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CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 5 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 5
If P(x) is a polynomial such that
P
(
x
2
+
1
)
=
{
P
(
x
2
)
}
2
+
1
and
P
(
0
)
=
0
then
P
′
(
0
)
is equal to
Report Question
0%
1
0%
0
0%
−
1
0%
none of these
Explanation
Given,
P
(
x
2
+
1
)
=
{
P
(
x
2
)
}
2
+
1
...(1)
Given
P
(
0
)
=
0
Put
x
=
0
in (1)
⇒
P
(
1
)
=
P
(
0
)
2
+
1
⇒
P
(
1
)
=
1
Also, put
x
=
1
in (1)
⇒
P
(
2
)
=
2
Also, since, P(x) is a polynomial so we have P(x)=x
⇒
P
′
(
x
)
=
1
P
′
(
0
)
=
1
If
y
x
=
x
sin
y
, find
d
y
d
x
.
Report Question
0%
y
x
[
x
log
y
−
sin
y
y
log
x
cos
y
−
x
]
0%
y
x
[
x
log
y
+
sin
y
y
log
x
cos
y
+
x
]
0%
−
y
x
[
x
log
y
−
sin
y
y
log
x
cos
y
−
x
]
0%
y
x
[
x
log
y
−
sin
y
y
log
x
cos
y
+
x
]
Explanation
Given
y
x
=
x
sin
y
.
Take log both sides.
x
log
y
=
sin
y
log
x
.
Differentiate w.r.t. x
log
y
+
x
.
1
y
d
y
d
x
=
1
x
sin
y
+
(
log
x
)
cos
y
.
d
y
d
x
(
log
y
−
sin
y
x
)
=
d
y
d
x
[
cos
y
log
x
−
x
y
]
∴
d
y
d
x
=
y
x
[
x
log
y
−
sin
y
y
log
x
cos
y
−
x
]
y
=
(
cot
x
)
sin
x
+
(
tan
x
)
cos
x
.Find dy/dx
Report Question
0%
sin
x
(
cot
x
)
sin
x
−
1
(
−
c
o
s
e
c
2
x
)
+
(
cot
x
)
sin
x
(
l
o
g
cot
x
)
cos
x
+
cos
x
(
tan
x
)
cos
x
−
1
sec
2
x
+
(
tan
x
)
c
o
s
x
(
log
t
a
n
x
)
(
−
s
i
n
x
)
0%
sin
x
(
cot
x
)
sin
x
−
1
(
−
c
o
sec
2
x
)
+
cos
x
(
tan
x
)
cos
x
−
1
sec
2
x
0%
sin
x
(
cot
x
)
sin
x
−
1
(
−
s
e
c
2
x
)
+
(
cot
x
)
sin
x
(
l
o
g
cot
x
)
cos
x
+
cos
x
(
tan
x
)
cos
x
+
1
c
o
sec
2
x
+
(
tan
x
)
c
o
s
x
(
log
t
a
n
x
)
(
s
i
n
x
)
0%
None of these
Explanation
Given
y
=
(
cot
x
)
sin
x
+
(
tan
x
)
cos
x
Let
u
=
(
cot
x
)
sin
x
Taking log on both sides
log
u
=
sin
x
log
(
cot
x
)
Differentiating w.r.t. x, we get,
1
u
d
u
d
x
=
sin
x
cot
x
(
−
c
o
s
e
c
2
x
)
+
cos
x
log
cot
x
⇒
d
u
d
x
=
sin
x
(
cot
x
)
sin
x
−
1
(
−
c
o
s
e
c
2
x
)
+
(
cot
x
)
sin
x
cos
x
log
cot
x
Now,
v
=
(
tan
x
)
cos
x
Taking log on both sides
log
v
=
cos
x
log
(
tan
x
)
Differentiating w.r.t. x, we get,
1
v
d
v
d
x
=
cos
x
tan
x
(
sec
2
x
)
−
sin
x
log
tan
x
⇒
d
v
d
x
=
cos
x
(
tan
x
)
cos
x
−
1
(
sec
2
x
)
+
(
tan
x
)
cos
x
(
−
sin
x
)
log
tan
x
Since,
y
=
u
+
v
⇒
d
y
d
x
=
d
u
d
x
+
d
v
d
x
⇒
d
y
d
x
=
sin
x
(
cot
x
)
sin
x
−
1
(
−
c
o
s
e
c
2
x
)
+
(
cot
x
)
sin
x
cos
x
log
cot
x
+
cos
x
(
tan
x
)
cos
x
−
1
(
sec
2
x
)
+
(
tan
x
)
cos
x
(
−
sin
x
)
log
tan
x
If
x
2
+
y
2
=
1
then
(
w
h
e
r
e
y
′
=
d
y
d
x
,
y
″
=
d
2
y
d
x
2
)
Report Question
0%
y
y
″
−
2
y
′
2
+
1
=
0
0%
y
y
″
+
y
′
2
+
1
=
0
0%
y
y
″
−
y
′
2
−
1
=
0
0%
y
y
″
+
2
y
′
2
+
1
=
0
Explanation
Given,
x
2
+
y
2
=
1
Differentiating w.r.t x,
2
x
+
2
y
y
′
=
0
⇒
x
+
y
y
′
=
0
Again differentiating w.r.t. x, we get
1
+
y
y
″
+
y
′
y
′
=
0
1
+
y
y
″
+
y
′
2
=
0
d
d
x
(
log
e
(
1
+
x
1
−
x
)
1
/
4
−
1
2
tan
−
1
x
.
)
Report Question
0%
x
2
1
−
x
4
.
0%
x
3
1
−
x
4
.
0%
x
4
1
−
x
4
.
0%
−
x
2
1
−
x
4
.
Explanation
Let
y
=
1
4
[
log
(
1
+
x
)
−
log
(
1
−
x
)
]
−
1
2
tan
−
1
x
.
⇒
d
y
d
x
=
1
4
[
1
1
+
x
−
1
1
−
x
(
−
1
)
]
−
1
2
.
1
1
+
x
2
=
1
2
(
1
−
x
2
)
−
1
2
(
1
+
x
2
)
=
x
2
1
−
x
4
.
If
x
=
e
y
+
e
y
+
e
y
+
.
.
.
∞
,
∀
x
>
0
, then
d
y
d
x
=
Report Question
0%
1
−
x
x
0%
1
x
0%
x
1
+
x
0%
1
+
x
x
Explanation
x
=
e
y
+
e
y
+
e
y
+
.
.
.
⇒
x
=
e
y
+
x
Take logarithm on both sides and differentiating with respect to
x
∴
1
x
=
1
+
d
y
d
x
⇒
d
y
d
x
=
1
−
x
x
If
f
is a real-valued differentiable function satisfying
|
f
(
x
)
−
f
(
y
)
|
≤
(
x
−
y
)
2
for all
x
,
y
ϵ
R
and
f
(
0
)
=
0
then
f
(
1
)
equals
Report Question
0%
0
0%
−
1
0%
1
0%
2
Explanation
Given
|
f
(
x
)
−
f
(
y
)
|
≤
|
x
−
y
|
2
x
≠
y
∴
|
f
(
x
)
−
f
(
y
)
x
−
y
|
≤
|
x
−
y
|
Taking limit as
y
→
x
, we get
lim
\displaystyle\Rightarrow\left|\lim _{ y\rightarrow x }{ \frac { f\left( x \right) -f\left( y \right) }{ x-y } } \right| \le \left| \lim _{ y\rightarrow x }{ \left( x-y \right) } \right|
\Rightarrow \left| f'\left( x \right) \right| \le 0\Rightarrow \left| f'\left( x \right) \right| =0\left[ \because \left| f'\left( x \right) \right| \ge 0 \right]
\therefore f'\left( x \right) =0\Rightarrow f\left( x \right) =c
\therefore h\left( x \right) =\int { f\left( x \right) } dx=\int { cdx } =cx+d
where
d
is constant of integraion
Therefore
h\left( x \right)
is a linear function of
x
which is constant for all
x.
f(0)=0
........given
\therefore f(1)=0
\displaystyle \frac{d}{dx}(\tan ^{-1}\frac{\sin x+\cos x}{\cos x-\sin x})
Report Question
0%
-1
0%
-2
0%
1
0%
2
Explanation
Let
\displaystyle y = \tan ^{-1}\frac{\cos x+\sin x}{\cos x-\sin x}=\tan ^{-1}\frac{1+\tan x}{1-\tan x}
\displaystyle \Rightarrow y =\tan ^{-1}\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}=\tan^{-1}\tan(\frac{\pi}{4}+x)=\frac{\pi}{4}+x
\therefore \cfrac{dy}{dx}=1
\displaystyle \frac{d}{dx}\tan ^{-1}\left(\frac{a \cos x-b\sin x}{b\cos x+a\sin x}\right)
Report Question
0%
-1
0%
-2
0%
1
0%
2
Explanation
Put
\displaystyle a= r\cos \alpha , b= r\sin \alpha
\displaystyle \therefore r= \sqrt{a^{2}+b^{2}},\tan \alpha = \dfrac{b}{a}
\displaystyle \therefore y= \tan ^{-1}\frac{r\cos\left ( x+\alpha \right )}{r\sin\left ( x+\alpha \right )}=\tan^{-1}\cot \left ( x+\alpha \right )
\displaystyle =\tan^{-1}\tan\left [ \dfrac{\pi}{2} -\left ( x+\alpha \right )\right ]
\displaystyle \therefore y=\dfrac{\pi} {2}-x-\alpha
\therefore \dfrac{dy}{dx}=-1
If
\displaystyle f\left( x \right) =\sqrt { 1+\sqrt { x } } , x > 0,
then
\displaystyle f\left ( x \right )\cdot f'\left ( x \right )
is equal to
Report Question
0%
\displaystyle \frac{1}{2\sqrt{x}}
0%
\displaystyle \frac{1}{2}
0%
\displaystyle \frac{1}{4\sqrt{x}}
0%
\displaystyle \frac{2\sqrt{x}+1}{4\sqrt{x}}
Explanation
Let
y=f\left( x \right) =\sqrt { 1+\sqrt { x } }
Then,
{ y }^{ 2 }=1+\sqrt { x }
Differentiating w.r.t. x, we get,
\displaystyle 2y\frac{dy}{dx}=\frac{1}{2\sqrt{x}}
\Rightarrow f(x)f'(x)=\displaystyle \frac{1}{4\sqrt{x}}
\displaystyle \dfrac{d}{dx}\tan ^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)
Report Question
0%
-\displaystyle \dfrac{1}{2}
0%
-\displaystyle \dfrac{1}{4}
0%
-\displaystyle \dfrac{1}{8}
0%
\displaystyle \dfrac{1}{2}
Explanation
Let
\displaystyle y = \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)
\Rightarrow \displaystyle y = \tan ^{-1}\left(\dfrac{(\cos^2 \dfrac{x}{2}-\sin^2\dfrac{x}{2})}{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin \dfrac{x}{2}.\cos\dfrac{x}{2}}\right)
\displaystyle y=\tan ^{-1}\left(\dfrac{\cos\left ( x/2 \right )-\sin \left ( x/2 \right )}{\cos \left ( x/2 \right )+\sin \left ( x/2 \right )}\right)
\displaystyle =\tan^{-1}\left(\frac{1-\tan\left ( x/2 \right )}{1+\tan \left ( x/2 \right )}\right)
\displaystyle =\tan^{-1}\left(\tan \left ( \pi /4-x/2 \right )\right)=\pi/4-x/2
\displaystyle \therefore \dfrac{dy}{dx}=-\dfrac{1}{2}
Differentiate
\displaystyle x^{\sin^{-1}x}
w.r.t.
\displaystyle \sin ^{-1}x.
Report Question
0%
\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]
0%
-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]
0%
\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]
0%
-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]
Explanation
Let
\displaystyle y = x^{\sin^{-1}x}
and
z = \displaystyle \sin ^{-1}x\Rightarrow \sin z = x
we have to find
\cfrac{dy}{dz}
\Rightarrow y = (\sin z)^z
taking
\log
both side
\log y = z\log \sin z
Differentiating w.r.t
z
\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}
\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]
If
y = \displaystyle (\tan x)^{\log x}
, then
\cfrac{dy}{dx} =
Report Question
0%
(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]
0%
\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx
0%
\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x }
0%
none of these
Explanation
Let
y = \displaystyle (\tan x)^{\log x}
\log y = \log x .\log \tan x
Differentiating both side w.r.t
x
\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)
\Rightarrow \cfrac{dy}{dx} = (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]
Differentiate
\displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.
Report Question
0%
\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1-x^{2n} \right ) }\right ]nx^{n-1}
0%
\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n}
0%
\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}
0%
\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}
Explanation
Let
y = \displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.
\Rightarrow \displaystyle y= \tan x^{n}+\tan^{n}x-\left ( \tan^{-1}a+\tan^{-1}x^{n} \right )
\displaystyle\therefore \frac{dy}{dx}= \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-0-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) } \right ]nx^{n-1}
= \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) } \right ]nx^{n-1}
If
\displaystyle y=x^{n} log x+x(log x)^{n}
, find
dy/dx.
Report Question
0%
x^{n-1}(1+n log x)+(log x)^{n-1}[n+log x]
0%
x^{n}(1+n log x)+(log x)^{n}[n+log x]
0%
x^{n-1}(1+(n-1) log x)+(log x)^{n-1}[n-1+log x]
0%
none of these
Explanation
\displaystyle y=x^{n} \log x+x (\log x)^{n}
\Rightarrow \cfrac{dy}{dx}=nx^{n-1} \log x+x^{n} .(1/x)+x.n(\log x)^{n-1}.(1.x)+1(\log x)^{n}
\Rightarrow \cfrac{dy}{dx} =x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]
If
\displaystyle x\sqrt{(1+y)}+y\sqrt{(1+x)}=0
, then
\displaystyle \frac{dy}{dx}
=
Report Question
0%
\displaystyle \frac{1}{(1+x)^{2}}
0%
\displaystyle -\frac{1}{(1+x)^{2}}
0%
\displaystyle -\frac{1}{(1-x)^{2}}
0%
\displaystyle \frac{1}{(1-x)^{2}}
Explanation
Given,
\displaystyle x\sqrt{(1+y)}+y\sqrt{(1+x)}=0
\Rightarrow x\sqrt{(1+y)}=-y\sqrt{(1+x)}
Squaring both side we get,
x^{2}(1+y)=y^{2}(1+x)
\displaystyle x+y+xy=0
or
y=x
(rejected)
\displaystyle y=-\frac{x}{1+x}=\frac{1}{1+x}-1
\displaystyle \therefore \frac{dx}{dy}=-\frac{1}{(1+x)^2}
If
x^{m}.y^{n}=\left ( x+y \right )^{m+n}
, then
\dfrac{dy}{dx}=
Report Question
0%
\displaystyle \frac{y}{x}
0%
\displaystyle \frac{-y}{x}
0%
\dfrac {my}{x}
0%
\dfrac {ny}{x}
Explanation
x^{m}\times y^{n}=\left ( x+y \right )^{m+n}
Taking log both sides we get
m\log x+n\log y=\left ( m+n \right )\log \left ( x+y \right )
Differentiating w.r.t. x we get
\displaystyle \dfrac{m}{x}+\dfrac{n}{y}\dfrac{dy}{dx}=\dfrac{m+n}{x+y}\left ( 1+\dfrac{dy}{dx} \right )
\Rightarrow
\displaystyle \dfrac{dy}{dx}\left ( \dfrac{n}{y}-\dfrac{m+n}{x+y} \right )=\dfrac{m+n}{x+y}-\dfrac{m}{x}
\Rightarrow
\displaystyle \dfrac{dy}{dx}\left ( \dfrac{nx+ny-my-ny}{y\left ( x+y \right )} \right )=\dfrac{mx+nx-mx-my}{x\left ( x+y \right )}
\Rightarrow
\displaystyle \dfrac{dy}{dx}=\left ( \dfrac{nx-my}{nx-my} \right )\dfrac{y}{x}=\dfrac{y}{x}
\Rightarrow
\displaystyle \dfrac{dy}{dx}=\dfrac{y}{x}
Differentiate the following:
\displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}
Report Question
0%
\displaystyle \frac{1}{2}.
0%
\displaystyle \frac{-1}{2}.
0%
\displaystyle \frac{1}{4}.
0%
\displaystyle \frac{-1}{4}.
Explanation
Let
y = \displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}
\displaystyle y= \cot^{-1}\frac{\left ( \cos\frac{x}{2}+\sin\frac{x}{2} \right )+\left ( \cos\frac{x}{2}-\sin\frac{x}{2} \right )}{\left ( \cos\frac{x}{2}+\sin \frac{x}{2} \right )-\left ( \cos \frac{x}{2}-\sin \frac{x}{2} \right )}=\cot ^{-1}\frac{2\cos \left ( x/2 \right )}{2\sin \left ( x/2 \right )}
or
\displaystyle y=\cot ^{-1}\cot \frac{x}{2}= \frac{x}{2},
\therefore \dfrac{dy}{dx}= \dfrac{1}{2}.
Note: Here assumption taken that
\tan\frac{x}{2}\geq 1
Find the differential equation of the family of curves whose equations are
\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}+\lambda }=1,
where
\displaystyle \lambda
is parameter.
Report Question
0%
\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{2}-x^2}{a^{2}}
0%
\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{2}+x^2}{a^{2}}
0%
\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{4}-x^2}{a^{2}}
0%
\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{4}+x^2}{a^{2}}
Explanation
Differentiating given equation w.r.t
x
we get
\displaystyle \frac{2x}{a^{2}}+\frac{2y}{a^{2}+\lambda }\frac{dy}{dx}=0
\displaystyle \therefore \left ( a^{2}+\lambda \right )=-\frac{a^{2}y{y}'}{x}
Putting for
\displaystyle a^{2}+\lambda
in the given equation , we get
\displaystyle\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}x}{a^{2}y{y}'}=1
\displaystyle\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}x}{a^{2}y{y}'}=1
\Rightarrow \displaystyle -\frac{xy}{a^{2}{y}'}=1-\frac{x^{2}}{a^{2}}=\frac{a^{2}-x^2}{a^{2}}
Let
\displaystyle f\left ( x \right )
be defined by
\displaystyle f\left ( x \right )=\left\{\begin{matrix}\sin 2x & \text{if } 0< x\leq \dfrac{\pi}6\\ ax+b& \text{if } \dfrac{\pi}6< x\leq 1\end{matrix}\right.
. The values of
a
and
b
such that
\displaystyle f
and
\displaystyle {f}'
are continuous, are
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\displaystyle a=1,b=\dfrac1{\sqrt{2}}+\dfrac{\pi}6
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\displaystyle a=\dfrac1{\sqrt{2}},b=\dfrac1{\sqrt{2}}
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\displaystyle a=1,b=\dfrac{\sqrt{3}}2-\dfrac{\pi}6
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None of these
Explanation
For
f
to be continuous
\displaystyle \frac { \sqrt { 3 } }{ 2 } =f\left( \frac { \pi }{ 6 } \right) =\displaystyle \lim_{ x\rightarrow \tfrac { \pi }{ 6 }^+ } f\left( x \right) =\displaystyle\lim_{ x\rightarrow \tfrac { \pi }{ 6 } + } \left( ax+b \right) =\frac { a\pi }{ 6 } +b
\displaystyle f'\left( x \right) =\begin{cases} 2\cos { 2x } \quad \quad \text{if }0<x<\dfrac { \pi }{ 6 } \\ a\quad \quad \quad \quad \quad \text{if }\dfrac { \pi }{ 6 } <x<1 \end{cases}
\displaystyle f'\left( \frac { \pi }{ 6 } + \right) =a
and
\displaystyle f'\left( \frac { \pi }{ 6 } - \right) =1
Thus
\displaystyle a=1,b=\frac { \sqrt { 3 } }{ 2 } -\frac { \pi }{ 6 }
Find the solution of
\displaystyle \frac{dy}{dx}= \frac{2x+2y-2}{3x+y-5}.
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\displaystyle \left ( 2x+y-3 \right )= k\left ( x-y-3 \right )^{4}
0%
\displaystyle \left ( 2x-y-3 \right )= k\left ( x-y-3 \right )^{4}
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\displaystyle \left ( 2x+y+3 \right )= k\left ( x-y-3 \right )^{4}
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\displaystyle \left ( 2x+y-3 \right )= k\left ( 2x-y-3 \right )^{4}
If
f\left( x \right)
is a polynomial of degree
n(>2)
and
f\left( x \right) =f\left( k-x \right) ,(
where
k
is a fixed real number
),
then degree of
f'(x)
is
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n
0%
n-1
0%
n-2
0%
None of these
Explanation
Clearly,
f(x)
must be of the from
f\left( x \right) ={ a }_{ 0 }\left[ { x }^{ n }+{ \left( k-x \right) }^{ n } \right] +{ a }_{ 1 }\left[ { x }^{ n-1 }+{ \left( k-x \right) }^{ n-1 } \right] +...+{ a }_{ n-1 }\left[ x+\left( k-x \right) \right] +{ a }_{ n. }
It may be noted that
n
must be even for otherwise
f(x)
will become a polynomial of degree
n-1.
Clearly,
f'(x)
is a polynomial of degree
n-1.
If
2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x
, then
\displaystyle \frac { d }{ dx } f\left( x \right)
is
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\sin{x}+\cos{x}
0%
2
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\displaystyle \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }
0%
none of these
Explanation
\displaystyle 2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x
...(1)
Replace
x
by
\displaystyle \frac { \pi }{ 2 } -x
\displaystyle 2f\left( \cos { x } \right) +f\left( \sin { x } \right) =\frac { \pi }{ 2 } -x
...(2)
Solving we get ,
\displaystyle 3f\left( \sin { x } \right) =\frac { \pi }{ 2 } +3x
\displaystyle \therefore f\left( x \right) =\frac { \pi }{ 6 } +\sin ^{ -1 }{ x } \quad \\\displaystyle\therefore \frac { d }{ dx } f\left( x \right) =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }
Obtain the differential equation whose solutions are
\displaystyle y=A\cos \left ( x+3 \right ),
A being constant.
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\displaystyle \frac{dy}{dx}+y\tan \left ( x+3 \right )=0
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\displaystyle \frac{dy}{dx}+y\tan \left ( x-3 \right )=0
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\displaystyle -\frac{dy}{dx}+y\tan \left ( x-3 \right )=0
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\displaystyle -\frac{dy}{dx}+y\tan \left ( x+3 \right )=0
Explanation
y = A\cos(x+3)
(i)
\displaystyle \frac{dy}{dx}=-A\sin \left ( x+3 \right )
..(ii)
by (ii)
\div
(i)
\displaystyle \frac{\frac{dy}{dx}}{y}=-\tan \left ( x+3 \right )
\displaystyle \frac{dy}{dx}+y\tan \left ( x+3 \right )=0
If
\displaystyle {f}'\left ( x \right )=g\left ( x \right )
and
\displaystyle {g}'\left ( x \right )=-f\left ( x \right )
and
\displaystyle f\left ( 2 \right )=4={f}'\left ( 2 \right )
then
\displaystyle f^{2}\left ( 16 \right )+g^{2}\left ( 16 \right )
is
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16
0%
32
0%
64
0%
None of these
Explanation
\displaystyle \frac { d }{ dx } \left( { f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right) \right) \\ =2\left[ f\left( x \right) f'\left( x \right) +g\left( x \right) g'\left( x \right) \right] \\ =2\left[ f\left( x \right) -g\left( x \right) f\left( x \right) \right] =0
Thus
{ f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right)
is constant.
Therefore
{ f }^{ 2 }\left( 16 \right) +{ g }^{ 2 }\left( 16 \right) ={ f }^{ 2 }\left( 2 \right) +{ g }^{ 2 }\left( 2 \right) \\ ={ f }^{ 2 }\left( 2 \right) +{ \left( f'\left( 2 \right) \right) }^{ 2 }=16+16=32
Let
f\left( x \right)=\sqrt { x-1 } +\sqrt { x+24-10\sqrt { x-1 } } ;1<x<26
be a real valued function. Then
f'(x)
for
1<x<26
is
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0
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\displaystyle \frac { 1 }{ \sqrt { x-1 } }
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2\sqrt { x-1 } -5
0%
none of these
Explanation
Let,
f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}
1< x< 26
be a real valued function ,
We have,
f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}
This can be written as
f(x) = \sqrt{x -1} + \sqrt{\left ( 5 - \sqrt{x-1} \right )^{2} }
\because 1< x< 26
f(x) = \sqrt{x-1} + 5 - \sqrt{x-1}
f(x) = 5
On differentiating wrt x , we get
f{}'(x) = 0
Hence ,Option A
A curve passing through the point
(1,1)
is such that the intercept made by a tangent to it on x-axis is three times the x co-ordinate of the point of tangency, then the equation of the curve is:
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\displaystyle y=\frac{1}{x^{2}}
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\displaystyle y=\sqrt{x}
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\displaystyle y=\frac{1}{\sqrt{x}}
0%
none
Explanation
Equation of tangent on any point on the curve is,
\displaystyle Y-y=\frac{dy}{dx}\left ( X-x \right )
Thus intercept on the x-axis is,
\displaystyle I_{x}=x-\frac{y}{dy/dx}
Now using given condition,
\displaystyle I_{x}=x-\frac{y}{dy/dx}=3x
\displaystyle \therefore y\frac{dx}{dy}+2x=0
or
\displaystyle \frac{dx}{2x}+\frac{dy}{y}=0
or
\displaystyle \frac{1}{2}\log x+\log y=k
or
\displaystyle \log y\sqrt{x}=e^{k}=constant
Now given this curve passing through (1,1)
\Rightarrow k=0
\therefore
Hence required curve is,
y=\cfrac{1}{\sqrt{x}}
Let
\displaystyle f
be a function satisfying
\displaystyle f\left ( x+y \right )=f\left ( x \right )f\left ( y \right )
for all
x
and
y
and
\displaystyle f\left ( 0 \right )={f}'\left ( 0 \right )=1
then
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\displaystyle f
is differentiable for all x
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\displaystyle {f}'\left ( x \right )=f\left ( x \right )
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\displaystyle f\left ( x \right )=e^{x}
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\displaystyle f
is continuous for alI x
Explanation
We have,
f(x+y)=f(x)f(y) ...........(1)
Now
f'(x) =\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\dfrac{f(x)f(h)-f(x)}{h}
using (1)
\Rightarrow f'(x) =\displaystyle\lim_{h\to 0}\dfrac{f(x)[f(h)-1)}{h}=f(x)\lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h}=f(x)f'(0)=f(x)
Integrating we get,
\log(f(x)) = x+c
Since at x
=0, f(x) =1\Rightarrow
we have,
c=0
Hence
f(x) =e^x
, which is continuous and differentiable for all
x
If
\displaystyle f\left ( 1 \right )=3
and
\displaystyle {f}'\left ( 1 \right )=-\dfrac13
then the derivative of
\displaystyle \left ( x^{11} +f\left ( x \right )\right )^{-2}
at
\displaystyle x=1
is
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-\dfrac12
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-1
0%
1
0%
\displaystyle {f}'\left ( 1 \right )
Explanation
The given equation is:
y = (x^{11} + f(x))^{-2}
Differentiating the equation w.r.t to x once we get,
\Rightarrow y' = -2(x^{11} + f(x))^{-3}.(11x^{10} + f'(x))
Value of the derivative at
x = 1
is given as,
\Rightarrow y' = \dfrac{-2}{(1+3)^3}.(11 - \dfrac{1}{3})
\Rightarrow y' = \dfrac{-2}{64}.\dfrac{32}{3}
\Rightarrow y' = \dfrac{-1}{3}
.....Answer
\Rightarrow y' = f'(1)
A polynomial
f(x)
leaves remainder
15
when divided by
(x-3)
and
(2x+1)
when divided by
(x-1)^2
. When
f
is divided by
(x-3)(x-1)^2,
the remainder is
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2x^2+2x+3
0%
2x^2-2x-3
0%
2x^2-2x+3
0%
none of these
Explanation
Since function
f(x)
leaves remainder
15
when divided by
x-3
, therefore
f(x)
can be written as
f(x)=(x-3)l(x)+15
...(1)
Also,
f(x)
leaves remainder
2x+1
when divided by
(x-1)^2.
Thus,
f(x)
can also be written as
f\left( x \right)={ \left( x-1 \right) }^{ 2 }m\left( x \right) +2x+1
...(2)
If
R(x)
be the remainder when
f(x)
is divided by
(x-3)(x-1)^2,
then we may write
f\left( x \right)=\left( x-3 \right) { \left( x-1 \right) }^{ 2 }n\left( x \right) +R\left( x \right)
...(3)
Since
(x-3)(x-1)^2
is a polynomial of degree three, the remainder has to be a polynomial of degree less than or equal to two.
Thus let
R(x)=ax^2+bx+c
From (1) and (3), we have
f(3)=15=R(3)\Rightarrow 9a+3b+c=15
...(4)
From (2) and (3), we have
f(1)=3=R(1)\Rightarrow a+b+c=3
...(5)
From (2) and (3), we have
f'(1)=2=R'(1)\Rightarrow 2a+b=2
...(6)
Solving equation (4),(5) and (6), we get
a=2,b=-2,c=3
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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