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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
Report Question
0%
|
tan
θ
|
>
1
0%
|
tan
θ
|
<
1
0%
|
tan
θ
|
≥
1
0%
|
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
.
Report Question
0%
x
+
y
=
1
0%
x
−
y
+
1
=
0
0%
x
+
y
=
2
0%
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:
Report Question
0%
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
∴
|
c
|
=
√
2
The equation of the normal to the curve
y
=
(
1
+
x
)
y
+
s
i
n
−
1
(
s
i
n
2
x
)
a
t
x
=
0
is
Report Question
0%
x+y=1
0%
x-y+1=0
0%
2x+y=2
0%
2x-y+1
Explanation
Given,
y
=
(
1
+
x
)
y
+
sin
−
1
(
sin
2
x
)
when
x
=
0
y
=
1
y
+
0
=
1
∴
(
x
1
,
y
1
)
=
(
0
,
1
)
d
y
d
x
=
(
1
+
x
)
y
[
ln
(
1
+
x
)
d
y
d
x
+
y
1
+
x
]
+
2
sin
x
cos
x
√
1
−
sin
4
x
d
y
d
x
(
0
,
1
)
=
1
[
1
]
+
0
∴
d
y
d
x
(
0
,
1
)
=
1
Therefore, slope of normal is
−
1
Equation of normal,
y
−
y
1
=
−
1
d
y
d
x
(
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
[
2
c
o
s
θ
−
l
o
g
(
c
o
s
e
c
θ
+
c
o
t
θ
)
]
,
y
=
c
s
i
n
2
θ
a
t
θ
=
π
3
i
s
Report Question
0%
c
2
0%
2
c
0%
3
c
2
0%
5
c
2
Explanation
Given,
x
=
c
[
2
cos
θ
−
log
(
csc
θ
+
cot
θ
)
]
⇒
d
x
d
θ
=
c
(
−
2
sin
θ
+
csc
θ
)
y
=
c
sin
2
θ
d
y
d
θ
=
2
c
cos
2
θ
Also,
y
1
=
c
sin
2
π
3
=
c
√
3
2
Slope of tangent at
θ
=
π
3
is
m
=
d
y
d
x
θ
=
π
3
=
√
3
Length of tangent
=
|
y
1
√
1
+
m
2
m
|
=
c
Length of subtangent
=
|
y
1
m
|
=
c
2
Required sum
=
c
+
c
2
=
3
c
2
If the tangent to the curve
x
=
a
t
2
,
y
=
2
a
t
is perpendicular to
x
-axis, then its point of contact is
Report Question
0%
(
a
,
a
)
0%
(
0
,
a
)
0%
(
0
,
0
)
0%
(
a
,
0
)
Explanation
Let the required point be
(
x
1
,
y
1
)
.
It is given that points lies on the curve.
∴
x
1
=
a
t
2
and
y
1
=
2
a
t
Now,
x
=
a
t
2
and
y
=
2
a
t
Differentiating w.r.t.
t
,
we get
⇒
d
x
d
t
=
2
a
t
and
d
y
d
t
=
2
a
⇒
d
y
d
x
=
d
y
d
t
d
x
d
t
⇒
d
y
d
x
=
2
a
2
a
t
⇒
d
y
d
x
=
1
t
⇒
d
y
d
x
=
2
a
y
[ Since,
y
=
2
a
t
]
Slope of the tangent
=
(
d
y
d
x
)
(
x
1
,
y
1
)
=
2
a
y
1
It is given that the tangent is perpendicular to the
y
−
a
x
i
s
.
It means theta it is parallel to the
x
−
a
x
i
s
.
∴
Slope of the tangent
=
Slope of the
x
−
a
x
i
s
.
⇒
2
a
y
1
=
0
⇒
a
=
0
Now,
x
1
=
a
t
2
=
(
0
)
t
2
=
0
y
1
=
2
a
t
=
2
(
0
)
t
=
0
∴
(
x
1
,
y
1
)
=
(
0
,
0
)
The slope of the tangent to the curve
x
=
t
2
+
3
t
−
8
,
y
=
2
t
2
−
2
t
−
5
at point
(
2
,
−
1
)
is
Report Question
0%
22
7
0%
6
7
0%
−
6
0%
7
6
Explanation
x
=
t
2
+
3
t
−
8
........
(
1
)
y
=
2
t
2
−
2
t
−
5
..........
(
2
)
Differentiate
(
1
)
, we get
d
x
d
t
=
2
t
+
3
Differentiate
(
2
)
, we get
d
y
d
x
=
4
t
−
2
m
=
d
y
d
x
=
4
t
−
2
2
t
+
3
Given point is
(
2
,
−
1
)
Put the point in original x and y, we get
⇒
x
=
t
2
+
3
t
−
8
2
=
t
2
+
3
t
−
8
t
2
+
3
t
−
10
=
0
(
t
−
2
)
(
t
+
5
)
=
0
t
=
2
,
t
=
−
5
⇒
y
=
2
t
2
−
2
t
−
5
(
−
1
)
=
2
t
2
−
2
t
−
5
2
t
2
−
2
t
−
4
=
0
(
t
+
1
)
(
t
−
2
)
=
0
t
=
−
1
,
t
=
2
Since,
t
=
2
common in both parts, so we take
d
y
d
x
=
4
t
−
2
2
t
−
3
at
t
=
2
At
t
=
2
d
y
d
x
=
4
(
2
)
−
2
2
(
2
)
−
3
=
8
−
2
4
+
3
=
6
7
m
=
d
y
d
x
=
6
7
.
The equation of tangent at those points where the curve
y
=
x
2
−
3
x
+
2
meets
x
-axis are
Report Question
0%
x
−
y
+
2
=
0
,
x
−
y
−
1
=
0
0%
x
+
y
−
1
=
0
,
x
−
y
−
2
=
0
0%
x
−
y
−
1
=
0
,
x
−
y
=
0
0%
x
−
y
=
0
,
x
+
y
=
0
Explanation
solution:
y
=
x
2
−
3
x
+
2
Let the tangent meet the x-axis at point
(
x
,
0
)
⇒
d
y
d
x
=
2
x
−
3
The tangent passes through point
(
x
,
0
)
∴
0
=
x
2
−
3
x
+
2
=
0
⇒
(
x
−
2
)
(
x
−
1
)
=
0
⇒
x
=
2
,
x
=
1
Case I: when
x
=
2
Slope of tangent,
d
y
d
x
|
(
2
,
0
)
=
2
(
2
)
−
3
=
4
−
3
=
1
∴
(
x
1
,
y
1
)
=
(
2
,
0
)
Equation of tangent
y
−
y
1
=
m
(
x
−
x
1
)
⇒
y
−
0
=
1
(
x
−
2
)
⇒
x
−
y
−
2
=
0
Case II: When
x
=
1
Slope of tangent
d
y
d
x
|
(
2
,
0
)
=
2
−
3
=
−
1
∴
(
x
1
,
y
1
)
=
(
1
,
0
)
Equation of tangent
y
−
y
1
=
m
(
x
−
x
1
)
y
−
0
=
−
1
(
x
−
1
)
⇒
x
+
y
−
1
=
0
.
The equation of the normal to the curve
y
=
x
+
sin
x
cos
x
at
x
=
π
2
is
Report Question
0%
x
=
2
0%
x
=
π
0%
x
+
π
=
0
0%
2
x
=
π
Explanation
y
=
x
+
sin
x
c
o
s
x
Differentiating both sides w.r.t.
x
,
⇒
d
y
d
x
=
1
+
cos
2
x
−
sin
2
x
Slope of the tangent
(
m
)
=
(
d
y
d
x
)
x
=
π
2
=
1
+
cos
2
π
2
−
sin
2
π
2
=
1
−
1
=
0
Slope of the normal
=
−
1
m
=
−
1
0
When,
x
=
π
2
,
y
=
π
2
+
cos
π
2
sin
π
2
∴
y
=
π
2
Now,
(
x
1
,
y
1
)
=
(
π
2
,
π
2
)
∴
Equation of the normal.
⇒
y
−
y
1
=
−
1
m
(
x
−
x
1
)
⇒
y
−
π
2
=
−
1
0
(
x
−
π
2
)
⇒
x
=
π
2
⇒
2
x
=
π
The point on the curve
y
=
12
x
−
x
2
, where the slope of the tangent is zero will be
Report Question
0%
(
0
,
0
)
0%
(
2
,
16
)
0%
(
3
,
9
)
0%
(
6
,
36
)
Explanation
Let the point be
P
(
x
,
y
)
y
=
12
x
−
x
2
d
y
d
x
=
12
−
2
x
(
d
y
d
x
)
(
x
1
,
y
1
)
=
12
−
2
x
1
since slope of tangent is zero
so
(
d
y
d
x
)
x
1
,
y
1
=
0
12
−
2
x
1
=
0
2
x
1
=
12
x
1
=
6
Also curve passing through tangent
y
1
=
12
x
1
−
x
2
1
y
1
=
12
×
6
−
36
y
1
=
72
−
36
y
1
=
36
The points are
(
6
,
36
)
.
The slope of the tangent to the curve
x
=
3
t
2
+
1
,
y
=
t
3
−
1
at
x
=
1
is
Report Question
0%
1
2
0%
0
0%
−
2
0%
∞
Explanation
Given curves are,
x
=
3
t
2
+
1
---- ( 1 )
y
=
t
3
−
1
---- ( 2 )
Substituting
x
=
1
in ( 1 ) we get,
⇒
3
t
2
+
1
=
1
⇒
3
t
2
=
0
⇒
t
=
0
Differentiate ( 1 ) w.r.t.
t
,
we get
⇒
d
x
d
t
=
6
t
Differentiate ( 2 ) w.r.t. $$t,$ we get
⇒
d
y
d
t
=
3
t
2
⇒
d
y
d
x
=
d
y
d
t
d
x
d
t
=
3
t
2
6
t
∴
d
y
d
x
=
t
2
Slope of the tangent
=
(
d
y
d
x
)
t
=
0
=
0
2
=
0
The point on the curve
y
=
x
2
−
3
x
+
2
where tangent is perpendicular to
y
=
x
is
Report Question
0%
(
0
,
2
)
0%
(
1
,
0
)
0%
(
−
1
,
6
)
0%
(
2
,
−
2
)
Explanation
y
=
x
2
−
3
x
+
2
[ Given ]
y
=
x
[ Given ]
Differentiating both sides w.r.t.
x
,
we get
⇒
d
y
d
x
=
1
Let
(
x
1
,
y
1
)
be the required point.
It is given that point lies on the curve.
∴
y
1
=
x
2
1
−
3
x
1
+
2
⇒
y
=
x
2
−
3
x
+
2
Differentiating both sides w.r.t.
x
,
we get
∴
d
y
d
x
=
2
x
−
3
Slope of the tangent
=
(
d
y
d
x
)
(
x
1
,
y
1
)
=
2
x
1
−
3
The tangent is perpendicular to the line.
∴
Slope of the tangent
=
−
1
s
l
o
p
e
o
f
t
h
e
l
i
n
e
=
−
1
1
=
−
1
Now,
2
x
1
−
3
=
−
1
⇒
2
x
1
=
2
⇒
x
1
=
1
⇒
y
1
=
x
2
1
−
3
x
1
+
2
⇒
y
1
=
(
1
)
2
−
3
(
1
)
+
2
⇒
y
1
=
1
−
3
+
2
∴
y
1
=
0
∴
(
x
1
,
y
1
)
=
(
1
,
0
)
The equation of the normal to the curve
y
=
x
(
2
−
x
)
at the point
(
2
,
0
)
is
Report Question
0%
x
−
2
y
=
2
0%
x
−
2
y
+
2
=
0
0%
2
x
+
y
=
4
0%
2
x
+
y
−
4
=
0
Explanation
The given curve is
y
=
2
x
−
x
2
i.e,
x
2
−
2
x
+
y
=
0
Now the equation of the tangent at the point (
x
1
,
y
1
)=
(
2
,
0
)
or,
x
x
1
−
(
x
+
x
1
)
+
1
2
(
y
+
y
1
)
=
0
i.e
2
x
+
(
−
1
)
(
x
+
2
)
+
1
2
(
y
+
0
)
=
0
i.e,
2
x
−
4
+
y
=
0
i.e,
y
=
−
2
x
+
4
So, the slope of the tangent is
(
−
2
)
Therefore the slope of the normal=
1
2
So,the equation of the normal at the point
(
2
,
0
)
is
y
−
0
=
1
2
(
x
−
2
)
i.e,
2
y
=
x
−
2
i.e,
x
−
2
y
=
2
At what points the slope of the tangent to the curve
x
2
+
y
2
−
2
x
−
3
=
0
is zero?
Report Question
0%
(
3
,
0
)
,
(
−
1
,
0
)
0%
(
3
,
0
)
,
(
1
,
2
)
0%
(
−
1
,
0
)
,
(
1
,
2
)
0%
(
1
,
2
)
,
(
1
,
−
2
)
Explanation
x
2
+
y
2
−
2
x
−
3
=
0
is zero.
Differentiate w.r.t. x
2
x
+
2
y
d
y
d
x
−
2
=
0
2
y
⋅
d
y
d
x
=
2
−
2
x
d
y
d
x
=
2
(
1
−
x
)
2
y
d
y
d
x
=
1
−
x
y
........
(
1
)
If line is parallel to x-axis
Angle with x-axis
=
θ
=
0
Slope of x-axis
=
tan
θ
=
tan
0
o
=
0
Slope of tangent
=
Slope of x-axis
d
y
d
x
=
0
1
−
x
y
=
0
x
=
1
Finding y when
x
=
1
x
2
+
y
2
−
2
x
−
3
=
0
(
1
)
2
+
y
2
−
2
(
1
)
−
3
=
0
1
+
y
2
−
2
−
3
=
0
y
=
±
2
Hence, the points are
(
1
,
2
)
and
(
1
,
−
2
)
.
Any tangent to the curve
y
=
2
x
7
+
3
x
+
5
Report Question
0%
Is parallel to
x
-axis
0%
Is parallel to
y
-axis
0%
Makes an acute angle with
x
-axis
0%
Makes an obtuse angle with
x
-axis
Explanation
We have,
y
=
2
x
7
+
3
x
+
5
d
y
d
x
=
14
x
6
+
3
⇒
d
y
d
x
>
3
(
∵
x
6
is always positive for any real value)
⇒
d
y
d
x
>
0
so,
tan
θ
>
0
Hence,
θ
lies in first quadrant.
Thus, the tangent to curve makes an acute angle.
The normal to the curve
x
2
=
4
y
passing through
(
1
,
2
)
is
Report Question
0%
x
+
y
=
3
0%
x
−
y
=
3
0%
x
+
y
=
1
0%
x
−
y
=
1
Explanation
Curve is
x
2
=
4
y
Diff wrt to x
2
x
=
4
d
y
d
x
⇒
d
y
d
x
=
x
2
Slope of normal
=
−
1
d
y
/
d
x
=
−
2
x
Let (h, k) be the point where normal and curve intersects
∴
Slope of normal at (h, k)
=
−
2
/
h
Equation of normal passes through (h, k)
y
−
y
1
=
m
(
x
−
x
1
)
y
−
k
=
−
2
h
(
x
−
h
)
Since normal passes through
(
1
,
2
)
2
−
k
=
−
2
h
(
1
−
h
)
k
=
2
+
2
h
(
1
−
h
)
..........
(
1
)
since (h, k) lies on the curve
x
2
=
4
y
h
2
=
4
k
k
=
h
2
4
.......
(
2
)
using
(
1
)
and
(
2
)
2
+
2
h
(
1
−
h
)
=
h
2
4
2
h
=
h
2
4
h
=
2
Putting
h
=
2
in
(
2
)
k
=
h
2
4
,
k
=
1
h
=
2
and
k
=
1
putting in equation of normal
⇒
y
−
k
=
−
2
(
k
−
h
)
h
⇒
y
−
1
=
−
2
(
x
−
2
)
2
=
y
−
1
=
−
1
(
x
−
2
)
⇒
y
−
1
=
−
x
+
2
⇒
x
+
y
=
2
+
1
⇒
x
+
y
=
3
.
The equation of the normal to the curve
x
=
a
cos
3
θ
,
y
=
a
sin
3
θ
at the point
θ
=
π
4
is
Report Question
0%
x
=
0
0%
y
=
0
0%
x
=
y
0%
x
+
y
=
a
Explanation
Slope of normal to the curve =
−
1
(
d
y
d
x
)
θ
=
π
/
4
x
=
a
cos
3
θ
,
y
=
a
sin
3
θ
d
x
d
θ
=
3
a
cos
2
θ
(
−
sin
θ
)
,
d
y
d
θ
=
3
a
sin
2
θ
cos
θ
∴
d
y
d
x
=
d
y
/
d
θ
d
x
/
d
θ
=
+
3
a
sin
2
θ
cos
θ
−
3
a
cos
2
θ
sin
θ
=
−
tan
θ
∴
(
d
y
d
x
)
θ
=
π
/
4
=
−
1
∴
Equation of normal
=
(
y
−
y
0
)
=
−
1
(
d
y
d
x
)
θ
=
π
/
4
[
x
−
x
0
]
y
−
a
2
√
2
=
−
1
−
1
(
x
−
a
2
√
2
)
[
x
0
=
a
cos
3
π
4
=
a
2
√
2
y
−
a
2
√
2
=
x
−
a
2
√
2
y
0
=
a
sin
3
π
4
=
a
2
√
2
]
y
=
x
The number of tangents to the cure
x
3
/
2
+
y
3
/
2
=
2
a
3
/
2
,
a
>
0
, which are equally inclined to the axes, is
Report Question
0%
2
0%
1
0%
0
0%
4
The curve given by
x
+
y
=
e
x
y
has a tangent parallel to the y-axis at the point
Report Question
0%
(
0
,
1
)
0%
(
1
,
0
)
0%
(
1
,
1
)
0%
None of these
Explanation
Differentiating w.r.t.x, we get
1
+
d
y
d
x
=
e
x
y
(
y
+
x
d
y
d
x
)
or
d
y
d
x
=
y
e
x
y
−
1
1
−
x
e
x
y
As the tangent is parallel along
y
−
a
x
i
s
d
y
d
x
=
∞
=
y
e
x
y
−
1
1
−
x
e
x
y
⇒
1
−
x
e
x
y
=
0
This holds for x = 1, y = 0
Mark the correct alternative of the following.
The point on the curve
9
y
2
=
x
3
, where the normal to the curve makes equal intercepts with the axes is?
Report Question
0%
(
4
,
±
8
/
3
)
0%
(
−
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
9
y
2
=
x
2
since
(
x
1
,
y
1
)
lies on the curve, therefore
9
y
2
1
=
x
3
1
.......
(
1
)
Now
9
y
2
=
x
3
⇒
d
y
d
x
=
x
2
6
y
⇒
(
d
y
d
x
)
(
x
1
,
y
1
)
=
x
2
6
y
1
since normal to the curve at
(
x
1
,
y
1
)
makes equal intersepts with the coordinate axis, therefore slope of the normal
=
±
1
⇒
1
−
(
d
y
/
d
x
)
(
x
1
,
y
1
)
=
±
1
⇒
(
d
y
/
d
x
)
(
x
1
,
y
1
)
=
±
1
→
x
2
1
6
y
1
=
±
1
⇒
x
4
1
=
36
y
2
1
=
36
(
x
3
1
9
)
(using
1
)
⇒
x
4
1
=
4
x
3
1
⇒
x
3
1
(
x
1
−
4
)
=
0
⇒
x
1
=
0
,
4
Putting
x
1
=
0
in
(
1
)
we get
9
y
2
1
=
0
⇒
y
1
=
0
Putting
x
1
=
4
in
(
1
)
we get
9
y
2
1
=
(
4
)
3
⇒
y
1
=
±
8
3
But the line making equal intersepts with the coordinate axes cannot pass through origin.
Hence, the required points are
(
4
,
8
3
)
and
(
4
,
−
8
3
)
.
Mark the correct alternative of the following.
The line
y
=
m
x
+
1
is a tangent to the curve
y
2
=
4
x
, if the value of m is?
Report Question
0%
1
0%
2
0%
3
0%
1
/
2
Explanation
Given equation of the tangent to the given curve
y
=
m
x
+
1
Now substituting the value of y in
y
2
=
4
x
, we get
⇒
(
m
x
+
1
)
2
=
4
x
⇒
m
2
x
2
+
1
+
2
m
x
−
4
x
=
0
⇒
m
2
x
2
+
x
(
2
m
−
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
−
4
a
c
=
0
(
2
m
−
4
)
2
−
4
(
m
2
)
(
1
)
=
0
⇒
4
m
2
−
16
m
+
16
−
4
m
2
=
0
⇒
−
16
m
+
16
=
0
⇒
m
=
1
.
The slope of the tangent to the curve
x
=
t
2
+
3
t
−
8
,
y
=
2
t
2
−
2
t
−
5
at the point
(
2
,
−
1
)
is
Report Question
0%
22
7
0%
6
7
0%
7
6
0%
−
6
7
Explanation
Given curve
x
=
t
2
+
3
t
−
8
and
y
=
2
t
2
−
2
t
−
5
slope of tangent to the curve
=
d
y
d
x
=
d
y
/
d
t
d
x
/
d
t
∴
d
x
d
t
=
2
t
+
3
,
d
y
d
t
=
4
t
−
2
d
y
d
x
=
d
y
/
d
t
d
x
/
d
t
=
4
t
−
2
2
t
+
3
(
d
y
d
x
)
t
=
2
4
(
2
)
−
2
2
(
2
)
+
3
=
6
7
The normal at the point
(
1
,
1
)
on the curve
2
y
+
x
2
=
3
is
Report Question
0%
x
+
y
=
0
0%
x
−
y
=
0
0%
x
+
y
+
1
=
0
0%
x
−
y
=
1
Explanation
f
(
x
)
=
x
(
x
−
4
)
2
,
x
∈
1
The equation of the givne curve is
2
y
+
x
2
=
3
Differentiating wrt to x, we have
2
d
y
d
x
+
2
x
=
0
⇒
d
y
d
x
=
−
x
∴
(
d
y
d
x
)
(
1
,
1
)
=
−
1
The slope of the normal to the given curve at point
(
1
,
1
)
is
−
1
(
d
y
/
d
x
)
(
1
,
1
)
=
1
Hence the equation of the normal to the given curve at
(
1
,
1
)
is given as
⇒
y
−
1
=
1
(
x
−
1
)
⇒
y
−
1
=
x
−
1
⇒
x
−
y
=
0
.
Consider the equation
x
y
=
e
x
−
y
What is
d
2
y
d
x
2
at
x
=
1
equal to ?
Report Question
0%
0
0%
1
0%
2
0%
4
Explanation
x
y
=
e
x
−
y
Taking
log
both sides, we have
y
log
x
=
(
x
−
y
)
log
e
y
log
x
=
x
−
y
.
.
.
.
.
(
1
)
y
x
=
1
+
log
x
At
x
=
1
⇒
y
=
1
Differentiating equation
(
1
)
w.r.t.
x
, we have
d
y
d
x
log
x
+
y
x
=
1
−
d
y
d
x
⇒
(
1
+
log
x
)
d
y
d
x
=
1
−
y
x
⇒
d
y
d
x
=
x
−
y
x
(
1
+
log
x
)
⇒
d
y
d
x
=
y
log
x
x
(
1
+
log
x
)
⇒
d
y
d
x
=
log
x
(
1
+
log
x
)
2
Again differentiating above equation w.r.t.
x
, we have
d
2
y
d
x
2
=
(
1
+
log
x
)
2
⋅
1
x
−
log
x
⋅
2
(
1
+
log
x
)
1
x
(
1
+
log
x
)
4
⇒
d
2
y
d
x
2
=
1
x
(
1
−
log
x
)
(
1
+
log
x
)
3
At
x
=
1
,
d
2
y
d
x
2
=
1
Consider the equation
x
y
=
e
x
−
y
What is
d
y
d
x
at
x
=
1
equal to ?
Report Question
0%
0
0%
1
0%
2
0%
4
Explanation
x
y
=
e
x
−
y
Taking
log
both sides, we have
y
log
x
=
(
x
−
y
)
log
e
y
log
x
=
x
−
y
.
.
.
.
.
(
1
)
At
x
=
1
⇒
y
=
1
Differentiating equation
(
1
)
w.r.t.
x
, we have
d
y
d
x
log
x
+
y
x
=
1
−
d
y
d
x
⇒
(
1
+
log
x
)
d
y
d
x
=
1
−
y
x
⇒
d
y
d
x
=
x
−
y
x
(
1
+
log
x
)
At
x
=
1
,
y
=
1
d
y
d
x
=
0
A curve
y
=
m
e
m
x
where
m
>
0
intersects y-axis at a point
P
.
How much angle does the tangent at
P
make with y-axis ?
Report Question
0%
tan
−
1
m
2
0%
cot
−
1
(
a
+
m
2
)
0%
sin
−
1
(
1
√
1
+
m
4
)
0%
sec
−
1
√
1
+
m
4
A curve
y
=
m
e
m
x
where
m
>
0
intersects y-axis at a point
P
.
What is the slope of the curve at the point of intersection
P
?
Report Question
0%
m
0%
m
2
0%
2
m
0%
2
m
2
Explanation
y
=
m
e
m
x
,
m
>
0
Therefore,
Slope
=
d
y
d
x
=
m
2
e
m
x
Substituting
x
=
0
, we have
S
l
o
p
e
=
m
2
e
m
⋅
0
=
m
2
For
x
>
1
,
y
=
log
e
x
satisfies the inequality
Report Question
0%
x
−
1
>
y
0%
x
2
−
1
>
y
0%
y
>
x
−
1
0%
x
−
1
x
<
y
The slope of the tangent to the curve
y
=
√
4
−
x
2
at the point, where the ordinate and the abscissa are equal , is
Report Question
0%
-1
0%
1
0%
0
0%
None of these
Explanation
Putting
y
=
x
in
y
=
√
4
−
x
2
, we get
x
=
√
2
,
−
√
2
.
So, the point is
(
√
2
,
√
2
)
.
Differentiating
y
2
+
x
2
=
4
w.r.t. x,
2
y
d
y
d
x
+
2
x
=
0
or
d
y
d
x
=
−
x
y
⇒
a
t
(
√
2
,
√
2
)
,
d
y
d
x
=
−
1
If m is the slope of a tangent to the curve
e
y
=
1
+
x
2
,
then
Report Question
0%
|
m
|
>
1
0%
m
>
1
0%
m
>
−
1
0%
|
m
|
≤
1
Explanation
Differentiating w.r.t.x, we get
e
y
d
y
d
x
=
2
x
⇒
d
y
d
x
=
2
x
1
+
x
2
(
∵
e
y
=
1
+
x
2
)
⇒
m
=
2
x
1
+
x
2
o
r
|
m
|
=
2
|
x
|
1
+
|
x
|
2
But
1
+
|
x
|
2
−
2
|
x
|
=
(
1
−
|
x
|
)
2
≥
0
⇒
1
+
|
x
|
2
≥
2
|
x
|
∴
|
m
|
≤
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
Report Question
0%
ln
(
√
3
+
√
7
7
)
and ln
(
√
3
+
√
5
2
)
0%
ln
(
√
3
+
√
7
2
)
0%
ln
(
√
7
+
√
3
2
)
0%
±
ln
(
√
3
+
√
7
2
)
If x=4 y = 14 is a normal to the curve
y
2
=
a
x
3
−
β
at (2,3) then the value of
α
+
β
is
Report Question
0%
9
0%
-5
0%
7
0%
-7
At what points of curve
y
=
2
3
x
3
+
1
2
x
2
, the tangent makes the equal with the axis?
Report Question
0%
(
1
2
,
5
24
)
and
(
−
1
,
−
1
6
)
0%
(
1
2
,
4
9
)
and
(
−
1
,
0
)
0%
(
1
3
,
1
7
)
and
(
−
3
,
1
2
)
0%
(
1
3
,
4
47
)
and
(
−
1
,
1
2
)
The curve represented parametrically by the equations x = 2 in
cot
t
+
1
and
y
=
tan
t
+
cot
t
Report Question
0%
tanfent and normal intersect at the point (2, 1)
0%
normal at
t
=
π
/
4
is parallel to the y-axis
0%
tangent at
t
=
π
/
4
is parallel to the line y = x
0%
tangent at
t
=
π
/
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
Report Question
0%
27
c
3
0%
4
27
c
3
0%
27
4
c
3
0%
4
9
c
3
The angle between the tangent to the curves
y
=
x
2
and
x
=
y
2
at (1, 1) is
Report Question
0%
cos
−
1
4
5
0%
sin
−
1
3
5
0%
tan
−
1
3
4
0%
tan
−
1
1
3
At the point
P
(
a
,
a
n
)
on the graph of
y
=
x
n
(
n
ϵ
n
)
in the first quadrant, a normal is drawn. the normal intersects the y-axis at the point (0, b) . if
l
i
m
a
→
b
b
=
1
2
, then n equals
Report Question
0%
1
0%
3
0%
2
0%
4
Point on the curve
f
(
x
)
=
x
1
−
x
2
where the tangent is inclined at an angle of
π
4
ot the x-axis are
Report Question
0%
(0, 0)
0%
(
√
3
,
−
√
3
2
)
0%
(
−
2
,
2
3
)
0%
(
−
√
3
,
√
3
2
)
The x-intercept of the tangent at any arbitrary point of the curve
a
x
2
+
b
y
2
=
1
is proportion to
Report Question
0%
square of the abscissa of the point of tangency
0%
square root of the abscissa of the point of tangency
0%
cube of the abscissa pf the point of tangency
0%
cube root of the abscissa of the point of tangency
If the tangent at any point
P
(
4
m
2
,
8
m
3
)
of
x
3
−
y
3
=
0
is also a normal to the curve
x
3
−
y
3
=
0
, then value of m is
Report Question
0%
m
=
√
2
3
0%
m
=
−
√
2
3
0%
m
=
3
√
2
0%
m
=
−
3
√
2
The slope of the tangent to the curve
y
=
f
(
x
)
at
[
x
,
f
(
x
)
]
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
Report Question
0%
5
6
0%
6
5
0%
1
6
0%
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
Report Question
0%
it makes a constant angle with x-axis
0%
it passes through the origin
0%
it is at a constant distance from the origin
0%
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
Report Question
0%
x
2
+
y
2
=
9
x
0%
4
x
2
+
y
2
=
9
x
0%
4
x
2
+
2
y
2
=
9
x
0%
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
Report Question
0%
(
y
+
√
x
2
−
y
2
)
x
k
−
1
=
c
0%
(
y
+
√
x
2
+
y
2
)
x
k
−
1
=
c
0%
(
y
−
√
x
2
−
y
2
)
x
k
−
1
=
c
0%
(
y
+
√
x
2
+
y
2
)
x
k
−
1
=
c
If
f
(
x
)
=
∫
x
1
e
t
2
/
2
(
1
−
t
2
)
d
t
,
then
d
d
x
f
(
x
)
at x=1 is
Report Question
0%
0
0%
1
0%
2
0%
−
1
Explanation
We have,
f
(
x
)
=
∫
x
1
e
t
2
/
2
(
1
−
t
2
)
d
t
f
′
(
x
)
=
[
e
x
2
/
2
(
1
−
x
2
)
]
2
f
′
(
1
)
=
e
1
/
2
.0
=
0
If
y
=
∫
x
2
x
t
2
d
t
,
then the equation of tangent at x=1 is
Report Question
0%
x
+
y
=
1
0%
y
=
x
−
1
0%
y
=
x
0%
y
=
x
+
1
Explanation
At
x
=
1
,
y
=
0
,
d
y
d
x
=
2
x
.
(
x
4
)
2
−
(
x
2
)
2
=
1
∴
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
)
Report Question
0%
on the left of x =c
0%
on the right of x = c
0%
at no point
0%
at all point
If the line ax +by + c = 0 is a normal to the curve xy = 1, then
Report Question
0%
a
>
0
,
b
>
0
0%
a
>
0
,
b
<
0
0%
a
<
0
,
b
>
0
0%
a
<
0
,
b
<
0
The equation of the curves through the point
(
1
,
0
)
and whose slope is
y
−
1
x
2
+
x
is
Report Question
0%
(
y
−
1
)
(
x
+
1
)
+
2
x
=
0
0%
2
x
(
y
−
1
)
+
x
+
1
=
0
0%
x
(
y
−
1
)
(
x
+
1
)
+
2
=
0
0%
None of these
Explanation
Slope
=
d
y
d
x
⟹
d
y
d
x
=
y
−
1
x
2
+
x
⟹
d
y
y
−
1
=
d
x
x
2
+
x
⟹
∫
1
y
−
1
d
y
=
∫
(
1
x
−
1
x
+
1
)
d
x
+
C
⟹
log
(
y
−
1
)
=
log
(
x
x
+
1
)
+
log
c
⟹
(
y
−
1
)
(
x
+
1
)
x
=
k
Putting
x
=
1
,
y
=
0
, we get
k
=
−
2
The equation is
(
y
−
1
)
(
x
+
1
)
+
2
x
=
0
The point(s) on the curve
y
3
+
3
x
2
=
12
y
,
where the tangent is vertical, is (are)
Report Question
0%
(
±
4
√
3
,
−
2
)
0%
(
±
√
11
3
,
1
)
0%
(0, 0)
0%
(
±
4
√
3
,
2
)
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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