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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 14 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 14
If
f
(
x
)
is a differentiable function in the interval
(
0
,
∞
)
such that
f
(
1
)
=
1
and
lim
t
→
x
t
2
f
(
x
)
−
x
2
f
(
t
)
t
−
x
=
1
, for each
x
>
0
, then
f
(
3
2
)
is equal to:
Report Question
0%
25
9
0%
13
6
0%
23
18
0%
31
18
Let
f
(
x
)
=
x
|
x
|
,
g
(
x
)
=
s
i
n
x
and
h
(
x
)
=
(
g
o
f
)
(
x
)
.
Then
Report Question
0%
h
′
(
x
)
is differentiable at x=0
0%
h
′
(
x
)
is continuous at x=0 but is not differentiable at x=0
0%
h
′
(
x
)
is differentiable at x=0 but
h
′
(
x
)
differentiable is not continuous at x=0
0%
h
′
(
x
)
is not differentiable at x=0
If
f
(
x
)
=
0
for
x
<
0
and
f
(
x
)
is differentiable at
x
=
0
, then for
x
≥
0
,
f
(
x
)
may be
Report Question
0%
x
2
0%
x
0%
−
x
0%
−
x
3
/
2
The value of
f
(
4
)
is
Report Question
0%
160
0%
240
0%
200
0%
N
o
n
e
o
f
t
h
e
s
e
If
y
=
sin
−
1
x
+
sin
−
1
√
1
−
x
2
, then
d
x
d
x
Report Question
0%
0
0%
1
0%
−
2
x
0%
y
x
y
=
x
20
,
find
d
2
y
d
x
2
Report Question
0%
381
x
18
0%
380
x
18
0%
389
x
18
0%
370
x
18
If
x
=
θ
−
1
θ
,
y
=
θ
+
1
θ
then
d
y
d
x
=
Report Question
0%
x
y
y
2
+
2
0%
y/x
0%
-x/y
0%
-y/x
Explanation
x
=
θ
−
1
θ
d
x
d
θ
=
1
−
(
−
1
θ
2
)
∴
y=\theta+\dfrac{1}{\theta}
\dfrac{dy}{d\theta}=1+\left(\dfrac{-1}{{\theta}^{2}}\right)
\therefore \dfrac{dy}{d\theta}=1-\dfrac{1}{{\theta}^{2}}
\therefore \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{1-\dfrac{1}{{\theta}^{2}}}{1+\dfrac{1}{{\theta}^{2}}}
We have
x=\theta-\dfrac{1}{\theta}
and
y=\theta+\dfrac{1}{\theta}
\Rightarrow xy={\theta}^{2}-\dfrac{1}{{\theta}^{2}}
and
{y}^{2}={\left(\theta+\dfrac{1}{\theta}\right)}^{2}
\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2\theta\dfrac{1}{\theta}
\Rightarrow {y}^{2}={\theta}^{2}+\dfrac{1}{{\theta}^{2}}-2
\Rightarrow {y}^{2}+2={\theta}^{2}+\dfrac{1}{{\theta}^{2}}
\therefore \dfrac{dy}{dx}=\dfrac{1-\dfrac{1}{{\theta}^{2}}}{1+\dfrac{1}{{\theta}^{2}}}
=\dfrac{xy}{{y}^{2}+2}
If
y=\cos ^{ -1 }{ (\frac { x-{ x }^{ -1 } }{ x+{ x }^{ -1 } } ) } \quad then\quad \frac { dy }{ dx } =
Report Question
0%
\frac { -2 }{ 1+{ x }^{ 2 } }
0%
\frac { 2 }{ 1+{ x }^{ 2 } }
0%
\frac { 1 }{ 1+{ x }^{ 2 } }
0%
\frac { -1 }{ 1+{ x }^{ 2 } }
dx\left\{ sin{ }^{ -1 }(\frac { 5x+12\sqrt { 1-x{ }^{ 2 } } }{ 13 } ) \right\} =
Report Question
0%
\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }
0%
\frac { -1 }{ \sqrt { 1+{ x }^{ 2 } } }
0%
\frac { 2 }{ \sqrt { 1-{ x }^{ 2 } } }
0%
0
Let fbe twice differentiable function such that
{ g }^{ 1 }\left( x \right) =-f\left( x \right) and\quad \quad \quad \quad \quad { f }^{ 1 }\left( x \right) =g\left( x \right) ,\\ h(x)={ \left( f\left( x \right) \right) }^{ 2 }+{ \left( g\left( x \right) \right) }^{ 2 }.\quad Ifh(5)=11,\quad thenh(10)\quad is
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0%
22
0%
11
0%
0
0%
8
If
y=\tan ^{ -1 }{ [x+\sqrt { 1+{ x }^{ 2 } } ] }
then
\frac { dy }{ dx } =
Report Question
0%
\frac { 1 }{ 1+{ x }^{ 2 } }
0%
\frac { 1 }{ 2(1+{ x }^{ 2 }) }
0%
\frac { 2 }{ 1+{ x }^{ 2 } }
0%
None of these
\frac { d[\sec ^{ -1 }{ (\sin { x } +{ x }^{ 2 }) } ] }{ dx } =
Report Question
0%
\frac { \sin { 2x } +{ x }^{ 2 } }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } }
0%
\frac { \cos { x } +2x }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } }
0%
\frac { 2x }{ \sin { x } \sqrt { \sin { x } -1 } }
0%
None of these
f(x) = \log_{1 - 2x}(1 + 2x)
for
x \ne 0
= k
for
x = 0
is continuous at
x = 0
, find
k.
Report Question
0%
1
0%
-1
0%
0
0%
\frac{1}{2}
The\, \, function\, \, g(x)=\left| { \begin{array} { *{ 20 }{ c } }{ x+b } & { x<0 } \\ { \cos \, x } & { x\ge 0 } \end{array} } \right. \, \, can\, \, be\, \, made\, differentiable\, at\, x=0
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0%
if\,b\,\,is\,\,equal\,to\,zero\,
0%
if\,b\,\,is\,\,not\,equal\,\,to\,zero\,
0%
if\,b\,\,is\,\,not\,equal\,\,to\,zero\,
0%
for\,no\,value\,of\,b
If
y={ tan }^{ -1 }\left( \dfrac { 1-{ cos }^{ 2 }x }{ 1+{ cos }^{ 2 } } \right) ,
, then
\dfrac { dy }{ xy } =
Report Question
0%
\dfrac { sinx }{ 1+{ cos }^{ 4 }x }
0%
\dfrac { sinx }{ 1+{ cos }^{ 2 }x }
0%
\dfrac { sin2x }{ 1+{ cos }^{ 4 }x }
0%
\dfrac { sin2x }{ 1+{ cos }^{ 2 }x }
Solve
\frac{d\tan ^{-1}}{dx} (\frac{5x + 1}{3 - x - 6x^2}) =
Report Question
0%
0
0%
1
0%
\frac{1}{1 + (3x + 2)^2} + \frac{1}{1 + (2x - 1)^2}
0%
\frac{3}{1 + (3x + 2)^2} + \frac{2}{1 + (2x - 1)^2}
If
\cos ^{-1}x + \cos ^{-1}y = \frac{\pi}{2}
then
\frac{dy}{dx}
Report Question
0%
\frac{x}{y}
0%
\frac{-2x}{y}
0%
\frac{-x}{y}
0%
\frac{-y}{x}
Let
f
be a function defined by
2f\left(\sin x\right) + f\left(\cos x\right) = x ~\forall x\in R
, then set of points where
f
is not differentiable is
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0%
set of all natural numbers
0%
set of all irrational number
0%
\left\{0,1,-1\right\}
0%
\left\{1,-1\right\}
If
f\left( x \right) =\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } }
, then
f(x)
is
Report Question
0%
continuous on
[-1,1]
0%
differentiable on
(-1,0)\cup (0,1)
0%
both (a) and (b)
0%
None of the above
y={ cos }^{ -1 }\cfrac { (9-{ x }^{ 2 }) }{ (9+{ x }^{ 2 }) }
then
Y'(-1)
is equal to:
Report Question
0%
\cfrac {3 }{ 5 }
0%
\cfrac { 3 }{ 5 }
0%
\cfrac { 2 }{ 7 }
0%
\cfrac { 3 }{ 8 }
The number of values of
x
at which
f(x) = \left| x - \frac { 3 } { 2 } \right| + | x - 2 | + ( x - 1 ) | x - 1 |
is not differentiable
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0%
2
0%
3
0%
4
0%
1
If
f(x)
is differentiable function and
f(1) = \sin 1, f(2) = \sin 4, f(3) = \sin 9
, then the minimum number of distinct solution of equation
f'(x) = 2x \cos x^2
in
(1, 3)
is
Report Question
0%
1
0%
2
0%
3
0%
4
Solve :
\dfrac { d } { d x } \left[ \tan ^ { - 1 } \sqrt { 1 + x ^ { 2 } } - \cot ^ { - 1 } \left( - \sqrt { 1 + x ^ { 2 } } \right) \right] =?
Report Question
0%
\pi
0%
1
0%
0
0%
\dfrac { 2 x } { \sqrt { 1 + x ^ { 2 } } }
The derivative of
\cos^{-1} x
w.r.t.
\sqrt{1 - x^2}
is
Report Question
0%
\dfrac{1}{\sqrt{1 - x^2}}
0%
\sqrt{1 - x^2}
0%
\dfrac{1}{2x}
0%
\dfrac{1}{x}
let f be the differeniable at x=0 and f'(0)=1 then
\underset { b\rightarrow 0 }{ lim } \frac { f(h)-f(-2h) }{ h }
Report Question
0%
3
0%
2
0%
1
0%
-1
If
f
is twice differentiable such that
f ^ { \prime \prime } ( x ) = - f ( x )
and
f ^ { \prime } ( x ) = g ( x ).
If
h ( x )
is a twice differentiable function that
h ^ { \prime } ( x ) = ( f ( x ) ) ^ { 2 } + ( g ( x ) ) ^ { 2 } .
If
h ( 0 ) = 2 , h ( 1 ) = 4 ,
then the equation
y = h ( x )
represents :
Report Question
0%
a curve of degree
2
0%
a curve passing through the origin
0%
a straight line with slope
2
0%
a straight line with
y
intercept equal to
2
f(x)=\begin{cases} x\left( \dfrac { a{ e }^{ \frac { 1 }{ \left| x \right| } }+3{ e }^{ \frac { -1 }{ x } } }{ \left( a+2 \right) { e }^{ \frac { 1 }{ \left| x \right| } }-{ e }^{ \frac { -1 }{ x } } } \right),\ \ x\neq 0 \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0 \end{cases}
is differentiable at
x=0
then
[a]=......
(
[\ ]
denotes greatest integer function)
Report Question
0%
0
0%
-1
0%
-2
0%
None\ of\ these
If
y = {\cot ^{ - 1}}\left( {\frac{{\sin x}}{{1 + \cos x}}} \right)
then
\frac{{dy}}{{dx}}
is
Report Question
0%
\frac{{ - 1}}{2}
0%
\frac{{ 1}}{2}
0%
0
0%
1
Number of points in
[0,\pi]
, where
f(x)=\left[\dfrac{x\tan{x}}{\sin{x}+\cos{x}}\right]
is non-differentiable is/are (where
[.]
denotes the greatest integer function)
Report Question
0%
0
0%
4
0%
9
0%
None\ of\ these
Let F(x) =
\left( f\left( x \right) \right) ^{ 2 }+\left( f\left( x \right) \right) ^{ 2 },F\left( 0 \right) -6
where f(x) is a differential function such that
\left| f\left( x \right) \right| \le 1\forall x\notin \left[ -1,1 \right]
then choose the correct statement (s)
Report Question
0%
There is atleast one point in each of the intervals (-1, 0) and (0, 1) where
\left| f'(x) \right| \le 2
0%
there is atleast one point in a each of the interval (-1, 0) and (0, 1) where f(x)
\le S
0%
there is no point of local maximum of F(x) in (-1, 1)
0%
For some
c\in \left( -1,1 \right)
,
F'\left( c \right) \ge 6,F"\left( c \right) \le 0
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Answered
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Not Answered
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Not Visited
Correct : 0
Incorrect : 0
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