MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 8 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 8
$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\cos (\tan x)-\cos x}{x^{4}} $$ is equal to
Report Question
0%
1/6
0%
-1/3
0%
1/2
0%
1
Explanation
$$ \cos (\tan x)-\cos x=2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)$$
$$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{\cos (\tan x)-\cos x}{x^{4}}=\lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}}$$
$$ =\displaystyle \lim _{x \rightarrow 0} \dfrac{2 \sin \left(\dfrac{x+\tan x}{2}\right) \sin \left(\dfrac{x-\tan x}{2}\right)}{x^{4}\left(\dfrac{x+\tan x}{2}\right)\left(\dfrac{x-\tan x}{2}\right)}\left(\dfrac{x^{2}-\tan ^{2} x}{4}\right) $$
$$ =\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\tan ^{2} x}{x^{4}} $$
$$ =\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{2}-\left(x+\dfrac{x^{3}}{3}+\dfrac{2}{15} x^{5}+\cdots\right)^{2}}{x^{4}} $$
$$ =\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x^{2}}\left(1-\left(1+\dfrac{x^{2}}{3}+\dfrac{2}{15} x^{4}+\dots\right)^{2}\right)=-\dfrac{1}{3} $$
The value of $$ \displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1}{n}+\dfrac{e^{1 / n}}{n}+\dfrac{e^{2 / n}}{n}+\ldots .+\dfrac{e^{(n-1) / n}}{n}\right] $$ is
Report Question
0%
1
0%
0
0%
e-1
0%
e+1
Explanation
$$ \quad \displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1}{n}+\dfrac{e^{1 / n}}{n}+\dfrac{e^{2 / n}}{n}+\dots+\dfrac{e^{(n-1) / n}}{n}\right] $$
$$=\displaystyle \lim _{n \rightarrow \infty}\left[\dfrac{1+e^{1 / n}+\left(e^{1 / n}\right)^{2}+\cdots+\left(e^{1 / n}\right)^{n-1}}{n}\right] $$
$$=\displaystyle \lim _{n \rightarrow \infty} \dfrac{1 \cdot\left[\left(e^{1 / n}\right)^{n}-1\right]}{n\left(e^{1 / n}-1\right)}=(e-1) \lim _{n \rightarrow \infty} \dfrac{1}{\left(\dfrac{e^{1 / n}-1}{1 / n}\right)} $$
$$=(e-1) \times 1=(e-1) $$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)} $$ is equal to
Report Question
0%
1
0%
0
0%
2
0%
None of these
Explanation
$$ \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)} $$
$$ =\dfrac{x^{4}\left(1+\tan ^{2} x+\tan ^{4} x\right)}{\tan ^{4} x\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\dfrac{x^{4}}{\tan ^{4} x}, x \neq 0 $$
$$ \Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}}{\tan ^{4} x}=1 $$
$$ \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{20} \cos ^{2 n}(x-10) $$ is equal to
Report Question
0%
0
0%
1
0%
19
0%
20
Explanation
$$ \because \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n} x=\begin{cases}1, x=r \pi, r \in I \\ 0, x \neq r \pi, r \in I\end{cases}$$
Here, for $$ x=10, \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n}(x-10)=1 $$
and in all other cases it is zero.
$$ \therefore \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{\infty} \cos ^{2 n}(x-10)=1 $$
The value of $$\displaystyle \underset { n\rightarrow \infty }{ lim } \left( \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 6n } \right) $$ is
Report Question
0%
$$\displaystyle \log { 2 } $$
0%
$$\displaystyle \log { 6 } $$
0%
$$\displaystyle 1$$
0%
$$\displaystyle \log { 3 } $$
A point where function $$f(x)$$ is not continuous where $$f(x)=\left[ \sin { \left[ x \right] } \right] $$ in $$\left( 0,2\pi \right) $$; is ($$\left[ \ast \right] $$ denotes greatest integer $$\le x$$)
Report Question
0%
$$(3,0)$$
0%
$$(2,0)$$
0%
$$(1,0)$$
0%
$$(4,-1)$$
$$\underset { x\rightarrow \pi /4 }{ Lim } \dfrac { 2\sqrt { 2 } \left( cosx+sinx \right) ^{ 3 } }{ 1-sin2x } =2$$ is equal to
Report Question
0%
$$\dfrac{3\sqrt{2}}{2}$$
0%
$$2\sqrt{2}$$
0%
$$\dfrac{4\sqrt{2}}{3}$$
0%
$does \not \exist$$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin \left(x^{2}\right)}{\ln \left(\cos \left(2 x^{2}-x\right)\right)} $$ is equal to
Report Question
0%
2
0%
-2
0%
1
0%
-1
Explanation
$$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{In(cos(2x^2-x))}$$
$$\displaystyle \
lim_{x \to 0}\dfrac{sin(x^2)}{log \left(1-2 sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}$$
$$=\displaystyle \lim_{x \to 0} \dfrac{sin(x^2)x^2}{\dfrac{x^2log \left(1-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}{-2sin^2 \left(\dfrac{2x^2-x}{2}\right)}\left[-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right]}$$
$$=\displaystyle \lim_{x \to 0} \dfrac{x^2}{\dfrac{2sin^2 \left(\dfrac{2x^2-x}{2}\right)}{\left(\dfrac{2x^2-x}{2}\right)^2}\left(\dfrac{2x^2-x}{2}\right)^2}$$
$$\displaystyle \lim_{x \to 0}-\dfrac{2x^2}{(2x^2-x)^2}=\displaystyle \lim_{x \to 0}-\dfrac{2}{(2x-1)^2}=-2$$
$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}} $$ is equal to
Report Question
0%
2
0%
-2
0%
1/2
0%
-1/2
Explanation
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{4 \sin ^{4} x}$$
$$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x}{4 \sin ^{4} x}\left[\dfrac{2 \tan x}{1-\tan ^{2} x}-2 \tan x\right]$$
$$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan ^{3} x}{2 \sin ^{4} x\left(1-\tan ^{2} x\right)}$$
$$=\dfrac{1}{2} \lim _{x \rightarrow 0} \dfrac{x}{\sin x} \dfrac{1}{\cos ^{3} x} \dfrac{1}{1-\tan ^{2} x}$$
$$=\dfrac{1}{2} \times 1 \times \dfrac{1}{1^{3}} \times \dfrac{1}{1-0}=\dfrac{1}{2}$$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$$ is equal to
Report Question
0%
$$e^{\sin ^{2} y}$$
0%
$$\sin 2 y e^{\sin ^{2} y}$$
0%
0
0%
None of these
Explanation
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t+\int_{a}^{x+y} e^{\sin ^{2} t} d t\right]=\lim _{x \rightarrow 0} \dfrac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t \\$$
$$\therefore\left(\dfrac{0}{0} \text { form }\right)\\$$
Apply L'Hopital Rule
$$\displaystyle =\lim _{x \rightarrow 0} \dfrac{e^{\sin ^{2}(x+y)}\left(1+\dfrac{d y}{d x}\right)-e^{\sin ^{2} y} \dfrac{d y}{d x}}{1} \\$$
$$=e^{\sin ^{2} y}\left[1+\dfrac{d y}{d x}-\dfrac{d y}{d x}\right]=e^{\sin ^{2} y}$$
If $$y^{r}=\dfrac{n !^{n+r-1} C_{r-1}}{r^{n}},$$ where $$n=k r(k \text { is constant }),$$ then $$\operatorname{lim}_{r\rightarrow\infty} y$$ is equal to
Report Question
0%
$$(k-1) \log _{e}(1+k)-k$$
0%
$$(k+1) \log _{e}(k-1)+k$$
0%
$$(k+1) \log _{e}(k-1)-k$$
0%
$$(k-1) \log _{e}(k-1)+k$$
Explanation
$$y^{r}=\left(1+\dfrac{1}{r}\right)\left(1+\dfrac{2}{r}\right)\left(1+\dfrac{3}{r}\right) \cdots\left(1+\dfrac{n-1}{r}\right)\\$$
$$\Rightarrow \log y=\dfrac{1}{r} \sum_{p=1}^{n-1} \log \left(1+\dfrac{p}{r}\right)\\$$
$$\Rightarrow \lim _{n \rightarrow \infty} y=\lim _{r \rightarrow \infty} y=\int_{-\infty}^{k} \log (1+x) d x=(k-1) \log _{e}(1+k)-k$$
The value of $$ \displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}} $$ is
Report Question
0%
$$\dfrac{2 a}{\pi} $$
0%
$$-\dfrac{2 a}{\pi} $$
0%
$$ \dfrac{4 a}{\pi} $$
0%
$$-\dfrac{4 a}{\pi} $$
Explanation
$$ \displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cdot \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$$
$$=\displaystyle \lim _{x \rightarrow a} \dfrac{\sqrt{a^{2}-x^{2}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}$$
$$=\dfrac{2}{\pi} \displaystyle \lim _{x \rightarrow a} \dfrac{\dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}(a+x)=\dfrac{4 a}{\pi}$$
If function $$f(x)=\dfrac{x^2-9}{x-3}$$ is continuous at $$x=3$$, then value of $$(3)$$ will be:
Report Question
0%
$$6$$
0%
$$3$$
0%
$$1$$
0%
$$0$$
Explanation
$$f(x)=\dfrac{x^2-9}{x-3}$$
Left hand limit
$$f(3+0)=\displaystyle \lim_{h \rightarrow 0}f(3+h)$$
$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{(3+h)^2-9}{3+h-3}$$
$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{h(6+h)}{h}$$
$$=\displaystyle \lim_{h \rightarrow 0} (6+h)$$
$$=6$$
Function is continous at $$x=3$$, so
$$f(3)=f(3+0)$$
$$f(3)=6$$
Hence, option $$(a)$$ is correct.
If $$f(x)=\begin{cases} \begin{matrix} \dfrac{\log (1+mx)- \log (1-nx)}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} k; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
is continuous at $$x=0$$ then the value of $$k$$ will be:
Report Question
0%
$$0$$
0%
$$m+n$$
0%
$$m-n$$
0%
$$m.n$$
Explanation
$$\therefore$$ Function is continous at $$x=0$$
$$\therefore \displaystyle \lim_{x \rightarrow 0} f(x)=f(0)$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{\log (1+mx)- \log (1-nx)}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{ [mx-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3} - ...]- [nx-\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} - ...]}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{x \left[ m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... \right]}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... =k$$
$$\Rightarrow m+n=k$$
$$\Rightarrow k=m+n$$
Hence, option $$(b)$$ is correct.
If function $$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
is continuous at $$x=2$$ then value of $$m$$ will be:
Report Question
0%
$$3$$
0%
$$1/3$$
0%
$$1$$
0%
$$0$$
Explanation
$$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
$$f(0)=m$$
$$f(0+0)=\displaystyle \lim_{h \rightarrow 0}f(0+h)$$
$$\displaystyle \lim_{h \rightarrow 0} \dfrac{\sin 3 (0+h)}{0+h}$$
$$\displaystyle \lim_{h \rightarrow 0}3 \dfrac{\sin 3 h}{3h}$$
$$=3 \times 1$$
$$=3$$
At $$x=a$$ function is continuous.
So, $$f(0)=f(0+0)$$
$$m=3$$
Hence, option $$(a)$$ is correct.
The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac {xe^{x} - \log_{e} (1 + x)}{x^{2}}$$ is
Report Question
0%
$$2/3$$
0%
$$1/3$$
0%
$$1/2$$
0%
$$3/2$$
Explanation
$$\displaystyle \lim_{x \rightarrow 0} \dfrac {xe^{x} - \log_{e} (1 + x)}{x^{2}}$$
$$x \left [1 + x + \dfrac {x^{2}}{2!} + \dfrac {x^{3}}{3!} + ....\right ]$$
$$= \displaystyle \lim_{x \rightarrow 0} \dfrac {-\left [x - \dfrac {x^{2}}{2} + \dfrac {x^{3}}{3} - \dfrac {x^{4}}{4} + ....\right ]}{x^{2}}$$
$$= \displaystyle \lim_{x \rightarrow 0} \dfrac {\left [x + x^{2} + \dfrac {x^{3}}{2!} + \dfrac {x^{4}}{3!} + .... - x + \dfrac {x^{2}}{2} - \dfrac {x^{3}}{3} + \dfrac {x^{4}}{4} - ....\right ]}{x^{2}}$$
$$= \displaystyle \lim_{x \rightarrow 0} \dfrac {x^{2} \left [1 + \dfrac {x}{2!} + \dfrac {x^{2}}{3!} + ... + \dfrac {1}{2} - \dfrac {x}{3} + \dfrac {x^{2}}{4} - .... \right ]}{x^{2}}$$
$$= \left [1 + 0 + 0 + ... + \dfrac {1}{2} - 0 + 0 ... \right ] = 1 + \dfrac {1}{2} = \dfrac {3}{2}$$
Hence, option (D) is correct.
If $$f(x)=\begin{cases} \dfrac { 1-\sqrt { 2 } \sin { x } }{ \pi -4x } ,\quad \quad ifx\neq \dfrac { \pi }{ 4 } \\ a\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ,\quad \quad ifx=\dfrac { \pi }{ 4 } \end{cases}$$ is continous at $$x=\dfrac {\pi}{4}$$ then $$a=$$
Report Question
0%
$$4$$
0%
$$2$$
0%
$$1$$
0%
$$1/4$$
The value of $$\displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin x}{x}$$ is
Report Question
0%
$$0$$
0%
$$\infty$$
0%
$$1$$
0%
$$-1$$
Explanation
$$\displaystyle \lim_{x \rightarrow \infty} \dfrac {\sin x}{x}$$
Let $$x = 1/y$$ or $$y = \dfrac {1}{x}$$
Som $$x\rightarrow \infty \Rightarrow y \rightarrow 0$$
$$\therefore \displaystyle \lim_{x \rightarrow \infty} \left (\dfrac {\sin x}{x}\right ) = \displaystyle \lim_{y \rightarrow 0} y \sin \left (\dfrac {1}{y}\right )$$
$$= \displaystyle \lim_{y \rightarrow 0} y \cdot \displaystyle \lim_{y \rightarrow 0} \sin \dfrac {1}{y}$$
$$= 0\times \text {(any variable quantity)}$$
$$= 0$$
Hence, option (A) is correct.
The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$$ is
Report Question
0%
$$0$$
0%
$$1/2$$
0%
$$-1/2$$
0%
$$-1$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$$
$$= \displaystyle \lim_{x\rightarrow 0} \dfrac {1 - 1 + 2\sin^{2} x/2}{x^{2}}$$
$$= \displaystyle \lim_{x\rightarrow 0} \dfrac {2\sin^{2} x/2}{x^{2}}$$
$$= \displaystyle \lim_{x\rightarrow 0} 2\left (\dfrac {\sin x/2}{x/2}\right )^{2} \times \dfrac {1}{4}$$
$$= \dfrac {1}{2}\times \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin x/2}{x/2}\right )^{2}$$
$$= \dfrac {1}{2} \times 1 = \dfrac {1}{2}$$
Hence option (B) is correct.
The value of $$\displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{\tan x}\right )^{4}$$ is
Report Question
0%
$$0$$
0%
$$81$$
0%
$$4$$
0%
$$1$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{\tan x}\right )^{4}$$
$$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\times 3x\right )^{4}\times \dfrac {1}{\tan^{4} x}$$
$$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\right )^{4} \times \dfrac {3^{4}}{\dfrac {\tan^{4}x}{x^{4}}}$$
$$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\right )^{4} \times \dfrac {1}{\left (\dfrac {\tan x}{x}\right )^{4}} \times 81$$
$$= 1\times 1\times 81 = 81$$
Hence, option (B) is correct.
The value of $$\displaystyle \lim_{x\rightarrow \infty} \sin \dfrac {\pi}{4x} \cos \dfrac {\pi}{4x}$$ is
Report Question
0%
$$\pi/4$$
0%
$$\pi/2$$
0%
$$0$$
0%
$$\infty$$
Explanation
$$\displaystyle \lim_{\rightarrow \infty} x \sin \dfrac {\pi}{4x}\cos \dfrac {\pi}{4x}$$
$$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {2\sin \dfrac {\pi}{4x} \cos \dfrac {\pi}{4x}}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {1}{2} \times \dfrac {\pi}{2x}$$
$$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (2\times \dfrac {\pi}{4x}\right )}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {\pi}{4}$$
$$= \dfrac {\pi}{4}\times \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (\dfrac {\pi}{2x}\right )}{\left (\dfrac {\pi}{2x}\right )} = \dfrac {\pi}{4}\times 1 = \dfrac {\pi}{4}$$
Hence, option (A) is correct.
Let $$f(x)=\displaystyle \frac{(256+ax)^{1/8}-2}{(32+bx)^{1/5}-2}$$. If $$f$$ is continuous at $$x = 0$$, then the value of $$a / b$$ is:
Report Question
0%
$$\displaystyle \frac{8}{5}f(0)$$
0%
$$\displaystyle \frac{32}{5}f(0)$$
0%
$$\displaystyle \frac{64}{5}f(0)$$
0%
$$\displaystyle \frac{16}{5}f(0)$$
Explanation
$$f(x)=\displaystyle \frac{(256+ax)^{1/8}-2}{(32+bx)^{1/5}-2}$$
Given , $$f$$ is continuous at $$x=0$$.
$$ \displaystyle \lim _{ x\rightarrow 0 } f(x)=f(0)$$
Now, $$\displaystyle \lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ \frac { (256+ax)^{ 1/8 }-2 }{ (32+bx)^{ 1/5 }-2 } } $$
It is of the form $$\displaystyle \frac{0}{0}$$, so applying L-Hospital's rule
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ 8 } a(256+ax)^{ -7/8 } }{ \frac { 1 }{ 5 } b(32+bx)^{ -4/5 } } } $$
$$\displaystyle = \frac { 5a }{ 64b } $$
Hence,$$\displaystyle \frac { 5a }{ 64b }=f(0)$$
$$\Rightarrow \displaystyle \frac { a }{ b }=\frac { 64 }{ 5 }f(0)$$
Let $$\alpha(a)$$ and $$\beta(a)$$ be the roots of the equation $$(\sqrt[3]{1+a}-1)x^{2}+(\sqrt{1+a}-1){x}+(\sqrt[6]{1+a}-1)=0$$ where $$a>-1$$.
Then $$ \underset{a\rightarrow 0^{+}}{\lim}\alpha(a)$$ and $$ \underset{a\rightarrow 0^{+}}{\lim}\beta(a)$$ are
Report Question
0%
$$-\displaystyle \frac{5}{2}$$ and 1
0%
$$-\displaystyle \frac{1}{2}$$ and $$-1$$
0%
$$-\displaystyle \frac{7}{2}$$ and 2
0%
$$-\displaystyle \frac{9}{2}$$ and 3
Explanation
Let $$1+a=y$$.
The equation becomes $$(y^{\frac {1}{3}}-1)x^2+(y^{\frac {1}{2}}-1)x+(y^{\frac {1}{6}}-1)=0$$
Dividing by $$y-1$$, we get
$$\displaystyle \left(\dfrac {y^{\frac {1}{3}}-1}{y-1}\right)x^2+\left(\dfrac {y^{\frac {1}{2}}-1}{y-1}\right)x+\left(\dfrac {y^{\frac {1}{6}}-1}{y-1}\right)=0$$
When $$a\rightarrow 0, y\rightarrow 1$$.
Hence, taking $$\underset {y\rightarrow 1}{\lim}$$ on both the sides,
$$\Rightarrow \displaystyle \underset {y\rightarrow 1}{\lim} \left(\dfrac {y^{\frac {1}{3}}-1}{y-1}\right)x^2+\left(\dfrac {y^{\frac {1}{2}}-1}{y-1}\right)x+\left(\dfrac {y^{\frac {1}{6}}-1}{y-1}\right)=0$$
$$\Rightarrow \dfrac {1}{3}x^2+\cfrac {1}{2}x+\cfrac {1}{6}=0$$ ...[Using $$\underset {x\rightarrow a}{\lim}\cfrac {x^n-a^n}{x-a}=na^{n-1}]$$
$$\Rightarrow 2x^2+3x+1=0$$
Solving the quadratic equation, we get
$$\Rightarrow x=-1$$ or $$x=\dfrac {-1}{2}$$.
Since $$\alpha (a)$$ and $$\beta (a)$$ are roots of the given equation, we get
$$\underset {a\rightarrow 0^+}{\lim}\alpha (a)=-1$$ and $$\underset {a\rightarrow 0^+}{\lim}\beta (a)=\dfrac {-1}{2}$$.
If $$f(x)=\left\{\begin{matrix}(cos x)^{1/sinx} &for &x\neq 0 \\ k & for & x=0 \end{matrix}\right.$$
Then the value of $$k$$, so that $$f$$ is continuous at $$x=0$$ is
Report Question
0%
$$0$$
0%
$$1$$
0%
$$\dfrac{1}{2}$$
0%
$$2$$
Explanation
$$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =k$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cos { x } \right) }^{ \cfrac { 1 }{ \sin { k } } } } =k$$
$${ 1 }^{ \infty }$$ form
$${ e }^{\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { \cos { x } -1 }{ \sin { x } } } }={ e }^{\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { -\sin { x } }{ \cos { x } } } }\quad \left( \cfrac { 0 }{ 0 } form \right) $$
$$={ e }^{ -\cfrac { 0 }{ 1 } }=1=k$$
The function $$f$$ : $$R/\{0\}\rightarrow R$$ given by $$f(x)=\displaystyle \frac{1}{x}-\frac{2}{e^{2x}-1}$$ can be made continuous at $$x=0$$ by defining $$f(0)$$ as -
Report Question
0%
$$2$$
0%
$$-1$$
0%
$$0$$
0%
$$1$$
Explanation
$$f(x) = \dfrac { 1 }{ x } -\dfrac { 2 }{ { e }^{ 2x }-1 }
= \dfrac { { e }^{ 2x }-1-2x }{ x{ (e }^{ 2x }-1) } $$
Using expansion of $$e^x$$
$$f(x) = \dfrac { \left(1+2x+\dfrac { { (2x) }^{ 2 } }{ 2! } +\dfrac { { (2x) }^{ 3 } }{ 3! } +...\right)-1-2x }{ x{ (e }^{ 2x }-1) }$$
$$f(x)=\dfrac { \dfrac { { (2x) }^{ 2 } }{ 2! } +\dfrac { { (2x) }^{ 3 } }{ 3! } }{ x{ (e }^{ 2x }-1) } =\quad \dfrac { 2x\left(\dfrac { { (2x) } }{ 2! } +\dfrac { { (2x) }^{ 2 } }{ 3! }\right) }{ x\times{ (e }^{ 2x }-1) }$$
$$f(x)=\dfrac { 2x }{ { (e }^{ 2x }-1) }\times\dfrac {\left(\dfrac { { (2x) } }{ 2! }
+\dfrac { { (2x) }^{ 2 } }{ 3! } \right) }{ x } $$
Now,
$$\displaystyle\lim_{ x\rightarrow 0}\dfrac { 2x }{ { (e }^{ 2x }-1) }\times\dfrac { \dfrac { { (2x) } }{ 2! } +\dfrac { { (2x) }^{ 2 } }{ 3! } +... }{ x } =1\times1=1$$
Hence, option D.
$$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} \frac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } -1 }{ \sqrt { { x }^{ 2 }+4 } -2} , & x\neq 0 \end{matrix} \\ \begin{matrix} a, & x=0 \end{matrix} \end{cases}$$ then the value of $$a$$ in order that $$f(x)$$ may be continuous at $$x=0$$ is
Report Question
0%
$$-8$$
0%
$$8$$
0%
$$-4$$
0%
$$4$$
Explanation
Given: $$\displaystyle f\left( x \right) =\begin{cases} \begin{matrix} \dfrac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x-1 } }{ \sqrt { { x }^{ 2 }+4 } -2 } , & \left( x\neq 0 \right) \end{matrix} \\ \begin{matrix} a, & \left( x\neq 0 \right) \end{matrix} \end{cases}$$
is continuous at $$x=0$$, then
$$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =\lim _{ x\rightarrow 0 }{ \dfrac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x-1 } }{ \sqrt { { x }^{ 2 }+4 } -2 } }$$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { \cos { 2x-1 } }{ \sqrt { { x }^{ 2 }+4 } -2 } } $$
As th function is of the $$\dfrac{0}{0}$$ form,applying L-Hospital's rule,
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \left[ \dfrac { \dfrac { -2\sin { 2x } }{ x } }{ \sqrt { { x }^{ 2 }+4 } } \right] } $$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ 2\frac { -\sin { 2x } \sqrt { { x }^{ 2 }+4 } }{ x } } $$
$$\displaystyle =-\lim _{ x\rightarrow 0 }{ 2\left[ 2\cos { 2x } \sqrt { { x }^{ 2 }+4 } +\sin { 2x } \frac { x }{ \sqrt { { x }^{ 2 }+4 } } \right] } $$
$$\displaystyle =-\left[ 4\sqrt { 4 } \right] =-8$$
Thus, $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =-8$$
For function to be continuous at $$x=0$$.
$$\displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ f\left( x \right) } $$
$$\displaystyle \Rightarrow a=-8$$
If $$\displaystyle f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^{5}};x\neq 0$$ is continuous at $$x=0$$ , then
Report Question
0%
$$\displaystyle {A}+{B}=2$$
0%
$$\displaystyle {A}+{B}=1$$
0%
$$\displaystyle {A}+{B}=0$$
0%
$$\displaystyle {A}-{B} =1$$
Explanation
$$f(x)=\displaystyle \frac { sin3x+Asin2x+Bsinx }{ x^{ 5 } } ;x\neq 0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ f(x)= } \lim _{ x\rightarrow 0 }{\displaystyle \frac { sin3x+Asin2x+Bsinx }{ x^{ 5 } } } $$
It is of the form $$\displaystyle \frac{0}{0}$$, so applying L-Hospital's rule
$$=\displaystyle \lim _{ x\rightarrow 0 }{ \displaystyle\frac { 3\cos { 3x } +2A\cos { 2x } +B\cos { x } }{ 5x^{ 4 } } } $$
As $$x\to 0 , D^{r} \to 0 \Rightarrow N^{r} \to 0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ 3\cos { 3x } +2A\cos { 2x } +B\cos { x } =0 } $$
$$\Rightarrow 3+2A+B=0$$ .....(i)
Again $$\displaystyle \lim _{ x\rightarrow 0 }{\displaystyle \frac { 3\cos { 3x } +2A\cos { 2x } +B\cos { x } }{ 5x^{ 4 } } } $$ is of the form $$\displaystyle \frac {0}{0}$$
$$=\displaystyle \lim _{ x\rightarrow 0 }{ \displaystyle\frac { -9\sin { 3x } -4A\sin { 2x } -B\sin { x } }{ 20x^{ 3 } } } $$
Again of the form $$\displaystyle \frac {0}{0}$$
$$=\displaystyle \lim _{ x\rightarrow 0 }{\displaystyle \frac { -27\cos { 3x } -8A\cos { 2x } -B\cos { x } }{ 60x^{ 2 } } } $$
As $$x\to 0 , D^{r} \to 0 \Rightarrow N^{r} \to 0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ -27\cos { 3x } -8A\cos { 2x } -B\cos { x } } =0$$
$$\Rightarrow 27+8A+B=0$$ .....(ii)
Solving (i) and (ii), we get
$$A=-4, B=5$$
Thus, $$A+B=1$$
$$f(x)=\left\{\begin{array}{ll}\dfrac{(x+bx^{2})^{1/_{2}}-x^{1/2}}{bx^{3/2}} & x>0\\c & x=0\\\dfrac{\sin(a+1)x+\sin x}{x} & x<0\end{array}\right.$$ is continuous at $${x}=0$$, then
Report Question
0%
$$a=\displaystyle \frac{-3}{2},b=0,c=\frac{1}{2}$$
0%
$$a=\displaystyle \frac{-3}{2},b\neq 0,c=\frac{1}{2}$$
0%
$$a=\displaystyle \frac{3}{2},b\neq 0,c=\frac{1}{2}$$
0%
$$a=\displaystyle \frac{3}{2},b\neq 0,c=-\dfrac{1}{2}$$
Explanation
Given: $$f(x)=\left\{\begin{array}{ll}\dfrac{(x+bx^{2})^{1/_{2}}-x^{1/2}}{bx^{3/2}} & x>0\\c & x=0\\\dfrac{\sin(a+1)x+\sin x}{x} & x<0\end{array}\right.$$ is continuous at $${x}=0$$
Multiply and divide by
$$((x+bx^{2})^{1/_{2}}+x^{1/2})$$
$$\displaystyle\lim_{x\rightarrow 0^{+}} =\lim_{x\to0}\dfrac{(x+bx^2) + (x)}{bx^{\frac{3}{2}}((x+bx^2)^{\frac{1}{2}} + x^{\frac{1}{2}})} $$
$$\displaystyle = \lim_{x\to 0}\dfrac{x^{\frac{1}{2}}}{{(x+bx^{2})^{\frac{1}{2}}+x^{\frac{1}{2}}}}$$
$$\displaystyle = \lim_{x\to0}\dfrac{1}{(1+bx)\dfrac{1}{2}+1} = \dfrac12$$
$$\displaystyle\lim_{x\rightarrow 0} \dfrac{(a+1)sin(a+1)x}{(a+1)x}+\dfrac{sin x}{x}$$
$$(a+1)+1 =\dfrac{1}{2}$$
$$a=\dfrac{-3}{2}$$
$$c=\dfrac{1}{2}$$ (By continuity)
The graph of the function $$y = f (x)$$ has a unique tangent at the point $$(e^{a} ,0)$$ through which the graph passes then $$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-sinf(x)}{3f(x)}$$ is
Report Question
0%
$$1$$
0%
$$2$$
0%
$$0$$
0%
$$-1$$
Explanation
Given $$y=f(x)$$ has a unique tangent at the point $$(e^a,0)$$.
So, as $$x\rightarrow e^{a}$$, $$f(x)\rightarrow 0$$
Now, $$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-\sin f(x)}{3f(x)}$$
$$\displaystyle \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 3f(x) } -\frac { \sin f(x) }{ 3f(x) } $$
$$\displaystyle =\frac { 7 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 7f(x) } -\frac { 1 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { \sin f(x) }{ f(x) } $$
$$=\dfrac{7-1}{3}=2$$
Assertion(A): $$f(x)=\left\{\begin{array}{ll}x^{2}\sin(\frac{1}{x}) , & x\neq 0\\0, & x=0\end{array}\right.$$ is continuous at $${x}=0$$
Reason(R): Both $$h(x)=x^{2},g(x)=
\left\{\begin{array}{ll}\sin(\frac{1}{x}) , & x\neq 0\\0, & x=0\end{array}\right.$$are continuous at $$x = 0$$
Report Question
0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true and R is not the correct explanation of A
0%
A is true but R is false
0%
R is true but A is false
Explanation
Assertion
$$f(0)=0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ 2 }\sin { \left( \cfrac { 1 }{ x } \right) } } ={ 0 }^{ 2 }\times $$ (finite value)
$$=0$$
$$\therefore $$ It is continuous at $$x=0$$
Reason: $$h\left( x \right) ={ x }^{ 2 }$$ is continuous
but $$g\left( x \right) $$ is not continuous
$$\displaystyle \lim _{ x\rightarrow 0 }{ \sin { \left( \cfrac { 1 }{ x } \right) } } =$$not defined (value oscillates)
$$\displaystyle \lim _{ x\rightarrow 0 }{ \sin { \left( \cfrac { 1 }{ x } \right) } } \neq 0$$
$$\therefore$$ not continuous.
Let $$a, b, c \epsilon R^+$$ and $$\displaystyle\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\displaystyle \frac{n}{(k+an)(k+bn)}=\displaystyle \frac{A}{a-b}ln\displaystyle \frac{a(b+B)}{b(a+C)}, a \neq b,$$ then $$(A + B + C)$$ is equal to
Report Question
0%
2
0%
3
0%
4
0%
5
Explanation
Let $$a, b, c$$ $$\epsilon R^+ $$and
$$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\dfrac{n}{n^{2}\left ( a+\dfrac{k}{n} \right )\left ( b+\dfrac{k}{n} \right )}$$
$$=\lim_{n\rightarrow \infty }\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\left ( a+\dfrac{k}{n} \right )\left ( b+\dfrac{k}{n} \right )}$$
$$=\int_{0}^{1}\dfrac{1}{(a+x)(b+x)}dx=\dfrac{1}{a-b}\int_{0}^{1}\dfrac{(a+x)-(b+x)}{(a+x)(b+x)}dx$$
$$=\dfrac{1}{a-b}\int_{0}^{1}\left ( \dfrac{1}{b+x}-\dfrac{1}{a+x} \right )dx$$
$$=\dfrac{1}{a-b}\left ( ln(b+x)-ln(a+x) \right )_{0}^{1}$$
$$=\dfrac{1}{a-b}\left ( ln\dfrac{(b+x)}{(a+x)} \right )_{0}^{1}=\dfrac{1}{a-b}ln\dfrac{(b+1)a}{(a+1)b}$$
$$\therefore A + B + C = 3$$
Hence, option 'B' is correct.
If f (x)$$=\left \{ \displaystyle \frac{|x+2|}{tan^{-1}(_{2}x+2)} \right.x\neq -2$$
$$x=-2,$$ then f(x) is
Report Question
0%
continuous at $$x = - 2$$
0%
not continuous at $$x = - 2$$
0%
differentiable at $$x = - 2$$
0%
continuous but not differentiable at $$ x = - 2$$
Explanation
L.H.L$$=\lim_{x\rightarrow -2^{-}}f(x)=\lim_{h\rightarrow 0^{+}}f(-2-h)$$
$$=\lim_{h\rightarrow 0^{+}}\dfrac{|-2-h+2|}{tan^{-1}(-2-h+2)}$$
$$\lim_{h\rightarrow 0^{+}}\dfrac{h}{tan^{-1}(-h)}$$
$$\lim_{h\rightarrow 0^{+}}\dfrac{h}{-tan^{-1}h}$$
& R.H.L. $$= \lim_{x\rightarrow -2^{+}}f(x)=\lim_{h\rightarrow 0^{+}}f_-2+h$$
$$=\lim_{h\rightarrow 0^{+}}\dfrac{|-2+h+2|}{tan^{-1}(-2+h+2)}$$
$$=\lim_{h\rightarrow 0^{+}}\dfrac{h}{tan^{-1}(h)}=1$$
$$\therefore \lim_{x\rightarrow -2^{-}}f(x)\neq \lim_{x\rightarrow -2^{+}}f(x)$$
so, f(x) is not continuous & not differentiable at x $$=- 2$$
The function $$f(x) = x - |x - x^2|, -1 \leq x \leq 1$$ is continuous on the interval
Report Question
0%
$$[-1, 1]$$
0%
$$[-1, 2]$$
0%
$$[-1, 1]- \{0\}$$
0%
$$(-1, 1)- \{0\}$$
Explanation
Given,
$$f(x)=x-|x-x^2|=x-|x(1-x)|$$
$$\therefore$$ continuity is to be checked at $$x=0$$ and $$x=1$$
At $$x=0$$
$$LHL =\displaystyle\lim_{h \rightarrow 0}f(0-h)=\displaystyle\lim_{h \rightarrow 0}[-h-|-h(1+h)|]=0$$
$$RHL =\displaystyle\lim_{h \rightarrow 0}f(0+h)=\displaystyle\lim_{h \rightarrow 0}[h-|h(1-h)|]=0$$
Also, $$f(0)=0$$
Since $$LHL = RHL=f(0)$$
$$\therefore f(x)$$ is continuous at $$x=0$$
At $$x=1$$
$$LHL = \displaystyle\lim_{h \rightarrow 0}f(1-h)=\displaystyle\lim_{h \rightarrow 0}[(1-h)-|(1-h)(1-1+h)|]=1$$
$$RHL =\displaystyle\lim_{h \rightarrow 0}f(1+h)=\displaystyle\lim_{h \rightarrow 0}[(1+h)-|(1+h)(1-1-h)|]=1$$
Also, $$f(1)=1$$
$$\therefore f(x)$$ is continuous at $$x=1$$
Hence $$f(x)$$ is continuous for all$$ x \in [-1,1]$$
$$\displaystyle \lim_{x\to\infty}{\displaystyle \frac{2\sqrt{x}+3\sqrt [\Large 3]{x}+4\sqrt [\Large 4]{x}+...+n\sqrt [\Large n]{x}}{\sqrt{(2x-3)}+\sqrt[\Large 3]{(2x-3)}+\sqrt[\Large 4]{(2x-3)}+...+\sqrt[\Large n]{(2x-3)}}}$$ is equal to
Report Question
0%
$$1$$
0%
$$\infty$$
0%
$$\sqrt{2}$$
0%
None of these
Explanation
$$\displaystyle \lim_{x\to\infty}{\displaystyle \dfrac{2\sqrt{x}+3\sqrt [\Large 3]{x}+4\sqrt [\Large 4]{x}+...+n\sqrt [\Large n]{x}}{\sqrt{(2x-3)}+\sqrt[\Large 3]{(2x-3)}+\sqrt[\Large 4]{(2x-3)}+...+\sqrt[\Large n]{(2x-3)}}}$$
Dividing numerator and denominator by $$\sqrt{x}$$
$$\displaystyle=\lim _{ x\to \infty }{ \displaystyle\dfrac { 2+3{ x }^{ -\frac { 1 }{ 6 } }+4{ x }^{ -\frac { 1 }{ 4 } }+...+n{ x }^{ \frac { (2-n) }{ 2n } } }{ \sqrt { (2-\dfrac { 3 }{ x } ) } +\dfrac { \sqrt [ 3 ]{ (2x-3) } }{ \sqrt { x } } +...+\dfrac { \sqrt [ n ]{ (2x-3) } }{ \sqrt { x } } } } $$
$$\displaystyle =\lim _{ x\to \infty }{ \dfrac { 2+\dfrac { 3 }{ { x }^{ 1/6 } } +\dfrac { 4 }{ { x }^{ 1/4 } } +...+\dfrac { n }{ { x }^{ \frac { (n-2) }{ 2n } } } }{ \sqrt { (2-\dfrac { 3 }{ x } ) } +\sqrt [ 6 ]{ \dfrac { { (2x-3) }^{ 2 } }{ { x }^{ 3 } } } +...+\sqrt [ 2n ]{ \dfrac { { (2x-3) }^{ 2 } }{ { x }^{ n } } } } } $$
$$\displaystyle=\dfrac { 2 }{ \sqrt { 2 } } =\sqrt { 2 } $$
If $$f(x) \displaystyle =\frac{3x^2 + ax + a + 1}{x^2 + x - 2}$$, then which of the following can be correct?
Report Question
0%
$$\displaystyle \lim_{x \rightarrow 1} f(x)$$ exists $$\Rightarrow a = - 2$$
0%
$$\displaystyle \lim_{x \rightarrow -2} f(x)$$ exists $$\Rightarrow a = 13$$
0%
$$\displaystyle \lim_{x \rightarrow 1} f(x) = \dfrac{4}{3}$$
0%
$$\displaystyle \lim_{x \rightarrow -2} f(x) = -\dfrac{1}{3}$$
Explanation
$$f(x) \displaystyle = \frac{3x^2 + ax + a + 1}{(x+2)(x-1)}$$
As $$x \rightarrow 1, $$ Denominator $$ \rightarrow 0$$. Hence as $$x \rightarrow 1, $$ Numerator $$ \rightarrow 0$$.
Therefore, $$3 + 2a + 1 = 0$$ or $$ a=-2$$
As $$x \rightarrow -2, $$Denominator $$ \rightarrow 0$$.Hence as $$x \rightarrow - 2, $$ Numerator $$ \rightarrow 0$$.
Therefore, $$12 - 2a + a + 1 = 0$$ or $$a = 13$$
Now, $$\lim_{x \rightarrow 1} f(x) = \displaystyle \lim_{x \rightarrow1} \frac{3x^2 - 2x - 1}{(x + 2)(x-1)}$$
$$\displaystyle = \lim_{x \rightarrow 1} \frac{(3x + 1)(x-1)}{(x+2)(x-1)} = \frac{4}{3}$$
Now,$$\displaystyle \lim_{x \rightarrow -2} \frac{3x^2 + 13x +14}{(x + 2)(x-1)}$$
$$\displaystyle = \lim_{x \rightarrow -2} \frac{(3x + 7)(x+2)}{(x + 2)(x-1)} = -\frac{1}{3}$$
If the function $$f(x)= \left\{\begin{matrix}
(1+\left | \tan x \right |)^{ \displaystyle \frac{p}{\left| \tan {x} \right|}} &, -\frac{\pi }{3}< x< 0 \\ \\
q& x=0\\ \\
e^{ \displaystyle \frac{\sin {3x}}{\sin {2x}}},& 0\: < \, x\, < \frac{\pi }{3}
\end{matrix}\right.$$
is continuous at $$x=0$$, then
Report Question
0%
$$\displaystyle \mathrm{p}=\frac{3}{2}$$
0%
$$\displaystyle \mathrm{p}=\frac{2}{3}$$
0%
$$\log_{e}\mathrm{q}=\mathrm{p}$$
0%
$$\mathrm{q}=2$$
Explanation
Since f(x) is continuous at $$x=0$$
$$ \displaystyle \lim_{x\rightarrow 0^{-}}f(x)=f(0)=\lim_{x\rightarrow 0^{+}}f(x)$$...(1)
Now ,
$$LHL= \displaystyle \lim _{ x\rightarrow 0^{ - } } f(x)=\lim _{ x\rightarrow 0 } (1-tan\, x)^{ -\frac { p }{ tan\, x } }$$
Since $$ \displaystyle \lim _{ x\rightarrow 0 } (1+x)^{ 1/x }=e$$
$$\Rightarrow LHL=e^{ p }$$...(2)
Now,
$$RHL \displaystyle =\lim_{x\rightarrow 0^{+}}f(x) =\lim_{x\rightarrow 0}e^{\displaystyle \frac{\sin {3x}}{\sin {2x}}}$$
$$ \displaystyle =\lim _{ x\rightarrow 0 } e^{ \displaystyle \frac { 3 }{ 2 } (\displaystyle \frac { \sin { 3x } }{ 3x } )(\frac { 2x }{ \sin { 2x } } ) }$$
$$RHL=e^{3/2}$$....(3)
From (1), (2) and (3)
$$ e^{ p }=e^{ 3/2 }=q$$
$$\Rightarrow p=\frac { 3 }{ 2 } =\log _{ e } q$$
Let $$\tan \alpha .x+\sin \alpha .y=\alpha $$ and $$\alpha \ \text{cosec} \alpha .x+\cos \alpha .y=1$$ be two variable straight line, $$\alpha $$ being the parameter. Let $$P$$ be the point of intersection of the lines. In the limiting position when $$\alpha \rightarrow 0$$, the point $$P$$ lies on the line
Report Question
0%
$$x=2$$
0%
$$x=-1$$
0%
$$y+1=0$$
0%
$$y=2$$
Explanation
Solving $$\tan \alpha .x+\sin \alpha .y=\alpha $$ and $$\alpha \ \text{cosec}\alpha .x+\cos \alpha .y=1$$, we get
$$x=\cfrac { \alpha \cos { \alpha } -\sin { \alpha } }{ \sin { \alpha } -\alpha } $$ and $$y=\cfrac { \alpha -x\tan { \alpha } }{ \sin { \alpha } } $$
$$\displaystyle \lim _{ \alpha \rightarrow 0 }{ x } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { \cos { \alpha } -\alpha \sin { \alpha } -\cos { \alpha } }{ \cos { \alpha } -1 } } \\ =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { \alpha \sin { \alpha } }{ 2\sin ^{ 2 }{ \cfrac { \alpha }{ 2 } } } } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { 4{ \left( \cfrac { \alpha }{ 2 } \right) }^{ 2 }\cfrac { \sin { \alpha } }{ \alpha } }{ { \left( \sin { \cfrac { \alpha }{ 2 } } \right) }^{ 2 }2 } } =2\\ \displaystyle \lim _{ \alpha \rightarrow 0 }{ y } =\displaystyle \lim _{ \alpha \rightarrow \alpha }{ \cfrac { \alpha -x\tan { \alpha } }{ \sin { \alpha } } } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \left( \cfrac { \alpha }{ \sin { \alpha } } -\cfrac { x }{ \cos { \alpha } } \right) } =1-2=-1$$
Hence $$P=\left( 2,-1 \right) $$
if $$f\left( x \right) = \left\{ {\matrix{ {\cos \left[ x \right],} & {x \ge 0} \cr {\left| x \right| + a,} & {x < 0} \cr
} } \right\}$$
Find
the value of a , given that $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$ exists,
where[.] denotes
Report Question
0%
-1
0%
2
0%
1
0%
0
$$f(x) = \left\{\begin{matrix}(3/x^{2})\sin 2x^{2} & if x M 0 \\\dfrac {x^{2} + 2x + c}{1 - 3x^{2}} & if\ x \geq 0, x \neq \dfrac {1}{\sqrt {3}}\\ 0 & x = 1/ \sqrt {3}\end{matrix}\right.$$ then in order that $$f$$ be continuous at $$x = 0$$, the value of $$c$$ is
Report Question
0%
$$2$$
0%
$$4$$
0%
$$6$$
0%
$$8$$
$$\displaystyle\underset{x\rightarrow 0}{Lt}\left(cosec x-\dfrac{1}{x}\right)=?$$
Report Question
0%
$$0$$
0%
$$1/2$$
0%
$$1$$
0%
Does not exits
For each $$t\in R$$, let [t] be the greatest integer less than or equal to t. Then,
$$\underset { { x\rightarrow 0 }^{ + } }{ lim } x([\frac { 1 }{ x } ]+[\frac { 2 }{ x } ]+...+[\frac { 15 }{ x } ])$$
Report Question
0%
is equal to 0
0%
is equal to 15
0%
is equal to 120
0%
does not exist (in R)
If $$\phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$$, then
Report Question
0%
$$\phi (x) = g(x)$$ for all x $$\in$$ R
0%
$$\phi (x) = f(x)$$ for all x $$\in$$ R
0%
$$\left\{\begin{matrix}g(x) & for -1 < x < 1\\ f(x) & for |x| \geq 1\end{matrix}\right.$$
0%
$$\left\{\begin{matrix}g(x) & for |x| < 1\\ f(x) & for |x| > 1 \\\displaystyle \frac{f(x) + g(x)}{2} & for |x| = 1\end{matrix}\right.$$
Explanation
We have,
$$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = \left\{\begin{matrix}0,& if |x| < 1\\ \infty, & if |x| > 1 \\ 1, & if |x| = 1\end{matrix}\right.$$
Thus, we have the following cases:
CASE I
When $$- 1 < x < 1$$
In this case, we have $$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = 0$$
$$\therefore \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}} = g(x)$$
CASE II
When $$|x| > 1$$
In this case, we have $$\displaystyle \lim_{n \rightarrow \infty} \displaystyle \frac{1}{x^{2n}} = 0$$
$$\therefore \phi (x) = \displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$$
$$\Rightarrow \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{f(x) + \displaystyle \frac{g(x)}{x^{2n}}}{1 + \displaystyle \frac{1}{x^{2n}}} = \displaystyle \frac{f(x) + 0}{1 + 0} = f(x)$$
CASE III
When $$|x| = 1$$
In this case, we have $$x^{2n} = 1\Rightarrow \displaystyle \lim_{n \rightarrow \infty } x^{2n}=1$$.
$$\therefore \phi (x) = \displaystyle \frac{f(x) + g(x)}{2}$$
If $$\displaystyle f(x) = \frac{x - e^x + cos 2x}{x^2}, x \neq 0$$, is continuous at $$x = 0$$,
where [x] and {x} denotes the greatest integer and fractional part functions, respectively.
Then which of the following is correct?
Report Question
0%
$$f(0) = 5/2$$
0%
$$[f(0)] = - 2$$
0%
$$\{ f(0) \} = - 0.5$$
0%
$$[f(0)]\{ f(0) \} = - 1.5$$
Explanation
$$\displaystyle \lim_{h \rightarrow 0} \frac{x- e^x + 1 - (1 - cos 2x)}{x^2}$$
$$\displaystyle = \lim_{h \rightarrow 0} \left [ \frac{x - e^x + 1}{x^2} - \frac{(1 - cos 2x)}{x^2} \right ]$$
$$= \displaystyle\lim_{h \rightarrow 0} \left [ \frac{x + 1 - \left (1 + x + \frac{x^2}{2} \right )}{x^2} - \frac{2sin^2 x}{x^2} \right ]$$ (Using expansion of $$e^x$$)
$$\displaystyle = - \frac{1}{2} - 2 = - \frac{5}{2}$$
Hence, for continuity, $$f(0) = - \displaystyle \frac{5}{2}$$
Now, $$[f(0)] = - 3; \{ f(0) \} = \displaystyle \left \{ - \frac{5}{2} \right \} = \frac{1}{2}$$.
Hence, $$[f(0)] \{ f(0) \} = \displaystyle - \frac{3}{2} = - 1.5$$
Hence, option D is correct.
Let $$f(x) \displaystyle = \frac{x^2 - 9x + 20}{x -[x]}$$ where [x] is the greatest integer not greater than $$x$$, then
Report Question
0%
$$\displaystyle \lim_{x \rightarrow 5^-} f(x) = 0$$
0%
$$\displaystyle \lim_{x \rightarrow 5^+} f(x) = 1$$
0%
$$\displaystyle \lim_{x \rightarrow 5} f(x)$$ does not exists
0%
$$none\ of\ these$$
Explanation
$$\displaystyle \lim_{x \rightarrow 5^-} f(x) $$
$$=\lim_{x \rightarrow 5^-} \displaystyle \dfrac{x^2 - 9x + 20}{x - [x]} $$
$$=\displaystyle \lim_{x \rightarrow 5^-} \dfrac{(x - 5)(x - 4)}{x - 4}$$
$$=\displaystyle \lim_{x \rightarrow 5^-} (x-5) = 0$$
Hence, option A is correct.
Now, $$\displaystyle \lim_{x \rightarrow 5^-} f(x) = 0$$
$$\displaystyle \lim_{x \rightarrow 5^+} \dfrac{x^2 - 9x + 20}{x - [x]}$$
$$=\displaystyle \lim_{x \rightarrow 5^+} \dfrac{(x- 5)(x-4)}{x - 5} $$
$$=\displaystyle \lim_{x \rightarrow 5^+} (x - 4) = 1$$
Hence, option B is correct.
Since, $$LHL \ne RHL$$
Hence, limit does not exist.
If $${ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },..,{ x }_{ n }$$ are the roots of the equation $$x^n+ax+b=0,$$ then the value of $$\left( { x }_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) \left( { x }_{ 1 }-{ x }_{ 4 } \right) ...\left( { x }_{ 1 }-{ x }_{ n } \right) $$ is equal to
Report Question
0%
$$n{ { x }_{ 1 } }^{ n-1 }+a$$
0%
$$n{ { x }_{ 1 } }^{ n-1 }$$
0%
$$nx-1+b$$
0%
$$n{ { x }_{ 1 } }^{ n-1 }+b$$
Explanation
Given:$${ x }_{ 1, }{ x }_{ 2, }{ x }_{ 3 }.....{ x }_{ n }$$ are the roots of $${ x }^{ n }+ax+b=0$$
To find: $$\left( { x }_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) ......\left( { x }_{ 1 }-{ x }_{ n } \right) =?$$
Sol: $$\left( { x }_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) ......\left( { x }_{ 1 }-{ x }_{ n } \right) $$
$$=\dfrac { \left( { x }_{ 1 }-{ x }_{ 1 } \right) \left( { x }_{ 1 }-{ x }_{ 2 } \right) }{ ({ x }_{ 1 }-{ x }_{ 1 } )} ........\left( { x }_{ 1 }-{ x }_{ 2 } \right) $$
$$=\lim _{ x\rightarrow { x }_{ 1 } }{ \dfrac { \left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right) .........\left( x-{ x }_{ n } \right) }{ \left( x-{ x }_{ 1 } \right) } } $$
$$=\lim _{ x\rightarrow { x }_{ 1 } }{ \dfrac { { x }^{ n }+ax+b }{ \left( x-{ x }_{ 1 } \right) } } $$
Using l'Hospital rule, we get
$$\lim _{ x\rightarrow { x }_{ 1 } }{ \dfrac { n.{ x }^{ n-1 }+a }{ 1 } } ={ nx }_{ 1 }^{ n-1 }+a$$
Hence, correct answer is $${ nx }_{ 1 }^{ n-1 }+a$$
STATEMENT-1 : $$\displaystyle \lim_{x \rightarrow 0} [x] \left \{ \frac{e^{1/x} - 1}{e^{1/x} + 1} \right \}$$ (where [.] represents the greatest integer function) does not exist.
STATEMENT-2 : $$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$$ does not exists.
Report Question
0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is False
0%
STATEMENT-1 is False, STATEMENT-2 is True
Explanation
I)
$$\displaystyle \lim_{x \rightarrow 0} [x] \left \{ \frac{e^{1/x} - 1}{e^{1/x} + 1} \right \}$$
$$RHL=\displaystyle \lim_{x \rightarrow 0^+} [x] \left ( \frac{e^{1/x} - 1}{e^{1/x}+1} \right) $$
$$=\lim _{ h\rightarrow 0 } [h]\left(\displaystyle \frac { e^{ 1/h }-1 }{ e^{ 1/h }+1 } \right) $$
$$= \lim_{h \rightarrow 0} [h] \left ( \displaystyle\frac{1 - e^{-1/h}}{1 + e^{-1/h}} \right) $$
$$= 0 \times 1 = 0$$
$$LHL=\displaystyle \lim_{x \rightarrow 0^-} [x] \left ( \frac{e^{1/x}- 1}{e^{1/x} + 1} \right )$$
$$= \lim_{h \rightarrow 0} [-h] \left ( \displaystyle\frac{e^{-1/h} - 1}{e^{-1/h} + 1} \right) $$
$$= -1 \times (-1) = 1$$
Here, $$LHL \ne RHL$$
Thus, given limit does not exist.
II)
$$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$$
$$RHL=\displaystyle \lim_{x \rightarrow 0^+} \left ( \frac{e^{1/x} - 1}{e^{1/x}+1} \right) $$
$$=\lim _{ h\rightarrow 0 }\left(\displaystyle \frac { e^{ 1/h }-1 }{ e^{ 1/h }+1 } \right) $$
$$= \lim_{h \rightarrow 0} \left ( \displaystyle\frac{1 - e^{-1/h}}{1 + e^{-1/h}} \right) $$
$$RHL= 1 $$
$$LHL=\displaystyle \lim_{x \rightarrow 0^-} \left ( \frac{e^{1/x}- 1}{e^{1/x} + 1} \right )$$
$$= \lim_{h \rightarrow 0} \left ( \displaystyle\frac{e^{-1/h} - 1}{e^{-1/h} + 1} \right) $$
$$LHL= -1$$
So, $$\displaystyle \lim_{x \rightarrow 0} \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right)$$ does not exist, but this cannot be taken as only reason for non-existence of $$\displaystyle \lim_{x \rightarrow 0} [x] \left ( \frac{e^{1/x} - 1}{e^{1/x} + 1} \right )$$.
If $$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$$ and $$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$$,
then the value of $$\displaystyle\lim_{x\rightarrow a}{f(x)g(x)}$$ is?
Report Question
0%
Does not exist
0%
Exists and is $$\displaystyle\frac{3}{4}$$
0%
Exists and is $$\displaystyle-\frac{3}{4}$$
0%
Exists and is $$\displaystyle\frac{4}{3}$$
Explanation
Given, $$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$$ and $$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$$
,
Lets, assume $$F(x) = (f(x) + g(x))$$ and $$ G(x) = (f(x) - g(x))$$
$$\because$$ Limits of both $$F(x)$$ and $$G(x)$$ exists as $$x\rightarrow a$$, we can say that
$$\displaystyle\lim_{x\rightarrow a}{(F(x)+G(x))}$$ and $$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))}$$ also exists.
$$\displaystyle\therefore\lim_{x\rightarrow a}{(F(x)+G(x))} = 3$$
$$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{f(x)} = 3$$
$$\Rightarrow \lim_{x\rightarrow a}{f(x)} = \dfrac{3}{2}$$
Similarly,
$$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))} = 1$$
$$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{g(x)} = 1$$
$$\Rightarrow \lim_{x\rightarrow a}{g(x)} = \dfrac{1}{2}$$
Hence,
$$\displaystyle\lim_{x\rightarrow a}{(f(x)\cdot g(x))} = \lim_{x\rightarrow a}{f(x)}\cdot\lim_{x\rightarrow a}{g(x)} = \dfrac{3}{2}\times \dfrac{1}{2} = \dfrac{3}{4}$$
Hence, option B.
$$x$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$f(x)$$
$$4$$
$$3$$
$$7$$
$$1$$
$$3$$
The function f is continuous on the closed interval $$[1, 5]$$ and values of the function are shown in the table above. If the values in the table are used to calculate a trapezoidal sum, the approximate value of $$\int_{1}^{5}f(x)dx$$ is
Report Question
0%
$$14$$
0%
$$14.5$$
0%
$$15$$
0%
$$29$$
$$\quad \lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \sum _{ r=1 }^{ 2n }{ \cfrac { r }{ \sqrt { { n }^{ 2 }+{ r }^{ 2 } } } } } $$ equal to:
Report Question
0%
$$1+\sqrt { 5 } $$
0%
$$-1+\sqrt { 5 } $$
0%
$$1+\sqrt { 2 } $$
0%
$$1+\sqrt { 2 } $$
The value of $$\lim _{ x\rightarrow 0 }{ \left( { \left( \sin { x } \right) }^{ 1/x }+{ \left( 1+x \right) }^{ \sin { x } } \right) } $$ whre $$x> 0$$ is
Report Question
0%
$$0$$
0%
$$-1$$
0%
$$1$$
0%
2
Explanation
Given $$\lim _{ x\rightarrow 0 }{ \left( { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+{ \left( 1+x \right) }^{ sinx } \right) } $$
$$=\lim _{ x\rightarrow 0 }{ { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } } $$
$$={ \left( sin0 \right) }^{ { \frac { 1 }{ 0 } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } $$
$$={ \left( 0 \right) }^{ { \infty } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } $$
$$=0+{ e }^{ \lim _{ x\rightarrow 0 }{ \left[ log{ \left( 1+x \right) }^{ sinx } \right] } }$$
$$\left( \because { e }^{ loga }=a \right) $$
$$={ e }^{ \lim _{ x\rightarrow 0 }{ \left[ sinx.log{ \left( 1+x \right) } \right] } }$$
$$={ e }^{ \lim _{ x\rightarrow 0 }{ sinx } \times \lim _{ x\rightarrow 0 }{ log\left( 1+x \right) } }$$
$$={ e }^{ sin\left( 0 \right) \times log\left( 1+0 \right) }$$
$$={ e }^{ 0\times log\left( 1 \right) }$$
$$={ e }^{ 0\times 0 }={ e }^{ 0 }$$
$$=1$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
