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CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 4 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 4
If
y
=
l
o
g
x
3
+
3
s
i
n
−
1
x
+
k
x
2
, then find
d
y
d
x
Report Question
0%
3
⋅
1
x
+
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
0%
3
⋅
1
x
3
+
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
0%
3
⋅
1
x
−
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
0%
3
⋅
1
x
+
3
⋅
1
√
1
−
x
2
+
2
x
Explanation
Here,
y
=
log
x
3
+
3
sin
−
1
x
+
k
x
2
On differentiating we get
d
y
d
x
=
d
d
x
[
log
x
3
]
+
d
d
x
[
3
sin
−
1
x
]
+
d
d
x
[
k
x
2
]
=
3
d
d
x
[
log
x
]
+
3
d
d
x
(
sin
−
1
x
)
+
k
d
d
x
(
x
2
)
=
3
⋅
1
x
+
3
⋅
1
√
1
−
x
2
+
k
(
2
x
)
If
y
=
5
3
−
x
2
+
(
3
−
x
2
)
5
, then
d
y
d
x
=
Report Question
0%
−
2
x
{
5
3
−
x
2
⋅
log
e
5
+
5
(
3
−
x
2
)
4
}
0%
−
x
{
5
3
−
x
2
⋅
log
e
5
+
5
(
3
−
x
2
)
4
}
0%
−
2
x
{
5
3
−
x
2
⋅
log
e
5
+
(
3
−
x
2
)
4
}
0%
−
2
x
{
5
3
−
x
2
+
5
(
3
−
x
2
)
4
}
Explanation
d
d
x
(
5
3
−
x
2
+
(
3
−
x
2
)
5
)
=
5
3
−
x
2
log
e
5
d
d
x
(
3
−
x
2
)
+
5
(
3
−
x
2
)
4
d
d
x
(
3
−
x
2
)
=
5
3
−
x
2
log
e
5
(
−
2
x
)
+
5
(
3
−
x
2
)
4
(
−
2
x
)
=
−
2
x
(
5
3
−
x
2
log
e
5
+
5
(
3
−
x
2
)
)
If
y
=
e
a
x
⋅
cos
(
b
x
+
c
)
, then find
d
y
d
x
Report Question
0%
a
e
a
x
cos
(
b
x
+
c
)
−
b
e
a
x
sin
(
b
x
+
c
)
0%
a
e
a
x
cos
(
b
x
+
c
)
+
b
e
a
x
sin
(
b
x
+
c
)
0%
e
a
e
a
x
cos
(
b
x
+
c
)
−
b
e
a
x
sin
(
b
x
+
c
)
0%
a
e
a
x
cos
(
b
x
+
c
)
Explanation
y
=
e
a
x
⋅
cos
(
b
x
+
c
)
On differentiating, we get
d
y
d
x
=
e
a
x
d
d
x
(
cos
(
b
x
+
c
)
)
+
cos
(
b
x
+
c
)
d
d
x
e
a
x
=
−
e
a
x
sin
(
b
x
+
c
)
d
d
x
(
b
x
+
c
)
+
cos
(
b
x
+
c
)
e
a
x
d
d
x
(
a
x
)
=
−
b
e
a
x
sin
(
b
x
+
c
)
+
a
cos
(
b
x
+
c
)
e
a
x
=
a
e
a
x
cos
(
b
x
+
c
)
−
b
e
a
x
sin
(
b
x
+
c
)
If
y
=
log
3
x
+
3
log
e
x
+
2
tan
x
, then
d
y
d
x
=
Report Question
0%
1
x
log
e
3
+
3
x
+
2
sec
2
x
0%
1
x
log
e
3
+
3
x
+
sec
2
x
0%
1
log
e
3
+
3
x
+
2
sec
2
x
0%
1
x
log
e
3
−
3
x
+
2
sec
2
x
Explanation
d
d
x
(
log
3
x
+
3
log
e
x
+
2
tan
x
)
=
d
d
x
(
log
e
x
log
3
+
3
log
e
x
+
2
tan
x
)
=
1
x
log
e
3
+
3
x
+
2
sec
2
x
If
y
=
e
x
log
a
+
e
a
log
x
+
e
a
log
a
, then
d
y
d
x
=
Report Question
0%
a
x
log
a
+
x
a
−
1
0%
a
x
log
a
+
a
x
0%
a
x
log
a
+
a
x
a
−
1
0%
a
x
log
a
+
a
x
a
Explanation
Let
y
=
e
x
l
o
g
a
+
e
a
l
o
g
x
+
e
a
l
o
g
a
=
e
log
e
a
x
+
e
log
e
x
a
+
e
log
e
a
a
=
a
x
+
x
a
+
a
a
......... Since
e
log
e
x
=
x
d
y
d
x
=
d
d
x
(
a
x
+
x
a
+
a
a
)
=
a
x
log
a
+
a
.
x
a
−
1
If
y
=
1
1
+
x
β
−
α
+
x
γ
−
α
+
1
1
+
x
α
−
β
+
x
γ
−
β
+
1
1
+
x
α
−
γ
+
x
β
−
γ
then
d
y
d
x
is equal to-
Report Question
0%
0
0%
1
0%
(
a
+
β
+
γ
)
X
α
+
β
+
γ
−
1
0%
None of these
If
2
x
+
2
y
=
2
x
+
y
, then
d
y
d
x
has the value equal to
Report Question
0%
−
2
y
2
x
0%
1
1
−
2
x
0%
1
−
2
y
0%
2
x
(
1
−
2
y
)
2
y
(
2
x
−
1
)
Explanation
2
x
+
2
y
=
2
x
+
y
Differentiating both the sides
⇒
d
d
x
(
2
x
+
2
y
)
=
d
d
x
(
2
x
+
y
)
⇒
2
x
log
2
+
2
y
log
2
d
y
d
x
=
2
x
+
y
log
2
(
1
+
d
y
d
x
)
⇒
log
2
(
2
x
+
2
y
d
y
d
x
)
=
log
2
(
2
x
+
y
+
2
x
+
y
d
y
d
x
)
⇒
(
2
y
−
2
x
+
y
)
d
y
d
x
=
2
x
+
y
−
2
x
⇒
d
y
d
x
=
2
x
2
y
−
2
x
2
y
−
2
x
2
y
=
2
x
(
2
y
−
1
)
2
y
(
1
−
2
x
)
=
2
x
(
1
−
2
y
)
2
y
(
2
x
−
1
)
If
f
′
(
x
)
=
sin
x
+
sin
4
x
⋅
cos
x
, then
f
′
(
2
x
2
+
π
2
)
is
Report Question
0%
4
x
{
cos
(
2
x
2
)
−
s
i
n
8
x
2
⋅
sin
2
x
2
}
0%
4
x
{
cos
(
2
x
2
)
+
sin
8
x
2
⋅
sin
2
x
2
}
0%
{
cos
(
2
x
2
)
−
sin
8
x
⋅
sin
2
x
2
}
0%
none of the above
Explanation
f
′
(
x
)
=
sin
x
+
sin
4
x
⋅
cos
x
f
′
(
2
x
2
+
π
2
)
=
d
d
x
(
2
x
2
+
π
2
)
[
sin
(
2
x
2
+
π
2
)
+
sin
4
(
2
x
2
+
π
2
)
.
cos
(
2
x
2
+
π
2
)
]
=
4
x
[
cos
2
x
2
−
sin
8
x
2
.
sin
2
x
2
]
If
y
=
|
cos
x
|
+
|
sin
x
|
, then
d
y
d
x
at
x
=
2
π
3
is
Report Question
0%
1
2
(
√
3
+
1
)
0%
2
(
√
3
−
1
)
0%
1
2
(
√
3
−
1
)
0%
none of these
Explanation
d
y
d
x
=
d
d
x
(
|
c
o
s
x
|
+
|
s
i
n
x
|
)
=
−
sin
x
+
cos
x
at
x
=
2
π
3
=
|
−
sin
2
π
3
|
+
|
cos
2
π
3
|
=
√
3
2
−
1
2
=
1
2
(
√
3
−
1
)
Find the derivative of
e
x
sin
x
Report Question
0%
e
x
sin
x
(
x
cos
x
−
sin
x
)
0%
e
x
sin
x
x
cos
x
0%
e
x
sin
x
(
−
x
cos
x
+
sin
x
)
0%
e
x
sin
x
(
x
cos
x
+
sin
x
)
Explanation
Find the derivative of
sec
−
1
(
x
+
1
x
−
1
)
+
sin
−
1
(
x
−
1
x
+
1
)
Report Question
0%
0
0%
1
0%
−
1
0%
x
+
1
x
−
1
Explanation
Let
y
=
sec
−
1
(
x
+
1
x
−
1
)
+
sin
−
1
(
x
−
1
x
+
1
)
y
=
cos
−
1
(
x
−
1
x
+
1
)
+
sin
−
1
(
x
−
1
x
+
1
)
=
π
2
∴
\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } }
If
y=\log_{10}x+\log_x 10+\log_xx+\log_{10} 10
, then
\displaystyle \frac{dy}{dx}=
Report Question
0%
\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}
0%
\displaystyle \frac {1}{\log_e 10}-\displaystyle \frac {\log_e 10}{x(\log_ex)^2}
0%
\displaystyle \frac {1}{x \log_e 10}-\displaystyle \frac {\log_e 10}{x^2(\log_ex)^2}
0%
None of these
Explanation
\displaystyle \dfrac { d }{ dx } \left( \log _{ 10 }{ x+\log _{ x }{ 10+\log _{ x }{ x+\log _{ 10 }{ 10 } } } } \right)
=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } +\displaystyle \dfrac { \left( -\log _{ e }{ 10\left( \displaystyle \dfrac { 1 }{ x } \right) } \right) }{ { \left( \log_e { x } \right) }^{ 2 } }
=\displaystyle \dfrac { 1 }{ x\log _{ e }{ 10 } } -\displaystyle \dfrac { \log _{ e }{ 10 } }{ x{ \left( \log_e { x } \right) }^{ 2 } }
If
\displaystyle y=\frac { x }{ a+\displaystyle\frac { x }{ b+\displaystyle\frac { x }{ a+\displaystyle\frac { x }{ b+.....\infty } } } }
, then
\cfrac{dy}{dx} =
Report Question
0%
\displaystyle\frac{a}{ab+2ay}
0%
\displaystyle\frac{b}{ab+2by}
0%
\displaystyle\frac{a}{ab+2by}
0%
\displaystyle\frac{b}{ab+2ay}
Explanation
\displaystyle y = \frac{x}{a+\frac{x}{b+y}}
\Rightarrow \displaystyle y =\frac{x(b+y)}{ab+ay+x}
\Rightarrow aby+ay^2+xy=bx+xy\Rightarrow aby+ay^2=bx
Differentiating both sides w.r.t
x
\Rightarrow\displaystyle (ab+2ay)\frac{dy}{dx}=b\therefore \frac{dy}{dx}=\frac{b}{ab+2ay}
If
\displaystyle y=\frac { \sin { x } }{ 1+\displaystyle \frac { \cos { x } }{ 1+\displaystyle \frac { \sin { x } }{ 1+\displaystyle\frac { \cos { x } }{ 1+\displaystyle \frac { \sin { x } }{ 1+ .....\infty} } } } }
,
then
y'(0)
is
Report Question
0%
equal to
0
0%
equal to
\frac{1}{2}
0%
equal to
1
0%
non existent
Explanation
\displaystyle y=\frac { \sin { x } }{ 1+\frac { \cos { x } }{ 1+\frac { \sin { x } }{ 1+\frac { \cos { x } }{ 1+\frac { \sin { x } }{ 1+.....\infty } } } } }
here,
y(0)=0
\displaystyle \Rightarrow y=\frac { \sin { x } }{ 1+\frac { \cos { x } }{ 1+y } }
\displaystyle \Rightarrow y\left( 1+y+\cos { x } \right) =\left( 1+y \right) \sin { x }
differentiating wrt x
y'\left( 1+y+\cos { x } \right) +y\left( y'-\sin { x } \right) =\left( 1+y \right) \cos { x } +y'\sin { x }
y'\left( 0 \right) \left( 1+y\left( 0 \right) +1 \right) +y\left( 0 \right) \left( y'\left( 0 \right) \right) =\left( 1+y\left( 0 \right) \right)
y'\left( 0 \right) =\dfrac { 1 }{ 2 }
Given :
f(x)=4x^3-6x^2\cos2a+3x \sin 2a.\sin 6a+\sqrt{\ln (2a-a^2)}
then
Report Question
0%
f(x)
is not defined at
x=\displaystyle \frac{1}{2}
0%
{f}'(\displaystyle \frac{1}{2})<0
0%
f'(x)
is not defined at
x=\displaystyle \frac{1}{2}
0%
{f}'(\displaystyle \frac{1}{2})>0
Explanation
f'(x)=12x^{2}-12xcos2a+3sin2a.sin6a+0
f'(1/2)=\frac{12}{4}-\frac{12}{2}cos2a+3sin2a.sin6a
=3-6cos2a+1.5(cos4a-cos8a)
So, it will alwyas be greater than
0
If
y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) + \sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)
, then
\displaystyle\frac{dy}{dx}
equals
Report Question
0%
1
0%
0
0%
\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}
0%
\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}
Explanation
y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) + \sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)
y=\cos ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) +\sin ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) } } =\displaystyle \frac { \pi }{ 2 }
.............
\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } }
\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0
Which of the following could be the sketch graph of
y=\displaystyle \frac{d}{dx}(x\ln x)
Report Question
0%
0%
0%
0%
Explanation
\displaystyle y=\frac { d }{ dx } \left( x\ln { x } \right) =\ln { x } +\frac { x }{ x } =\ln { x } +1
At
x=0 ,y=1
At
y=0, x={ e }^{ -1 }
The solution set of
{f}'(x)>{g}'(x)
where
f(x)=\displaystyle \frac{1}{2}(5^{2x+1})
&
g(x)= 5^x+4x(\ln 5)
is
Report Question
0%
x>1
0%
0< x< 1
0%
x \leq 0
0%
x>0
Explanation
Given,
f(x)= \dfrac {1}{2}(5^{2x+1})
f'(x)= 5^{2x+1}\times \log 5
g'(x)=5^{x}l\log 5 +4\ln 5=\ln 5(5^{x}+4)
f'(x)>g'(x)
=>5\times 5^{2x}>(5^{x}+4)
Let
t=5^{x}
=> 5t^{2}-t-4>0
=>(t-1)(t+\frac{4}{5})>0
=>t>1
or
t>-\frac{4}{5}
=>5^{x}>5^{0} =>x>0
so,
x>0
The equation
y^2e^{xy} =9e^{-3}.x^2
defines
y
as a differentiable function of x. The value of
\displaystyle \frac{dy}{dx}
for
x=-1
and
y= 3
is
Report Question
0%
-\displaystyle \frac{15}{2}
0%
-\displaystyle \frac{9}{5}
0%
3
0%
15
Explanation
y^2e^{xy} =9e^{-3}.x^2
Differentiate both sides
\displaystyle \frac { d }{ dx } \left( { y }^{ 2 }{ e }^{ xy } \right) =\displaystyle \frac { d }{ dx } \left( 9{ e }^{ -3 }{ x }^{ 2 } \right)
2y\displaystyle \frac { dy }{ dx } { e }^{ xy }+{ y }^{ 2 }{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right) =9{ e }^{ -3 }\left( 2x \right)
On solving, we get
\displaystyle \frac { dy }{ dx } =\displaystyle \frac { 18x{ e }^{ -3 }-{ y }^{ 3 }{ e }^{ xy } }{ 2y{ e }^{ xy }+{ y }^{ 2 }{ e }^{ xy }x }
putting
x=-1,\quad y=3
, we get
\displaystyle \frac { dy }{ dx } =\displaystyle \frac { -18{ e }^{ -3 }-27{ e }^{ -3 } }{ 6{ e }^{ -3 }-9{ e }^{ -3 } } =\displaystyle \frac { -45{ e }^{ -3 } }{ -3{ e }^{ -3 } } =15
f:R\rightarrow R
and
\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}
, then
f(x)
is
Report Question
0%
one-one ito
0%
many-one onto
0%
one-one onto
0%
many-one into
Explanation
\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}
\displaystyle \quad \quad =\frac {x(x^4+1)(x+1)+(x^4+1)+1}{x^2+x+1}
\displaystyle \quad \quad =\frac {(x^4+1)(x^2+x+1)+1}{x^2+x+1}
\displaystyle \quad \quad =(x^4+1)+\frac {1}{x^2+x+1}
\displaystyle f'(x) =4x^3-\frac {2x+1}{(x^2+x+1)^2}=
not always positive or negative
Thus,
f
is many one.
Also range and co-domain of
f
are not same,
Hence is many-one into function
Suppose the function
f(x)-f(2x)
has the derivative
5
at
x=1
and derivative
7
at
x=2
.The derivative of the function
f(x)-f(4x)
at
x=1
, has the value equal to
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0%
19
0%
9
0%
17
0%
14
If
y=x^{1/x}
, the value of
\displaystyle \frac{dy}{dx}
at
x=e
is
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0%
1
0%
0
0%
-1
0%
none of these
Explanation
Given,
y=x^{1/x}
\displaystyle \log { y } =\frac { 1 }{ x } \log { x }
\Rightarrow \displaystyle x \log { y }= \log { x }
\displaystyle \frac { x }{ y } \frac { dy }{ dx } +\log { y } =\frac { 1 }{ x }
\Rightarrow \displaystyle \frac { dy }{ dx } =\frac { y(1-x\log { y } ) }{ { x }^{ 2 } }
\displaystyle \left( \frac { dy }{ dx } \right) _{ x=e }{ =\frac { e^{ 1/e }(1-e\log {e}^\frac {1}{e}) }{ { e }^{ 2 } } }=0
If for all
x, y
the function
f
is defined by
f(x)+f(y)+f(x).f(y)=1
and
f(x)>0
then
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f^{'}(x)
does not exist
0%
f^{'}(x)=0
for all
x
0%
f^{'}(0)< f^{'}(1)
0%
none of these
Explanation
f(x)+f(y)+f(x).f(y)=1
put
y = 0
\Rightarrow f(x)+f(0)+f(x).f(0)=1
\Rightarrow f(x) = \cfrac{1-f(0)}{1+f(0)} = C
(constant)
\therefore f'(x) = 0\forall x
If
{ S }_{ n }
denotes the sum of
n
terms of a G.P. whose common ratio is
r
, then
\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr }
is equal to
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\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }
0%
\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }
0%
\left( n-1 \right) { S }_{ n }
0%
None of these
Explanation
We have,
\displaystyle { S }_{ n }=\frac { a\left( { r }^{ n }-1 \right) }{ r-1 }
\Rightarrow \left( r-1 \right) { S }_{ n }={ ar }^{ n }-a
Differentiating both sides with respect to
r,
we get
\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr } +{ S }_{ n }=na{ r }^{ n-1 }-0
\displaystyle \Rightarrow \left( r-1 \right) \frac { d{ S }_{ n } }{ dr } =na{ r }^{ n-1 }-{ S }_{ n }
=n[n
th term of
g.p.]-{ S }_{ n }
=n\left( { S }_{ n }-{ S }_{ n-1 } \right) -{ S }_{ n }
=\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }.
If
x^{y}=e^{x+y}
then
\displaystyle \frac{dy}{dx}
at
x=1
is equal to
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0%
0
0%
-2
0%
1
0%
none of these
Explanation
Given,
x^y= e^{x+y}
Taking
\log
on both sides, we get
y\log x=x+y
Taking derivative,
\displaystyle \frac{y}{x}+logx.y'=1+y'
At,
x=1 => y=-1
So,
y'=-2
Let
\displaystyle f\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\frac { f\left( { x }_{ 1 } \right) +f\left( { x }_{ 2 } \right) +...+f\left( { x }_{ n } \right) }{ n }
where all
{ x }_{ i }\in R
are independent to each other and
n\in N
. if
f(x)
is differentiable and
f'\left( 0 \right) =a,f\left( 0 \right) =b
and
f'\left( x \right)
is equal to
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0%
a
0%
0
0%
b
0%
None of these
Explanation
Differentiating the given equation w.r.t.
{ x }_{ 1 },
we get
\displaystyle \dfrac { 1 }{ n } f'\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n } }{ n } \right) =\dfrac { f'\left( { x }_{ 1 } \right) }{ n }
[
Since all
{ x }_{ i }'s
are independent to each other,
\displaystyle \therefore \dfrac { d{ x }_{ i } }{ { dx }_{ j } } =0
if
i\neq j
and
\displaystyle \dfrac { { dx }_{ i } }{ { dx }_{ j } } =1
if
(i=j)]
On putting
{ x }_{ 1 }={ x }_{ 2 }=...={ x }_{ n-1 }=0
and
{ x }_{ n }=x,
we get
\displaystyle f'\left( \dfrac { x }{ n } \right) =f'\left( 0 \right) =a.
On integrating, we get
\displaystyle nf'\left( \dfrac { x }{ n } \right) =ax+c
Since
f(0)=b,
we have
c=nb
\displaystyle \therefore nf'\left( \dfrac { x }{ n } \right) =ax+nb\Rightarrow nf'\left( x \right) =nax+nb\Rightarrow f\left( x \right) =ax+b.
\displaystyle \therefore f'\left( x \right) =a,\forall x\in R
If
5f(x)+3f\left ( \displaystyle \frac{1}{x} \right )=x+2
and
y=xf(x)
then
\left (\displaystyle \frac{dy}{dx} \right )_{x=1}
is equal to ?
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14
0%
\displaystyle \frac{7}{8}
0%
1
0%
none of these
Explanation
5f(x)+3f\left ( \displaystyle \frac{1}{x} \right )=x+2\Rightarrow (1)
Replace
x
by
\dfrac{1}{x}
\Rightarrow 5f\left ( \displaystyle \frac{1}{x} \right )+3f(x)=\dfrac{1}{x}+2\Rightarrow (2)
\Rightarrow 5\times(1)-3\times (2)\Rightarrow 16f(x) =5x-\dfrac{3}{x}+4=\dfrac{5x^2+4x-3}{x}
\therefore y = xf(x) =\dfrac{5x^2+4x-3}{16}
\Rightarrow \dfrac{dy}{dx}=\dfrac{10x+4}{16}
\therefore \left(\dfrac{dy}{dx}\right)_{x=1}=\dfrac{10+4}{16}=\dfrac{7}{8}
y=\sqrt{\sin x+\sqrt{\sin x +\sqrt{\sin x+-\infty }}}
then
\displaystyle \frac{dy}{dx}
equals:
(\sin x> 0)
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\displaystyle \frac{\cos x}{2y-1}
0%
\displaystyle \frac{y}{2\tan x+y\sec x}
0%
\displaystyle \frac{1}{\sqrt{1+4\sin x}}
0%
\displaystyle \frac{2\cos x}{\sin x+2y}
Explanation
y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}
Taking square on both the sides, we get
{ y }^{ 2 }=\sin { x } +\sqrt { \sin x\sqrt { \sin x\sqrt { \sin x+-\infty } } }
{ y }^{ 2 }=\sin { x } +y
............ Since
y=\sqrt{\sin x\sqrt{\sin x \sqrt{\sin x+-\infty }}}
2y\displaystyle \frac { dy }{ dx } =\cos { x } +\displaystyle \frac { dy }{ dx }
\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x } }{ 2y-1 }
If
xe^{xy}-y=\sin ^{2}x
then
\displaystyle \frac{dy}{dx}
at
x=0
is
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0%
0
0%
1
0%
-1
0%
None of these
Explanation
Given
x{ e }^{ xy }=y+\sin ^{ 2 }{ x }
...(1)
on putting
x=0
, we get
0.{ e }^{ 0 }=y+0\Rightarrow y=0
on differentiating (1) both sides w.r.t
x
, we get
\displaystyle 1.{ e }^{ xy }+x.{ e }^{ xy }\left( x.\frac { dy }{ dx } +y \right) =\frac { dy }{ dx } +2\sin { x } \cos { x }
On putting
x=0,y=0
, we get
\displaystyle { e }^{ 0 }+0\left( 0+0 \right) =\left[ \frac { dy }{ dx } \right] +2\sin { 0 }
\Rightarrow \dfrac { dy }{ dx } =1
If
x^{y}.y^{x}=16
then
\frac{dy}{dx}
at (2, 2) is
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0%
1
0%
-1
0%
0
0%
none of these
Explanation
x^{y}.y^{x}=16
\log { x^{ y } } +\log { y^{ x } } =\log { 16 }
y\log { x } +x\log { y } =\log { 16 }
Differentiating w.r.t x , we get
\displaystyle \frac { y }{ x } +\log { x } \frac { dy }{ dx } +\frac { x }{ y } \frac { dy }{ dx } +\log { y } =0
So, at x=2, y=2
\displaystyle 1+\log { 2 } \left( \frac { dy }{ dx } \right) _{ (2,2) }+1\left( \frac { dy }{ dx } \right) _{ (2,2) }+\log { 2 } =0
\Rightarrow \left( \frac { dy }{ dx } \right) _{ (2,2) }=-1
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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