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CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Tangents And Its Equations
Quiz 3
Equation of the tangent to the parabola
y
2
=
4
x
+
5
which is parallel to the line
y
=
2
x
+
7
is
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y
=
2
x
+
3
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y
=
2
x
−
3
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y
=
2
x
+
5
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y
=
2
x
−
5
Explanation
Differentiating the equation given,we obtain
2
y
′
y
=
2
y
′
=
2
2
y
=
2
y
=
1
x
=
−
1
(
y
−
1
)
=
2
(
x
+
1
)
y
=
2
x
+
3
For the parabola
y
2
=
8
x
, tangent and normal are drawn at
P
(
2
,
4
)
which meet the axis of the parabola in
A
and
B
, then the length of the diameter of the circle through
A
,
P
,
B
is
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2
0%
4
0%
8
0%
6
Explanation
Considering the equation of the parabola
y
2
=
8
x
The axis of symmetry is the positive
x
−
axis.
By, differentiating with respect to
x
, we get
2
y
.
y
′
=
8
y
′
=
4
y
Hence,
y
′
(
2
,
4
)
=
1
Thus slope of the tangent at the point of contact
P
(
2
,
4
)
is
1
while that of normal is
−
1
.
Hence, equation of the tangent will be
y
−
4
x
−
2
=
1
−
x
+
y
=
2
...(i)
Hence, the tangent meets the x axis at
(
−
2
,
0
)
Therefore,
A
=
(
−
2
,
0
)
.
Equation of normal will be
y
−
4
x
−
2
=
−
1
x
+
y
=
6
...(ii)
Hence the normal meets the x axis at
(
6
,
0
)
.
Thus,
B
=
(
6
,
0
)
Therefore the three points through which the circle passes are
(
−
2
,
0
)
,
(
6
,
0
)
,
(
2
,
4
)
Now let the equation of the circle be
(
x
−
h
)
2
+
(
y
−
k
)
2
=
r
2
Therefore
(
−
2
−
h
)
2
+
k
2
=
r
2
→
(
2
+
h
)
2
+
k
2
=
r
2
(
6
−
h
)
2
+
k
2
=
r
2
Subtracting {ii} from {i}, we get
2
h
−
4
=
0
h
=
2
Therefore the equation of the circle reduces to
(
x
−
2
)
2
+
(
y
−
k
)
2
=
r
2
Now
(
2
−
2
)
2
+
(
4
−
k
)
2
=
r
2
→
(
4
−
k
)
2
=
r
2
(
6
−
2
)
2
+
k
2
=
r
2
→
16
+
k
2
=
r
2
Subtracting i from ii, we get
16
+
k
2
−
(
k
−
4
)
2
=
0
16
+
(
2
k
−
4
)
(
4
)
=
0
4
+
2
k
−
4
=
0
k
=
0
Thus the centre of the circle lies at
C
=
(
2
,
0
)
.
Hence the radius of the circle is
C
A
=
C
P
=
C
B
Now
C
A
=
√
(
2
−
(
−
2
)
)
2
+
0
2
=
2
−
(
−
2
)
=
4
=
r
Hence, the diameter of the circle is
8
units.
If the curves
y
=
x
2
−
1
,
y
=
8
x
−
x
2
−
9
touch each other at (2, 3) then equation of the common tangent is
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4
x
−
y
=
5
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4
x
+
y
=
5
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x
−
4
y
=
5
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x
+
4
y
=
14
Explanation
y
1
1
=
2
x
y
1
2
=
8
−
2
x
at the tangent point slope will be
some.
2
x
=
8
−
2
x
4
x
=
8
x
=
2
y
=
3
y
1
=
4
(
y
−
3
)
=
4
(
x
−
2
)
y
=
4
x
−
5
The point on the hyperbola
y
=
x
−
1
x
+
1
at which the tangents are parallel to
y
=
2
x
+
1
are
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(
0
,
−
1
)
only
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(
−
2
,
3
)
only
0%
(
0
,
−
1
)
,
(
−
2
,
3
)
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(
−
2
,
3
)
,
(
5
,
4
)
Explanation
y
′
=
(
x
+
1
)
−
(
x
−
1
)
(
x
+
1
)
2
y
′
=
2
(
∴
parallel to
y =2x+1)
\dfrac{2}{(x+1)^{2}}=2
(x+1)^{2}=1
x=0
or
x=-2
Using the equation of hyperbola, when
x=0
,
y=-1
.
when
x=-2
,
y=3
Assertion(A): The tangent to the curve
y=x^{3}-x^{2}-x+2
at (1, 1) is parallel to the x axis.
Reason(R): The slope of the tangent to the above curve at (1, 1) is zero.
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Both A and R are true R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
y^{1}=3x^{2}-2x-1
y^{1}|_{x=1}=0=m
m=0
\therefore
the curve is parallel to x-axis.
Assertion A: The curves
x^{2}=y,\ x^{2}=-y
touch each other at (0, 0).
Reason R: The slopes of the tangents at (0, 0) for both the curves are equal.
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Both A and R are true R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
y'=2x
y'|_{x=0}=0=m_{1}
y'=-2x
y'|_{x=0}=0=m_{2}
m_{1}=m_{2}
\therefore
both the curve touches each other
Observe the following statements for the curve
y=2.e^{\frac{-x}{3}}
I : The slope of the tangent to the curve where it meets y-axis is
\displaystyle \frac{-2}{3}
II:The equation of normal to the curve where it meets y-axis is
3x+2y+4=0
.
Which of the above statement is correct
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only I
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only II
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both I and II
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neither I nor II
Explanation
y'=\cfrac{-2}{3}e^{-x}
y'|_{x=0}=\frac{-2}{3}
y=2,\ x=0
(y-2)=\dfrac{3}{2}(x-0)
2y-3x-4=0
only 2 is true
Match the points on the curve
2y^{2}=x+1
with the slope of normals at those points and choose
the correct answer.
Point
Slope of normal
I :
(7, 2)
a){-4\sqrt{2}}
II:
(0, \displaystyle \frac{1}{\sqrt{2}})
b) -8
III :
(1, 1)
c) -4
IV:
(3, \sqrt{2})
d){-2\sqrt{2}}
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i-b,ii- d,iii- c,iv- a
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i-b,ii- a,iii- d,iv- c
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i-b,ii- c,iii- d,iv- a
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i-b,ii- d,iii- a,iv- c
Explanation
\displaystyle y'=\frac{1}{4y}=m
Slope of normal
m'=-4y
For given points slope is
(1)
m'_{1}=-4\times 2 ==-8
(2)
m'_{2}=-4\times \dfrac{1}{\sqrt 2}=-2\sqrt{2}
(3)
m'_{3}=-4\times 1=-4
(4)
m'_{4}=-4\times {\sqrt 2}=-4\sqrt{2}
Option A is correct answer
The equation of the tangent to the curve
y=e^{-|x|}
at the point where the curve cuts the line
x=1
is
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x+y=e
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e(x+y)=1
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y+ ex=1
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x+ey=2
Explanation
y=e^{-x}
y'=\dfrac{-1e^{-x}}{e^{-x}}
y'=-e^{-1} \ \ at \ \ x=1
y=e^{-1} \ \ at \ \ x=1
(y-e^{-1})=-e^{-1}(x-1)
x+ey=2
The equation of the normal at
x
=
2a
for the curve
\displaystyle y=\frac{8a^{3}}{4a^{2}+x^{2}}
is
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2x-y=3a
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2x+y=2a
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x+2y=6a
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x+y=a
Explanation
Equation of normal at
x=2a
for
y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ x }^{ 2 } }
\Rightarrow x=2a
\Rightarrow y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ (2a) }^{ 2 } } =a
y'=\dfrac { dy }{ dx } =\dfrac { -8{ a }^{ 3 }(2x) }{ { \left( 4{ a }^{ 2 }+{ x }^{ 2 } \right) }^{ 2 } } ,x=2a\Rightarrow y'=\dfrac { -8{ a }^{ 3 }\times 2\times 2a }{ { \left( 4{ a }^{ 2 }+4{ a }^{ 2 } \right) }^{ 2 } }
=\dfrac { -8{ a }^{ 3 }\times 4a }{ 64{ a }^{ 4 } }
=-\dfrac { 1 }{ 2 }
Slope of normal
=\dfrac { -1 }{ y' } =2
\Rightarrow y-a=2(x-2a)
2x-4a=y-a
\therefore 2x-y=3a
is a equation of normal.
The distance of the origin from the normal to the curve
y=e^{2x}+x^{2}
at
x=0
is
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\displaystyle \frac{2}{5}
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\displaystyle \frac{2}{\sqrt{5}}
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2\sqrt{5}
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5\sqrt{2}
Explanation
y'=2e^{2x}+2x
y=1
at
x=0
y'=2
m
of normal
=\dfrac{-1}{2}
(y-1)=\dfrac{-1}{2}(x-0)
x+2y-2=0
distance from origin
d=\left |\dfrac{x_{1}+2y_{1}-2}{\sqrt{1^{2}+2^{2}}} \right | (x_{1}y_{1})=(0,0)
d=\dfrac{2}{\sqrt{5}}
Equation of the tangent line to
y=be^{\frac{-x}{a}}
where it crosses y-axis is
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ax+ by=1
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\displaystyle \frac{x}{a}+\frac{y}{b}=1
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\displaystyle \frac{x}{b}+\frac{y}{a}=1
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ax- by=1
Explanation
y=b e^{\frac{-x}{a}} at x=0
y=b
y'=\dfrac{-b e^{-x}}{a}
y'=\dfrac{-b }{a} \ \ (at \ \ x=0)
(y-b )=\dfrac{-b }{a}(x-0)
\dfrac{x}{a}+\dfrac{y}{b }=1
The portion of the tangent to xy
=a^{2}
at any point on it between the axes is
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Trisected at that point
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bisected at that point
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constant
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with ratio
1 : 4
at the point
Explanation
Let xy tangent at
(a,a)
xy'+y=0
y^{1}=\cfrac{-y}{x}
y'=-1
(y-a)=-1(x-a)
Bisected by that point.
y' \ \ at \ \ B =\dfrac{-y}{x}
=\dfrac{-ay^{2}}{ty^{2}}
(y-\dfrac{a^{2}}{t})=\dfrac{-a^{2}}{t^{2}}(x-t)
x=0 , y=\dfrac{2a^{2}}{t}
y=0, x=2t
(t,\dfrac{a^{2}}{t})
to the mid point of (2+10) and
(0,\dfrac{2a^{2}}{t})
.
H.P.
Area of the triangle formed by the normal to the curve
x=e^{\sin y}
at (1, 0) with the coordinate axes is
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\displaystyle \frac{1}{4}
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\displaystyle \frac{1}{2}
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\displaystyle \frac{3}{4}
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1
The arrangment of the slopes of the normals to the curve
y=e^{\log(cosx)}
in the ascending order at the points given below.
A) \displaystyle x=\frac{\pi}{6}, B) \displaystyle x=\frac{7\pi}{4}, C)x=\frac{11\pi}{6}, D)x=\frac{\pi}{3}
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C, B, D, A
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B, C, A, D
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A, D, C, B
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D, A, C, B
Explanation
We have,
y = e^{\log \cos x} =(\cos x)^{\log e}[\because a^{\log_bc} =c^{\log_ba}] =\cos x
\Rightarrow \dfrac{dy}{dx} =-\sin x=m
(say)
Thus slope of normal to the given curve at any point is,
m'=- \dfrac{1}{m}=\dfrac{1}{\sin x}
Now,
m'_{\frac{\pi}{6}}=2, m'_{\frac{7\pi}{4}}=-\sqrt{2}, m'_{\frac{\pi}{6}}=-2,m'_{\frac{\pi}{6}}=\dfrac{\sqrt{3}}{2}
Clearly ascending order is,
C,B,D,A
Note: Given function is not defined where
\cos x<0
lf the normal at the point
p(\theta)
of the curve
x^{\tfrac{2}{3}}+y^{\tfrac{2}{3}}=a^{\tfrac{2}{3}}
passes through the origin then
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\theta =\dfrac {\pi}3
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\theta =\dfrac {\pi}6
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\theta =\dfrac {\pi}4
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\theta =\dfrac {\pi}2
Explanation
x=a\ \cos^{3}\theta
y=a\ \sin^{3}\theta
y^{1}=\dfrac{a\ 3\ \sin^{2}\theta\ \cos\ \theta}{a\ 3\ \cos^{2}\theta-\cos\ \theta}
\dfrac{-\sin\ \theta}{\cos\ \theta}(y-a\ \sin^{3}\theta)=\dfrac{\cos\ \theta}{\sin\ \theta}(x-a\ \cos^{3}\ \theta)
(0, 0)
satisfies the eq
^{n}
\sin^{4}\ \theta=\cos^{4}\ \theta
\theta=\dfrac {\pi}4
The equations of the tangents at the origin to the curve
y^{2}=x^{2}(1+x)
are
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y=\pm x
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y=\pm 2x
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y=\pm 3x
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x=\pm 2y
Explanation
y=+_-\sqrt{x^{2}+x^{3}}
y'=+_-\dfrac{2x+3x^{2}}{2\sqrt{x^{2}+x^{3}}}
y'=+_-\dfrac{2+3x}{2\sqrt{1+x}}
y'|_{x=0}=+_-1
y=+_-x
The equation of the common normal at the point of contact of the curves
x^{2}=y
and
x^{2}+y^{2}-8y=0
Report Question
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x=y
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x=0
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y=0
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x+y=0
Explanation
y^{2}+y-8y=0
y=0.\ y=7
x=0,\ x=\sqrt{7}
y^{1}=2x=m_{1}
yy^{1}-4y^{1}=-x
y^{1}=\frac{x}{4-y}=m_{2}
m_{1},\ m_{2}\ at\ (0,0)
m_{1}=m_{2}=0
Slope of the normal
=\alpha
\therefore x=0
is the required line
cl. y-axis
lf the chord joining the points where
x= p,\ x =q
on the curve
y=ax^{2}+bx+c
is parallel to the tangent drawn to the curve at
(\alpha, \beta)
then
\alpha=
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2pq
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\sqrt{pq}
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\displaystyle \frac{p+q}{2}
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\displaystyle \frac{p-q}{2}
Explanation
y'=2ax+b
y'|_{x=b}=(2a\alpha +b)
2a\alpha+b=\left ( \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )
2a\alpha+b=\dfrac{ap^{2}+bp-aq^{2}-bq}{(p-q)}
=\dfrac{a(p-q)(p+q)+b(p-a)}{(p-q)}
2a\alpha+b=a(p+q)+b
\alpha=\left ( \dfrac{p+q}{2} \right )
The arrangement of the following curves in the ascending order of slopes of their tangents at the given points.
A) \displaystyle y=\frac{1}{1+x^{2}}
at
x=0
B) y=2e^{\frac{-x}{4}},
where it cuts the y-axis
C) y= cos(x)
at
\displaystyle x=\frac{-\pi}{4}
D) y=4x^{2}
at
x=-1
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DCBA
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ACBD
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ABCD
0%
DBAC
Explanation
(A)
y^{1}=-\frac{2x}{(1+x^{2})}
y^{1}|_{x=0}=0
(B)
y^{1}=-\frac{2x}{(4}
y^{1}|_{x=0}=-\frac{1}{2}
(C)
y^{1}=-sin\ x
y^{1}|_{x=^{-pi}/_4}=\frac{1}{\sqrt{2}}
(D)
y^{1}=8x
y^{1}|_{x=-1}=-8
(C)>(A)>(B)>(D)
Observe the following lists for the curve
y=6+x-x^{2}
with the slopes of tangents at the given points; I, II, III, IV
Point
Tangent slope
I:
(1, 6)
a)
3
II:
(2, 4)
b)
5
III:
(-1, 4)
c)
-1
IV:
(-2, 0)
d)
-3
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a ,b, c ,d
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b, c, d ,a
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c, d ,b ,a
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c, d ,a ,b
Explanation
y^{'}=\ \ \ 1-2x
(A)
m_{1}=-1
(B)
m_{2}=-3
(C)
m_{3}=3
(D)
m_{4}=5
lf the tangent to the curve
2y^{3}=ax^{2}+x^{3}
at the point
(a,\ a)
cuts off intercepts
\alpha
and
\beta
on the coordinate axes such that
\alpha^{2}+\beta^{2}=61
, then
a
is equal to
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\pm 30
0%
\pm 5
0%
\pm 6
0%
\pm 61
Explanation
2\times 3\times y^{2}y'=2ax+3x^{2}
y' = \dfrac{2ax+3x^2}{6y^2}
m = \left.y'\right|_{(a,a)}=\dfrac{5}{6}
Equation of the tangent at
(a,a)
with slope,
m = \dfrac56
(y-a)=\dfrac{5}{6}(x-a)
Intersection points with coordinate axes are,
\left(0,\dfrac{a}{6}\right)
and
\left(-\dfrac{a}{5},0\right)
So,
\alpha = \dfrac a6
and
\beta = -\dfrac a5
Given that,
{\alpha}^2 + {\beta}^2 = 61
\Rightarrow \dfrac{a^2}{36} + \dfrac{a^2}{25} = 61
\Rightarrow a^2 = {30}^2
\Rightarrow a = \pm 30
Assertion(A): If the tangent at any point
P
on the curve
xy = a^{2}
meets the axes at
A
and
B
then
AP : PB = 1 : 1
Reason(R): The tangent at
P(x, y)
on the curve
X^{m}.Y^{n}=a^{m+n}
meets the axes at
A
and
B
. Then the ratio of
P
divides
\overline{AB}
is
n : m
.
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Both A and R are true R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
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A is true but R is false
0%
A is false but R is true
Explanation
y=\dfrac{a^{2}}{x}
y'=-\dfrac{a^{2}}{x^{2}}
y'=-\dfrac{a^{2}}{t^{2}}
\left ( y-\dfrac{a^{2}}{t} \right )=\dfrac{-a^{2}}{t^{2}}(n-t)
B=(2+10)
A=\left ( 0,\ \dfrac{2a^{2}}{t} \right )
\dfrac{AP}{BP}=1
The point on the curve
\sqrt{x}+\sqrt{y}=2a^{2}
, where the tangent is equally inclined to the axes, is
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\left ( a^{4}, a^{4} \right )
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\left ( 0, 4a^{4} \right )
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\left ( 4a^{4}, 0 \right )
0%
none of these
Explanation
y=x
is the equation of line as it is equally inclined to axes
So let point
(k,k)
is the point of tangency
Therefore,
\sqrt { k } +\sqrt { k } =2{ a }^{ 2 }\\ k={ a }^{ 4 }
Hence point is
\left( { a }^{ 4 },{ a }^{ 4 } \right)
Match List-I with List-II and select the correct answer using the code given below. A B C D
List-I
List-II
a) Equation of tangent to the curve
y=be^{-x/a}
at
x=0
1)
x-2y=2
b) Equation of tangent to the curve
y=x^{2}+1
at
(1, 2)
2)
y = 2x
c) Equation of normal to the curve
y=2x-x^{2}
at
(2, 0)
3)
x-y =\pi
d) Equation of normal to the curve
y= \sin x
at
x=\pi
4)
\displaystyle {\frac{x}{a}+\frac{y}{b}=1}
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4 1 2 3
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4 2 1 3
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1 2 3 4
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1 4 3 2
Explanation
(A)
\displaystyle y'=\frac{-b}{a}e^{-x/a}
\displaystyle m=-\frac{b}{a}
\displaystyle (y-b)=\frac{-b}{a}(x-0)
\displaystyle \frac{x}{a}+\frac{y}{b}=1
(B)
y'=2x
y'|_{x=1}=2
(y-2)=2(x-1)
y=2x
(C)
y'=2-2x
y'|_{x=2}=-2
Slope of normal
\displaystyle =\frac{1}{2}
\displaystyle (y-0)=\frac{1}{2}(x-2)
2y=x-2
(D)
y'=\cos x
y'|_{x={\pi}}=-1
Slope of normal =1
(y-0)=1(x-\pi)
x-y=\pi
Area of the triangle formed by the tangent, normal at
(1, 1)
on the curve
\sqrt{x}+\sqrt{y}=2
and the y axis is (in sq. units)
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1
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2
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\displaystyle \frac{1}{2}
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4
Explanation
\dfrac{1}{\sqrt{x}}+\dfrac{y^{1}}{\sqrt{y}}=0
y'=-\sqrt{\dfrac{y}{x}}
y'=-1\ \ at\ (n=1,\ y=1)
(y-1)=-1(x-1)
y=2
(y-1)=1(x-1)
y=0
Area=\dfrac{1}{2}\times 2 \times 1
=1
A curve with equation of the form
y=ax^{4}+bx^{3}+c+cx+d
has zero gradient at the point (0,1) and also touches the x-axis at the point (-1, 0) then the values of the x for which the curve has a negative gradient are :
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x > -1
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x < 1
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x < -1
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-1\leq x\leq 1
Explanation
\dfrac{dy}{dx}|_{x=0}=4ax^{3}+3bx^{2}+c=0
\therefore c=0
\dfrac{dy}{dx}<0
4ax^{3}+3bx^{2}<0
x^{2}(4ax+3b)<0
x<-\dfrac{3b}{4a}
Assertion (A): The points on the curve
y=x^{3}-3x
at which the tangent is parallel to
x
-axis are
(1, -2)
and
(-1, 2).
Reason (R): The tangent at
(x_{1}, y_{1})
on the curve
y=f(x)
is vertical then
\displaystyle \frac{dy}{dx}
at
(x_{1}, y_{1})
is not defined.
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Both A and R are true and R is the correct explanation for A
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Both A and R are true but R is not the correct explanation for A
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A is true but R is false
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A is false but R is true
Explanation
y'=3x^{2}-3
y'=0
\therefore
tangent is parallel to x-axis
x^{2}=1
x =^+_-1
y=-2
y=2
y tangent is vertical then
\dfrac{dy}{dx}
is not defined.
lf the tangent at any point on the curve
x^{4}+y^{4}=c^{4}
cuts off intercepts
a
and
b
on the coordinate axes, the value of
a^{-4/3}+b^{-4/3}
is
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c^{-4/3}
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c^{-1/2}
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c^{1/2}
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c^{{4}/{3}}
Explanation
Given equation of curve is
x^4+y^4=c^4
\dfrac{dy}{dx}=-\dfrac{x^3}{y^3}
Slope of tangent at point
(x_1,y_1)
on the curve is
-\dfrac{x_1^3}{y_1^3}
Let the equation of tangent to curve be
\dfrac{x}{a}+\dfrac{y}{b}=1
Since, this line intersects X-axis at
(a,0)
and
(0,b)
\dfrac{x}{c^4/x_1^3}+\dfrac{y}{c^4/y_1^3}=1
\Rightarrow a=\dfrac{c^4}{x_1^3},b=\dfrac{c^4}{y_1^3}
\Rightarrow a^{-4/3}=\dfrac{c^{-16/3}}{x_1^{-4}}, b^{-4/3}=\dfrac{c^{-16/3}}{y_1^{-4}}
\Rightarrow a^{-4/3}+b^{-4/3}=c^{-16/3}(x_1^4+y_1^4)
\Rightarrow a^{-4/3}+b^{-4/3}=c^{-4/3}
I. lf the curve
y=x^{2}+ bx +c
touches the straight line
y=x
at the point
(1, 1)
then
b
and
c
are given by
1, 1.
II. lf the line
Px+ my +n=0
is a normal to the curve
xy=1
, then
P> 0,\ m <0
.
Which of the above statements is correct
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only I
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only II
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both I and II
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Neither I nor II
Explanation
(I)
1+b+c=1
b+c=0
y'=2x+b
y'=1
2+b=1
(b=-1)
c=1
(II)
y'=-\dfrac{1}{x^{2}}
m=-\dfrac{1}{x^{2}}
-\dfrac{P}{m}=\dfrac{x^{2}}{x}
-P=x^{2}m
p and m will be of opposite num.
At origin the curve
y^{2}=x^{3}+x
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Touches the x-axis
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Touches the y-axis
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Bisects the angle between the axes
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touches both the axes
Explanation
\dfrac{dy}{dx}=\dfrac{3x^{2}+1}{2y}
At origin
x=0,y=0
\tan\ \theta=\infty
\theta =\dfrac{\pi}{2}
\therefore
Curve touches the y-axis
Observe the following statements
I: If
p
and
q
are the lengths of perpendiculars from the origin on the tangent and normal at any point on the curve
x^{\frac{2}{3}}+y^{\frac{2}{3}}=1
then
4p^{2}+q^{2}=1
.
II: If the tangent at any point
P
on the curve
x^{3}.y^{2}=a^{5}
cuts the coordinate axes at
A
and
B
then
AP : PB = 3 : 2
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only I
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only II
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both I and II
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neither I nor II
Explanation
x=a\ \cos^{3}\theta
y=a\ \sin^{3}\theta
y'=-\tan\ \theta
(y-sin^{3}\theta)=-\tan\ \theta(x-\cos^{3}\theta)
y+n\ \tan\ \theta-\tan\ \theta\ \cos^{3}\theta-\sin^{3}\theta=0
Perpendicular from origin
\left | \dfrac{\tan\ \theta\ \cos^{3}\theta+\sin^{3}\theta}{\sqrt{1+\tan^{2}\theta}} \right |=p
p_{2}\ \ \sin\ \theta\ \cos\ \theta
p=\dfrac{\sin^{2}\theta}{2}
(y-\sin 3\theta)=\frac{1}{\tan\ \theta}(x-\cos3\theta)
y\ \tan\ \theta-\sin3\theta\ \tan\ \theta-x+\cos3\ \theta=0
Perpendicular frm the origin
\left | \dfrac{\cos3\theta-\sin3\theta\ \tan\ \theta}{\sqrt{1+\tan^{2}\theta}} \right |=q
q=\cos2\theta
4p^{2}+q^{2}=1
Observe the following statements for the curve
x = at^{3}
,
y = at^{4}
at
t = 1
.
I : The equation of the tangent to the curve is
4x-3y- a = 0
II : The equation of the normal to the curve is
3x +4y- 7a = 0
III: Angle between tangent and normal at any point on the curve is
\displaystyle \frac{\pi}{2}
Which of the above statements are correct.
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I and II
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II and III
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I and III
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I, II, III
Explanation
\dfrac{dy}{dx}=\dfrac{4at^{3}}{3at^{2}}
\dfrac{dy}{dx}=\dfrac{4t}{3}
\dfrac{dy}{dx}|_{t=1}=\dfrac{4}{3}
(y-a)=\dfrac{4}{3}(x-a)
(tangent)
4x-3y-a=0
(y-a)=-\dfrac{3}{4}(x-a)
(Normal)
4y+3x-7a=0
Tangent and normal are always at an angle of
{\pi}/{2}
to each other
Assertion (A): The normal to the curve
ay^{2}=x^{3}(a\neq 0, x\neq 0)
at a point
(x, y)
on it makes equal intercepts on the axes, then
\displaystyle x=\frac{4a}{9}
.
Reason (R): The normal at
(x_{1}, y_{1})
on the curve
y=f(x)
makes equal intercepts on the coordinate axes, then
\left.\displaystyle \frac{dy} {dx}\right|_{(x_{1},y_{1})}=1
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Both (A) and (R) are true and (R) is the correct explanation for (A).
0%
Both (A) and (R) are true but (R) is not the correct explanation for (A).
0%
(A) is true but (R) is false.
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(A) is false but (R) is true.
Explanation
ay^{2}=x^{3}
2ay\dfrac{dy}{dx}=3x^{2}\Rightarrow\dfrac{dy}{dx}=\dfrac{3x^{2}}{2ay}
Now, slope of normal
= -\dfrac{dx}{dy}=-\dfrac{2ay}{3x^{2}}=-\dfrac{2a\times x^{\frac{3}{2}}}{3x^{2}\sqrt{a}}=-\dfrac{2\sqrt{a}}{3\sqrt{x}}
For the normal to make equal intercepts on the axes, the slope has to be
\pm1
.
Hence,
\dfrac{-2\sqrt{a}}{3\sqrt{x}}=\pm1
On squaring both sides,
\dfrac{4a}{9x}=1\Rightarrow x=\dfrac{4a}{9}
.
Given the curves
y=f(x)
passing through the point
(0, 1)
and
y=\displaystyle \int_{-\infty}^{x}{f(t)}
passing through the point
\left( 0, \dfrac{1}{2}\right).
The tangents drawn to both the curves at the points with equal abscissae intersect on the
x
- axis. Then the curve
y=f(x)
is
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f(x)=x^2+x+1
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f(x)=\dfrac{x^2}{e^x}
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f(x)=e^{2x}
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f(x)=x-e^x
The point of intersection of the tangents drawn to the curve
x^{2}y=1-y
at the points where it is met by the curve
xy=1-y
is given by
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(0, -1)
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(1, 1)
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(0, 1)
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(1, \infty)
Explanation
x^{2}y=xy
y=1,\ x=0
y=^{1}/_{2},\ x=1
2x\dfrac{dx}{dy}=-\dfrac{-1}{y^{2}}
\dfrac{dy}{dx}=-2xy^{2}
(x=0)
\dfrac{dy}{dx}=0
(y-1)=x-0
y=1
------- (1)
\left ( y-\dfrac{1}{2} \right )=\dfrac{-1}{2}(x-1)
-----(2)
x=0
(By (1) & (2))
(0, 1)
The number of tangents to the curve
x^{3/2}+y^{3/2}=a^{3/2}
, where the tangents are equally inclined to the axes, is
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2
0%
1
0%
0
0%
4
Explanation
\dfrac{dy}{dx}=-1
(where x is equally inclined)
\dfrac{3}{2}x^{^{1}/_{2}}+^{3}/_{2}y^{^{1}/_{2}}\dfrac{dy}{dx}=0
\dfrac{dy}{dx}=-\sqrt{\dfrac{x}{y}}=1
-\sqrt{\dfrac{x}{y}}=-1
\sqrt{x}=\sqrt{y}
x=y
x^{^{3}/_{2}}+x^{^{3}/_{2}}=a^{^{3}/_{2}}
2x^{^{3}/_{2}}=a^{^{3}/_{2}}
x=\dfrac{a}{2^{^{2}/_{3}}}
y=\dfrac{a}{2^{^{2}/_{3}}}
\therefore
only one tangent.
The points on the hyperbola
x^{2}-y^{2}=2
closest to the point (0, 1) are
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(\displaystyle \pm\frac{3}{2},\frac{1}{2})
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(\displaystyle \frac{1}{2},\pm\frac{3}{2})
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(\displaystyle \frac{1}{2},\frac{1}{2})
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(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})
Explanation
Let point be
(\sqrt 2 sec\theta,\sqrt 2 tan\theta)
closest to (0, 1)
So distance
=\sqrt {2 sec^2\theta+(\sqrt 2 tan\theta-1)^2}
=\sqrt {2 sec^2\theta+2tan^2\theta-2\sqrt 2 tan\theta+1}
=\sqrt {4 tan^2\theta-2\sqrt 2 tan\theta+3}
=2\sqrt {tan^2\theta-\frac {tan\theta}{\sqrt 2}+\frac {3}{4}}
=2\sqrt {(tan\theta-\frac {1}{2\sqrt 2})^2+\frac {3}{4}-\frac {1}{8}}
So distance is min when
tan\theta=\frac {1}{2\sqrt 2}
sec\theta=\pm \frac {3}{2\sqrt 2}
lf the tangent to the curve
f(x)=x^{2}
at any point
(c, f(c))
is parallel to the line joining points
(a, f(a))
and
(b,f(b))
on the curvel then
a,\ c,\ b
are in
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AP
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GP
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HP
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AGP
Explanation
\partial^{'}M=2s
at
(c,\ \partial(c))=2c
\dfrac{\partial(a)-\partial(L)}{a-b}=2c
\dfrac{a^{2}=b^{2}}{a-b}=2c
a+b=2c
\therefore\ a,c,b
are in AP
A
curve is given by the equations
x=\sec^{2}\theta, y= \cot\theta
. If the tangent at
P
where
\displaystyle \theta=\frac{\pi}{4}
meets the curve again at
Q
, then length
PQ
, is
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\displaystyle \frac{\sqrt{5}}{2}
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\displaystyle \frac{3\sqrt{5}}{2}
0%
10
0%
20
Explanation
y = \cot\theta
\Rightarrow \dfrac{dy}{d\theta}=-\text{cosec}^{2}\theta
x = \sec^2\theta
\Rightarrow \dfrac{dx}{d\theta}=2\sec\theta\cdot \sec\theta\ \tan\theta
\left.\dfrac{dy}{dx}\right|_{\theta = \frac{\pi}4}=-\dfrac{\text{cosec}^{2}\theta}{2\ \sec^{2}\theta\ \tan\theta}
=\dfrac{-2}{2\times 2\times 1}=-\dfrac{1}{2}
y=\dfrac{1}{\tan\theta}
x=1+\tan^{2}\theta
x=1+\dfrac{1}{y^{2}}
\Rightarrow y^{2}x=y^{2}+1
\Rightarrow x=2,\ y=1
Now, at
(x,y) = (2,1)
, equation of a line with slope
-\dfrac12
, is
(y-1)=-\dfrac{1}{2}(x-2)
\Rightarrow 2y+x=4
\Rightarrow y^{2}(4-2y)=y^{2}+1
\Rightarrow 4y^{2}-2y^{3}=y^{2}+1
\Rightarrow 2y^{3}-3y^{2}+1=0
\Rightarrow y=1,\ -\dfrac{1}{2}
\Rightarrow x=2,\ 5
PQ = \sqrt{(2-5)^{2}+\left (1+\dfrac{1}{2} \right )^{2}}
=\left ( 3^{2}+\dfrac{3^{2}}{4} \right )^{0.5}
=\dfrac{3\sqrt{5}}{2}
Hence, option B.
lf the parametric equation of a curve given by
x=e^{t}\cos t,\ y=e^{t}\sin t
, then the tangent to the curve at the point
t=\dfrac{\pi}{4}
makes with axis of
x
the angle.
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0
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\dfrac{\pi}{4}
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\dfrac{\pi}{3}
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\dfrac{\pi}{2}
Explanation
at\ t={\pi}/{4}
y=e^{t}sint
\dfrac{dy}{dl}=e^{t}(sin\ t+cos\ t)
\dfrac{dx}{dl}=d^{t}(cos\ t+sin\ t)
\dfrac{dy}{dx}=\dfrac{sin \ t+cos\ t}{cos\ t-sin\ t}
= \alpha
\therefore tan^{-1}(\infty)={\pi}/{2}
The normal to the curve
\mathrm{x}=\mathrm{a}(1 +\cos\theta ),\ \mathrm{y}=a\sin\theta
at any point
\theta
always passes through the fixed point:
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(a, 0)
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(0, a)
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(0, 0)
0%
(a, a)
Explanation
x=a\left( 1+\cos { \theta } \right) ,y=a\sin { \theta }
Differentiating w.r.t
\theta
, we get
\displaystyle \frac { dx }{ d\theta } =a\left( -\sin { \theta } \right) ,\frac { dy }{ d\theta } =a\cos { \theta }
\displaystyle \therefore \frac { dy }{ dx } =-\frac { \cos { \theta } }{ \sin { \theta } }
Therefore equation of normal at
\left( a\left( 1+\cos { \theta } \right) ,\sin { \theta } \right)
is
\displaystyle \left( y-a\sin { \theta } \right) =\frac { \sin { \theta } }{ \cos { \theta } } \left( x-a\left( 1+\cos { \theta } \right) \right)
It is clear that in the given options normal passes through the point
\left( a,0 \right)
lf the curve
y=px^{2}+qx+r
passes through the point (1, 2) and the line
y=x
touches it at the origin, then the values of
p,\ q
and
r
are
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p=1,q=-1,r=0
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p=1,q=1,r=0
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p=-1,q=1,r=0
0%
p=1,q=2,r=3
Explanation
2=p+q+r
r=0
\dfrac{dx}{dy}|_{x=0}=1
\dfrac{dy}{dx}=2px+q
q=1
2=p+1+0
p=1
For the curve
y=3\sin \theta\cos\theta, x=e^{\theta}\sin \theta, 0\leq \theta\leq\pi
; the tangent is parallel to
x
-axis when
\theta
is
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0
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\displaystyle \frac{\pi}{2}
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\displaystyle \frac{\pi}{4}
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\displaystyle \frac{\pi}{6}
Explanation
\dfrac{dy}{dx}=0
\dfrac{dy}{d\theta}=d\dfrac{3}{2}(sin2\theta)
\dfrac{dy}{d\theta}=d\dfrac{3}{4}(cos2\theta)
---- (1)
\dfrac{dy}{d\theta}=e^{\theta}sin\theta+e^{\theta}cos\theta
---- (2)
\dfrac{dx}{dy}=\dfrac{3}{4}\dfrac{cos2\theta}{e^{\theta}(sin\ \theta+ cos\ \theta)}=0
cos2\theta=0
2\theta={\pi}/{2}
\theta={\pi}/{4}
If the circle
x^{2}+y^{2}+2gx+2fy+c=0
is touched by
y=x
at
P
such that
OP=6\sqrt{2},
then the value of
c
is
Report Question
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36
0%
144
0%
72
0%
None of these
Explanation
Let the point of contact be
x_{1},y_{1}
However it lies on the line
y=x
Hence
x_{1}=y_{1}
Applying distance formula, we get
\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}
2x_{1}^2=72
x_{1}=6
...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
2x+2yy'+2g+2fy'=0
Now
y'=1
since
y'
is the slope of the line
y=x
.
By substituting, we get
x+y=-(g+f)
Now
x_{1}=y_{1}=6
Hence
12=-(g+f)
...(i)
Substituting
x=y=6
in the equation of the circle, we get
36+36+2(6)(g+f)+c=0
72+12(-12)+c=0
c=144-72
c=72
If
x+ 4y=14
is a normal to the curve
y^2=\alpha x ^3-\beta
at
(2,3)
, then the value of
\alpha+\beta
is
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9
0%
-5
0%
7
0%
-7
Explanation
Given equation of curve is
y^2=\alpha x ^3-\beta
Since, it passes through (2,3)
4=8\alpha -\beta
\displaystyle \frac{dy}{dx}=\frac{3\alpha x^2}{2y}
Slope of tangent at (2,3) is
2\alpha
Slope of normal at
\displaystyle (2,3)= -\frac{1}{2\alpha}
Given equation of normal is
x+4y=14
Slope of normal is
-\cfrac{1}{4}
\Rightarrow \alpha =2
\Rightarrow \beta=7
Hence
\alpha+\beta =9
The number of tangents to the curve
x^{3/2} + y^{3/2}= 2a^{3/2}
,
a>0
, which are equally inclined to the axes, is
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2
0%
1
0%
0
0%
4
Explanation
For a tangent to be equally inclined to the axes, the slope of the tangent should be 1.
Thus
y'=1
.
Now
x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}}
.
\dfrac{3}{2}(\sqrt{x})+\dfrac{3}{2}(\sqrt{y})=0
\sqrt{x}=-\sqrt{y}
x=y
x_{1}=y_{1}
Substituting int he above equation, we get
2x^{\frac{3}{2}}=2a^{\frac{3}{2}}
x=y=a
.
Hence the tangent touches the curve at
(a,a)
.
Thus we get only one tangent, satisfying that criteria.
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
0%
m>-1
0%
\left | m \right | \leq 1
Explanation
e^{y}=1+x^{2}
\dfrac{dy}{dx}=\dfrac{2x}{1+x^{2}}
So
|m|=|\dfrac{2x}{x^{2}+1}|
Now, since
x^{2}\ge 0
\Rightarrow x^{2}+1 \ge 1
\Rightarrow \dfrac{1}{ x^{2}+1} \le 1
\Rightarrow \dfrac{2x}{ x^{2}+1} \le 1
|m| \le 1
If the circle
x^2 + y^2 + 2gx + 2fy + c =0
is touched by y = x at P in the first quadrant, such that
OP = 6 \sqrt2
, then the value of
c
is
Report Question
0%
36
0%
144
0%
72
0%
None of these
Explanation
Let the point of contact be
x_{1},y_{1}
However it lies on the line
y=x
Hence
x_{1}=y_{1}
Applying distance formula, we get
\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}
2x_{1}^2=72
x_{1}=6
...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
2x+2yy'+2g+2fy'=0
Now
y'=1
since
y'
is the slope of the line
y=x
.
By substituting, we get
x+y=-(g+f)
Now
x_{1}=y_{1}=6
Hence
12=-(g+f)
...(i)
Substituting
x=y=6
in the equation of the circle, we get
36+36+2(6)(g+f)+c=0
72+12(-12)+c=0
c=144-72
c=72
The equations of the tangents to the curve
y = x^4
from the point (2, 0) not on the curve, are given by
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y=0
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y-1=5(x-1)
0%
\displaystyle y - \frac{{4098}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)
0%
\displaystyle y - \frac{{32}}{{243}} = \frac{{80}}{{81}}\left( {x - \frac{2}{3}} \right)
Explanation
y=4\left ( \frac{8}{3} \right )^3 (x-2)
y-0 = m(x-2)
y=m(x-2)
m(x_1 -2) = x_1^4
4x_1^4 - 8 x_1^3 = x_1^4
(
m=4x_1^3
)
3x_1^4 = 8x_1^3
x_1 =\frac{8}{3}
\therefore \left ( y -\frac{4098}{81} \right ) = \frac{2048}{27} \left ( x- \frac{8}{3} \right )
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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