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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 15 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 15
if the function
f
(
x
)
=
{
a
+
sin
−
1
(
x
+
b
)
,
x
≥
1
x
,
x
<
1
is differentiable at
x
=
1
, then
a
b
is equal to
Report Question
0%
1
0%
0
0%
−
1
0%
2
The function
f
(
x
)
=
x
1
+
|
x
|
is differentiable at which of the following?
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0%
Every where
0%
Everywhere except at
x
=
1
0%
Everywhere except at
x
=
0
0%
Everywhere except at
x
=
0
or
1
If
y
=
e
sin
−
1
(
t
2
−
1
)
&
x
=
e
sec
−
1
(
1
t
2
−
1
)
, then
d
y
d
x
is equal to
Report Question
0%
x
y
0%
−
y
x
0%
y
x
0%
−
x
y
If
f
(
x
)
=
s
i
n
−
1
[
2
x
1
+
x
2
]
,then
f
(
x
)
is differentiable on
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0%
[-1,1]
0%
R-{-1,1}
0%
R-(-1,1)
0%
None of these
f(X)=|x|+|x-1| is continuous at
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0%
'0' only
0%
0,1 only
0%
Every where
0%
No where
Explanation
Step-1: Finding values of
f
(
x
)
f
(
x
)
=
|
x
|
+
|
x
−
1
|
f
(
x
)
=
−
x
−
|
x
−
1
|
at
x
≤
0
=
x
−
(
x
−
1
)
at
0
≤
x
<
1
=
x
+
(
x
−
1
)
at
x
≥
1
Hence, we get
f
(
x
)
=
1
−
2
x
at
x
≤
0
=
1
at
0
≤
x
<
1
=
2
x
−
1
at
x
≥
1
Step-2: Finding limits of the function at constraint values
At
x
=
0
,
lim
x
→
0
f
(
x
)
=
1
Hence,
f
(
x
)
is continuous at
x
=
0
At
x
=
1
,
lim
x
→
1
f
(
x
)
=
1
Hence,
f
(
x
)
is continuous at
x
=
1
Hence,
f
(
x
)
is continuous everywhere
Hence,Correct option is (C)
If
f
n
(
x
)
=
e
f
n
−
1
(
x
)
for all
n
ϵ
N
and
f
0
(
x
)
=
x
then
d
d
x
{
f
n
(
x
)
}
is equal to
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0%
f
n
(
x
)
.
d
d
x
{
f
n
−
1
(
x
)
}
0%
f
n
(
x
)
.
f
n
−
1
(
x
)
0%
f
n
(
x
)
.
f
n
−
1
(
x
)
.
.
.
.
.
f
2
(
x
)
.
f
1
(
x
)
0%
n
∏
i
=
1
f
i
(
x
)
Explanation
Hence, Option (A,C,D) is the correct answer.
The order of the differential equation of all circles whose radius is
4
, is?
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0%
1
0%
2
0%
3
0%
4
Explanation
Circle with radius
4
is represented as follows:-
(
x
−
h
)
2
+
(
y
−
k
)
2
=
4
2
.....
(
1
)
Differentiating w.r.t
x
2
(
x
−
h
)
+
2
(
y
−
k
)
d
y
d
x
=
0
⇒
(
x
−
h
)
+
(
y
−
k
)
d
y
d
x
=
0
.....
(
2
)
Again differentiating w.r.t
x
1
+
(
y
−
k
)
d
2
y
d
x
2
+
(
d
y
d
x
)
2
=
0
⇒
(
y
−
k
)
=
−
(
d
y
d
x
)
2
−
1
d
2
y
d
x
2
....
(
3
)
From
(
2
)
we get
⇒
(
x
−
h
)
=
−
(
y
−
k
)
d
y
d
x
⇒
(
x
−
h
)
=
−
{
(
d
y
d
x
)
2
−
1
d
2
y
d
x
2
}
d
y
d
x
∴
....
(4)
Substituting
(4)
and
(3)
in eq.
(1)
\left (\dfrac{\left(\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx}}{\dfrac{d^2y}{dx^2}} \right )^2 + \left (\dfrac{-\left (\dfrac{dy}{dx} \right )^2 - 1}{\dfrac{d^2y}{dx^2}} \right )^2 = 4^2
\left [\left (\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx} \right ]^2 + \left [\left (\dfrac{dy}{dx} \right )^2 + 1 \right ]^2 = 16 \dfrac{d^2y}{dx^2}
\therefore
Order of D.E of circle whose radius is
4
is
2
.
If
f\left( x \right) = {\left| x \right|^{\left| {\sin x} \right|}}
, then
{f'}\left( { - \dfrac{\pi }{4}} \right)
is equals
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{\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( { - \dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi }} \right)
0%
{\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } + \dfrac{{2\sqrt 2 }}{\pi }} \right)
0%
{\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{\pi }{4} + \dfrac{{2\sqrt 2 }}{\pi }} \right)
0%
None of these.
Let
f:(-1,1)\rightarrow R
be a differentiable function satisfying
(f'(x))^4=16(f(x))^2
for all
x\in (-1,1)
f(0)=0
The number of such functions is
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0%
2
0%
3
0%
4
0%
more than
4
Explanation
Given:
{\left({f}^{\prime}{\left(x\right)}\right)}^{4}=16{\left(f{\left(x\right)}\right)}^{2}
for all
x\in\,\left(–1 , 1\right)
f\left(0\right)=0
\Rightarrow\, {\left({f}^{\prime}{\left(x\right)}\right)}^{2}=\pm\,4f\left(x\right)
\Rightarrow\, {f}^{\prime}{\left(x\right)}=\pm\,2\sqrt{\pm\,f\left(x\right)}
(i)
Case
1
{f}^{\prime}{\left(x\right)}=2\sqrt{\pm\,f\left(x\right)}
\displaystyle\int{\dfrac{d\left(f\left(x\right)\right)}{f\left(x\right)}}=\displaystyle\int{2\,dx}
\Rightarrow\,2\sqrt{f\left(x\right)}=2x+c
\Rightarrow\,f\left(0\right)=0\Rightarrow\,c=0
\Rightarrow\,\sqrt{f\left(x\right)}=x
\Rightarrow\,x\ge 0
f\left(x\right)={x}^{2},\,\,0\le x<1
(ii)
Case
2
{f}^{\prime}{\left(x\right)}=-2\sqrt{f\left(x\right)}
\Rightarrow\,\sqrt{f\left(x\right)}=-x\Rightarrow\,x\le 0
f\left(x\right)={x}^{2};\,\,\,-1<x\le 0
(iii)
Case
3
{f}^{\prime}{\left(x\right)}=2\sqrt{-f\left(x\right)}
\sqrt{-f\left(x\right)}=x
-f\left(x\right)={x}^{2}
f\left(x\right)=-{x}^{2};\,\,\,0\le x< 1
(iv)
Case
4
{f}^{\prime}{\left(x\right)}=-2\sqrt{-f\left(x\right)}
\sqrt{-f\left(x\right)}=-x
f\left(x\right)=-{x}^{2};\,\,\,-1< x\le 0
(v)
Also, one singular solution of given differential equation is
f\left(x\right)=0,\,\,-1<x<1
Hence, there are more than
4
function possible
{f}_{1}\left( x \right) =\begin{cases} {x}^{2};0\le x< 1 \\ {-x}^{2}; -1<x<0 \end{cases}
{f}_{2}\left( x \right) =\begin{cases} -{x}^{2};0\le x< 1 \\ {x}^{2}; -1<x<0 \end{cases}
{f}_{3}\left( x \right) = {x}^{2},\,-1<x<1
{f}_{4}\left( x \right) =-{x}^{2},\,-1<x<1
{f}_{5}\left( x \right) =0,\,-1<x<1
...
Hence there are more than
4
solutions
If
\displaystyle y = \dfrac{\sqrt{(1 + t^{2})} - \sqrt{(1 - t^{2})}}{\sqrt{(1 + t^{2})} +\sqrt{(1 - t^{2})}}
and
\displaystyle x = \sqrt {(1 - t^{4})}
, then
\dfrac{dy}{dx}
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\displaystyle \dfrac{-1}{t^{2}\left \{1 +\sqrt{1 - t^{4}} \right \}}
0%
\displaystyle \dfrac{\left \{\sqrt{(1 - t^{4})} - 1\right \}}{t^{6}}
0%
\displaystyle \dfrac{1}{t^{2}\left \{1 + \sqrt{(1 - t^{4})}\right \}}
0%
\displaystyle \dfrac{1 - \sqrt{( 1 - t^{4})}}{t^{6}}
f(x) is diffrentiable function and (f(x). g(x)) is differentiable a x=a , then
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0%
g(x) must be differentiable at x=a
0%
if g(x) is discontinuous , then f(a) =0
0%
f(a)
\neq
0 then g(x) must be differentiable
0%
nothing can be said
Give that f(x) =xg(x) /
\left | x \right |
, g(0) = 0 and f(x) is continous at x=Then the value of f' (0)
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0%
Does not exist
0%
is -1
0%
is 1
0%
is 0
Let f(x)=
Report Question
0%
f' is differentiable
0%
f is differentiable
0%
f' is continuous
0%
f is continuous
If
f(x) = \left\{\begin{matrix} \dfrac{x\log \cos x}{\log(1+x^2)}, & x \neq 0\\ 0, & x=0\end{matrix}\right.
then
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0%
f(x) is not continuous at x=0.
0%
f(x) is continuous at x=0.
0%
f(x) is continuous at x=0 but not differentiable at x=0.
0%
f(x) is differentiable at x=0.
Let f:
R\rightarrow R
be a function such that f(x+y)= f(x)+f(y),
\forall
x,y
\epsilon R
. If f(x) is differentiable at x=0, then
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0%
f(x) is differentiable only in a finite interval containing zero
0%
f(x) is continuous
\forall x\epsilon R
0%
f(x) is constant
\forall x\epsilon R
0%
f(x) is differentiable except at finitely many points
Let
f(x)
be a function satisfying
f(x+y) = f(x)+f(y)
and
f(x) = xg(x)
\forall x, ~y \in
R, where
g(x)
is a continuous function then, which of the following is true?
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0%
f'(x)=g'(0)
0%
f'(x)=g (0)
0%
f(x)=g (0)
0%
f(x)=g'(0)
Which of the following is differentiable at x= 0
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0%
cos
\left ( \left | x \right | \right )+\left | x \right |
0%
cos
\left ( \left | x \right | \right )-\left | x \right |
0%
sin
\left ( \left | x \right | \right )+\left | x \right |
0%
sin \left ( \left | x \right | \right )-\left | x \right |
\cos |x|
is differentiable everywhere.
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0%
True
0%
False
If the function
f:[0,8] \rightarrow R
is differentiable, then for
0<a, b<2, \int_{0}^{8} f(t) d t
is equal to
Report Question
0%
3\left[\alpha^{3} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]
0%
3\left[\alpha^{3} f(\alpha)+\beta^{3} f(\beta)\right]
0%
{3}\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]
0%
3\left[\alpha^{2} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]
Explanation
\operatorname{Let} g(x)=\int_{0}^{x^{3}} f(t) d t\\
\text { Now } \int_{0}^{8} f(t) d t=g(2) =\dfrac{g(2)-g(1)}{2-1}+\dfrac{g(1)-g(0)}{1-0} \\
=g^{\prime}(\alpha)+g^{\prime}(\beta) \\
=3\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]
Let f(x) and g(x) be differentiable for
0\leq x\leq 1
, such that f(0) such that f'(c)=2g'(c), then the value of g(1) must be
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0%
1
0%
3
0%
-2
0%
-1
f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)
and
g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)
The value of
g(0)
is
Report Question
0%
0
0%
-3
0%
2
0%
None of these
Explanation
f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)
and
g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)
\text { Here put } g^{\prime}(1)=a, g^{\prime \prime}(2)=b \quad\quad\quad(1)
\text { Then } f(x)=x^{2}+a x+b, f(1)=1+a+b \Rightarrow f^{\prime}(x)=2 x+a
\quad f^{\prime \prime}(x)=2
\therefore g(x)=(1+a+b) x^{2}+(2 x+a) x+2=x^{2}(3+a+b)+a x\quad+2
\Rightarrow g^{\prime}(x)=2 x(3+a+b)+a \text { and } g^{\prime \prime}(x)=2(3+a+b)
\text { Hence, } g^{\prime}(1)=2(3+a+b)+a \quad\quad\quad(2)
g^{\prime \prime} (2)=2(3+a+b)\quad\quad\quad(3)
From (1),(2) and (3) we have
a=2(3+a+b)+a
\text { and } b=2(3+a+b)
\Rightarrow 3+a+b=0 \text { and } b+2 a+6=0
Hence,
b=0
and
a=-3 .
So,
f(x)=x^{2}-3 x
and
g(x)=-3 x+2
g(x)=-3 x+2 \Rightarrow g(0)=2
f(x)
is not invertible for
Report Question
0%
x \in\left[-\dfrac{\pi}{2}-\tan ^{-1} 2, \dfrac{\pi}{2}-\tan ^{-1} 2\right]
0%
x \in\left[\tan ^{-1} \dfrac{1}{2}, \pi+\tan ^{-1} \dfrac{1}{2}\right]
0%
x \in\left[\pi+\cot ^{-1} 2,2 \pi+\cot ^{-1} 2\right]
0%
None of these
Explanation
\begin{array}{l}f(x)=\sin x+\sin x \int_{-\pi / 2}^{\pi / 2} f(t) d t+\cos x \int_{-\pi / 2}^{\pi / 2} t f(t) d t \\ \qquad \begin{aligned}=\sin x\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos x \int_{-\pi / 2}^{\pi / 2} t(t) d t & \\=& A \sin x+B \cos x \\ \text { Thus, } A=& 1+\int_{-\pi / 2}^{\pi / 2} f(t) d t \\=& 1+\int_{-\pi / 2}^{\pi / 2}(A \sin t+B \cos t) d t \\=& 1+2 B \int_{0}^{\pi / 2} \cos t d t \\ &=\int_{-\pi / 2}^{\pi / 2} t(A \sin t+B \cos t) d t \\ &=1+2 B \\ = & 2 A \int_{0}^{\pi / 2} t f(t) d t \end{aligned}\end{array} \\
\begin{aligned}&=2 A[-t \cos t+\sin t]_{0}^{\pi / 2} \\\Rightarrow \quad B &=2 A\end{aligned}\\
From equations (1) and
(2),
we get
A=-1 / 3, B=-2 / 3 \\
\Rightarrow f(x)=-\dfrac{1}{3}(\sin x+2 \cos x)\\
Thus, the range of
f(x)
is
\left[-\dfrac{\sqrt{5}}{3}, \dfrac{\sqrt{5}}{3}\right]\\
\begin{aligned}f(x) &=-\dfrac{1}{3}(\sin x+2 \cos x) \\&=-\dfrac{\sqrt{5}}{3} \sin \left(x+\tan ^{-1} 2\right) \\&=-\dfrac{\sqrt{5}}{3} \cos \left(x-\tan ^{-1} \dfrac{1}{2}\right)\end{aligned}\\
f(x)
is invertible if
-\dfrac{\pi}{2} \leq x+\tan ^{-1} 2 \leq \dfrac{\pi}{2}\\
\Rightarrow-\dfrac{\pi}{2}-\tan ^{-1} 2 \leq x \leq \dfrac{\pi}{2}-\tan ^{-1} 2\\
or
0 \leq x-\tan ^{-1} \dfrac{1}{2} \leq \pi\\
\Rightarrow \tan ^{-1} \dfrac{1}{2} \leq x \leq \pi+\tan ^{-1} \dfrac{1}{2}\\
or
\pi \leq x-\tan ^{-1} \dfrac{1}{2} \leq 2 \pi\\
\Rightarrow x \in\left[\pi+\cot ^{-1} 2,2 \pi+\cot ^{-1} 2\right]\\
Let
f(0,\infty)\rightarrow R
be a differentiable function such that
f'(x)=2-\dfrac{f(x)}{x}
for all
x\epsilon (0,\infty)
and
f(1)\neq 1
Then
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\underset{x\rightarrow 0+}{lim}f'(\dfrac{1}{x})=1
0%
\underset{x\rightarrow 0+}{lim}xf'(\dfrac{1}{x})=2
0%
\underset{x\rightarrow 0+}{lim}x^2f'(\dfrac{1}{x})=0
0%
|f(x)|\le 2
for all
x\epsilon(0,2)
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