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CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Tangents And Its Equations
Quiz 10
If tangent at any point on the curve
y
2
=
1
+
x
2
m
a
k
e
s
a
n
a
n
g
l
e
θ
with positive direction of the x-axis then
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0%
|
tan
θ
|
>
1
0%
|
tan
θ
|
<
1
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|
tan
θ
|
≥
1
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|
tan
θ
|
≤
1
The equation of the normal to the curve
y
=
(
1
+
x
)
y
+
sin
−
1
(
sin
2
x
)
a
t
x
=
0
i
s
.
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x
+
y
=
1
0%
x
−
y
+
1
=
0
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x
+
y
=
2
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2
x
−
y
+
1
=
0
If the line
a
x
+
y
=
c
, touches both the curves
x
2
+
y
2
=
1
and
y
2
=
4
√
2
x
, then
|
c
|
is equal to:
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1
/
2
0%
2
0%
√
2
0%
1
√
2
Explanation
Tangent to
y
2
=
4
√
2
x
is
y
=
m
x
+
√
2
m
is also tangent to
x
2
+
y
2
=
1
⇒
|
√
2
/
m
√
1
+
m
2
|
=
1
⇒
m
=
±
1
Tangent will be
y
=
x
+
√
2
or
y
=
−
x
−
√
2
compare with
y
=
−
a
x
+
c
⇒
a
=
±
1
;
c
=
±
√
2
∴
The equation of the normal to the curve
y=\left( 1+x \right) ^{ y }+{ sin }^{ -1 }\left( { sin }^{ 2 }x \right) at\quad x=0
is
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x+y=1
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x-y+1=0
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2x+y=2
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2x-y+1
Explanation
Given,
y=(1+x)^y+\sin ^{-1}(\sin ^2x)
when
x=0
y=1^y+0=1
\therefore (x_1,y_1)=(0,1)
\dfrac{dy}{dx}=(1+x)^y\left [ \ln (1+x)\dfrac{dy}{dx}+\dfrac{y}{1+x} \right ]+\dfrac{2\sin x\cos x}{\sqrt{1-\sin ^4x}}
\dfrac{dy}{dx}_{(0,1)}=1[1]+0
\therefore \dfrac{dy}{dx}_{(0,1)}=1
Therefore, slope of normal is
-1
Equation of normal,
y-y_1=\dfrac{-1}{\frac{dy}{dx}}(x-x_1)
y-1=(-1)(x-0)
y-1=-x
x+y=1
is the equation of normal
The sum of the length of sub tangent of sub tangent and tangent to the curve
x=c\left[ 2cos\theta -log\left( cos\quad ec\theta +cot\theta \right) \right] ,y=csin2\theta \quad at\quad \theta =\frac { \pi }{ 3 } is
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\dfrac { c }{ 2 }
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2c
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\dfrac { 3c }{ 2 }
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\dfrac { 5c }{ 2 }
Explanation
Given,
x=c[2\cos \theta −\log (\csc \theta +\cot \theta )]
\Rightarrow \dfrac {dx}{d\theta }=c(-2\sin \theta +\csc \theta )
y=c\sin 2\theta
\dfrac{dy}{d\theta }=2c\cos 2\theta
Also,
y_1=c\sin \dfrac{2\pi }{3}=\dfrac{c\sqrt 3}{2}
Slope of tangent at
\theta =\dfrac{\pi }{3}
is
m=\dfrac{dy}{dx}_{\theta =\frac{\pi }{3}}=\sqrt 3
Length of tangent
=\left | \dfrac{y_1\sqrt{1+m^2}}{m} \right |=c
Length of subtangent
=\left | \dfrac{y_1}{m} \right |=\dfrac{c}{2}
Required sum
=c+\dfrac{c}{2}=\dfrac{3c}{2}
If the tangent to the curve
x=at^2, y=2at
is perpendicular to
x
-axis, then its point of contact is
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(a, a)
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(0, a)
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(0, 0)
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(a, 0)
Explanation
Let the required point be
(x_1,y_1).
It is given that points lies on the curve.
\therefore
x_1=at^2
and
y_1=2at
Now,
x=at^2
and
y=2at
Differentiating w.r.t.
t,
we get
\Rightarrow
\dfrac{dx}{dt}=2at
and
\dfrac{dy}{dt}=2a
\Rightarrow
\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}
\Rightarrow
\dfrac{dy}{dx}=\dfrac{2a}{2at}
\Rightarrow
\dfrac{dy}{dx}=\dfrac{1}{t}
\Rightarrow
\dfrac{dy}{dx}=\dfrac{2a}{y}
[ Since,
y=2at
]
Slope of the tangent
=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{2a}{y_1}
It is given that the tangent is perpendicular to the
y-axis.
It means theta it is parallel to the
x-axis.
\therefore
Slope of the tangent
=
Slope of the
x-axis.
\Rightarrow
\dfrac{2a}{y_1}=0
\Rightarrow
a=0
Now,
x_1=at^2=(0)t^2=0
y_1=2at=2(0)t=0
\therefore
(x_1,y_1)=(0,0)
The slope of the tangent to the curve
x=t^2+3t-8, y=2t^2-2t-5
at point
(2, -1)
is
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\dfrac {22}{7}
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\dfrac {6}{7}
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-6
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\dfrac {7}{6}
Explanation
x=t^2+3t-8
........
(1)
y=2t^2-2t-5
..........
(2)
Differentiate
(1)
, we get
\dfrac{dx}{dt}=2t+3
Differentiate
(2)
, we get
\dfrac{dy}{dx}=4t-2
m=\dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}
Given point is
(2, -1)
Put the point in original x and y, we get
\Rightarrow x=t^2+3t-8
2=t^2+3t-8
t^2+3t-10=0
(t-2)(t+5)=0
t=2, t=-5
\Rightarrow y=2t^2-2t-5
(-1)=2t^2-2t-5
2t^2-2t-4=0
(t+1)(t-2)=0
t=-1, t=2
Since,
t=2
common in both parts, so we take
\dfrac{dy}{dx}=\dfrac{4t-2}{2t-3}
at
t=2
At
t=2
\dfrac{dy}{dx}=\dfrac{4(2)-2}{2(2)-3}=\dfrac{8-2}{4+3}=\dfrac{6}{7}
m=\dfrac{dy}{dx}=\dfrac{6}{7}
.
The equation of tangent at those points where the curve
y=x^2-3x+2
meets
x
-axis are
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x-y+2=0,x-y-1=0
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x+y-1=0,x-y-2=0
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x-y-1=0,x-y=0
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x-y=0,x+y=0
Explanation
solution:
y=x^2-3x+2
Let the tangent meet the x-axis at point
(x, 0)
\Rightarrow \dfrac{dy}{dx}=2x-3
The tangent passes through point
(x, 0)
\therefore 0=x^2-3x+2=0
\Rightarrow (x-2)(x-1)=0
\Rightarrow x=2, x=1
Case I: when
x=2
Slope of tangent,
\left.\dfrac{dy}{dx}\right|_{(2, 0)}=2(2)-3=4-3=1
\therefore (x_1, y_1)=(2, 0)
Equation of tangent
y-y_1=m(x-x_1)
\Rightarrow y-0=1(x-2)
\Rightarrow x-y-2=0
Case II: When
x=1
Slope of tangent
\left.\dfrac{dy}{dx}\right|_{(2, 0)}=2-3=-1
\therefore (x_1, y_1)=(1, 0)
Equation of tangent
y-y_1=m(x-x_1)
y-0=-1(x-1)
\Rightarrow x+y-1=0
.
The equation of the normal to the curve
y=x+\sin x\cos x
at
x=\dfrac {\pi}{2}
is
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x=2
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x=\pi
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x+\pi =0
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2x=\pi
Explanation
y=x+\sin x cos x
Differentiating both sides w.r.t.
x,
\Rightarrow
\dfrac{dy}{dx}=1+\cos^2x-\sin^2x
Slope of the tangent
(m)=\left(\dfrac{dy}{dx}\right)_{x=\frac{\pi}{2}}=1+\cos^2\dfrac{\pi}{2}-\sin^2\dfrac{\pi}{2}
=1-1
=0
Slope of the normal
=\dfrac{-1}{m}=\dfrac{-1}{0}
When,
x=\dfrac{\pi}{2},
y=\dfrac{\pi}{2}+\cos\dfrac{\pi}{2}\sin\dfrac{\pi}{2}
\therefore
y=\dfrac{\pi}{2}
Now,
(x_1,y_1)=\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)
\therefore
Equation of the normal.
\Rightarrow
y-y_1=\dfrac{-1}{m}(x-x_1)
\Rightarrow
y-\dfrac{\pi}{2}=\dfrac{-1}{0}\left(x-\dfrac{\pi}{2}\right)
\Rightarrow
x=\dfrac{\pi}{2}
\Rightarrow
2x=\pi
The point on the curve
y=12x-x^2
, where the slope of the tangent is zero will be
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(0, 0)
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(2, 16)
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(3, 9)
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(6, 36)
Explanation
Let the point be
P(x, y)
y=12x-x^2
\dfrac{dy}{dx}=12-2x
\left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=12-2x_1
since slope of tangent is zero
so
\left(\dfrac{dy}{dx}\right)_{x_1, y_1}=0
12-2x_1=0
2x_1=12
x_1=6
Also curve passing through tangent
y_1=12x_1-x^2_1
y_1=12\times 6-36
y_1=72-36
y_1=36
The points are
(6, 36)
.
The slope of the tangent to the curve
x=3t^2+1, y=t^3-1
at
x=1
is
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\dfrac {1}{2}
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0
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-2
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\infty
Explanation
Given curves are,
x=3t^2+1
---- ( 1 )
y=t^3-1
---- ( 2 )
Substituting
x=1
in ( 1 ) we get,
\Rightarrow
3t^2+1=1
\Rightarrow
3t^2=0
\Rightarrow
t=0
Differentiate ( 1 ) w.r.t.
t,
we get
\Rightarrow
\dfrac{dx}{dt}=6t
Differentiate ( 2 ) w.r.t. $$t,$ we get
\Rightarrow
\dfrac{dy}{dt}=3t^2
\Rightarrow
\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3t^2}{6t}
\therefore
\dfrac{dy}{dx}=\dfrac{t}{2}
Slope of the tangent
=\left(\dfrac{dy}{dx}\right)_{t=0}=\dfrac{0}{2}=0
The point on the curve
y=x^2-3x+2
where tangent is perpendicular to
y=x
is
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(0, 2)
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(1, 0)
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(-1, 6)
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(2, -2)
Explanation
y=x^2-3x+2
[ Given ]
y=x
[ Given ]
Differentiating both sides w.r.t.
x,
we get
\Rightarrow
\dfrac{dy}{dx}=1
Let
(x_1,y_1)
be the required point.
It is given that point lies on the curve.
\therefore
y_1=x_1^2-3x_1+2
\Rightarrow
y=x^2-3x+2
Differentiating both sides w.r.t.
x,
we get
\therefore
\dfrac{dy}{dx}=2x-3
Slope of the tangent
=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=2x_1-3
The tangent is perpendicular to the line.
\therefore
Slope of the tangent
=\dfrac{-1}{slope\,of\,the\,line}=\dfrac{-1}{1}=-1
Now,
2x_1-3=-1
\Rightarrow
2x_1=2
\Rightarrow
x_1=1
\Rightarrow
y_1=x_1^2-3x_1+2
\Rightarrow
y_1=(1)^2-3(1)+2
\Rightarrow
y_1=1-3+2
\therefore
y_1=0
\therefore
(x_1,y_1)=(1,0)
The equation of the normal to the curve
y=x(2-x)
at the point
(2, 0)
is
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x-2y=2
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x-2y+2=0
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2x+y=4
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2x+y-4=0
Explanation
The given curve is
y=2x-x^2
i.e,
x^2-2x+y=0
Now the equation of the tangent at the point (
x_{1},y_{1}
)=
(2,0)
or,
xx_1-(x+x_1)+\dfrac{1}{2}(y+y_1)=0
i.e
2x+(-1)(x+2)+\dfrac{1}{2}(y+0)=0
i.e,
2x-4+y=0
i.e,
y=-2x+4
So, the slope of the tangent is
(-2)
Therefore the slope of the normal=
\dfrac{1}{2}
So,the equation of the normal at the point
(2,0)
is
y-0=\dfrac{1}{2}(x-2)
i.e,
2y=x-2
i.e,
x-2y=2
At what points the slope of the tangent to the curve
x^2+y^2-2x-3=0
is zero?
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(3, 0), (-1, 0)
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(3, 0), (1, 2)
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(-1, 0), (1, 2)
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(1, 2), (1, -2)
Explanation
x^2+y^2-2x-3=0
is zero.
Differentiate w.r.t. x
2x+2y\dfrac{dy}{dx}-2=0
2y\cdot \dfrac{dy}{dx}=2-2x
\dfrac{dy}{dx}=\dfrac{2(1-x)}{2y}
\dfrac{dy}{dx}=\dfrac{1-x}{y}
........
(1)
If line is parallel to x-axis
Angle with x-axis
=\theta =0
Slope of x-axis
=\tan\theta =\tan 0^o=0
Slope of tangent
=
Slope of x-axis
\dfrac{dy}{dx}=0
\dfrac{1-x}{y}=0
x=1
Finding y when
x=1
x^2+y^2-2x-3=0\\
(1)^2+y^2-2(1)-3=0\\
1+y^2-2-3=0\\
y=\pm 2
Hence, the points are
(1, 2)
and
(1, -2)
.
Any tangent to the curve
y=2x^7+3x+5
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Is parallel to
x
-axis
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Is parallel to
y
-axis
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Makes an acute angle with
x
-axis
0%
Makes an obtuse angle with
x
-axis
Explanation
We have,
y=2x^7+3x+5
\dfrac{dy}{dx}=14x^6+3
\Rightarrow \dfrac{dy}{dx} > 3
(
\because x^6
is always positive for any real value)
\Rightarrow \dfrac{dy}{dx} > 0
so,
\tan \theta > 0
Hence,
\theta
lies in first quadrant.
Thus, the tangent to curve makes an acute angle.
The normal to the curve
x^2=4y
passing through
(1, 2)
is
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x+y=3
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x-y=3
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x+y=1
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x-y=1
Explanation
Curve is
x^2=4y
Diff wrt to x
2x=\dfrac{4dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}
Slope of normal
=\dfrac{-1}{dy/dx}=\dfrac{-2}{x}
Let (h, k) be the point where normal and curve intersects
\therefore
Slope of normal at (h, k)
=-2/h
Equation of normal passes through (h, k)
y-y_1=m(x-x_1)
y-k=\dfrac{-2}{h}(x-h)
Since normal passes through
(1, 2)
2-k=\dfrac{-2}{h}(1-h)
k=2+\dfrac{2}{h}(1-h)
..........
(1)
since (h, k) lies on the curve
x^2=4y
h^2=4k
k=\dfrac{h^2}{4}
.......
(2)
using
(1)
and
(2)
2+\dfrac{2}{h}(1-h)=\dfrac{h^2}{4}
\dfrac{2}{h}=\dfrac{h^2}{4}
h=2
Putting
h=2
in
(2)
k=\dfrac{h^2}{4}, k=1
h=2
and
k=1
putting in equation of normal
\Rightarrow y-k=\dfrac{-2(k-h)}{h}
\Rightarrow y-1=\dfrac{-2(x-2)}{2}=y-1=-1(x-2)
\Rightarrow y-1=-x+2\\
\Rightarrow x+y=2+1\\
\Rightarrow x+y=3
.
The equation of the normal to the curve
x=a\cos^3\theta, y=a\sin^3\theta
at the point
\theta =\dfrac {\pi}{4}
is
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x=0
0%
y=0
0%
x=y
0%
x+y=a
Explanation
Slope of normal to the curve =
- \dfrac{1}{\left(\dfrac{dy}{dx} \right)_{\theta = \pi/4}}
x = a \cos^3 \theta, \, y = a \sin^3 \theta
\dfrac{dx}{d \theta} = 3 a \cos^2 \theta (- \sin \theta) , \dfrac{dy}{d \theta} = 3 a \sin^2 \theta \cos \theta
\therefore \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx / d \theta} = \dfrac{+3a \sin^2 \theta \cos \theta}{-3 a \cos^2 \theta \sin \theta} = -\tan \theta
\therefore \left(\dfrac{dy}{dx}\right)_{\theta = \pi/4} = -1
\therefore
Equation of normal
= (y - y_0) = \dfrac{-1}{\left(\dfrac{dy}{dx}\right)_{\theta = \pi/4}} [ x - x_0]
y - \dfrac{a}{2 \sqrt{2}} = \dfrac{-1}{-1} \left(x - \dfrac{a}{2 \sqrt{2}} \right)
[x_0 = a \cos^3 \dfrac{\pi}{4} = \dfrac{a}{2 \sqrt{2}}
y - \dfrac{a}{2 \sqrt{2}} = x- \dfrac{a}{2 \sqrt{2}}
y_0 = a \sin^3 \dfrac{\pi}{4} = \dfrac{a}{2 \sqrt{2}}]
y = x
The number of tangents to the cure
x^{3/2}+y^{3/2}=2a^{3/2}, a> 0
, which are equally inclined to the axes, is
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0%
2
0%
1
0%
0
0%
4
The curve given by
x + y = e^{xy}
has a tangent parallel to the y-axis at the point
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(0,1)
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( 1, 0 )
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(1, 1)
0%
None of these
Explanation
Differentiating w.r.t.x, we get
1 + \dfrac{dy}{dx} = e^{xy}\left ( y + x\dfrac{dy}{dx} \right )
or
\dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}
As the tangent is parallel along
y-axis
\dfrac{dy}{dx} = \infty=\dfrac{ye^{xy}-1}{1-xe^{xy}}
\Rightarrow 1-xe^{xy}=0
This holds for x = 1, y = 0
Mark the correct alternative of the following.
The point on the curve
9y^2=x^3
, where the normal to the curve makes equal intercepts with the axes is?
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(4, \pm 8/3)
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(-4, 8/3)
0%
(-4, -8/3)
0%
(8/3, 4)
Explanation
Let the required point be
(x_1, y_1)
equation of the curve is
9y^2=x^2
since
(x_1, y_1)
lies on the curve, therefore
9y^2_1=x^3_1
.......
(1)
Now
9y^2=x^3
\Rightarrow \dfrac{dy}{dx}=\dfrac{x^2}{6y}
\Rightarrow \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=\dfrac{x^2}{6y^1}
since normal to the curve at
(x_1, y_1)
makes equal intersepts with the coordinate axis, therefore slope of the normal
=\pm 1
\Rightarrow \dfrac{1}{-(dy/dx)_{(x_1, y_1)}}=\pm 1
\Rightarrow (dy/dx)_{(x_1, y_1)}=\pm 1
\rightarrow \dfrac{x^2_1}{6y^1}=\pm 1
\Rightarrow x^4_1=36y^2_1=36\left(\dfrac{x^3_1}{9}\right)
(using
1
)
\Rightarrow x^4_1=4x^3_1\Rightarrow x^3_1(x_1-4)=0
\Rightarrow x_1=0, 4
Putting
x_1=0
in
(1)
we get
9y^2_1=0\Rightarrow y_1=0
Putting
x_1=4
in
(1)
we get
9y^2_1=(4)^3\Rightarrow y_1=\pm \dfrac{8}{3}
But the line making equal intersepts with the coordinate axes cannot pass through origin.
Hence, the required points are
\left(4, \dfrac{8}{3}\right)
and
\left(4, \dfrac{-8}{3}\right)
.
Mark the correct alternative of the following.
The line
y=mx+1
is a tangent to the curve
y^2=4x
, if the value of m is?
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0%
1
0%
2
0%
3
0%
1/2
Explanation
Given equation of the tangent to the given curve
y=mx+1
Now substituting the value of y in
y^2=4x
, we get
\Rightarrow (mx+1)^2=4x
\Rightarrow m^2x^2+1+2mx-4x=0
\Rightarrow m^2x^2+x(2m-4)+1=0
..........
(1)
Since, a tangent touches the curve at one point, the root of equation
(1)
must be equal.
Thus, we get
Discriminant,
D=b^2-4ac=0
(2m-4)^2-4(m^2)(1)=0
\Rightarrow 4m^2-16m+16-4m^2=0
\Rightarrow -16m+16=0
\Rightarrow m=1
.
The slope of the tangent to the curve
x=t^2+3t-8, y=2t^2-2t-5
at the point
(2, -1)
is
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\dfrac {22}{7}
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\dfrac {6}{7}
0%
\dfrac {7}{6}
0%
-\dfrac {6}{7}
Explanation
Given curve
x = t^2 + 3t - 8
and
y = 2t^2 - 2t - 5
slope of tangent to the curve
= \dfrac{dy}{dx} = \dfrac{dy/dt}{dx / dt}
\therefore \dfrac{dx}{dt} = 2t + 3 , \, \dfrac{dy}{dt} = 4t - 2
\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4t - 2}{2t + 3}
\left(\dfrac{dy}{dx} \right)_{t = 2}
\dfrac{4(2) - 2}{2(2) + 3} = \dfrac{6}{7}
The normal at the point
(1, 1)
on the curve
2y+x^2=3
is
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x+y=0
0%
x-y=0
0%
x+y+1=0
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x-y=1
Explanation
f(x)=x(x-4)^2, x\in 1
The equation of the givne curve is
2y+x^2=3
Differentiating wrt to x, we have
\dfrac{2dy}{dx}+2x=0
\Rightarrow \dfrac{dy}{dx}=-x
\therefore \left(\dfrac{dy}{dx}\right)_{(1, 1)}=-1
The slope of the normal to the given curve at point
(1, 1)
is
\dfrac{-1}{(dy/dx)_{(1, 1)}}=1
Hence the equation of the normal to the given curve at
(1, 1)
is given as
\Rightarrow y-1=1(x-1)
\Rightarrow y-1=x-1
\Rightarrow x-y=0
.
Consider the equation
x^y=e^{x-y}
What is
\dfrac{d^2y}{dx^2}
at
x=1
equal to ?
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Explanation
{x}^{y} = {e}^{x - y}
Taking
\log
both sides, we have
y \log{x} = \left( x - y \right) \log{e}
y \log{x} = x - y ..... \left( 1 \right)
\dfrac{y}{x}=1+\log x
At
x = 1 \Rightarrow y = 1
Differentiating equation
\left( 1 \right)
w.r.t.
x
, we have
\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}
\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}
\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}
\Rightarrow \cfrac{dy}{dx} = \cfrac{y \log{x}}{x \left( 1 + \log{x} \right)}
\Rightarrow \cfrac{dy}{dx} = \cfrac{\log{x}}{{\left( 1 + \log{x} \right)}^{2}}
Again differentiating above equation w.r.t.
x
, we have
\cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{{\left( 1 + \log{x} \right)}^{2} \cdot \frac{1}{x} - \log{x} \cdot 2 \left( 1 + \log{x} \right) \frac{1}{x}}{{\left( 1 + \log{x} \right)}^{4}}
\Rightarrow \cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{\frac{1}{x} \left( 1 - \log{x} \right)}{{\left( 1 + \log{x} \right)}^{3}}
At
x = 1
,
\cfrac{{d}^{2}y}{d{x}^{2}} = 1
Consider the equation
x^y=e^{x-y}
What is
\dfrac{dy}{dx}
at
x=1
equal to ?
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Explanation
{x}^{y} = {e}^{x - y}
Taking
\log
both sides, we have
y \log{x} = \left( x - y \right) \log{e}
y \log{x} = x - y ..... \left( 1 \right)
At
x = 1 \Rightarrow y = 1
Differentiating equation
\left( 1 \right)
w.r.t.
x
, we have
\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}
\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}
\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}
At
x = 1, \; y = 1
\cfrac{dy}{dx} = 0
A curve
y=me^{mx}
where
m > 0
intersects y-axis at a point
P
.
How much angle does the tangent at
P
make with y-axis ?
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\tan^{-1}m^2
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\cot^{-1}(a+m^2)
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\sin^{-1}(\dfrac{1}{\sqrt{1+m^4}})
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\sec^{-1}\sqrt{1+m^4}
A curve
y=me^{mx}
where
m > 0
intersects y-axis at a point
P
.
What is the slope of the curve at the point of intersection
P
?
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m
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m^2
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2m
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2m^2
Explanation
y = m {e}^{mx}, \; m > 0
Therefore,
Slope
= \cfrac{dy}{dx} = {m}^{2} {e}^{mx}
Substituting
x = 0
, we have
Slope={m}^{2} {e}^{m \cdot 0} = {m}^{2}
For
x > 1, y=\log_e x
satisfies the inequality
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x-1 > y
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x^2 -1 >y
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y > x-1
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\dfrac {x-1}{x} < y
The slope of the tangent to the curve
y = \sqrt{4-x^{2}}
at the point, where the ordinate and the abscissa are equal , is
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-1
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1
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0
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None of these
Explanation
Putting
y=x
in
y = \sqrt{4-x^{2}}
, we get
x = \sqrt{2}, -\sqrt{2}
.
So, the point is
(\sqrt{2}, \sqrt{2})
.
Differentiating
y^{2}+x^{2} = 4
w.r.t. x,
2y \dfrac{dy}{dx}+ 2x = 0
or
\dfrac{dy}{dx}= -\dfrac{x}{y}
\Rightarrow at(\sqrt{2}, \sqrt{2}), \dfrac{dy}{dx} = -1
If m is the slope of a tangent to the curve
e^{y}=1+x^{2},
then
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\left | m \right |> 1
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m> 1
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m> -1
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\left | m \right |\leq 1
Explanation
Differentiating w.r.t.x, we get
e^{y} \dfrac{dy}{dx} = 2x
\Rightarrow \dfrac{dy}{dx} = \dfrac{2x}{1+x^{2}} (\because e^{y} = 1 +x^{2})
\Rightarrow m = \dfrac{2x}{1 + x^{2}} or \left | m \right |= \dfrac{2\left | x \right |}{1 + \left | x \right |^{2}}
But
1 + \left | x \right |^{2} - 2\left | x \right |=(1-\left | x \right |)^{2}\geq 0
\Rightarrow 1 + \left | x \right |^{2} \geq 2\left | x \right |
\therefore \left | m \right |\leq 1
The abscissa of points P and Q in the curve
y = e^{x}+e^{-x}
such that tangents at P and Q make
60^{o}
with the x-axis
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ln
\left ( \dfrac{\sqrt{3}+\sqrt{7}}{7} \right )
and ln
\left ( \dfrac{\sqrt{3}+\sqrt{5}}{2} \right )
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ln
\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )
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ln
\left ( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right )
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\pm
ln
\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )
If x=4 y = 14 is a normal to the curve
y^{2}=ax^{3}-\beta
at (2,3) then the value of
\alpha +\beta
is
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9
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-5
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7
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-7
At what points of curve
y = \dfrac{2}{3}x^{3}+\dfrac{1}{2}x^{2}
, the tangent makes the equal with the axis?
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(\dfrac{1}{2},\dfrac{5}{24})
and
\left ( -1,\dfrac{-1}{6} \right )
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(\dfrac{1}{2},\dfrac{4}{9})
and
( -1,0)
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\left ( \dfrac{1}{3},\dfrac{1}{7} \right )
and
\left ( -3, \dfrac{1}{2} \right )
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\left ( \dfrac{1}{3},\dfrac{4}{47} \right )
and
\left ( -1, \dfrac{1}{2} \right )
The curve represented parametrically by the equations x = 2 in
\cot t+1
and
y=\tan t+\cot t
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tanfent and normal intersect at the point (2, 1)
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normal at
t = \pi /4
is parallel to the y-axis
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tangent at
t = \pi /4
is parallel to the line y = x
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tangent at
t = \pi /4
is parallel to the x-axis
If a variable tangent to the curve
x^{2}y=c^{3}
makes intercepts a, b on x-and y-axes, respectively, then the value of
a^{2}b
is
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27c^{3}
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\dfrac{4}{27}c^{3}
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\dfrac{27}{4}c^{3}
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\dfrac{4}{9}c^{3}
The angle between the tangent to the curves
y = x^{2}
and
x = y^{2}
at (1, 1) is
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\cos ^{-1}\dfrac{4}{5}
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\sin ^{-1}\dfrac{3}{5}
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\tan ^{-1}\dfrac{3}{4}
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\tan ^{-1}\dfrac{1}{3}
At the point
P(a, a^{n})
on the graph of
y = x^{n}(n \epsilon n)
in the first quadrant, a normal is drawn. the normal intersects the y-axis at the point (0, b) . if
\underset{a\rightarrow b}{lim}b=\dfrac{1}{2}
, then n equals
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Point on the curve
f(x)=\dfrac{x}{1-x^{2}}
where the tangent is inclined at an angle of
\dfrac{\pi }{4}
ot the x-axis are
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(0, 0)
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\left ( \sqrt{3},\dfrac{-\sqrt{3}}{2} \right )
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\left ( -2 ,\dfrac{2}{3}\right )
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\left (- \sqrt{3},\dfrac{\sqrt{3}}{2} \right )
The x-intercept of the tangent at any arbitrary point of the curve
\dfrac{a}{x^{2}}+\dfrac{b}{y^{2}}=1
is proportion to
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square of the abscissa of the point of tangency
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square root of the abscissa of the point of tangency
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cube of the abscissa pf the point of tangency
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cube root of the abscissa of the point of tangency
If the tangent at any point
P(4m^{2}, 8m^{3})
of
x^{3}-y^{3}=0
is also a normal to the curve
x^{3}-y^{3}=0
, then value of m is
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m = \dfrac{\sqrt{2}}{3}
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m = -\dfrac{\sqrt{2}}{3}
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m = \dfrac{3}{\sqrt{2}}
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m = -\dfrac{3}{\sqrt{2}}
The slope of the tangent to the curve
y = f(x)
at
\left [ x, f(x) \right ]
is 2x +If the curve passes through the point (1, 2)then the area bounded by the curve, the x-axis and the line x = 1 is
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\dfrac{5}{6}
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\dfrac{6}{5}
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\dfrac{1}{6}
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6
The normal to the curve
x = a (\cos 0 + 0\sin 0), y= a (\sin 0- 0\cos 0)
at any point 0 is such that
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it makes a constant angle with x-axis
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it passes through the origin
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it is at a constant distance from the origin
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none of these
A curve passes through
(2,1)
and is such that the square of the ordinate is twice the contained by the abscissa and the intercept of the normal. Then the equation of curve is
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x^2 +y^2=9x
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4x^2 +y^2=9x
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4x^2 +2y^2=9x
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None of these
The curve for which the ratio of the length of the segment by any tangent on the
Y-
axis to the length of the radius vector is constant
(K)
, is
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(y+\sqrt {x^2 -y^2})x^{k-1}=c
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(y+\sqrt {x^2 +y^2})x^{k-1}=c
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(y-\sqrt {x^2 -y^2})x^{k-1}=c
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(y+\sqrt {x^2 +y^2})x^{k-1}=c
If
f(x) = \displaystyle \int_{1}^{x} e^{t^2/2}(1-t^2)dt,
then
\dfrac{d}{dx} f(x)
at x=1 is
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-1
Explanation
We have,
f(x)=\displaystyle \int_{1}^{x}e^{t^2/2}(1-t^2)dt
f'(x)=[e^{x^2/2}(1-x^2)]^2
f'(1)=e^{1/2}.0=0
If
y=\displaystyle \int_{x}^{x^2}t^2dt ,
then the equation of tangent at x=1 is
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x+y=1
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y=x-1
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y=x
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y=x+1
Explanation
At
x=1,y=0,
\dfrac{dy}{dx}=2x.(x^4)^2-(x^2)^2=1
\,\therefore
Equation of tangent is
y=x-1.
The tangent to the curve
y = e^{x}
drawn at the point
(c, e^{c})
intersects the line joining the points
(c-1, e^{c-1})
and
(c+1, e^{c+1})
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on the left of x =c
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on the right of x = c
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at no point
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at all point
If the line ax +by + c = 0 is a normal to the curve xy = 1, then
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a > 0, b> 0
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a > 0, b < 0
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a < 0, b > 0
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a < 0, b < 0
The equation of the curves through the point
(1,0)
and whose slope is
\dfrac{y -1}{x^{2} + x}
is
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(y - 1)(x + 1) + 2x = 0
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2x(y - 1) + x + 1 = 0
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x(y - 1)(x + 1) + 2 = 0
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None of these
Explanation
Slope
= \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{y - 1}{x^{2} + x}
\implies \dfrac{dy}{y-1} = \dfrac{dx}{x^{2} + x}
\implies \int\dfrac{1}{y-1}dy = \int(\dfrac{1}{x} - \dfrac{1}{x+1})dx + C
\implies \log(y-1)=\log(\dfrac{x}{x+1})+\log{c}
\implies \dfrac{(y-1)(x+1)}{x} = k
Putting
x=1, y=0
, we get
k=-2
The equation is
(y-1)(x+1) + 2x = 0
The point(s) on the curve
y^{3} + 3x^{2} = 12y,
where the tangent is vertical, is (are)
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\left ( \pm \dfrac{4}{\sqrt{3}}, -2 \right )
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\left ( \pm \sqrt{\dfrac{11}{3}}, 1 \right )
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(0, 0)
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\left ( \pm \dfrac{4}{\sqrt{3}}, 2 \right )
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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