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CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Application Of Derivatives
Quiz 4
Let the parabolas
y
=
x
2
+
a
x
+
b
and
y
=
x
(
c
−
x
)
touch each other at the point
(
1
,
0
)
. Then
Report Question
0%
a
=
−
3
0%
b
=
1
0%
c
=
2
0%
b
+
c
=
3
Explanation
As point
(
1
,
0
)
lies on
y
=
x
2
+
a
x
+
b
⇒
a
+
b
=
−
1
...(1)
And also lies on
y
=
x
(
c
−
x
)
⇒
c
=
1
...(2)
They touch each other, so slope of tangents drawn from
(
1
,
0
)
will be same, then
d
y
d
x
=
2
x
+
a
=
c
−
2
x
⇒
a
=
c
−
4
⇒
a
=
−
3
...(3)
Using (1) and (3) we get
−
3
+
b
=
1
⇒
b
=
2
...(4)
Now using (2) and (4) we get
b
+
c
=
2
+
1
=
3
Hence, options 'A' and 'D' are correct.
The curve
x
n
a
n
+
y
n
b
n
=
2
touches the line
x
a
+
y
b
=
2
at the point
Report Question
0%
(
b
,
a
)
0%
(
a
,
b
)
0%
(
1
,
1
)
0%
(
1
a
,
1
b
)
Explanation
Slope of curve
x
n
a
n
+
y
n
b
n
=
2
is
d
y
d
x
=
−
b
n
a
n
x
n
−
1
y
n
−
1
Slope of tangent
x
a
+
y
b
=
2
is
d
y
d
x
=
−
b
a
Equating both slopes, we get
−
b
n
a
n
x
n
−
1
y
n
−
1
=
−
b
a
Only
x
=
a
y
=
b
satisfy the above equation.
Hence, option 'B' is correct.
If the line joining the points
(
0
,
3
)
and
(
5
,
−
2
)
is the tangent to the curve
y
=
c
x
+
1
then the value of
c
is
Report Question
0%
1
0%
−
2
0%
4
0%
none of these
Explanation
Slop of line joining the points
(
0
,
3
)
a
n
d
(
5
,
2
)
is given by
3
+
2
0
−
5
=
−
1
This is equal to the slop of tangent on the curve and that is given by
d
y
d
x
=
−
c
(
x
+
1
)
2
d
y
d
x
=
−
1
⇒
c
=
(
x
+
1
)
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
1
)
\
Equation of line joining the points
(
0
,
3
)
a
n
d
(
5
,
2
)
is given by
(
y
−
3
)
=
−
1
(
x
−
0
)
Solving equation of tangent and the curve for point of intersection
c
x
+
1
+
x
=
3
.............(2)
Solving (1) and (2)
x
=
1
Putting this in (2), we get c=4
A point on the ellipse
4
x
2
+
9
y
2
=
36
where the tangent is equally inclided to the axes is
Report Question
0%
(
9
√
13
,
4
√
13
)
0%
(
−
9
√
13
,
4
√
13
)
0%
(
9
√
13
,
−
4
√
13
)
0%
(
4
√
13
,
−
9
√
13
)
Explanation
Slope of tangent to the curve
4
x
2
+
9
y
2
=
36
is
1
. since it is equally inclined to the axes.
d
y
d
x
=
4
x
9
y
=
1
⇒
x
=
9
y
4
On solving obtained equation with the equation of curve, we get
y
=
±
4
√
13
Substituting value of y we get
(
x
,
y
)
=
(
9
√
13
,
4
√
13
)
,
(
−
9
√
13
,
4
√
13
)
,
(
9
√
13
,
−
4
√
13
)
Hence, options 'A', 'B' and 'C' are correct.
If error in measuring the edge of a cube is
k
% then the percentage error in estimating its volume is
Report Question
0%
k
0%
3
k
0%
k
3
0%
none of these
Explanation
Let the actual length of the cube be a.
Therefore the measured length of the cube will be
=
a
(
1
±
0.0
k
)
=
a
(
1
±
k
100
)
Considering positive error,
a
′
=
a
(
1
+
k
100
)
V
′
=
a
3
(
1
+
k
100
)
3
=
a
3
(
1
+
3
(
k
100
)
+
3
(
k
100
)
2
+
(
k
100
)
3
)
Since
k
100
<<
1
, hence we neglect the higher order terms.
Thus
V
′
=
a
3
(
1
+
3
(
k
100
)
)
Actual volume V
V
=
a
3
Therefore
V
′
−
V
=
a
3
(
1
+
3
k
100
)
−
a
3
=
a
3
(
3
k
100
)
V
′
−
V
V
=
a
3
3
k
100
a
3
=
3
k
100
=
3
k
100
V
′
−
V
V
×
100
=
3
k
Therefore percentage error in volume is
3
k
.
The angle between two tangents to the ellipse
x
2
16
+
y
2
9
=
1
at the points where the line
y
=
1
cuts the curve is
Report Question
0%
π
4
0%
tan
−
1
6
√
2
7
0%
π
2
0%
none of these
Explanation
Substituting
y
=
1
in
x
2
16
+
y
2
9
=
1
We get
x
=
±
8
√
2
3
And slope of tangents
d
y
d
x
=
9
x
16
y
=
m
1
,
m
2
=
±
3
√
2
2
Therefore
t
a
n
(
θ
)
=
m
1
−
m
2
1
+
m
1
m
2
=
|
−
6
√
2
7
|
⇒
θ
=
t
a
n
−
1
(
6
√
2
7
)
A tangent to the curve
y
=
∫
x
0
|
t
|
d
t
, which is parallel to the line y=x, cuts off an intercept from the y-axis equal to
Report Question
0%
1
0%
−
1
2
0%
1
2
0%
−
1
Explanation
Since the integral is in the first quadrant
|
t
|
=
t
Thus
∫
x
x
=
0
|
t
|
.
d
t
=
∫
x
x
=
0
t
.
d
t
=
x
2
2
Or
2
y
=
x
2
is the equation of the curve.
Now
2
y
′
=
2
x
Or
y
′
=
x
...(i)
Now the tangent is parallel to
y
=
x
Hence slope
=
y
′
=
1
Thus the x-coordinate of point of contact is 1.
Hence
y
=
1
2
.
Hence equation of the tangent will be
y
−
1
2
=
(
x
−
1
)
.
Or
x
−
y
=
1
2
Or
y
=
x
−
1
2
.
Hence intercept cut on the y axis is
−
1
2
.
Angle between the tangents to the curve
y
=
x
2
−
5
x
+
6
at the points
(
2
,
0
)
and
(
3
,
0
)
is
Report Question
0%
π
2
0%
π
3
0%
π
6
0%
π
4
Explanation
Given equation of curve
y
=
x
2
−
5
x
+
6
⇒
d
y
d
x
=
2
x
−
5
Slope of tangent to the curve at
(
2
,
0
)
is
(
d
y
d
x
)
(
2
,
0
)
=
2
(
2
)
−
5
=
−
1
=
m
1
Slope of tangent to the curve at
(
3
,
0
)
is
(
d
y
d
x
)
(
3
,
0
)
=
2
(
3
)
−
5
=
1
=
m
2
Since
m
1
m
2
=
−
1
∴
Angle between the tangents to the curve at
(
2
,
0
)
and
(
3
,
0
)
is
π
2
The number of tangents to the curve
y
=
e
|
x
|
at the point
(
0
,
1
)
is
Report Question
0%
2
0%
1
0%
4
0%
0
Explanation
For
x
>
0
,
y
=
e
x
.
d
y
d
x
=
e
x
.
Now
d
y
d
x
x
=
0
=
1
.
Hence
y
−
1
=
1
(
x
−
0
)
Or
x
−
y
+
1
=
0
is the required equation of tangent.
Similarly for
x
<
0
y
=
e
−
x
.
d
y
d
x
=
−
e
−
x
.
Now
d
y
d
x
x
=
0
=
−
1
.
Hence
y
−
1
=
−
1
(
x
−
0
)
Or
x
+
y
−
1
=
0
is the required equation of tangent.
Hence at
x
=
0
we will have 2 tangents to the curve
y
=
e
|
x
|
which are mutually perpendicular.
The curve
y
+
e
x
y
+
x
=
0
has a tangent parellel to y-axis at a point
Report Question
0%
(
−
1
,
0
)
0%
(
1
,
0
)
0%
(
1
,
1
)
0%
(
0
,
0
)
Explanation
d
y
d
x
+
e
x
y
[
x
d
y
d
x
+
y
]
+
1
=
0
Or
d
y
d
x
[
1
+
x
.
e
x
y
]
+
1
+
y
.
e
x
y
=
0
Or
d
y
d
x
=
−
(
1
+
y
.
e
x
y
)
1
+
x
.
e
x
y
Now
1
+
x
e
x
y
=
0
since the slope of the tangent is
90
0
.
Hence
x
e
x
y
=
−
1
Or
x
(
−
x
−
y
)
=
−
1
Or
x
2
+
x
y
=
1
Or
y
=
1
−
x
2
x
Or
y
=
1
x
−
x
...(i)
Considering
y
=
0
,
x
2
=
1
x
=
±
1
.
Hence we get 2 points
(
1
,
0
)
and
(
−
1
,
0
)
.
Out of these 2, only
(
−
1
,
0
)
lies on the curve.
Hence the required point is
(
−
1
,
0
)
.
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
)
(
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 points
Explanation
Equation of straight line joining
A
(
c
+
1
,
e
c
+
1
)
and
B
(
c
−
1
,
e
c
−
1
)
is
y
−
e
c
+
1
=
e
c
+
1
−
e
c
−
1
2
(
x
−
c
−
1
)
(1)
Equation of tangent at
(
c
,
e
c
)
is
y
−
e
c
=
e
c
(
x
−
c
)
(2)
Subtracting (1) from (2),we get
e
c
(
e
−
1
)
=
e
c
[
(
x
−
c
)
−
1
2
(
e
−
e
−
1
)
(
x
−
c
)
+
1
2
(
e
−
e
−
1
)
]
⇒
1
2
(
e
+
e
−
1
)
−
1
=
(
x
−
c
)
[
1
−
1
2
(
e
−
e
−
1
)
]
⇒
x
−
c
=
e
+
e
−
1
−
2
2
−
e
+
e
−
1
<
0
[
∵
e
+
e
−
1
>
2
and
2
+
e
−
1
−
e
<
0
]
⇒
x
<
c
Thus the two lines meet to the left of
x
=
c
For
a
∈
[
π
,
2
π
]
and
n
∈
Z
, the critical points of
f
(
x
)
=
1
3
sin
a
tan
3
x
+
(
sin
a
−
1
)
tan
x
+
√
a
−
2
8
−
a
are
Report Question
0%
x
=
n
π
0%
x
=
2
n
π
0%
x
=
(
2
n
+
1
)
π
0%
None of these
Explanation
Given,
f
(
x
)
=
1
3
sin
a
tan
3
x
+
(
sin
a
−
1
)
tan
x
+
√
a
−
2
8
−
a
f
′
(
x
)
=
sin
a
tan
2
x
sec
2
x
+
(
sin
a
−
1
)
sec
2
x
=
(
sin
a
tan
2
x
+
sin
a
−
1
)
sec
2
x
At critical points, we must have
f
′
(
x
)
=
0
⇒
sin
a
tan
2
x
+
sin
a
−
1
=
0
(
∵
sec
2
x
≠
0
f
o
r
a
n
y
x
∈
R
)
⇒
tan
2
x
=
1
−
sin
a
sin
a
Since
a
∈
[
π
,
2
π
]
,
1
−
sin
a
sin
a
<
0
∴
tan
2
x
=
1
−
sin
a
sin
a
has no solution in
R
⇒
f
(
x
)
has no critical points
Report Question
0%
Assertion is true and Reason is true; Reason is a correct explanation for Assertion.
0%
Assertion is True, Reason is true; Reason is not a correct explanation for Assertion.
0%
Assertion is true, Reason is false
0%
Assertion is false, Reason is true
Explanation
We have
y
=
x
2
+
b
x
+
c
⇒
d
y
d
x
=
2
x
+
b
Since the curve touches the line
y
=
x
at the point
(
1
,
1
)
[
2
x
+
b
]
(
1
,
1
)
=
1
⇒
2
+
b
=
1
⇒
b
=
−
1
Also, the curve passes through the point
(
1
,
1
)
∴
1
=
1
+
b
+
c
⇒
c
=
−
b
=
1
∴
y
=
x
2
−
x
+
1
⇒
d
y
d
x
=
2
x
−
1
Now,
d
y
d
x
<
0
⇒
2
x
−
1
<
0
⇒
x
<
1
2
y
=
4
x
2
and
y
=
x
2
.
The two curves
Report Question
0%
intersect each other
0%
touch each other
0%
do not meet
0%
represent parabola
Explanation
y
=
4
x
2
and
y
=
x
2
represent two parabola concave upward and symmetric about positive y-axis. Both the curves do not intersect, however they touch each other at
x
=
0
.
y
′
=
8
x
and
y
′
=
2
x
Now
8
x
=
2
x
Or
x
=
0
Hence both curves have equal slope at
x
=
0
and are horizontal (parallel to x axis)at
x
=
0
If the normal to the curve
y
=
f
(
x
)
at the point
(
3
,
4
)
makes an angle
3
π
4
with the positive x-axis then
f
′
(
3
)
is equal to
Report Question
0%
−
1
0%
−
3
4
0%
4
3
0%
1
Explanation
Slope of normal
=
−
1
f
′
(
x
)
=
t
a
n
(
3
π
4
)
=
−
1
Hence
f
′
(
x
)
=
1
Or
d
y
d
x
=
1
Or
y
=
x
+
c
Hence
y
=
f
(
x
)
is an equation of a straight line parallel to
y
=
x
.
Hence
f
′
(
x
)
is independent of x and its value is 1.
Find the slopes of the tangents of the curve
y
=
(
x
+
1
)
(
x
−
3
)
at the points where it cuts the X-axis.
Report Question
0%
4
0%
−
4
0%
2
0%
−
2
Explanation
f
(
x
)
=
(
x
+
1
)
(
x
−
3
)
Now
f
(
x
)
=
0
⇒
x
=
−
1
,
x
=
3
Differentiating
f
(
x
)
with respect to x
d
y
d
x
=
x
+
1
+
x
−
3
=
2
x
−
2
=
2
(
x
−
1
)
Now slope of the tangent at
(
h
,
k
)
will be
d
y
d
x
h
,
k
Hence slopes of the tangent at
x
=
−
1
and
x
=
3
, will be
d
y
d
x
x
=
−
1
=
2
(
x
−
1
)
x
=
−
1
=
−
4
And
d
y
d
x
x
=
3
=
2
(
x
−
1
)
x
=
3
=
4
Find the points on the curve
y
=
x
3
, the tangents at which are inclined at an angle of
60
∘
to x-axis.
Report Question
0%
x
=
±
1
√
√
3
,
y
=
1
√
3
.
1
√
√
3
.
0%
x
=
1
√
√
3
,
y
=
1
√
3
.
1
√
√
3
.
0%
x
=
±
1
√
√
3
,
y
=
±
1
√
3
.
1
√
√
3
.
0%
x
=
−
1
√
√
3
,
y
=
±
1
√
3
.
1
√
√
3
.
Explanation
Hence slope of tangents
=
t
a
n
60
0
=
√
3
=
d
y
d
x
Hence
d
y
d
x
=
3
x
2
=
√
3
Or
x
2
=
1
√
3
Hence
x
=
±
1
√
√
3
.
Thus y
=
1
√
3
.
1
√
√
3
.
Find the points on the curve
y
=
x
/
(
1
−
x
2
)
where the tangents makes an angle of
π
/
4
with x-axis
Report Question
0%
(
√
3
,
−
√
2
3
)
,
(
−
√
2
,
√
2
3
)
0%
(
√
3
,
−
√
3
4
)
,
(
−
√
3
,
√
3
4
)
0%
(
√
3
,
−
√
3
2
)
,
(
−
√
3
,
√
3
2
)
0%
none of these
Explanation
d
y
d
x
=
t
a
n
45
0
Or
d
y
d
x
=
1
Or
(
1
−
x
2
)
−
x
(
−
2
x
)
(
1
−
x
2
)
2
=
1
Or
1
−
x
2
+
2
x
2
=
(
1
−
x
2
)
2
Or
1
+
x
2
=
1
−
2
x
2
+
x
4
Or
x
4
−
3
x
2
=
0
Or
x
2
[
x
2
−
3
]
=
0
x
=
0
or
x
=
±
√
3
Hence
y
=
√
3
1
−
3
=
−
√
3
2
y
=
−
√
3
1
−
3
=
√
3
2
Hence
(
√
3
,
−
√
3
2
)
,
(
−
√
3
,
√
3
2
)
.
Find the condition that the line
A
x
+
B
y
=
1
may be a normal to the curve
a
n
−
1
y
=
x
n
.
Report Question
0%
a
n
B
(
B
2
+
n
A
2
)
n
=
A
n
n
n
.
0%
a
n
−
1
B
(
B
2
+
n
A
2
)
n
−
1
=
A
n
n
n
.
0%
a
n
B
(
B
2
+
n
A
2
)
n
−
1
=
A
n
n
n
.
0%
a
n
−
1
B
(
B
2
−
n
A
2
)
n
−
1
=
A
n
n
n
.
Explanation
Given,
a
n
−
1
y
=
x
n
∴
d
y
d
x
=
n
x
n
−
1
a
n
−
1
=
n
x
n
−
1
a
n
−
1
⋅
1
x
=
n
y
x
.
∴
Normal is:
Y
−
y
=
−
1
d
y
/
d
x
(
X
−
x
)
=
−
x
n
y
(
X
−
x
)
∴
X
x
+
Y
n
y
=
n
y
2
+
x
2
.
Compare with
A
X
+
B
Y
=
1
∴
X
A
=
n
y
B
=
n
y
2
+
x
2
1
=
k
,
say.
∴
x
=
A
k
,
y
=
(
B
k
/
n
)
and
n
y
2
+
x
2
=
k
or
k
2
[
(
B
2
/
n
+
A
2
)
]
=
k
∴
k
=
n
B
2
+
n
A
2
..(1)
Now
a
n
−
1
y
=
x
n
.
Put for
x
and
y
.
a
n
−
1
y
⋅
B
k
n
=
A
n
k
n
⇒
a
n
−
1
B
=
n
A
n
k
n
−
1
⇒
a
n
−
1
B
=
n
A
n
⋅
(
n
B
2
+
n
A
2
)
n
−
1
,
by (1)
⇒
a
n
−
1
B
(
B
2
+
n
A
2
)
n
−
1
=
A
n
n
n
.
Above is the required condition.
If the normal to the curve
y
=
f
(
x
)
at the point
(
3
,
4
)
makes an angle
3
π
/
4
with the positive x-axis, then
f
′
(
3
)
=
Report Question
0%
−
1
0%
0
0%
1
0%
√
3
Explanation
Tangent being perpendicular to given line of slope 2, will have its slope as -
1
2
.
Slope of tangent=
−
f
x
f
y
=
−
6
x
+
1
2
(
y
+
1
)
=
−
1
2
∴
y
=
6
x
.
Sloping with the given curve, we have
3
x
2
+
36
x
2
+
x
+
12
x
=
0
or
13
x
(
3
x
+
1
)
=
0
∴
x
=
0
,
−
1
/
3
∴
y
=
0
,
−
2
Hence the two points are
(
0
,
0
)
,
(
−
1
3
,
−
2
)
∴
y
=
−
1
2
x
and
y
+
2
=
−
1
2
(
x
+
1
3
)
or
2
y
+
x
=
0
and
2
y
+
x
+
13
3
=
0
Ans: D
The curve
y
−
e
x
y
+
x
=
0
has a vertical tangent at the point
Report Question
0%
(
1
,
1
)
0%
n
o
p
o
i
n
t
0%
(
0
,
1
)
0%
(
1
,
0
)
Explanation
d
y
d
x
−
e
x
y
[
x
d
y
d
x
+
y
]
+
1
=
0
Or
d
y
d
x
[
1
−
x
.
e
x
y
]
+
1
−
y
.
e
x
y
=
0
Or
d
y
d
x
=
−
(
1
−
y
.
e
x
y
)
1
−
x
.
e
x
y
Now
1
−
x
e
x
y
=
0
since the slope of the tangent is
90
0
.
Hence
x
e
x
y
=
1
Or
x
(
x
+
y
)
=
1
Or
x
2
+
x
y
=
1
Or
y
=
1
−
x
2
x
Or
y
=
1
x
−
x
...(i)
Considering
y
=
0
,
x
2
=
1
x
=
±
1
.
Hence we get 2 points
(
1
,
0
)
and
(
−
1
,
0
)
.
Out of these 2, only
(
1
,
0
)
lies on the curve.
Hence the required point is
(
1
,
0
)
.
The set of all values of x for which the function
f
(
x
)
=
(
k
2
−
3
k
+
2
)
(
cos
2
x
4
−
sin
2
x
4
)
+
(
k
−
1
)
x
+
sin
1
does not posses critical points is
Report Question
0%
(
−
4
,
4
)
0%
(
0
,
4
)
0%
(
0
,
1
)
∪
(
1
,
4
)
0%
(
0
,
2
)
∪
(
2
,
4
)
Explanation
f
(
x
)
=
(
k
2
−
3
x
+
2
)
cos
x
2
+
(
k
−
1
)
x
+
sin
1
f
′
(
x
)
=
(
k
−
1
)
(
k
−
2
)
(
−
1
2
sin
x
2
)
+
(
k
−
1
)
=
(
k
−
1
)
[
1
−
k
−
2
2
sin
x
2
]
Since f(x) does not possess critical points therefore f'(x) is not equal to zero.
i.e.,
k
≠
1
or
1
−
k
−
2
2
sin
x
2
=
0
does not posses a solution or
sin
x
2
=
2
k
−
2
does not have a solution.
Hence we must
have
|
2
k
−
2
|
>
1
as
|
sin
x
2
|
<
1.
Above implies that
|
k
−
2
|
2
≤
4
or
−
2
<
(
k
−
2
)
<
2
∵
x
2
<
a
2
⇒
(
x
2
−
a
2
)
=
−
i
v
e
or
−
a
<
x
<
a
∴
0
<
k
<
4
. Also
k
≠
1.
∴
k
ϵ
(
0
,
1
)
∪
(
1
,
4
)
Determine the intervals of monotonicity of
f
(
x
)
=
log
|
x
|
.
Report Question
0%
increasing for
x
>
0
0%
increasing for
x
<
0
0%
decreasing for
x
>
0
0%
decreasing for
x
<
0
Explanation
f
′
(
x
)
=
1
x
,
x
≠
0
, at
x
=
0
,
f
(
x
)
is not defined.
Hence
f
(
x
)
is increasing for
x
>
0
since
f
′
(
x
)
>
0
∀
x
∈
(
0
,
∞
)
and
f
(
x
)
is decreasing for
x
<
0
since
f
′
(
x
)
<
0
∀
x
∈
(
−
∞
,
0
)
If
x
cos
α
+
y
sin
α
=
p
touches
x
2
+
a
2
y
2
=
a
2
,
then
Report Question
0%
p
2
=
a
2
sin
2
α
+
cos
2
α
0%
p
2
=
a
2
cos
2
α
+
sin
2
α
0%
1
/
p
2
=
sin
2
α
+
α
2
cos
2
α
0%
1
/
p
2
=
cos
2
α
+
α
2
sin
2
α
Explanation
Solving for y, we have
y
=
p
sin
α
−
x
cot
α
Putting this value in
x
2
+
a
2
y
2
=
a
2
, we have
x
2
+
a
2
(
p
sin
α
−
x
cot
α
)
2
=
a
2
⇒
x
2
(
1
+
a
2
cot
2
α
)
−
2
a
2
x
p
cot
α
sin
α
+
a
2
p
2
sin
2
α
−
a
2
=
0
The discriminant of this equation must be zero. So
a
4
p
2
cot
2
α
sin
2
α
=
(
1
+
a
2
cot
2
α
)
(
a
2
p
2
sin
2
α
−
a
2
)
⇒
a
2
p
2
cot
2
α
=
(
1
+
a
2
cot
2
α
)
(
p
2
−
sin
2
α
)
⇒
p
2
(
a
2
cot
2
α
−
1
−
a
2
cot
2
α
)
=
sin
2
α
−
a
2
cos
2
α
⇒
p
2
=
a
2
cos
2
α
+
sin
2
α
If the line
a
x
+
b
y
+
c
=
0
is a normal to the curve
x
y
=
1
, then
Report Question
0%
a
>
0
,
b
>
0
0%
a
>
0
,
b
<
0
0%
a
<
0
,
b
>
0
0%
a
<
0
,
b
<
0
Explanation
Given equation of curve
x
y
=
1
x
d
y
d
x
+
y
=
0
⇒
d
y
d
x
=
−
y
x
Slope of tangent at
P
(
x
1
,
y
1
)
=
−
1
x
2
1
Slope of normal at P is
=
x
2
....(1)
Given equation of normal
a
x
+
b
y
+
c
=
0
Slope of the normal at P is
−
a
b
.....(2)
From (1) and (2),
x
2
=
−
a
b
Since,
x
2
>
0
−
a
b
>
0
⇒
a
b
<
0
i.e.
a
b
should be negative.
⇒
a
<
0
,
b
>
0
or
a
>
0
,
b
<
0
Find the co-ordinates of the points on the curve
y
=
x
/
(
1
+
x
2
)
where the tangent to the curve has greatest slope.
Report Question
0%
(
√
3
,
√
3
4
)
0%
(
0
,
0
)
0%
(
−
√
3
,
−
√
3
4
)
0%
(
1
,
1
2
)
Explanation
Given curve
y
=
x
(
1
+
x
2
)
Here slope
S
=
d
y
d
x
=
{
1.
(
1
+
x
2
)
−
2
x
.
x
}
(
1
+
x
2
)
2
⇒
S
=
(
1
−
x
2
)
(
1
+
x
2
)
2
Now,
d
S
d
x
=
{
−
2
x
(
1
+
x
2
)
2
−
2
(
1
+
x
2
)
.2
x
(
1
−
x
2
)
}
(
1
+
x
2
)
4
=
−
2
x
(
1
+
x
2
)
(
3
−
x
2
)
(
1
+
x
2
)
4
d
S
d
x
=
2
x
[
x
−
(
−
√
3
)
]
[
x
−
√
3
]
(
1
+
x
2
)
3
.
For maximum or minimum of S,
d
S
d
x
=
0
.
⇒
x
=
−
√
3
,
0
,
√
3
.
Now, at
x
=
0
,
d
s
d
x
changes from +ive to -ive
At
x
=
±
√
3
. it changes from -ive to +ive .
Hence slope S is maximum when x=0 and min. when
x
=
±
√
3
, thus for greatest slope, we have x=0 and y=0.
Hence the required point is (0, 0), that is, the origin.
The line
y
=
x
is a tangent to the parabola
y
=
a
x
2
+
b
x
+
c
at the point
x
=
1
.If the parabola passes through the point
(
−
1
,
0
)
, then determine
a
,
b
,
c
.
Report Question
0%
a
=
1
2
,
b
=
1
4
,
c
=
1
3
.
0%
a
=
1
4
,
b
=
1
2
,
c
=
1
4
.
0%
a
=
2
,
b
=
1
,
c
=
4.
0%
a
=
4
,
b
=
2
,
c
=
4.
Explanation
Given equation of parabola is
y
=
a
x
2
+
b
x
+
c
d
y
d
x
=
2
a
x
+
b
Slope of tangent to the curve at
x
=
1
is
2
a
+
b
Given tangent is
y
=
x
. Slope of this tangent is
1.
So,
2
a
+
b
=
1
...(1)
Since, the parabola passes through
(
−
1
,
0
)
∴
a
−
b
+
c
==
0
...(2)
Given
y
=
x
is a tangent at
x
=
1
∴
y
=
1.
Hence
(
1
,
1
)
lies both on tangent and parabola
∴
a
+
b
+
c
=
1
...(3)
Solving (1), (2) and (3), we get
a
=
1
4
,
b
=
1
2
,
c
=
1
4
.
Find
d
y
d
x
if
y
=
[
x
+
√
x
+
√
x
]
1
/
2
, at
x
=
1
Report Question
0%
3
+
4
√
2
8
√
2
(
√
1
+
√
2
)
0%
Not defined
0%
0
0%
e
Explanation
y
=
[
x
+
√
x
+
√
x
]
1
2
y
2
=
x
+
√
x
+
√
x
y
2
−
x
=
√
x
+
√
x
y
4
−
2
x
y
2
+
x
2
=
x
+
√
x
Differentiating with respect to x gives us
4
y
3
y
′
−
2
y
2
−
4
x
y
y
′
+
2
x
=
1
+
1
2
√
x
At
x
=
1
4
y
3
x
=
1
y
′
−
2
y
2
x
=
1
−
4
y
y
′
x
=
1
+
2
=
1
+
1
2
4
y
3
x
=
1
y
′
−
2
y
2
x
=
1
−
4
y
y
′
x
=
1
=
−
1
2
Now at
x
=
1
y
=
√
1
+
√
1
+
√
1
=
√
1
+
√
2
Substituting in the above equation gives us
4
y
3
x
=
1
y
′
−
2
y
2
x
=
1
−
4
y
y
′
x
=
1
=
−
1
2
4
(
√
1
+
√
2
)
(
1
+
√
2
)
y
′
−
2
(
1
+
√
2
)
−
4
(
√
1
+
√
2
)
y
′
=
−
1
2
4
(
√
1
+
√
2
)
y
′
(
1
+
√
2
−
1
)
=
−
1
2
+
2
(
1
+
√
2
)
4
√
2
(
√
1
+
√
2
)
y
′
=
3
2
+
2
√
2
8
√
2
(
√
1
+
√
2
)
y
′
=
3
+
4
√
2
y
′
=
3
+
4
√
2
8
√
2
(
√
1
+
√
2
)
Hence
d
y
d
x
x
=
1
=
3
+
4
√
2
8
√
2
(
√
1
+
√
2
)
A and B are points
(
−
2
,
0
)
and
(
1
,
3
)
on the curve
y
=
4
−
x
2
. If the tangent at P on the curve be parallel to chord AB, then co-ordinates of point P are
Report Question
0%
(
−
1
3
,
5
3
)
0%
(
1
2
,
−
15
4
)
0%
(
−
1
2
,
15
4
)
0%
(
−
1
3
,
1
5
)
Explanation
Given equation of curve
y
=
4
−
x
2
...(i)
d
y
d
x
=
−
2
x
Given points
A
(
−
2
,
0
)
and
(
1
,
3
)
Slope of AB
=
3
3
=
1
⇒
−
2
x
=
1
⇒
x
=
−
1
2
So, by equation (i), we get
y
=
15
4
Hence, the point is
(
−
1
2
,
15
4
)
The line
x
a
+
y
b
=
1
touches the curve
y
=
b
e
−
x
/
a
at the point
Report Question
0%
(
a
,
b
/
a
)
0%
(
−
a
,
b
/
a
)
0%
(
a
,
a
/
b
)
0%
None of these
Explanation
Simplifying the equation of the line we get
b
x
+
a
y
=
a
b
a
y
=
−
b
x
+
a
b
Or
y
=
−
b
a
.
x
+
b
Hence
d
y
d
x
=
−
b
a
Or
−
b
a
.
e
−
x
/
a
.
=
−
b
a
Or
e
−
x
/
a
=
1
Or
x
=
0
Hence
y
=
b
.
Therefore the point is
(
0
,
b
)
If the line,
a
x
+
b
y
+
c
=
0
is a normal to the curve
x
y
=
2
,
then
Report Question
0%
a
<
0
,
b
>
0
0%
a
>
0
,
b
<
0
0%
a
>
0
,
b
>
0
0%
a
<
0
,
b
<
0
Explanation
x
y
′
+
y
=
0
Or
y
′
=
−
y
x
Hence slope of normal
=
x
y
=
2
y
2
Hence the slope is always positive.
Now
slope of the line is
−
a
b
Hence we are left with 2 options.
Either
a
<
0
and
b
>
0
Or
a
>
0
and
b
<
0
.
The function
f
(
x
)
=
2
log
(
x
−
2
)
−
x
2
+
4
x
+
1
increases in the interval
Report Question
0%
(
1
,
2
)
0%
(
2
,
3
)
0%
(
5
/
2
,
3
)
0%
(
2
,
4
)
Explanation
f
(
x
)
=
2
log
(
x
−
2
)
−
x
2
+
4
x
+
1
f
′
(
x
)
=
2
/
(
x
−
2
)
−
2
x
+
4
=
2
⋅
1
−
(
x
−
2
)
2
x
−
2
f
′
(
x
)
=
−
2
⋅
(
x
−
1
)
(
x
−
3
)
x
−
2
For
f
(
x
)
to be increasing,
f
′
(
x
)
>
0
−
2
⋅
(
x
−
1
)
(
x
−
3
)
x
−
2
>
0
⇒
x
ϵ
(
2
,
3
)
(since domain of f is
(
2
,
∞
)
Hence option B, C are correct.
The critical points of the function
f
(
x
)
=
|
x
−
1
|
x
2
are
Report Question
0%
0
0%
1
0%
2
0%
-1
Explanation
for
x
≥
1
f
(
x
)
=
x
−
1
x
2
&
f
′
(
x
)
=
−
x
2
+
2
x
x
4
for
x
<
1
f
(
x
)
=
−
x
+
1
x
2
&
f
′
(
x
)
=
x
2
−
2
x
x
4
for critical points:
f
′
(
x
)
=
0
⇒
x
=
0
,
2
now,
f
′
(
1
+
)
=
1
&
f
′
(
1
−
)
=
−
1
Since,
f
′
(
x
)
changes sign at
x
=
1
Therefore,
x
=
1
is also a critical point
Thus critical points of
f
(
x
)
are
x
=
0
,
1
,
2
Ans: A,B,C
If
f
(
0
)
=
0
and
f
″
(
x
)
>
0
for all
x
>
0
, then
f
(
x
)
x
Report Question
0%
decreases on
(
0
,
∞
)
0%
increases on
(
0
,
∞
)
0%
decreases on
(
1
,
∞
)
0%
neither increases nor decreases on
(
0
,
∞
)
Explanation
Given,
f
(
0
)
=
0
f
″
(
x
)
>
0
, it means
f
′
(
x
)
is also increasing
Let
g
(
x
)
=
f
(
x
)
x
g
′
(
x
)
=
x
f
′
(
x
)
−
f
(
x
)
x
2
f
′
(
x
)
is increasing and
x
ϵ
(
0
,
∞
)
thus
f
′
(
x
)
=
positive
f
(
x
)
=
positive
x
2
=
positive
Therefore,
g
(
x
)
=
positive
>
0
Thus,
f
(
x
)
x
increases for
x
ϵ
(
0
,
∞
)
The interval(s) of decrease of of the function
f
(
x
)
=
x
2
log
27
−
6
x
log
27
+
(
3
x
2
−
18
x
+
24
)
log
(
x
2
−
6
x
+
8
)
is
Report Question
0%
(
3
−
√
1
+
1
/
3
e
,
2
)
0%
(
4
,
3
+
√
1
+
1
/
3
e
)
0%
(
3
,
4
+
√
1
+
1
/
3
e
)
0%
none of these
Explanation
f
′
(
x
)
=
2
x
log
27
+
(
6
x
−
18
)
log
(
x
2
−
6
x
+
8
)
+
(
3
x
2
−
18
+
24
)
(
2
x
−
6
)
x
2
−
6
x
+
8
=
6
(
x
−
3
)
log
(
3
(
x
2
−
6
x
+
8
)
e
)
For
f
(
x
)
to be define
x
2
−
6
x
+
8
>
0
⇒
x
>
4
or
x
<
2
If
x
>
4
then
f
′
(
x
)
<
0
if
log
3
(
x
2
−
6
x
+
8
)
e
<
0
i
.
e
.3
(
x
2
−
6
x
+
8
)
e
<
1
i
.
e
.
x
2
−
6
x
+
(
8
−
1
/
3
e
)
<
0
i
.
e
.
(
x
−
(
3
+
√
1
+
1
/
3
e
)
)
(
x
−
(
3
−
√
1
+
1
/
3
e
)
)
<
0
⇔
3
−
√
1
+
1
/
3
e
<
x
<
3
+
√
1
+
1
/
3
e
Hence
x
ϵ
(
4
,
3
+
√
1
+
1
/
3
e
)
Similarly if
x
<
2
then
f
′
(
x
)
<
0
if
log
3
(
x
2
−
6
x
+
8
)
e
>
0
i
.
e
x
<
3
−
√
1
+
1
/
3
e
or
x
>
3
+
√
1
+
1
/
3
e
Hence
x
ϵ
(
3
−
√
1
+
1
/
3
e
,
2
)
The slope of the tangent to the curve represented by
x
=
t
2
+
3
t
−
8
and
y
=
2
t
2
−
2
t
−
5
at the point
M
(
2
,
−
1
)
is
Report Question
0%
7/6
0%
2/3
0%
3/2
0%
6/7
Explanation
We first determine the value of
t
corresponding to the given values ofx and
y
. From
t
2
+
3
t
−
8
=
2
, we get
t
=
2
,
−
5
, and from
2
t
2
−
2
t
−
5
=
2
we get
t
=
2
,
−
1
. Hence to the given point there corresponds the value
t
=
2
. Therefore, the slope of the tangent at
(
2
,
−
1
)
is
|
y
′
|
t
=
2
=
|
d
y
/
d
t
d
x
/
d
t
|
t
=
2
=
|
4
t
−
2
2
t
+
3
|
t
=
2
=
6
7
The number of critical points of the fuction
f
′
(
x
)
,
where
f
′
(
x
)
=
|
x
−
2
|
x
3
are
Report Question
0%
0
0%
1
0%
3
0%
4
Explanation
f
(
x
)
=
{
x
−
2
x
3
,
x
>
1
2
−
x
x
3
,
x
<
1
,
x
≠
0
f
′
(
x
)
=
{
2
(
3
−
x
)
x
4
,
x
∈
(
−
∞
,
0
)
∪
(
0
,
1
)
2
(
x
−
3
)
x
4
,
x
∈
(
1
,
∞
)
f
″
(
x
)
=
{
6
(
x
−
4
)
x
5
,
x
∈
(
−
∞
,
0
)
∪
(
0
,
1
)
6
(
4
−
x
)
x
5
,
x
∈
(
1
,
3
)
∪
(
3
,
∞
)
f
″
(
x
)
doesn't exits at
x
=
3
Thus critical point of
f
′
(
x
)
is 3.
The value of a for which the function
f
(
x
)
=
(
4
a
−
3
)
(
x
+
log
5
)
+
2
(
a
−
7
)
cot
(
x
/
2
)
sin
2
(
x
/
2
)
does not possess critical points is
Report Question
0%
(
−
∞
,
−
4
/
3
)
0%
(
−
∞
,
−
1
)
0%
(
1
,
∞
)
0%
(
2
,
∞
)
Explanation
f
(
x
)
=
(
4
a
−
3
)
(
x
+
log
5
)
+
2
(
a
−
7
)
cot
(
x
/
2
)
sin
2
(
x
/
2
)
⇒
f
(
x
)
=
(
4
a
−
3
)
(
x
+
log
5
)
+
(
a
−
7
)
sin
x
f
(
x
)
posses critical points when
f
′
(
x
)
=
(
4
a
−
3
)
+
(
a
−
7
)
cos
x
=
0
⇒
cos
x
=
−
4
a
−
3
a
−
7
⇒
−
1
≤
−
4
a
−
3
a
−
7
≤
1
⇒
−
4
3
≤
a
≤
2
Therefore,
f
(
x
)
does not have critical points when
a
∈
(
−
∞
,
−
4
/
3
)
∪
(
2
,
∞
)
Ans: A,D
The critical points of the function
f
(
x
)
=
(
x
−
2
)
2
/
3
(
2
x
+
1
)
are
Report Question
0%
−
1
,
2
0%
1
0%
1
,
−
1
2
0%
1
,
2
Explanation
Given,
f
(
x
)
=
(
x
−
2
)
2
/
3
(
3
x
+
1
)
f
′
(
x
)
=
2
3
(
x
−
2
)
−
1
3
(
2
x
+
1
)
+
2
(
x
−
2
)
2
3
=
10
(
x
−
1
)
(
x
−
2
)
−
1
3
f
′
(
x
)
=
0
⇒
x
=
1
Also
f
′
(
x
)
does not exits at
x
=
2
Hence the critical points are
x
=
1
,
2
The coordinates of the point on the curve
(
x
2
+
1
)
(
y
−
3
)
=
x
where a tangent to the curve has the greatest slope are given by
Report Question
0%
(
√
3
,
3
+
√
3
/
4
)
0%
(
−
√
3
,
3
−
√
3
/
4
)
0%
(
0
,
3
)
0%
none of these
Explanation
Solving for y the given equation we have
y
=
3
=
x
x
2
+
1
⇒
d
y
d
x
=
1
−
x
2
(
1
+
x
2
)
2
=
f
(
x
)
Now
f
′
(
x
)
=
−
2
x
(
3
−
x
2
)
(
1
+
x
2
)
2
For extremum of
f
(
x
)
, we have
f
′
(
x
)
=
0
⇒
x
=
0
,
x
=
±
√
3
At
x
=
0
,
f
′
(
x
)
changes sign from positive to negative
⇒
f
(
x
)
has maxima at
x
=
0
.
Thus the required point is
(
0
,
3
)
The angle at which the curve
y
=
k
e
k
x
intersects the
y
-axis is
Report Question
0%
tan
−
1
(
k
2
)
0%
cot
−
1
(
k
2
)
0%
sin
−
1
(
1
/
√
1
+
k
4
)
0%
sec
−
1
(
1
/
√
1
+
k
4
)
Explanation
d
y
d
x
=
k
2
e
k
x
.
The curve intersects
y
-axis at
(
0
,
k
)
So,
d
y
d
x
|
(
0
,
k
)
=
k
2
.
If
θ
is the angle at which the given
curve intersects the
y
-axis then
tan
(
π
/
2
−
θ
)
=
k
2
−
0
1
+
0.
k
2
=
k
2
.
Hence
θ
=
cot
−
1
k
2
The lines tangent to the curves
y
3
−
x
2
y
+
5
y
−
2
x
=
0
and
x
4
−
x
3
y
2
+
5
x
+
2
y
=
0
at the origin intersect at an angle
θ
equal to
Report Question
0%
π
6
0%
π
4
0%
π
3
0%
π
2
Let
f
(
x
)
=
x
3
+
a
x
+
b
with
a
≠
b
and suppose the tangent lines to the graph of
f
at
x
=
a
and
x
=
b
have the same gradient Then the value of
f
(
1
)
is equal to
Report Question
0%
0
0%
1
0%
−
1
3
0%
2
3
Explanation
f
(
x
)
=
x
3
+
a
x
+
b
f
′
(
x
)
=
3
x
2
+
a
Given gradient at
x
=
a
and at
x
=
b
are same
⇒
3
a
2
+
a
=
3
b
2
+
a
⇒
b
2
=
a
2
But given
a
≠
b
⇒
a
+
b
=
0
.
.
(
1
)
Hence
f
(
1
)
=
1
+
a
+
b
=
1
using (1)
A curve with equation of the form
y
=
a
x
4
+
b
x
3
+
c
x
+
d
has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1, 0) then the values of x for which the curve has a negative gradient are
Report Question
0%
x > -1
0%
x < 1
0%
x < -1
0%
−
1
≤
×
≤
1
Explanation
Given,
y
=
a
x
4
+
b
x
3
+
c
x
+
d
⇒
y
′
=
4
a
x
3
+
3
b
x
2
+
c
Using given conditions,
y
(
0
)
=
1
⇒
d
=
1
y
′
(
0
)
=
0
⇒
c
=
0
y
(
−
1
)
=
0
⇒
a
−
b
=
−
1
.
.
(
1
)
and
y
′
(
−
1
)
=
0
⇒
4
a
−
3
b
=
0
.
.
(
2
)
Solving equation (1) and (2) we get,
a
=
3
,
b
=
4
Hence the polynomial is,
y
=
3
x
4
+
4
x
3
+
1
y
′
=
12
x
2
(
1
+
x
)
Now for negative gradient
y
′
<
0
⇒
12
x
2
(
1
+
x
)
<
0
⇒
x
<
−
1
If
f
′
(
x
)
=
g
(
x
)
(
x
−
a
)
2
, where
g
(
a
)
≠
0
and
g
is continuous at
x
=
a
then
Report Question
0%
f
is increasing near a if
g
(
a
)
>
0
0%
f
is increasing near a if
g
(
a
)
<
0
0%
f
is decreasing near a if
g
(
a
)
>
0
0%
f
is decreasing near a if
g
(
a
)
<
0
Explanation
Since
g
is continuous at
x
=
a
if
g
(
a
)
>
0
, there exist an open interval
I
containing
a
so that
g
(
x
)
>
0
∀
x
∈
I
⇒
f
′
(
x
)
≥
0
∀
x
∈
I
. Therefore,
f
is increasing near
a
.
Similarly
f
is decreasing near
a
if
g
(
a
)
<
0
.
Ans: A,D
The curve
y
=
a
x
3
+
b
x
2
+
c
x
+
8
touches
x
-axis at
P
(
−
2
,
0
)
and cuts the
y
-axis at a point
Q
(
0
,
8
)
where its gradient isThe values of
a
,
b
,
c
are respectively
Report Question
0%
−
1
2
,
−
3
4
,
3
0%
3
,
−
1
2
,
−
4
0%
−
1
2
,
−
7
4
,
2
0%
none of these
Explanation
Given,
y
=
a
x
3
+
b
x
2
+
c
x
+
8
∴
d
y
d
x
=
3
a
x
2
+
2
b
x
+
c
Since the curve touches
x
-axis at
(
−
2
,
0
)
so
d
y
d
x
|
(
−
2
,
0
)
=
0
⇒
12
a
−
4
b
+
c
=
0
(
i
)
The curve cut the
y
-axis at
(
0
,
8
)
so
d
y
d
x
|
(
0
,
8
)
=
3
⇒
c
=
3
Also the curve passes through
(
−
2
,
0
)
so
0
=
−
8
a
+
4
b
−
2
c
+
8
⇒
−
8
a
+
4
b
−
2
=
0
(
i
i
)
Solving
(
i
)
and
(
i
i
)
a
=
−
1
/
4
,
b
=
0
Suppose
f
′
(
x
)
exists for each
x
and
h
(
x
)
=
f
(
x
)
−
(
f
(
x
)
)
2
+
(
f
(
x
)
)
3
for every real number
x
. Then
Report Question
0%
h
is increasing whenever f is increasing
0%
h
is increasing whenever f is decreasing
0%
h
is decreasing whenever f is decreasing
0%
nothing can be said in general.
Explanation
h
(
x
)
=
f
(
x
)
−
(
f
(
x
)
)
2
+
(
f
(
x
)
)
3
h
′
(
x
)
=
f
′
(
x
)
−
2
f
′
(
x
)
f
(
x
)
+
3
f
′
(
x
)
f
(
x
)
2
=
3
f
′
(
x
)
[
f
(
x
)
2
−
2
3
f
(
x
)
+
1
3
]
=
3
f
′
(
x
)
[
(
f
(
x
)
−
1
/
3
)
2
+
2
/
9
]
Thus,
h
′
(
x
)
>
0
if
f
′
(
x
)
>
0
and
h
′
(
x
)
<
0
if
f
′
(
x
)
<
0
Therefore,
h
increases whenever
f
increases
and
h
decreases whenever
f
decreases
Ans: A,C
The critical points of the function
f
(
x
)
=
(
x
−
2
)
2
/
3
(
2
x
+
1
)
are
Report Question
0%
1
and
2
0%
1
and
−
1
2
0%
−
1
and
2
0%
1
Explanation
f
(
x
)
=
(
x
−
2
)
2
/
3
(
2
x
+
1
)
f
′
(
x
)
=
2
3
(
x
−
2
)
−
1
/
3
(
2
x
+
1
)
(
x
−
2
)
2
/
3
=
0.2
Clearly
f
′
(
x
)
is not defined at
x
=
2
.
∴
x
=
2
is a critical point.
Another critical point is given by
f
′
(
x
)
=
0
,
i.e.,
2
3
2
x
+
1
(
x
−
2
)
1
/
3
+
2
(
x
−
2
)
2
/
3
=
0
⇒
2
3
(
2
x
+
1
)
+
2
(
x
−
2
)
=
0
⇒
4
x
+
2
+
6
x
−
12
=
0
⇒
x
=
1
The graph a function
f
is given.
On what interval is
f
increasing ?
Report Question
0%
(
−
1
,
3
]
0%
(
−
3
,
1
)
0%
(
−
3
,
1
]
0%
none of these
Explanation
A function
f
(
x
)
is said to be increasing if as
x
increases
f
(
x
)
increases as well
It can be Clearly observed from the graph that for the interval
(
−
1
,
3
]
function f is increasing
The points of contact of the vertical tangents
x
=
2
−
3
sin
θ
,
y
=
3
+
2
cos
θ
are
Report Question
0%
(
2
,
5
)
,
(
2
,
1
)
0%
(
−
1
,
3
)
,
(
5
,
3
)
0%
(
2
,
5
)
,
(
5
,
3
)
0%
(
−
1
,
3
)
,
(
2
,
1
)
Explanation
For the tangents to be vertical,
d
y
d
x
=
∞
Or
d
x
d
y
=
0
Or
d
x
d
θ
=
0
Or
−
3
cos
θ
=
0
Or
θ
=
2
n
−
1
2
π
Hence
θ
=
π
2
,
3
π
2
.
Now
x
π
2
=
−
1
y
π
2
=
3
.
Similarly
x
3
π
2
=
5
y
3
π
2
=
3
.
Hence the points are
(
−
1
,
3
)
and
(
3
,
5
)
.
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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