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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
lim
x
→
0
cos
(
tan
x
)
−
cos
x
x
4
is equal to
Report Question
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1/6
0%
-1/3
0%
1/2
0%
1
Explanation
cos
(
tan
x
)
−
cos
x
=
2
sin
(
x
+
tan
x
2
)
sin
(
x
−
tan
x
2
)
⇒
lim
x
→
0
cos
(
tan
x
)
−
cos
x
x
4
=
lim
x
→
0
2
sin
(
x
+
tan
x
2
)
sin
(
x
−
tan
x
2
)
x
4
=
lim
x
→
0
2
sin
(
x
+
tan
x
2
)
sin
(
x
−
tan
x
2
)
x
4
(
x
+
tan
x
2
)
(
x
−
tan
x
2
)
(
x
2
−
tan
2
x
4
)
=
1
2
lim
x
→
0
x
2
−
tan
2
x
x
4
=
1
2
lim
x
→
0
x
2
−
(
x
+
x
3
3
+
2
15
x
5
+
⋯
)
2
x
4
=
1
2
lim
x
→
0
1
x
2
(
1
−
(
1
+
x
2
3
+
2
15
x
4
+
…
)
2
)
=
−
1
3
The value of
lim
n
→
∞
[
1
n
+
e
1
/
n
n
+
e
2
/
n
n
+
…
.
+
e
(
n
−
1
)
/
n
n
]
is
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0%
1
0%
0
0%
e-1
0%
e+1
Explanation
lim
n
→
∞
[
1
n
+
e
1
/
n
n
+
e
2
/
n
n
+
⋯
+
e
(
n
−
1
)
/
n
n
]
=
lim
n
→
∞
[
1
+
e
1
/
n
+
(
e
1
/
n
)
2
+
⋯
+
(
e
1
/
n
)
n
−
1
n
]
=
lim
n
→
∞
1
⋅
[
(
e
1
/
n
)
n
−
1
]
n
(
e
1
/
n
−
1
)
=
(
e
−
1
)
lim
n
→
∞
1
(
e
1
/
n
−
1
1
/
n
)
=
(
e
−
1
)
×
1
=
(
e
−
1
)
lim
x
→
0
x
4
(
cot
4
x
−
cot
2
x
+
1
)
(
tan
4
x
−
tan
2
x
+
1
)
is equal to
Report Question
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1
0%
0
0%
2
0%
None of these
Explanation
x
4
(
cot
4
x
−
cot
2
x
+
1
)
(
tan
4
x
−
tan
2
x
+
1
)
=
x
4
(
1
+
tan
2
x
+
tan
4
x
)
tan
4
x
(
tan
4
x
−
tan
2
x
+
1
)
=
x
4
tan
4
x
,
x
≠
0
⇒
lim
x
→
0
x
4
(
cot
4
x
−
cot
2
x
+
1
)
(
tan
4
x
−
tan
2
x
+
1
)
=
lim
x
→
0
x
4
tan
4
x
=
1
lim
n
→
∞
20
∑
x
=
1
cos
2
n
(
x
−
10
)
is equal to
Report Question
0%
0
0%
1
0%
19
0%
20
Explanation
∵
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
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\displaystyle \log { 2 }
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\displaystyle \log { 6 }
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\displaystyle 1
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\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
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(3,0)
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(2,0)
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(1,0)
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(4,-1)
\underset { x\rightarrow \pi /4 }{ Lim } \dfrac { 2\sqrt { 2 } \left( cosx+sinx \right) ^{ 3 } }{ 1-sin2x } =2
is equal to
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\dfrac{3\sqrt{2}}{2}
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2\sqrt{2}
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\dfrac{4\sqrt{2}}{3}
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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
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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
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2
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-2
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1/2
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-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
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e^{\sin ^{2} y}
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\sin 2 y e^{\sin ^{2} y}
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0
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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
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(k-1) \log _{e}(1+k)-k
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(k+1) \log _{e}(k-1)+k
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(k+1) \log _{e}(k-1)-k
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(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
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\dfrac{2 a}{\pi}
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-\dfrac{2 a}{\pi}
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\dfrac{4 a}{\pi}
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-\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
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6
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3
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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:
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0
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m+n
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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
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3
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1/3
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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
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2/3
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1/3
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1/2
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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=
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4
0%
2
0%
1
0%
1/4
The value of
\displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin x}{x}
is
Report Question
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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
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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
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\pi/4
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\pi/2
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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:
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\displaystyle \frac{8}{5}f(0)
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\displaystyle \frac{32}{5}f(0)
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\displaystyle \frac{64}{5}f(0)
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\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
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-\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
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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 -
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2
0%
-1
0%
0
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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
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-8
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8
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-4
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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
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\displaystyle {A}+{B}=2
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\displaystyle {A}+{B}=1
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\displaystyle {A}+{B}=0
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\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
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a=\displaystyle \frac{-3}{2},b=0,c=\frac{1}{2}
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a=\displaystyle \frac{-3}{2},b\neq 0,c=\frac{1}{2}
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a=\displaystyle \frac{3}{2},b\neq 0,c=\frac{1}{2}
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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
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1
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2
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0
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-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
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Both A and R are true and R is the correct explanation of A
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Both A and R are true and R is not the correct explanation of A
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A is true but R is false
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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
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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
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continuous at
x = - 2
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not continuous at
x = - 2
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differentiable at
x = - 2
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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
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[-1, 1]
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[-1, 2]
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[-1, 1]- \{0\}
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(-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
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1
0%
\infty
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\sqrt{2}
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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?
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\displaystyle \lim_{x \rightarrow 1} f(x)
exists
\Rightarrow a = - 2
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\displaystyle \lim_{x \rightarrow -2} f(x)
exists
\Rightarrow a = 13
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\displaystyle \lim_{x \rightarrow 1} f(x) = \dfrac{4}{3}
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\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
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\displaystyle \mathrm{p}=\frac{3}{2}
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\displaystyle \mathrm{p}=\frac{2}{3}
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\log_{e}\mathrm{q}=\mathrm{p}
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\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
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x=2
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x=-1
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y+1=0
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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
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-1
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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
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2
0%
4
0%
6
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8
\displaystyle\underset{x\rightarrow 0}{Lt}\left(cosec x-\dfrac{1}{x}\right)=?
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0
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1/2
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1
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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 } ])
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is equal to 0
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is equal to 15
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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
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\phi (x) = g(x)
for all x
\in
R
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\phi (x) = f(x)
for all x
\in
R
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\left\{\begin{matrix}g(x) & for -1 < x < 1\\ f(x) & for |x| \geq 1\end{matrix}\right.
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\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?
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f(0) = 5/2
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[f(0)] = - 2
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\{ f(0) \} = - 0.5
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[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
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\displaystyle \lim_{x \rightarrow 5^-} f(x) = 0
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\displaystyle \lim_{x \rightarrow 5^+} f(x) = 1
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\displaystyle \lim_{x \rightarrow 5} f(x)
does not exists
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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
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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.
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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?
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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
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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:
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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
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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
0:0:2
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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