \ldots . .( ii )$
Adding $(i)$ and $(ii),$ we get
$6 a=3$
$\therefore a=\frac{1}{2}$
Substituting $a=\frac{1}{2}$ in (ii), we get
$4\left(\frac{1}{2}\right)+b=5 $
$\therefore 2+b=5 $
$\therefore b=3$
$\therefore a=\frac{1}{2}, b=3$
View full question & answer→MCQ 71 Mark
$f(x)=\frac{x^2-7 x+10}{x^2+2 x-8}$, for $x \in[-6,-3]$
- A
$f$ is discontinuous at $x=2$
- ✓
$f$ is discontinuous at $x=-4$
- C
$f$ is discontinuous at $x=0$
- D
$f$ is discontinuous at $x=2$ and $x=-4$
AnswerCorrect option: B. $f$ is discontinuous at $x=-4$
(B) $f$ is discontinuous at $x =-4$
Hint:
$f(x)=\frac{x^2-7 x+10}{x^2+2 x-8}, \text { for } x \in[-6,-3] $
$=\frac{x^2-7 x+10}{(x+4)(x-2)}$
Here $f(x)$ is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when $x=-4$ or $x=2$
But $x=2$ does not lie in the given interval.
$\therefore x =-4$ is the point of discontinuity.
View full question & answer→MCQ 81 Mark
If $f(x)=\frac{(\sin 2 x) \tan 5 x}{\left(e^{2 x}-1\right)^2}$, for $x \neq 0$ is continuous at $x=0$, then $f(0)$ is
- A
$\frac{10}{e^2}$
- B
$\frac{10}{e^4}$
- C
$\frac{5}{4}$
- ✓
$\frac{5}{2}$
AnswerCorrect option: D. $\frac{5}{2}$
(D) $\frac{5}{2}$
Hint:
$f (x)$ is continuous at $x=0$
$f (0)$
$=\lim _{x \rightarrow 0} \frac{(\sin 2 x)(\tan 5 x)}{\left( e ^{2 x}-1\right)^2} $
$=\frac{\lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \times \lim _{x \rightarrow 0} \frac{\tan 5 x}{5 x} \times 2 \times 5}{\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{2 x}\right)^2 \times(2)^2} $
$=\frac{1 \times 1 \times 2 \times 5}{(1)^2 \times 4} \cdots$
$[\because x \rightarrow 0,2 x \rightarrow 0,5 x \rightarrow 0 , \text { and } \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1, \lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1]$
$=\frac{5}{2}$
View full question & answer→MCQ 91 Mark
If $f(x)=\frac{1-\sqrt{2} \sin x}{\pi-4 x}$, for $x \neq \frac{\pi}{4}$ is continuous at $x=\frac{\pi}{4}$, then $f\left(\frac{\pi}{4}\right)=$
- A
$\frac{1}{\sqrt{2}}$
- B
$-\frac{1}{\sqrt{2}}$
- C
$-\frac{1}{4}$
- ✓
$\frac{1}{4}$
AnswerCorrect option: D. $\frac{1}{4}$
(D) $\frac{1}{4}$
Hint:
$f(x)$ is continuous at $x=\frac{\pi}{4}$
$\therefore \quad f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x) $
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x} $
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}\left(\sin x-\frac{1}{\sqrt{2}}\right)}{4\left(x-\frac{\pi}{4}\right)} $
$=\frac{\sqrt{2}}{4} \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\sin \frac{\pi}{4}}{x-\frac{\pi}{4}} $
$=\frac{\sqrt{2}}{4} \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \cos \left(\frac{\left.x+\frac{\pi}{4}\right)}{2}\right) \cdot \sin \left(\frac{\left.x-\frac{\pi}{4}\right)}{2}\right)}{x-\frac{\pi}{4}}$
$=\frac{\sqrt{2}}{4} \cdot \lim _{x \rightarrow \frac{\pi}{4}} \cos \left(\frac{x}{2}+\frac{\pi}{8}\right) \cdot \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin \left(\frac{x-\frac{\pi}{4}}{2}\right)}{\frac{x-\frac{\pi}{4}}{2}}$
$=\frac{\sqrt{2}}{4} \cdot \cos \left(\frac{x}{8}+\frac{\pi}{8}\right) \times 1\ldots$
$\left[\because x \rightarrow \frac{\pi}{4}, x-\frac{\pi}{4} \rightarrow 0, \frac{x-\frac{\pi}{4}}{2} \rightarrow 0 ,\text { and } \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\frac{\sqrt{2}}{4} \times \cos \frac{\pi}{4}$
$=\frac{1}{4} \quad$
View full question & answer→MCQ 101 Mark
$\begin{aligned}
& f(x)=\frac{2^{\mathrm{cotx}}-1}{\pi-2 x}, \text { for } x \neq \frac{\pi}{2} \\
& =\log \sqrt{2}, \text { for } x=\frac{\pi}{2}
\end{aligned}$
- ✓
$f$ is continuous at $x=\frac{\pi}{2}$
- B
$f$ has a jump discontinuity at $x=\frac{\pi}{2}$
- C
f has a removable discontinuity
- D
$\lim _{x \rightarrow \frac{\pi}{2}} f(x)=2 \log 3$
AnswerCorrect option: A. $f$ is continuous at $x=\frac{\pi}{2}$
(A) $f$ is continuous at $x=\frac{\pi}{2}$
Hint:
$
\begin{aligned}
& f\left(\frac{\pi}{2}\right)=\log \sqrt{2} \\
& \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2^{\cot x}-1}{\pi-2 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2^{tan(\frac{\pi}{2}-xt)}-1}{2(\frac{\pi}{2}-x)}
\end{aligned}
$
Put $\frac{\pi}{2}-x=h$
As $x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0$
$
\begin{aligned}
& \therefore \quad \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} \frac{2^{\text{tan h}}-1}{2 \mathrm{~h}} \\
& =\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{2^{\text{tan h}}-1}{\tan h} \times \frac{\tan h}{h}\right) \\
& \ldots(\because \mathrm{h} \rightarrow 0, \tan h \rightarrow 0, \tan h \neq 0) \\
& =\frac{1}{2} \lim _{h \rightarrow 0} \frac{2^{\tanh }-1}{\tanh } \times \lim _{h \rightarrow 0} \frac{\tan h}{h} \\
& =\frac{1}{2} \cdot \log 2 \cdot(1) \\
& =\log \sqrt{2}=f\left(\frac{\pi}{2}\right) \\
&
\end{aligned}
$
$\therefore \quad \mathrm{f}(x)$ is continuous at $x=\frac{\pi}{2}$
View full question & answer→MCQ 112 Marks
The number of discontinuities of the greatest integer function $f(x)=[x], x \in\left(-\frac{7}{2}, 100\right)$ is
Answer(d) : Given $f(x)=[x], x \in\left(\frac{-7}{2}, 100\right)$ i.e.. $x \in(-3 \cdot 5,100)$
As we know that greatest integer function is discontinuous on integer value.
In interval $(-3.5,100)$, the integer values are $-3,-2,-1,0$, 99 i.e., 103 in number
$\therefore \quad f(x)$ is discontinuous at 103 points.
View full question & answer→MCQ 122 Marks
The function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x)=\left\{\begin{array}{cc}\frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right) & , x \neq 0 \\ k & , \quad x=0\end{array}\right.$ is continuous at $x=0$, then $k$ is
AnswerSo, $\lim _{x \rightarrow 0} f(x)=f(0)=k$
$\Rightarrow k=\lim _{x \rightarrow 0} \frac{\log \frac{(1+3 x)}{(1-2 x)}}{x}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{(1+3 x)}{(1-2 x)}-1\right)}{\left(\frac{3 x+1}{1-2 x}-1\right) x} \times\left[\left(\frac{3 x+1}{1-2 x}\right)-1\right]$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{5 x}{1-2 x}\right)}{\frac{5 x}{1-2 x}} \times \frac{5}{1-2 x}$
$[$As we know that $\lim _{f(x) \rightarrow 0} \frac{\log (1+f(x))}{f(x)}=1]$
$\Rightarrow \lim _{x \rightarrow 0} \frac{5}{1-2 x}=5$
$\Rightarrow k=5$
View full question & answer→MCQ 132 Marks
The function $f(x)=[x] \cdot \cos \left(\frac{2 x-1}{2}\right) \pi$, where $[.]$ denotes the greatest integer function, is discontinuous at
AnswerCorrect option: B. no $x$.
(b) : As, $f(x)=[x] \cos \left(\frac{2 x-1}{2}\right) \pi$
For a function to be continuous LHL $=$ RHL $=f(x)$
So, let us check continuity for $x=a$ where $a$ is an integer
L.HL at $x=a$ is given by
$
\begin{aligned}
& \lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{-}}[x] \cos \left(\frac{2 x-1}{2}\right) \pi \\
& =0 \\
&
\end{aligned}
$
RHL at $x=a$ is given by
$
\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}}[x] \cos \left(\frac{2 x-1}{2}\right) \pi=0
$
Also $f(a)=[a] \cos \left(\frac{2 a-1}{2}\right) \pi=0$
$
\Rightarrow LHL = RHL =f(x), \text { for } x=a
$
So, $f(x)$ will be continuous for all $x$.
View full question & answer→MCQ 142 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{x-4}{|x-4|}+a, & \text { for } x<4 \\ a+b, & \text { for } x=4 \\ \frac{x-4}{|x-4|}+b, & \text { for } x>4\end{array}\right.$
is continuous at $x=4$, then
- A
$a=1, b=1$
- ✓
$a=1, b=-1$
- C
$a=0, b=0$
- D
$a=-1, b=1$
AnswerCorrect option: B. $a=1, b=-1$
(b) : Since $f(x)$ is continuous at $x=4$ So, $\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(x)$ at $x=4-1+a=a+b=1+b$ Now, $a-1=a+b \Rightarrow b=-1$ and $a+b=1+b \Rightarrow a=1$ Hence, $a=1, b=-1$.
View full question & answer→MCQ 152 Marks
If $f(x)\left\{\begin{array}{rr}\log \left(\sec ^2 x\right)^{\cot 2 x}, & \text { for } x \neq 0 \\ K+1, & \text { for } x=0\end{array}\right.$ is continuous at $x=0$, then value of $K$ is
Answer(a) : $f(x)=\left\{\begin{array}{cc}\log \left(\sec ^2 x\right)^{\cot ^2 x}, & x \neq 0 \\ K+1 & , x=0\end{array}\right.$ is continuous at $x=0$.
$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \log \left(\sec ^2 x\right)^{\cot ^2 x}=K+1 \\
& \Rightarrow \lim _{x \rightarrow 0} \cot ^2 x \log \left(1+\tan ^2 x\right)=K+1 \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x}=K+1 \\
& \Rightarrow K+1=1 \Rightarrow K=0 \quad \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]
\end{aligned}
$
View full question & answer→MCQ 162 Marks
If $f(x)=\left\{\begin{array}{ll}x, & \text { if } x \leq 1 \\ 7, & \text { if } x>1\end{array}\right.$ then at $x=1$
AnswerCorrect option: D. $f$ is discontinuous
(d) $: f(1)=1$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x=1, \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 7=7$
Since, $f(1) \neq \lim _{x \rightarrow 1^{+}} f(x)$
$\therefore \quad f$ is discontinuous at $x=1$
View full question & answer→MCQ 172 Marks
Determine the value of $k$ for which the function $f(x)$ is continuous at $x=4$.$f(x)= \begin{cases}\frac{x^2-16}{x-4}, & x \neq 4 \\ k, & x=4\end{cases}$
Answer(d) : Since $f(x)$ is continuous at $x=4$. Therefore, $\lim _{x \rightarrow 4} f(x)=f(4)$
$\Rightarrow \lim _{x \rightarrow 4} f(x)=k \quad[\because f(4)=k]$
$\Rightarrow \lim _{x \rightarrow 4} \frac{x^2-16}{x-4}=k \Rightarrow \lim _{x \rightarrow 4} \frac{(x-4)(x+4)}{x-4}=k$
$\Rightarrow \lim _{x \rightarrow 4}(x+4)=k \Rightarrow k=8$
View full question & answer→MCQ 182 Marks
Which of the following function is not continuous at $x=0$ ?
- A
$f(x)= \begin{cases}(1+2 x)^{1 / x}, & x \neq 0 \\ e^2, & x=0\end{cases}$
- B
$f(x)= \begin{cases}\sin x-\cos x, & x \neq 0 \\ -1, & x=0\end{cases}$
- ✓
$f(x)= \begin{cases}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ -1, & x=0\end{cases}$
- D
$f(x)= \begin{cases}\frac{e^{5 x}-e^{2 x}}{\sin 3 x}, & x \neq 0 \\ 1, & x=0\end{cases}$
AnswerCorrect option: C. $f(x)= \begin{cases}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ -1, & x=0\end{cases}$
(c) $f(x)= \begin{cases}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ -1, & x=0\end{cases}$
View full question & answer→MCQ 192 Marks
If the function $f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}$, $x \neq 0$ is continuous at $x=0$, then $f(0)=$
- A
$\log a-\log b$
- ✓
$a+b$
- C
$\log a+\log b$
- D
$a-b$
Answer(b) : If $f(x)$ is continuous at $x=0$, then $\lim _{x \rightarrow 0} f(x)=f(0)$
$
\therefore f(0)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}
$
$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\log (1+a x)}{x}-\lim _{x \rightarrow 0} \frac{\log (1-b x)}{x} \\
& =\lim _{x \rightarrow 0} \frac{a}{1+a x}+\lim _{x \rightarrow 0} \frac{b}{1+b x} \quad \text { [By L.H. Rule] } \\
& =a+b
\end{aligned}
$
View full question & answer→MCQ 202 Marks
If function $f(x)=\left\{\begin{array}{ll}x-\frac{|x|}{x}, & x<0 \\ x+\frac{|x|}{x}, & x>0 \\ 1, & x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{-}} f(x)$ does not exist
- B
$\lim _{x \rightarrow 0^{+}} f(x)$ does not exist
- ✓
$f(x)$ is continuous at $x=0$
- D
$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$
AnswerCorrect option: C. $f(x)$ is continuous at $x=0$
(c) : We have, $f(0)=1$
$
\begin{aligned}
& \begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(x-\frac{|x|}{x}\right)=\lim _{x \rightarrow 0}\left(x+\frac{x}{x}\right) \\
&=\lim _{x \rightarrow 0}(x+1)=1 \\
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(x+\frac{|x|}{x}\right)=\lim _{x \rightarrow 0}\left(x+\frac{x}{x}\right) \\
&=\lim _{x \rightarrow 0}(x+1)=1
\end{aligned}
\end{aligned}
$
Thus, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\therefore f(x)$ is continuous at $x=0$.
View full question & answer→MCQ 212 Marks
If the function $f(x)=\left\{\begin{array}{l}\frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}, x \neq 0 \\ 16, x=0\end{array}\right.$ is continuous at $x=0$, then $k=$
- A
$\pm \frac{1}{8}$
- B
$\pm 4$
- C
$\pm 2$
- ✓
$\pm 8$
AnswerCorrect option: D. $\pm 8$
Since, $f(x)$ is continuous at $x=0$.
$\therefore \lim _{x \rightarrow 0} f(x)=f(0) $
$\Rightarrow \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}=16$
$ \Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right)}{k x} \times \frac{\tan k x}{k x} \times k^2=16$
$ \Rightarrow \frac{k^2}{4} \lim _{x \rightarrow 0} \frac{e^{k x}-1}{k x} \times \lim _{x \rightarrow 0} \frac{\tan k x}{k x}=16 $
$\Rightarrow \frac{k^2}{4} \times 1 \times 1=16$
$ \Rightarrow k^2=64 $
$\Rightarrow k= \pm 8$
View full question & answer→MCQ 222 Marks
If $f(x)=\frac{e^{x^2}-\cos x}{x^2}$, for $x \neq 0$ is continuous at $x=0$, then value of $f(0)$ is
- A
$\frac{2}{3}$
- B
$\frac{5}{2}$
- C
- ✓
$\frac{3}{2}$
AnswerCorrect option: D. $\frac{3}{2}$
(d) : Given, $f(x)=\frac{e^{x^2}-\cos x}{x^2}$ is continuous at $x=0$
$\Rightarrow$ L.H.L. $=$ R.H.L $=f(0)$ at $x=0$
L.H.L. $($ at $x=0)=\lim _{h \rightarrow 0} \frac{e^{(0-h)^2}-\cos (0-h)}{(0-h)^2}$
$=\lim _{h \rightarrow 0} \frac{e^{h^2}-\cos h-1+1}{h^2}=\lim _{h \rightarrow 0}\left(\frac{e^{h^2}-1}{h^2}+\frac{1-\cos h}{h^2}\right)$
$=\lim _{h \rightarrow 0} \frac{e^{h^2}-1}{h^2}+\lim _{h \rightarrow 0} \frac{2 \sin ^2\left(\frac{h}{2}\right)}{h^2}$
$=1+2 \lim _{h \rightarrow 0} \frac{\sin ^2(h / 2)}{(h / 2)^2} \times \frac{1}{4}=1+\frac{2 \times 1}{4}=\frac{3}{2}$
Hence, $f(0)=\frac{3}{2}$
View full question & answer→MCQ 232 Marks
If $f(x)=x^2+\alpha$ for $x \geq 0$ $=2 \sqrt{x^2+1}+\beta$ for $x<0$ is continous at $x=0$ and $f\left(\frac{1}{2}\right)=2$, then $\alpha^2+\beta^2$ is
- A
- B
$\frac{8}{25}$
- ✓
$\frac{25}{8}$
- D
$\frac{1}{3}$
AnswerCorrect option: C. $\frac{25}{8}$
(c) : We have $f(x)=\left\{\begin{array}{c}x^2+\alpha \text { for } x \geq 0 \\ 2 \sqrt{x^2+1}+\beta \text { for } x<0\end{array}\right.$
$\therefore \quad$ L.H.L. $($ at $x=0)=\lim _{h \rightarrow 0} 2 \sqrt{(0-h)^2+1}+\beta$
$=\lim _{h \rightarrow 0} 2 \sqrt{h^2+1}+\beta=2+\beta$
and R.H.L. $($ at $x=0)=\lim _{h \rightarrow 0}(0+h)^2+\alpha=\lim _{h \rightarrow 0} h^2+\alpha=\alpha$
Now, $f(0)=(0)^2+\alpha=\alpha$
Since $f(x)$ is continuous at $x=0$.
$\Rightarrow$ L.H.L. at $x=0=$ R.H.L at $x=0=f(0)$
$\Rightarrow 2+\beta=\alpha$
Also $f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+\alpha \Rightarrow \frac{1}{4}+\alpha=2 \Rightarrow \alpha=\frac{7}{4}$.
Put $\alpha=\frac{7}{4}$ in (i), we get $2+\beta=\frac{7}{4} \Rightarrow \beta=\frac{-1}{4}$
Now, $\alpha^2+\beta^2=\left(\frac{7}{4}\right)^2+\left(\frac{-1}{4}\right)^2=\frac{49}{16}+\frac{1}{16}=\frac{25}{8}$
View full question & answer→MCQ 242 Marks
If $f(x)=\left\{\begin{array}{cc}\log \left(\sec ^2 x\right)^{\cot ^2 x} & \text { for } x \neq 0 \\ K & \text { for } x=0\end{array}\right.$ is continuous at $x=0$ then $K$ is
Answer(b) : Since $f(x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} \log \left(\sec ^2 x\right)^{\cot ^2 x}$
$\Rightarrow K=\lim _{x \rightarrow 0} \cot ^2 x \cdot \log \left(1+\tan ^2 x\right) \Rightarrow K=\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x}$
$\Rightarrow K=1 \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
View full question & answer→MCQ 252 Marks
If the function $f(x)=\left\{\left[\begin{array}{cc}{\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{1 / x}} & \text { for } x \neq 0 \\ K & \text { for } x=0\end{array}\right.\right.$ is continuous at $x=0$, then $K=$ ?
- A
$e$
- B
$e^{-1}$
- ✓
$e^2$
- D
$e^{-2}$
Answer(c) : We are given that $f(x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow K=\lim _{x \rightarrow 0}\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\left[(1+\tan x)^{\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}{\left.(1-\tan x)^{-\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}$ $=\frac{e}{e^{-1}}=e^2$
View full question & answer→MCQ 262 Marks
For what value of $k$, the function defined by $f(x)=\left\{\begin{array}{cc}\frac{\log (1+2 x) \sin x^{\circ}}{x^2} & \text { for } x \neq 0 \\ k & \text { for } x \neq 0\end{array}\right.$ is continuous at $x=0$ ?
- A
- B
$\frac{1}{2}$
- ✓
$\frac{\pi}{90}$
- D
$\frac{90}{\pi}$
AnswerCorrect option: C. $\frac{\pi}{90}$
(c) : $f(x)= \begin{cases}\frac{\log (1+2 x) \sin x^0}{x^2} & , x \neq 0 \\ k & , x=0\end{cases}$
Since, $f(x)$ is continuous at $x=0$
$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\
& \therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left[2 \cdot \frac{\log (1+2 x)}{2 x} \frac{\sin \left(\frac{\pi x}{180}\right)}{x}\right] \\
& =2 \times \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x} \frac{\sin \left(\frac{\pi x}{180}\right)}{\left(\frac{\pi x}{180}\right)} \times \frac{\pi}{180} \\
& \pi
\end{aligned}
$
$
=2 \times \frac{\pi}{180}=\frac{\pi}{90}\left[\because \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]
$
Also $f(0)=k$
$\therefore k=\frac{\pi}{90}$
View full question & answer→MCQ 272 Marks
If the function $f(x)$ defined by$f(x)= \begin{cases}x \sin \frac{1}{x} & \text { for } x \neq 0 \\ k & \text { for } x \neq 0\end{cases}$ is continuous at $x=0$, then $k=$
AnswerGiven $f(x)=\left\{\begin{array}{ll}x \sin \frac{1}{x} & , x \neq 0 \\ k & , x=0\end{array}\right.$
Also, $f(x)$ is continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left[x \sin \frac{1}{x}\right]...(i)$
We know, $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$
$\Rightarrow-x \leq x \sin \left(\frac{1}{x}\right) \leq x$
$\Rightarrow \lim _{x \rightarrow 0}(-x) \leq \lim _{x \rightarrow 0}\left(x \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0}(x)$
$\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left[x \sin \frac{1}{x}\right] \leq 0 $
$\Rightarrow \lim _{x \rightarrow 0} x \sin \frac{1}{x}=0$
So, from $ (i) \lim _{x \rightarrow 0^{-}} f (x)=0 \quad$ and $f(0)=k ...(ii)$
$\therefore$ From $(i)$ and $(ii),$ we get $k=0$
View full question & answer→MCQ 282 Marks
For every integer $n$, let $a_n$ and $b_n$ be real number
Let function $f : R \rightarrow R$ be given by
$f(x)=\left\{\begin{array}{lll}a_n+\sin \pi x & \text { for } & x \in[2 n, 2 n+1] \\b_n+\cos \pi x & \text { for } & x \in(2 n-1,2 n)\end{array}\right.$
for all integers $n$. If $f$ is continuous, then which of the following does not hold for all $n$ ?
- A
$a_n-b_{n+1}=-1$
- ✓
$a_{n-1}-b_{n-1}=0$
- C
$a_n-b_n=1$
- D
$a_{n-1}-b_n=-1$
AnswerCorrect option: B. $a_{n-1}-b_{n-1}=0$
(B)
Since $f$ is continuous at every point in $R$.
$\therefore \quad f$ is continuous at $x=2 n$.
$\therefore \quad \lim _{x \rightarrow(2 n )^{-}} f (x)=\lim _{x \rightarrow(2 n )^{+}} f (x)= f (2 n )$
$\Rightarrow \lim _{x \rightarrow(2 \pi)^{-}}\left( b _{ n }+\cos \pi x\right)=\lim _{x \rightarrow(2 \pi)^{+}}\left( a _{ n }+\sin \pi x\right)$
$\Rightarrow b_n+\cos 2 n \pi=a_n+\sin 2 n \pi$
$\Rightarrow b _{ n }+ l = a _{ n } \Rightarrow a _{ n }- b _{ n }=1$
So, option (C) is correct.
Also, f is continuous at $x=2 n +1$.
$\lim _{x \rightarrow(2 n +1)^{-}} f (x)=\lim _{x \rightarrow(2 n +1)^{+}} f (x)= f (2 n +1)$
$\Rightarrow \lim _{x \rightarrow(2 n+1)^{-}}\left(a_n+\sin \pi x\right)=\lim _{x \rightarrow(2 n+1)^{+}}\left(b_{n+1}+\cos \pi x\right)$
$\Rightarrow a_n+\sin (2 n+1) \pi=b_{n+1}+\cos (2 n+1) \pi$
$\ldots .\left[\because f (x)= b _{ n +1}+\cos \pi x, x \in(2 n +1,2 n +2)\right]$
$\Rightarrow a _{ n }= b _{ n +1}-1$
$\Rightarrow a _{ n }- b _{ n +1}=-1$
Replacing $n$ by $n-1$, we get
$a _{ n -1}- b _{ n }=-1$
So, options (A) and (D) are correct.
Hence, option (B) does not hold.
View full question & answer→MCQ 292 Marks
The value of $p$ and $q$ for which the function
$f(x)=\left\{\begin{array}{ll}\frac{\sin (p+1) x+\sin x}{x}, & x<0 \\q, & x=0 \\\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}}, & x>0\end{array}\right.$
is continuous for all x in R, are
- A
$p=\frac{1}{2}, q=-\frac{3}{2}$
- B
$p=\frac{5}{2}, q=\frac{1}{2}$
- ✓
$p=-\frac{3}{2}, q=\frac{1}{2}$
- D
$p=\frac{1}{2}, q=\frac{1}{2}$
AnswerCorrect option: C. $p=-\frac{3}{2}, q=\frac{1}{2}$
(C)
Since $f (x)$ is continuous for all $x$ in R .
$\therefore f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0^{-}} f(x)$
$\Rightarrow q =\lim _{x \rightarrow 0} \frac{\sin ( p +1) x+\sin x}{x}$
$\Rightarrow q =\lim _{x \rightarrow 0}\left[( p +1) \times \frac{\sin ( p +1) x}{( p +1) x}+\frac{\sin x}{x}\right]$
$\Rightarrow q =( p +1)+1$
$\Rightarrow q=p+2$
The values of $p$ and $q$ in option (C) satisfies this condition.
View full question & answer→MCQ 302 Marks
If the function
$f(x)=\left\{\begin{array}{c}1+\sin \frac{\pi x}{2}, \text { for }-\infty < x \leq 1 \\a x+b, \text { for } 1 < x < 3 \\6 \tan \frac{\pi x}{12}, \text { for } 3 \leq x < 6\end{array}\right.$
is continuous in the interval $(-\infty, 6)$, then the values of a and b are respectively
Answer(C)
Since $f(x)$ is continuous in $(-\infty, 6)$.
$\therefore $ it is continuous at $x=1$ and $x=3$.
$\therefore \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}\left(1+\sin \frac{\pi x}{2}\right)=\lim _{x \rightarrow 1^{+}}(a x+b)$
$\Rightarrow 1+\sin \frac{\pi}{2}=a+b$
$\Rightarrow a+b=2$ ...(i)
Also, $\lim _{x \rightarrow 3^{-}} f (x)=\lim _{x \rightarrow 3^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 3^{-}}(a x+b)=\lim _{x \rightarrow 3^{+}}\left(6 \tan \frac{\pi x}{12}\right)$
$\Rightarrow 3 a+b=6 \tan \frac{3 \pi}{12}$
$\Rightarrow 3 a+b=6$ ...(ii)
From (i) and (ii), we get $a=2, b=0$
View full question & answer→MCQ 312 Marks
If $f(x)$ is continuous in $[0, \pi]$, where
$f(x)=\left\{\begin{array}{ll}x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\2 x \cot x+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2}, \text { then } \\a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi\end{array}\right.$
- A
$a=\frac{\pi}{6}, b=\frac{\pi}{12}$
- B
$a=-\frac{\pi}{6}, b=\frac{\pi}{12}$
- ✓
$a=\frac{\pi}{6}, b=-\frac{\pi}{12}$
- D
$a=-\frac{\pi}{6}, b=-\frac{\pi}{12}$
AnswerCorrect option: C. $a=\frac{\pi}{6}, b=-\frac{\pi}{12}$
(C)
Sincc $f(x)$ is continuous in $[0, \pi]$.
$\therefore $ it is continuous at $x=\frac{\pi}{4}$ and $x=\frac{\pi}{2}$.
$\therefore \lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}} f (x)=\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}}(x+ a \sqrt{2} \sin x)=\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}}(2 x \cot x+ b )$
$\Rightarrow \frac{\pi}{4}+ a \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)=2\left(\frac{\pi}{4}\right)(1)+ b$
$\Rightarrow \frac{\pi}{4}+ a =\frac{\pi}{2}+ b$
$\Rightarrow a-b=\frac{\pi}{4}$ $\quad\ldots(i)$
Also, $\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} f (x)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(2 x \cot x+ b )=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}}( a \cos 2 x- b \sin x)$
$\Rightarrow 2\left(\frac{\pi}{2}\right)(0)+ b = a (-1)- b (1)$
$\Rightarrow b =- a - b$
$\Rightarrow a+2 b=0$ $\quad\ldots(ii)$
From (i) and (ii), we get
$a=\frac{\pi}{6}$ and $b=\frac{-\pi}{12}$
View full question & answer→MCQ 322 Marks
Let $f (x)=\left\{\begin{array}{cc}\frac{x^4-5 x^2+4}{|(x-1)(x-2)|} ; & x \neq 1,2 \\ 6 ; & x=1 \\ 12 ; & x=2\end{array}\right.$
then $f (x)$ is continuous on the set
- A
- B
$R -\{1\}$
- C
$R -\{2\}$
- ✓
$R -\{1, 2\}$
AnswerCorrect option: D. $R -\{1, 2\}$
(D)
$f (x)=\frac{(x-1)(x+1)(x-2)(x+2)}{|x-1||x-2|}$
Since $\lim _{x \rightarrow 1} \frac{x-1}{|x-1|}$ does not exist.
Also, $\lim _{x \rightarrow 2} \frac{x-2}{|x-2|}$ does not exist
$\therefore f (x)$ is discontinuous at $x=1,2$.
For any $x \neq 1,2, f (x)$ is the quotient of two polynomials and a polynomial is everywhere continuous. Therefore, $f (x)$ is continuous for all $x \neq 1,2$.
$\therefore f (x)$ is continuous on $R -\{1,2\}$.
View full question & answer→MCQ 332 Marks
A function $f (x)$ is defined by $f (x)=\frac{ e ^x+ e ^{-x}-2}{x \sin x}$ for $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$. The value of $f(0)$ so that $f$ will be continuous in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is
Answer(B)
$f (x)$ is continuous in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ except at $x=0$.
For $f (x)$ to be continuous in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,
$f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{ e ^x+ e ^{-x}-2}{x \sin x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{ e ^x- e ^{-x}}{x \cos x+\sin x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{ e ^x+ e ^{-x}}{-x \sin x+\cos x+\cos x}$
$=\frac{ e ^0+ e ^0}{0+2 \cos 0}=\frac{1+1}{2(1)}=1$
View full question & answer→MCQ 342 Marks
The value of $f(0)$ so that the function
$f(x)=\frac{\sqrt{1+x}-(1+x)^{\frac{1}{3}}}{x}$
becomes continuous is equal to
- ✓
$\frac{1}{6}$
- B
$\frac{1}{4}$
- C
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{1}{6}$
(A)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{2}}-(1+x)^{\frac{1}{3}}}{x}$
$=\lim _{x \rightarrow 0} \frac{\left(1+\frac{1}{2} x-\frac{1}{8} x^2+\ldots .\right)-\left(1+\frac{1}{3} x-\frac{1}{9} x^2+\ldots .\right)}{x}$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{1}{2}-\frac{1}{3}\right) x+\left(\frac{1}{9}-\frac{1}{8}\right) x^2+\ldots .}{x}$
$=\lim _{x \rightarrow 0}\left[\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{9}-\frac{1}{8}\right) x+\ldots\right]$
$\therefore f(0)=\frac{1}{6}$
View full question & answer→MCQ 352 Marks
The values of $a$ and $b$ such that the function defined by
$f(x)=\left\{\begin{array}{cl}7 & , \text { if } x \leq 2 \\ax+b & , \text { if } 2 < x < 9 \text { is continuous on its } \\21 & , \text { if } x \geq 9\end{array}\right.$
domain are
Answer(B)
Since $f (x)$ is continuous on its domain.
$\therefore $ it is continuous at $x=2$ and $x=9$.
$\therefore \lim _{x \rightarrow 2^{+}} f (x)=\lim _{x \rightarrow 2^{-}} f (x)$
$\Rightarrow \lim _{x \rightarrow 2^{+}}( ax + b )=7$
$\Rightarrow 2 a+b=7$ $\quad\ldots(i)$
Also, $\lim _{x \rightarrow 9^{-}} f (x)=\lim _{x \rightarrow 9^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 9^{-}}( ax + b )=21$
$\Rightarrow 9 a+b=21$ $\quad\ldots(ii)$
Solving (i) and (ii), we get $a=2, b=3$
View full question & answer→MCQ 362 Marks
If $f(x)$ is continuous in $[-2,2]$, where
$f(x)=\left\{\begin{array}{ll}\frac{\sin a x}{x}-2, & \text { for }-2 \leq x < 0 \\2 x+1, & \text { for } 0 \leq x \leq 1 \\2 b \sqrt{x^2+3}-1, & \text { for } 1 < x \leq 2\end{array}\right.$
then the value of $(a+b)$ is
Answer(B)
Since $f (x)$ is continuous in $[-2,2]$.
∴ $\quad$ it is continuous at $x=0$ and $x=1$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(\frac{\sin a x}{x}-2\right)=\lim _{x \rightarrow 0^{+}}(2 x+1)$
$\Rightarrow a-2=0+1 \Rightarrow a=3$
Also, $\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}(2 x+1)=\lim _{x \rightarrow 1^{+}}\left(2 b \sqrt{x^2+3}-1\right)$
$\Rightarrow 2(1)+1=2 b \sqrt{1+3}$
$\Rightarrow 3=4 b-1$
$\Rightarrow b =1$
$\therefore \quad a+b=3+1=4$
View full question & answer→MCQ 372 Marks
If the function $f (x)$ is continuous in $[0,8]$, where
$f(x)$ $=\left\{\begin{array}{ll}x^2+a x+6, & 0 \leq x< 2 \\3 x+2, & 2 \leq x \leq 4 \\2 a x+5 b, & 4< x \leq 8\end{array}\right. \text {, then }$
- A
$a=1, b=\frac{22}{5}$
- ✓
$a=-1, b=\frac{22}{5}$
- C
$a=1, b=\frac{-22}{5}$
- D
$a=-1, b=\frac{-22}{5}$
AnswerCorrect option: B. $a=-1, b=\frac{22}{5}$
(B)
Since $f(x)$ is continuous in $[0,8]$.
$\therefore \quad$ it is continuous at $x=2$ and $x=4$.
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 2^{-}}\left(x^2+a x+6\right)=\lim _{x \rightarrow 2^{+}}(3 x+2)$
$\Rightarrow(2)^2+2 a+6=3(2)+2$
$\Rightarrow 10+2 a =8$
$\Rightarrow a=-1$ ...(i)
Also, $\lim _{x \rightarrow 4^{-}} f (x)=\lim _{x \rightarrow 4^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 4^{-}}(3 x+2)=\lim _{x \rightarrow 4^{+}}(2 a x+5 b)$
$\Rightarrow 3(4)+2=2 a (4)+5 b$
$\Rightarrow 14=8 a+5 b$
$\Rightarrow b=\frac{22}{5} \quad\ldots[From (i)]$
View full question & answer→MCQ 382 Marks
If $f (x)=\left\{\begin{array}{ll}\sin x & \text { if } x \leq 0 \\ x ^ 2+ a ^2 & \text { if } 0 < x<1 \\ b x+2 & \text { if } 1 \leq x \leq 2 \\ 0 & \text { if } x>2\end{array}\right.$
is continuous on R , then $a + b + ab =$
Answer(D)
$f (x)=\sin x$
$\therefore \quad f(0)=\sin 0=0$
$\lim _{x \rightarrow 0^{+}} x^2+a^2=0^2+a^2=a^2$
Since the function is continuous at $x=0$,
$\lim _{x \rightarrow 0^{+}} f (x)= f (0)$
$\Rightarrow 0= a ^2$
$\Rightarrow a =0$
$\lim _{x \rightarrow 1^{-}} x^2+a^2=1^2+a^2=1$
$f (x)= b x+2$
$\therefore f (1)= b +2$
Now, $\lim _{x \rightarrow 1^{-}} f (x)= f (1)$
$\Rightarrow 1=b+2$
$\Rightarrow b =-1$
$a+b+a b=0-1+0(-1)=-1$
View full question & answer→MCQ 392 Marks
The function $f (x)=\sin |x|$ is
- ✓
- B
Continuous only at certain points
- C
Differentiable at all points
- D
Answer(A)
Let $g (x)=|x|$ and $h (x)=\sin x$.
Then, $f (x)=(\operatorname{hog})(x)$ for all $x \in R$.
As both $g$ and $h$ are continuous functions on $R$.
$\therefore \quad f (x)$ is also continuous for all $x \in R$.
View full question & answer→MCQ 402 Marks
The function $f(x)=\frac{x^2-4}{\sin x-2}$ is
- ✓
continuous for all real values of x
- B
- C
discontinuous when sin x = 2
- D
AnswerCorrect option: A. continuous for all real values of x
(A)
For all $x \in R ,-1 \leq \sin x \leq 1$
$\therefore f (x)$ is continuous for all real values of $x$.
View full question & answer→MCQ 412 Marks
The function $f (x)=[x]$, where $[x]$ the greatest function is continuous at
Answer(C)
Since $f (x)=[x]$ is continuous at every non integer points.
∴ option (C) is the correct answer.
View full question & answer→MCQ 422 Marks
The function $f (x)=\left\{\begin{array}{ll}x+2, & 1 \leq x<2 \\ 4, & x=2 \\ 3 x-2, & x>2\end{array}\right.$ is continuous at
- A
$x=2$ only
- B
$x \leq 2$
- ✓
$x \geq 2$
- D
AnswerCorrect option: C. $x \geq 2$
(C)
The given function is defined only in the interval $[1, \infty)$. For $x>2, y=3 x-2$ which is a straight line, hence continuous. Also, the given function is continuous at $x=2$.
∴ option (C) is the correct answer.
View full question & answer→MCQ 432 Marks
Let $f (x)=\left\{\begin{aligned} \frac{x^3+x^2-16 x+20}{(x-2)^2} & ; \text { if } x \neq 2 \\ k & ; \text { if } x=2\end{aligned}\right.$. If $f (x)$ is continuous for all $x$, then $k =$
Answer(A)
Since $f (x)$ is continuous for all $x$.
$\therefore f (x)$ is continuous at $x=2$.
$\therefore f (2)=\lim _{x \rightarrow 2} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 2} \frac{x^3+x^2-16 x+20}{(x-2)^2}$
$=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^2+3 x-10\right)}{(x-2)^2}$
$=\lim _{x \rightarrow 2} \frac{(x-2)^2(x+5)}{(x-2)^2}$
$=7$
View full question & answer→MCQ 442 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x<0 \\ \frac{2 x+1}{x-2}, & 0 \leq x \leq 1\end{array}\right.$ is continuous in $[-1,1]$, then $p$ is equal to
- A
- ✓
$\frac{-1}{2}$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: B. $\frac{-1}{2}$
(B)
Since $f (x)$ is continuous in $[-1,1]$.
∴ it is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+ p x}-\sqrt{1- p x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{(1+ p x)-(1- p x)}{x(\sqrt{1+ p x}+\sqrt{1- p x})}=\frac{-1}{2}$
$\Rightarrow p =\frac{-1}{2}$
View full question & answer→MCQ 452 Marks
Let $f (x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f (x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f \left(\frac{\pi}{4}\right)$ is
- A
- B
$\frac{1}{2}$
- ✓
$-\frac{1}{2}$
- D
AnswerCorrect option: C. $-\frac{1}{2}$
(C)
Since $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$
$\therefore \quad$ it is continuous at $x=\frac{\pi}{4}$
$\therefore f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}$
Applying L'Hospital rule on R.H.S., we get
$f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^2 x}{4}$
$\Rightarrow f \left(\frac{\pi}{4}\right)=\frac{-2}{4}=\frac{-1}{2}$
View full question & answer→MCQ 462 Marks
$\begin{aligned}\text {If } f (x) & =\frac{\left(2^x-1\right)^2}{\tan x \cdot \log (1+x)}, \text { for } x \neq 0 \\ & =\log 4 \quad, \text { for } x=0, \text { then }\end{aligned}$
- A
$f (x)$ is continuous at $x=0$
- ✓
$f (x)$ has removable discontinuity at $x=0$
- C
$f (x)$ has irremovable discontinuity at $x=0$
- D
AnswerCorrect option: B. $f (x)$ has removable discontinuity at $x=0$
(B)
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^2}{\tan x \cdot \log (1+x)}$
$=\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^2}{x^2} \times \frac{1}{\frac{\tan x}{x}} \times \frac{1}{\frac{\log (1+x)}{x}}$
$=(\log 2)^2 \times 1=(\log 2)^2$
and $f(0)=\log 4$
$\therefore f (x)$ is discontinuous at $x=0$.
Here, $\lim _{x \rightarrow 0} f(x)$ exists but not equal to $f(0)$.
∴ the discontinuity at $x=0$ is removable.
View full question & answer→MCQ 472 Marks
If the function is defined as
$\begin{aligned}f(x) & =\frac{5^{\cos x}-1}{\frac{\pi}{2}-x}, \text { when } x \neq \frac{\pi}{2} \\& =2 \log 5, \text { when } x=\frac{\pi}{2}, \text { then }\end{aligned}$
- A
$f (x)$ is continuous at $x=\frac{\pi}{2}$
- ✓
$f (x)$ has removable discontinuity at $x=\frac{\pi}{2}$
- C
$f (x)$ has irremovable discontinuity at $x=\frac{\pi}{2}$
- D
AnswerCorrect option: B. $f (x)$ has removable discontinuity at $x=\frac{\pi}{2}$
(B)
Applying L'Hospital rule, we get
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{5^{\cos x}-1}{\frac{\pi}{2}-x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{5^{\cos x} \cdot \log 5(-\sin x)}{-1}$
$=5^{\cos \frac{\pi}{2}} \cdot \log 5 \sin \frac{\pi}{2}=\log 5$
and $f\left(\frac{\pi}{2}\right)=2 \log 5$
$\therefore f (x)$ is discontinuous at $x=\frac{\pi}{2}$.
Here, $\lim _{x \rightarrow \frac{\pi}{2}} f (x)$ exists but not equal to $f \left(\frac{\pi}{2}\right)$.
$\therefore $ the discontinuity at $x=\frac{\pi}{2}$ is removable.
View full question & answer→MCQ 482 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{5^x- e ^{ z }}{\sin 2 x} \quad ; \quad x \neq 0 \\ \frac{1}{2}(\log 5+1) \quad ; \quad x=0\end{array}\right.$ then
- A
$f (x)$ is continuous at $x=0$
- ✓
$f (x)$ is discontinuous at $x=0$
- C
$\lim _{x \rightarrow 0} f (x)$ does not exist
- D
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=0$
(B)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{5^x- e ^x}{\sin 2 x}=\lim _{x \rightarrow 0} \frac{5^x-1+1- e ^x}{\sin 2 x}$
$=\lim _{x \rightarrow 0} \frac{\frac{5^x-1}{x}-\frac{ e ^x-1}{x}}{\frac{\sin 2 x}{2 x} \times 2}$
$=\frac{\log 5-\log e }{2}=\frac{1}{2}(\log 5-1)$
$\therefore \lim _{x \rightarrow 0} f (x) \neq f (0)$
$\therefore f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 492 Marks
$\begin{aligned}\text {If } f(x) & =\frac{x \cos x-3 \tan x}{x^2+2 \sin x}, x \neq 0 \\ & =1, \quad x=0, \text { then }\end{aligned}$
- ✓
$f ( x )$ is discontinuous at $x=0$
- B
$f(x)$ is continuous at $x=0$
- C
$\lim _{x \rightarrow 0} f(x)$ does not exist
- D
AnswerCorrect option: A. $f ( x )$ is discontinuous at $x=0$
(A)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{x \cos x-3 \tan x}{x^2+2 \sin x}$
$=\lim _{x \rightarrow 0} \frac{\frac{x \cos x-3 \tan x}{x}}{\frac{x^2+2 \sin x}{x}}$
$=\lim _{x \rightarrow 0} \frac{\cos x-\frac{3 \tan x}{x}}{x+\frac{2 \sin x}{x}}$
$=\frac{1-3}{0+2}=-1$
$\therefore \quad \lim _{x \rightarrow 0} f (x) \neq f (0)$
$\therefore \quad f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 502 Marks
$\begin{aligned}\text {If } f(x) & =\frac{\sin 2 x}{\sqrt{1-\cos 2 x}}, 0 < x \leq \frac{\pi}{2} \\ & =\frac{\cos x}{\pi-2 x}, \frac{\pi}{2}< x< \pi, \text { then }\end{aligned}$
- A
$\lim _{x\rightarrow\frac{\pi^{-}}{2}} f(x)=1$
- B
$\lim _{x\rightarrow{\frac{\pi^+}{2}}} f(x)=1$
- C
$f ( x )$ is continuous at $x=\frac{\pi}{2}$
- ✓
$f ( x )$ is discontinuous at $x=\frac{\pi}{2}$
AnswerCorrect option: D. $f ( x )$ is discontinuous at $x=\frac{\pi}{2}$
(D)
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin 2 x}{\sqrt{1-\cos 2 x}}=\frac{\sin \pi}{\sqrt{1-\cos \pi}}=0$
$\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\pi-2 x}$
$=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}$
$=\lim _{h \rightarrow 0} \frac{-\sin h}{-2 h}=\frac{1}{2}(1)=\frac{1}{2}$
$\therefore \lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x) \neq \lim _{x \rightarrow \frac{\pi^{+}}{2}} f (x)$
$\therefore f (x)$ is discontinuous at $x=\frac{\pi}{2}$.
View full question & answer→