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CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Maths
Application Of Derivatives
Quiz 10
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 function f(x) is
Report Question
0%
increasing for all x
0%
non-monotonic
0%
decreasing for all x
0%
None of these
Explanation
We have the equations of the tangents of the curve
y
=
∫
x
−
∞
f
(
t
)
d
t
and
y
=
f
(
x
)
at arbitrary points on them are
Y
−
∫
x
−
∞
f
(
t
)
d
t
=
f
(
x
)
(
X
−
x
)
and
Y
−
f
(
x
)
=
f
′
(
x
)
(
X
−
x
)
As Eqs.(i) and (ii) intersect at the same point on the X-axis
Putting Y=0 and equating x-coordinates, we have
x
−
f
(
x
)
f
′
(
x
)
=
x
−
∫
x
−
∞
f
(
t
)
d
t
f
(
x
)
⇒
f
(
x
)
∫
x
−
∞
f
(
t
)
d
t
=
f
′
(
x
)
f
(
x
)
⇒
∫
x
−
∞
f
(
t
)
d
t
=
c
f
(
x
)
As, f(0)=1
⇒
∫
x
−
∞
f
(
t
)
d
t
=
c
×
1
⇒
c
=
1
2
⇒
∫
x
−
∞
f
(
t
)
d
t
=
1
2
f
(
x
)
;
differentiating both the sides and integrating and using boundary conditions, we get
f
(
x
)
=
e
2
x
;
y
=
2
e
x
is tangent to
y
=
e
2
x
∴
Number of solutions=1
Clearly, f(x) is increasing for all x.
∴
l
i
m
x
→
∞
(
e
2
x
)
e
−
2
x
=
1
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
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
Number of critical point for
y
=
f
(
x
)
for
x
∈
[
0
,
2
]
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
Integrating,
d
2
y
d
x
2
=
6
x
−
4
,
we get
d
y
d
x
=
3
x
2
−
4
x
+
A
When
x
=
1
,
d
y
d
x
=
0
so that
A
=
1
.
Hence,
d
y
d
x
=
3
x
2
−
4
x
+
1
Integrating we get
y
=
x
3
−
2
x
2
+
x
+
B
.
When
x
=
1
,
y
=
5
so that
B
=
5
Thus, we have
y
=
x
3
−
2
x
2
+
x
+
5
From Eq. (i) we get the critical points
x
=
1
3
,
x
=
1
At the critical point
x
=
1
3
,
d
2
y
d
x
2
is negative.
Therefore, at
x
=
1
3
,
y
has a local maximum.
At
x
=
1
,
d
2
y
d
x
2
is positive. THerefore, at
x
=
1
,
y
has a local
minimum.
Also,
f
(
1
)
=
5
,
f
(
1
3
)
=
157
27
,
f
(
0
)
=
5
,
f
(
2
)
=
7
Hence, the global value
=
7
And the global minimum value
=
5
The point of the curve
y
2
=
x
where the tangent makes an angle of
π
4
with x-axis is
Report Question
0%
(
1
2
,
1
4
)
0%
(
1
4
,
1
2
)
0%
(
4
,
2
)
0%
(
1
,
1
)
Explanation
Given curve is
y
2
=
x
Therefore as per given condition
d
y
d
x
=
1
2
y
=
t
a
n
π
4
=
1
⇒
y
=
1
2
⇒
x
=
1
4
Therefore, correct answer is
B
.
The abscissa of the point on the curve
3
y
=
6
x
−
5
x
3
the normal at which passes through origin is :
Report Question
0%
1
0%
1
3
0%
2
0%
1
2
Explanation
Let
(
x
1
,
y
1
)
be the point on the given curve
3
y
=
6
x
−
5
x
3
at which the normal passes through the origin. Then we have
(
d
y
d
x
)
(
x
1
,
y
1
)
=
2
−
5
x
2
1
.
Again the equation of the normal at
(
x
1
,
y
1
)
passing through the origin gives
2
−
5
x
2
1
=
−
x
1
y
1
=
−
3
6
−
5
x
2
1
Since.
x
1
=
1
satisfies the equation, therefore, correct answer is
(
A
)
.
The curve
y
=
x
1
5
has at
(
0
,
0
)
Report Question
0%
a vertical tangent (parallel to y-axis)
0%
a horizontal tangent (parallel to x-axis)
0%
an oblique tangent
0%
no tangent
Explanation
We have,
y
=
x
1
/
5
⇒
d
y
d
x
=
1
5
x
1
5
−
1
=
1
5
x
−
4
/
5
∴
(
d
y
d
x
)
(
0
,
0
)
=
1
5
×
(
0
)
−
4
/
5
=
∞
So, the curve
y
=
x
1
/
5
has vertical tangent at
(
0
,
0
)
, which is parallel to y-axis.
The equation of the curve satisfying the differential equation
y
2
(
x
2
+
1
)
=
2
x
y
1
passing through the point
(
0
,
1
)
and having slope of tangent at
x
=
0
as
3
(where
y
2
and
y
1
represents 2nd and 1st order derivative), then
Report Question
0%
y
=
f
(
x
)
is a strictly increasing function
0%
y
=
f
(
x
)
is non-monotomic finction
0%
y
=
f
(
x
)
has three distinct real roots
0%
y
=
f
(
x
)
has only one negative root
Explanation
The given differential equation is
y
2
(
x
2
+
1
)
=
2
x
y
1
or,
y
2
y
1
=
2
x
x
2
+
1
Integrating both sides, we get
log
y
1
=
log
(
x
2
+
1
)
+
log
C
y
1
=
C
(
x
2
+
1
)
(1)
It is given that
y
1
=
3
at
x
=
0
Putting
x
=
0
,
y
1
=
3
in equation (1), we get
3
=
C
Substituting the value of
C
in (1), we obtain
y
1
=
3
(
x
2
+
1
)
(2)
Integrating both sides w.r.t. to
x
, we get
y
=
x
3
+
3
x
+
C
2
This passes through the point
(
0
,
1
)
. Therefore,
1
=
C
2
Hence, the requiredequation of the curve is
y
=
x
3
+
3
x
+
1
Obviously it is strictly increasing from equation (2)
Also
f
(
0
)
=
1
>
0
, then the only root is negative.
The tangent to the curve
y
=
e
2
x
at the point
(
0
,
1
)
meets x-axis at:
Report Question
0%
(
0
,
1
)
0%
(
−
1
2
,
0
)
0%
(
2
,
0
)
0%
(
0
,
2
)
Explanation
The equation of curve is
y
=
e
2
x
Since, it passes through the point
(
0.1
)
.
∴
d
y
d
x
=
e
2
x
.2
=
2.
e
2
x
⇒
(
d
y
d
x
)
x
=
0
=
2
e
2
×
0
=
2
=
slope of tangent to the curve
∴
Equation of tangent is
y
−
1
=
2
(
x
−
0
)
⇒
y
=
2
x
+
1
Since, tangent to curv
y
=
e
2
x
at the point
(
0
,
1
)
meets X-axis i.e.,
y
=
0
.
∴
0
=
2
x
+
1
⇒
x
=
−
1
2
So, the required point is
(
−
1
2
,
0
)
.
The slope of 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%
−
6
7
0%
−
6
Explanation
Equation of curve is given is given by
x
=
t
2
+
3
t
−
8
and
y
=
2
t
2
−
2
t
−
5.
∴
d
x
d
t
=
2
t
+
3
and
d
y
d
x
=
4
t
−
2
⇒
d
y
d
x
=
d
y
d
t
d
x
d
t
=
4
t
−
2
2
t
+
3
.
.
.
.
(
i
)
Since, the curve passes through the point
(
2
,
−
1
)
.
∴
2
=
t
2
+
3
t
−
8
and
−
1
=
2
t
2
−
2
t
−
5
⇒
t
2
+
3
t
−
10
=
0
and
2
t
2
−
2
t
−
4
=
0
⇒
t
2
+
5
t
−
2
t
−
10
=
0
and
2
t
2
+
2
t
−
4
t
−
4
=
0
⇒
t
(
t
+
5
)
−
2
(
t
+
5
)
=
0
and
2
t
(
t
+
1
)
−
4
(
t
+
1
)
=
0
⇒
(
t
−
2
)
(
t
+
5
)
=
0
and
(
2
t
−
4
)
(
t
+
1
)
=
0
⇒
t
=
2
,
−
5
and
t
=
−
1
,
2
⇒
t
=
2
∴
Slope of tangent,
(
d
y
d
x
)
a
t
t
−
2
=
4
×
2
−
2
2
×
2
+
3
=
6
7
[using Eq.(i)]
The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is :
Report Question
0%
an ellipse
0%
parabola
0%
circle
0%
rectangular hyperbola
Explanation
Step -1: Differentiate slope
.
Given: slope of tangent is equal to ratio of abscissa to the ordinate of the point
.
Let slope of the tangent be
d
y
d
x
.
According to the question
d
y
d
x
=
x
y
Step -2: Solve further to find the equation
.
Separating the variables
.
y
d
y
=
x
d
x
Upon integration we get,
∫
y
d
y
=
∫
x
d
x
⇒
y
2
2
=
x
2
2
+
C
⇒
y
2
=
x
2
+
2
C
⇒
x
2
−
y
2
+
2
C
=
0
⇒
x
2
−
y
2
=
k
[ k = - 2 C]
Hence, the given equation is of a rectangular hyperbola
.
The line
5
x
−
2
y
+
4
k
=
0
is tangent to
4
x
2
−
y
2
=
36
, then k is:
Report Question
0%
9
4
0%
81
16
0%
4
9
0%
2
3
Explanation
The equation of the hyperbola is
4
x
2
−
y
2
=
36
4
x
2
36
−
y
2
36
=
1
4
x
2
9
−
y
2
36
=
1
............(1)
a
2
=
9
;
b
2
=
36
The equation of the line is
5
x
−
2
y
+
4
k
=
0
2
y
=
5
x
+
4
k
y
=
5
2
x
+
2
k
............(2)
From equation (2)
m
=
5
2
,
c
=
2
k
c
2
=
a
2
m
2
−
b
2
(
2
k
)
2
=
9
(
5
2
)
2
−
36
4
k
2
=
9
×
25
4
−
36
4
k
2
=
225
−
144
4
k
2
=
81
16
⇒
k
=
9
4
The function
f
(
x
)
=
t
a
n
x
−
x
Report Question
0%
always increases
0%
always decreases
0%
never increases
0%
sometime increase and sometimes decreases
Which of the following function is decreasing on
(
0
,
π
2
)
Report Question
0%
s
i
n
2
x
0%
t
a
n
x
0%
c
o
s
x
0%
c
o
s
3
x
If the tangent at
(
1
,
1
)
on
y
2
=
x
(
2
−
x
)
2
meets the curve again at
P
, then
P
is
Report Question
0%
(
4
,
4
)
0%
(
−
1
,
2
)
0%
(
3
,
6
)
0%
(
9
4
,
3
8
)
Explanation
(
9
4
,
3
8
)
[Hint:
y
2
=
x
(
2
−
x
)
2
=
x
(
4
−
4
x
+
x
2
)
=
x
3
−
4
x
2
−
4
x
∴
2
y
d
y
d
x
=
3
x
2
−
8
x
+
4
∴
d
y
d
x
=
3
x
2
−
8
x
+
4
2
y
∴
(
d
y
d
x
)
a
t
(
1
,
1
)
=
3
(
1
)
2
−
8
(
1
)
+
4
2
(
1
)
=
1
2
= slope of the tangent at
(
1
,
1
)
y
−
1
=
−
1
2
(
x
−
1
)
∴
2
y
−
12
=
−
x
+
1
∴
x
+
2
y
=
3
Only the coordinates
(
9
4
,
3
8
)
satisfy both the
equation
y
2
=
x
(
2
−
x
)
2
and
x
+
2
y
=
3
∴
P
is
(
9
4
,
3
8
)
].
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
S
l
o
p
e
o
f
t
a
n
g
e
n
t
t
o
c
u
r
v
e
y
=
f
(
x
)
i
s
d
y
d
x
U
s
i
n
g
c
h
a
i
n
r
u
l
e
d
y
d
x
=
(
d
y
d
t
)
(
d
x
d
t
)
G
i
v
e
n
x
=
t
2
+
3
t
−
8
⟹
d
x
d
t
=
2
t
+
3
G
i
v
e
n
y
=
2
t
2
−
2
t
−
5
⟹
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
∴
Answer (B)
6
7
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,
y
2
=
4
x
⇒
2
y
d
y
d
x
=
4
⇒
d
y
d
x
=
2
y
but slope of tangent is
m
=
2
y
⇒
y
=
2
m
When
y
=
2
m
,
x
=
y
2
4
=
4
/
m
2
/
4
=
1
m
2
x
=
1
m
2
,
y
=
2
m
lie on the line
y
=
m
x
+
1
2
m
=
m
×
1
m
2
+
1
⇒
2
m
=
1
m
+
1
2
m
=
1
⇒
m
=
1
Correct answer (A).
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%
x
−
y
=
1
Explanation
Given,
2
y
+
x
2
=
3
2
d
y
d
x
+
2
x
=
0
⇒
d
y
d
x
=
−
x
∴
d
y
d
x
=
−
1
Slope of tangent =
−
1
∴
equation of normal is
y
−
1
=
1
(
x
−
1
)
⇒
y
−
x
=
0
⇒
x
−
y
=
0
∴
Answer (B).)
The slope of the normal to the curve
y
=
2
x
2
+
3
sin
x
at
x
=
0
is
Report Question
0%
3
0%
1
/
3
0%
−
3
0%
−
1
/
3
Explanation
Given,
y
=
2
x
2
+
3
sin
x
slope
=
d
y
d
x
=
4
x
+
3
cos
x
d
y
d
y
|
x
=
0
=
4
(
0
)
=
3
slope of normal =
−
1
/
d
y
d
x
=
−
1
/
3
The normal to the curve
x
2
=
4
y
passing
(
1
,
2
)
is
Report Question
0%
x
+
y
=
3
0%
x
−
y
=
3
0%
x
+
y
=
1
0%
x
−
y
=
1
Explanation
Given,
4
y
=
x
2
⇒
4
d
y
d
x
=
2
x
⇒
d
y
d
x
|
x
=
x
1
=
x
1
2
slope of tangent is
=
x
1
2
slope of normal is
=
−
2
x
1
equation of normal
y
−
y
1
=
−
2
x
1
(
x
−
x
1
)
but this passes they
(
1
,
2
)
2
−
y
1
=
−
2
x
(
1
−
x
)
⇒
2
−
y
1
=
−
2
x
1
+
2
y
1
=
2
x
1
but
x
2
=
4
y
x
2
1
=
4
y
1
x
2
1
=
4
×
2
x
1
=
8
⇒
x
1
=
2
when
x
1
=
2
y
1
=
2
2
=
1
∴
equation of normal is
y
−
1
=
−
2
2
×
(
x
−
2
)
⇒
y
−
1
=
(
x
−
2
)
x
+
y
=
3
The line
y
=
x
+
1
is a tangent to the curve
y
2
=
4
x
at the point
Report Question
0%
(
1
,
2
)
0%
(
2
,
1
)
0%
(
1
,
−
2
)
0%
(
−
1
,
2
)
Explanation
Given,
y
2
=
4
x
⇒
2
y
d
x
d
x
=
4
⇒
d
x
d
y
=
2
/
y
but given that
y
=
x
+
1
is a tangent to the curve its slope =
1
(
y
=
m
x
+
c
)
2
y
=
1
⟹
y
=
2
x
=
2
−
1
=
1
The line
y
=
x
+
1
is a tangent to the curve
y
2
=
4
x
at the point
(
1
,
2
)
.
The points on the curve
9
y
2
=
x
3
, where the normal to the curve makes equal intercepts with the axes are ...........
Report Question
0%
(
4
,
±
8
3
)
0%
(
4
,
−
8
3
)
0%
(
4
,
+
8
3
)
0%
(
±
4
,
8
3
)
For
a
∈
[
π
,
2
π
]
and
n
∈
I
, the critical points of
f
(
x
)
=
1
3
sin
a
tan
3
x
+
(
sin
a
−
1
)
tan
x
+
√
a
−
2
8
−
a
is
Report Question
0%
x
=
n
π
0%
x
=
2
n
π
0%
x
=
(
2
n
+
1
)
π
0%
no critical points
Explanation
f
(
x
)
=
1
3
sin
a
tan
3
(
x
)
+
(
sin
a
−
1
)
tan
x
+
√
a
−
2
8
−
a
f
′
(
x
)
=
sec
2
x
(
sin
a
tan
2
x
+
sin
a
−
1
)
For critical points,
f
′
(
x
)
=
0
⇒
sec
2
x
(
sin
a
tan
2
x
+
sin
a
−
1
)
=
0
⇒
tan
2
x
=
1
−
sin
a
sin
a
[
∵
sec
2
x
=
0
(not possible)]
Since,
a
∈
[
π
,
2
π
]
⇒
1
−
sin
a
sin
a
<
0
⇒
tan
2
x
<
0
which is false.
Hence, no critical points.
Let
f
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
d
, where
a
,
b
,
c
,
d
are real and
3
b
2
<
c
2
, is an increasing function and
g
(
x
)
=
a
f
′
(
x
)
+
b
f
″
(
x
)
+
c
2
. lf
G
(
x
)
=
∫
x
α
g
(
t
)
d
t
,
α
∈
R
, then for
α
<
x
<
α
+
1
,
Report Question
0%
G(x) is a decreasing function
0%
G(x) is an increasing function
0%
G(x) is neither increasing nor decreasing
0%
G(x) is a one-one function
Explanation
f
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
d
f
′
(
x
)
=
3
a
x
2
+
2
b
x
+
c
Since,
f
(
x
)
is increasing.
⇒
f
′
(
x
)
=
3
a
x
2
+
2
b
x
+
c
>
0
⇒
a
>
0
and
4
b
2
−
12
a
c
<
0
⇒
a
>
0
and
b
2
<
3
a
c
Also,
g
(
x
)
=
a
f
′
(
x
)
+
b
f
″
(
x
)
+
c
2
⇒
g
(
x
)
=
a
(
3
a
x
2
+
2
b
x
+
c
)
+
b
(
6
a
x
+
2
b
)
+
c
2
⇒
g
(
x
)
=
3
a
2
x
2
+
8
a
b
x
+
2
b
2
+
c
2
+
a
c
D
=
64
a
2
b
2
−
4
(
3
a
2
)
(
2
b
2
+
c
2
+
a
c
)
⇒
D
=
4
a
2
(
10
b
2
−
3
c
2
−
3
a
c
)
<
4
a
2
(
10
b
2
−
3
c
2
−
b
2
)
⇒
D
<
4
a
2
(
9
b
2
−
3
c
2
)
⇒
D
<
12
a
2
(
3
b
2
−
c
2
)
Since
⇒
3
b
2
−
c
2
<
0
,
⇒
D
<
0
⇒
g
(
x
)
>
0
for all
x
∈
R
.
G
(
x
)
=
∫
x
α
g
(
t
)
d
t
is a strictly increasing function.
Hence,
G
(
x
)
is one-one in the given interval.
Let
f
′
(
sin
x
)
<
0
and
f
″
(
sin
x
)
>
0
,
∀
x
∈
(
0
,
π
2
)
and
g
(
x
)
=
f
(
sin
x
)
+
f
(
cos
x
)
,
then
g
(
x
)
is decreasing in
Report Question
0%
(
π
4
,
π
2
)
0%
(
0
,
π
4
)
0%
(
0
,
π
2
)
0%
(
π
6
,
π
2
)
Explanation
Given
g
(
x
)
=
f
(
sin
x
)
+
f
(
cos
x
)
Differentiating w.r.t
x
, we get
g
′
(
x
)
=
f
′
(
sin
x
)
.
cos
x
−
f
′
(
cos
x
)
.
sin
x
Again differentiating w.r.t
x
, we get
g
″
(
x
)
=
−
f
′
(
sin
x
)
.
sin
x
+
f
″
(
sin
x
)
.
cos
2
x
−
f
′
(
cos
x
)
.
cos
x
+
f
″
(
cos
x
)
.
sin
2
x
...(2)
Now, given
f
′
(
sin
x
)
<
0
⇒
f
′
(
sin
(
π
2
−
x
)
)
<
0
(
∵
x
∈
(
0
,
π
2
)
⇒
π
2
−
x
∈
(
0
,
π
2
)
)
⇒
f
′
(
cos
x
)
<
0
...(2)
Similarly
f
″
(
sin
x
)
>
0
⇒
f
″
(
sin
(
π
2
−
x
)
)
>
0
⇒
f
″
(
cos
x
)
>
0
...(3)
∴
g
″
(
x
)
>
0
(
Using (1),(2) and (3)
)
⇒
g
′
(
x
)
is increasing in
(
0
,
π
2
)
.
Also
g
′
(
π
4
)
=
0
⇒
g
′
(
x
)
>
0
∀
x
∈
(
π
4
,
π
2
)
and
g
′
(
x
)
<
0
∀
x
∈
(
0
,
π
4
)
Thus
g
(
x
)
is decreasing in
(
0
,
π
4
)
The point of contact of vertical tangent to the curve given by the equations
x
=
3
−
2
cos
θ
,
y
=
2
+
3
sin
θ
is
Report Question
0%
(1, 5)
0%
(1, 2)
0%
(5, 2)
0%
(2, 5)
Explanation
x
=
3
−
2
c
o
s
θ
,
y
=
2
+
3
s
i
n
θ
The slope of the tangent at
′
θ
′
is
d
y
d
x
=
d
y
d
θ
d
x
d
θ
∴
for vertical tangent
d
x
d
θ
=
0
i
.
e
.
,
2
s
i
n
θ
=
0
⇒
θ
=
0
or
π
o
r
2
π
Then
x
=
1
,
y
=
2
;
x
=
5
,
y
=
2
;
x
=
1
,
y
=
2
∴
the points of contact are
(
1
,
2
)
,
(
5
,
2
)
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
)
∪
(
2
,
∞
)
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
)
(
1
+
0
)
+
2
(
a
−
7
)
[
cot
x
2
×
2
sin
x
2
cos
x
2
×
1
2
+
sin
2
x
2
×
(
−
csc
2
x
2
)
×
1
2
]
f
′
(
x
)
=
(
4
a
−
3
)
+
2
(
a
−
7
)
[
cos
2
x
2
−
1
2
]
f
′
(
x
)
=
(
4
a
−
3
)
+
2
(
a
−
7
)
×
(
cos
x
)
2
[
∵
1
+
cos
x
=
2
cos
2
x
2
]
f
′
(
x
)
=
(
4
a
−
3
)
+
(
a
−
7
)
cos
x
Now, for no critical points
f
′
(
x
)
≠
0
(
4
a
−
3
)
≠
(
7
−
a
)
cos
x
cos
x
≠
4
a
−
3
7
−
a
∴
4
a
−
3
7
−
a
>
1
4
a
−
3
>
7
−
a
5
a
>
10
a
>
2
and
4
a
−
3
7
−
a
<
−
1
4
a
−
3
<
a
−
7
3
a
<
−
4
a
<
−
4
/
3
∴
a
ϵ
(
−
∞
,
−
4
/
3
)
⋃
(
2
,
∞
)
is correct.
The greatest inclination between the tangents is
Report Question
0%
tan
−
1
(
a
+
b
2
√
a
b
)
0%
tan
−
1
(
a
−
b
2
√
a
b
)
0%
tan
−
1
√
a
b
0%
tan
−
1
√
b
a
Explanation
The slope of the tangent to the ellipse
=
−
b
a
cot
θ
The slope of tangent to the circle is
−
cot
θ
∴
the angle
ϕ
between the tangents is given by
tan
ϕ
=
(
−
b
a
+
1
)
cot
θ
1
+
b
a
cot
2
θ
=
(
a
−
b
)
a
tan
θ
+
b
cot
θ
=
a
−
b
(
√
a
tan
θ
−
√
b
cot
θ
)
2
+
2
√
a
b
This is maximum when
√
a
tan
θ
−
√
b
cot
θ
=
0
∴
the maximum angle
=
tan
−
1
(
a
−
b
2
√
a
b
)
A function
y
=
f
(
x
)
has a second order derivative
f
″
(
x
)
=
6
(
x
−
1
)
.
If its graph passes through the point
(
2
,
1
)
and at that point the tangent to the graph is
y
=
3
x
−
5
, then the function is
Report Question
0%
(
x
−
1
)
2
0%
(
x
+
1
)
2
0%
(
x
+
1
)
3
0%
(
x
−
1
)
3
Explanation
f
″
(
x
)
=
6
(
x
−
1
)
⇒
f
′
(
x
)
=
3
x
2
−
6
x
+
k
...(1)
Slope of tangent at
(
2
,
1
)
is
3
.
⇒
3
=
3
×
2
2
−
6
×
2
+
k
⇒
k
=
3
...(2)
From (1) and (2)
f
(
x
)
=
x
3
−
3
x
2
+
3
x
+
c
Using
f
(
x
)
=
y
=
1
when
x
=
2
, we get
1
=
2
+
c
⇒
c
=
−
1
Hence, the function is
f
(
x
)
=
x
3
−
3
x
2
+
3
x
−
1
⇒
f
(
x
)
=
(
x
−
1
)
3
If
f
(
x
)
=
x
sin
x
and
g
(
x
)
=
x
tan
x
where
0
<
x
≤
1
then in the interval
Report Question
0%
Both f(x) and g(x) are increasing functions
0%
Both f(x) and g(x) are decreasing functions
0%
f(x) is an increasing function
0%
g(x) is an increasing function
Explanation
We know that if
f
′
(
x
)
>
0
then
f
is increasing on that interval and if
f
′
(
x
)
<
0
then
f
is decreasing on the interval.
Given
f
(
x
)
=
x
s
i
n
x
⟹
f
′
(
x
)
=
s
i
n
x
−
x
c
o
s
x
s
i
n
2
x
=
0
⟹
s
i
n
x
−
x
c
o
s
x
=
0
⟹
x
=
s
i
n
x
c
o
s
x
=
t
a
n
x
In the interval
(
0
,
1
)
there is no solution for
x
=
t
a
n
x
Therefore,
f
′
(
x
)
>
0
Hence,
f
(
x
)
is an increasing function.
Given
g
(
x
)
=
x
t
a
n
x
⟹
g
′
(
x
)
=
t
a
n
x
−
x
s
e
c
2
x
t
a
n
2
x
=
0
⟹
t
a
n
x
−
x
s
e
c
2
x
=
0
⟹
x
=
s
e
c
2
x
t
a
n
x
=
1
s
i
n
x
c
o
s
x
In the interval
(
0
,
1
)
,
g
(
x
)
is not an increasing function.
A function
y
=
f
(
x
)
is given by
x
=
cos
2
θ
&
y
=
cot
θ
sec
2
θ
for all
θ
>
0
, then
f
is :
Report Question
0%
increasing in
x
∈
(
0
,
3
2
)
& decreasing in
x
∈
(
3
2
,
∞
)
0%
increasing in
x
∈
(
0
,
1
)
0%
increasing in
x
∈
(
0
,
2
)
0%
decreasing in
x
∈
(
2
,
∞
)
Suppose
a
,
b
,
c
are such that the curve
y
=
a
x
2
+
b
x
+
c
is tangent to
y
=
3
x
−
3
at
(
1
,
0
)
and is also tangent to
y
=
x
+
1
at
(
3
,
4
)
then the value of
(
2
a
−
b
−
4
c
)
equals
Report Question
0%
7
0%
8
0%
9
0%
10
Explanation
y
=
a
x
2
+
b
x
+
c
Since
(
1
,
0
)
lies on the curve
⇒
a
+
b
+
c
=
0
...... (i)
Slope of curve
=
d
y
d
x
=
2
a
x
+
b
Slope of curve at (1,0)
=
2
a
+
b
Slope of curve at (3,4)
=
6
a
+
b
Slope of tangent
y
=
3
x
−
3
is 3
Slope of tangent
y
=
x
+
1
is 1
(
d
y
d
x
)
x
=
1
=
3
and
(
d
y
d
x
)
x
=
3
=
1
2
a
+
b
=
3
.... (ii)
6
a
+
b
=
1
.... (iii)
on solving we get a
=
−
1
2
,
b
=
4
,
c
=
−
7
2
∴
2
a
−
b
−
4
c
=
−
1
−
4
+
14
=
9
For the curve
y
=
3
sin
θ
cos
θ
,
x
=
e
θ
sin
θ
,
0
≤
θ
≤
π
, the tangent is parallel to x-axis when
θ
is :
Report Question
0%
π
4
0%
π
2
0%
3
π
4
0%
π
6
Explanation
y
=
3
sin
θ
cos
θ
⇒
d
y
d
θ
=
3
(
cos
θ
−
sin
θ
)
(
cos
θ
+
sin
θ
)
x
=
e
θ
sin
θ
d
x
d
θ
=
3
e
θ
(
cos
θ
−
sin
θ
)
d
y
d
x
=
3
e
−
θ
(
cos
θ
−
sin
θ
)
d
y
d
x
=
0
⇒
sin
θ
=
cos
θ
⇒
θ
=
π
4
If
f
(
x
)
=
x
2
/
3
then
Report Question
0%
(
0
,
0
)
is a point of maxima
0%
(
0
,
0
)
is a point of minima
0%
(
0
,
0
)
is a critical point
0%
There is no critical point
Explanation
f
(
x
)
=
x
2
/
3
f
′
(
x
)
=
2
3
x
1
/
3
Here,
f
′
(
x
)
is undefined at
x
=
0
. So,
f
′
(
x
)
has singularity at
x
=
0
.
If
f
(
x
)
=
{
x
2
+
2
x
<
0
3
x
=
0
x
+
2
x
>
0
, then which of the following statement(s) is/are false ?
Report Question
0%
f
(
x
)
has a local maximum at
x
=
0
0%
f
(
x
)
is strictly decreasing on the left of
x
=
0
0%
f
′
(
x
)
is strictly increasing on the left of
x
=
0
0%
f
′
(
x
)
is strictly increasing on the right of
x
=
0
Explanation
f
(
0
)
=
3
lim
x
→
0
−
f
(
x
)
=
2
lim
x
→
0
+
f
(
x
)
=
2
Therefore there is a local maximum at
x
=
0
f
′
(
x
)
=
2
x
for
x
<
0
therefore
f
(
x
)
is decreasing for
x
<
0
f
″
(
x
)
=
2
for
x
<
0
, hence
f
′
(
x
)
is increasing for
x
<
0
f
′
(
x
)
=
1
>
0
f
o
r
x
>
0
therefore
f
(
x
)
is increasing for
x
>
0
f
′
(
x
)
is constant for
x
>
0
, hence option D is false
For the curve represented parametrically by the equations,
x
=
2
l
n
cot
(
t
)
+
1
&
y
=
tan
(
t
)
+
cot
(
t
)
Report Question
0%
tangent at
t
=
π
/
4
is parallel to x - axis
0%
normal at
t
=
π
/
4
is parallel to y - axis
0%
tangent at
t
=
π
/
4
is parallel to the line
y
=
x
0%
tangent and normal intersect at the point
(
2
,
1
)
Explanation
x
=
2
l
n
c
o
t
(
t
)
+
1
d
x
d
t
=
2
c
o
t
(
t
)
(
−
c
o
s
e
c
2
t
)
=
−
2
s
i
n
2
(
t
)
c
o
t
(
t
)
(
d
x
d
t
)
t
=
π
4
=
−
4
y
=
t
a
n
(
t
)
+
c
o
t
(
t
)
d
y
d
t
=
s
e
c
2
t
−
c
o
s
e
c
2
t
=
s
i
n
2
t
−
c
o
s
2
t
s
i
n
2
t
c
o
s
2
t
(
d
y
d
t
)
t
=
π
4
=
0
⇒
d
y
d
x
=
0
Slope of tangent
=
0
⇒
Tangent is parallel to x-axis.
Slope of normal
=
−
d
x
d
y
=
−
∞
⇒
Normal is parallel to y-axis.
A curve passes through
(
2
,
0
)
and the slope of the tangent at any point
(
x
,
y
)
is
x
2
−
2
x
for all values of
x
. The point of minimum ordinate on the curve where
x
>
0
is
(
a
,
b
)
'
Then find the value of
a
+
6
b
.
Report Question
0%
2
0%
4
0%
−
2
0%
−
4
Explanation
d
y
d
x
=
x
2
−
2
x
y
=
x
3
3
−
x
2
+
C
Passing through (2, 0)
⇒
C
=
4
3
∴
y
=
x
3
3
−
x
2
+
4
3
y
′
=
x
2
−
2
x
=
x
(
x
−
2
)
For maxima or minima,
y
′
=
0
⇒
x
=
0
,
2
y
″
>
0
at
x
=
2
At x = 2, y takes the minimum value.
∴
minimum value of y is
=
8
3
−
4
+
4
3
=
0
∴
a
=
2
,
b
=
0
a
+
6
b
=
2
The value of
x
at which tangent to the curve
y
=
x
3
−
6
x
2
+
9
x
+
4
,
0
≤
x
≤
5
has maximum slope is
Report Question
0%
0
0%
2
0%
5
2
0%
5
Explanation
y
=
x
3
−
6
x
2
+
9
x
+
4
d
y
d
x
=
3
x
2
−
12
x
+
9
Slope of tangent,
m
=
f
(
x
)
=
3
x
2
−
12
x
+
9
For maxima or minima,
f
′
(
x
)
=
0
⇒
6
x
−
12
=
0
⇒
x
=
2
.
Now,
m
(
0
)
=
9
,
m
(
2
)
=
−
3
and
m
(
5
)
=
24
Hence,
The maximum value is attained at
x
=
5
.
Hence, option D.
The point on the curve
y
2
=
x
,
the tangent at which makes an angle of
45
0
with positive direction of
x
−
axis will be given by
Report Question
0%
(
1
2
,
1
4
)
0%
(
1
2
,
1
2
)
0%
(
2
,
4
)
0%
(
1
4
,
1
2
)
Explanation
Step 1: Find the slope of tangent to given curve at any point
Given equation of curve is
y
2
=
x
...(1)
Differentiating w.r.t. x we get
⇒
2
y
d
y
d
x
=
1
⇒
d
y
d
x
=
1
2
y
Hence slope of tangent to given curve at any point
(
x
,
y
)
is
d
y
d
x
=
1
2
y
Step 2: Find the required point
Given slope of tangent
=
tan
45
∘
Hence
1
2
y
=
1
⇒
y
=
1
2
⇒
x
=
1
4
[From eq.(1)]
Hence, the required point is
(
1
4
,
1
2
)
Hence, Option 'D' is correct
A function
y
=
f
(
x
)
has a second-order derivative
f
″
(
x
)
=
6
(
x
−
1
)
. If its graph passes through the point
(
2
,
1
)
and at the point tangent to the graph is
y
=
3
x
−
5
, then the value of
f
(
0
)
is
Report Question
0%
1
0%
−
1
0%
2
0%
0
Explanation
f
″
(
x
)
=
6
x
−
6
∫
f
″
(
x
)
d
x
=
∫
(
6
x
−
6
)
d
x
⇒
f
′
(
x
)
=
3
x
2
−
6
x
+
C
Since, it passes through (2,1) and tangent is
y
=
3
x
−
5
⇒
C
=
3
So,
f
′
(
x
)
=
3
x
2
−
6
x
+
3
∫
f
′
(
x
)
d
x
=
∫
(
3
x
2
−
6
x
+
3
)
d
x
⇒
f
(
x
)
=
x
3
−
3
x
2
+
3
x
+
C
1
Since, it passes through
(
2
,
1
)
⇒
C
1
=
−
1
Hence,
f
(
x
)
=
x
3
−
3
x
2
+
3
x
−
1
f
(
0
)
=
−
1
The period of oscillation
T
of a pendulum of length
l
at a place of acceleration due to gravity
g
is given by
T
=
2
π
√
l
g
. If the calculated length is
0.992
times the actual length and if the value assumed for
g
is
1.002
times its actual value, the relative error in the computed value of
T
is
Report Question
0%
0.005
0%
−
0.005
0%
0.003
0%
−
0.003
Explanation
Relative error will calculate by
Δ
T
T
=
1
2
[
Δ
L
L
−
Δ
g
g
]
Δ
T
T
=
1
2
[
0.992
L
L
−
1.002
g
g
]
Δ
T
=
−
0.005
T
The focal length of a mirror is given by
1
v
−
1
u
=
2
f
. If equal errors (
α
) are made in measuring
u
and
v
, then the relative error in
f
is
Report Question
0%
2
α
0%
α
(
1
u
+
1
v
)
0%
α
(
1
u
−
1
v
)
0%
none of these
Explanation
Given,
1
v
−
1
u
=
2
f
⇒
−
Δ
v
v
2
+
Δ
u
u
2
=
−
2
Δ
f
f
2
Since, given equal errors in measuring u and v i.e.
Δ
u
=
Δ
v
=
α
⇒
α
(
1
u
−
1
v
)
(
1
u
+
1
v
)
=
−
2
Δ
f
f
2
But,
1
v
−
1
u
=
2
f
⇒
Δ
f
f
=
α
(
1
u
+
1
v
)
Hence, correct option is B
The tangent of the acute angle between the curves
y
=
|
x
2
−
1
|
and
y
=
√
7
−
x
2
at their points of intersection is
Report Question
0%
5
√
3
2
0%
3
√
5
2
0%
5
√
3
4
0%
3
√
5
4
Explanation
The given curves are
y
=
|
x
2
−
1
|
and
y
=
√
7
−
x
2
To find the point of intersection, let us equate both the curves.
|
x
2
−
1
|
=
√
7
−
x
2
Squaring on both sides gives us
x
4
−
x
2
−
6
=
0
(
x
2
−
3
)
(
x
2
+
2
)
=
0
On solving above quadratic equation, we get
x
2
=
3
The point of intersection is
(
±
√
3
,
2
)
The slope of the tangent to
y
=
|
x
2
−
1
|
at the point
(
√
3
,
2
)
is
m
1
=
d
y
d
x
=
2
x
=
2
√
3
The slope of tangent to
y
=
√
7
−
x
2
at the point
(
√
3
,
2
)
is
m
2
=
d
y
d
x
=
−
x
√
7
−
x
2
=
−
√
3
2
Let
θ
be the acute angle between the two tangents.
tan
θ
=
m
1
−
m
2
1
+
m
1
m
2
⇒
tan
θ
=
−
5
√
3
4
∴
The tangent of the acute angle between the curves is
5
√
3
4
.
Hence, option C is correct.
The angle made by the tangent of the curve
x
=
a
(
t
+
sin
t
cos
t
)
,
y
=
a
(
1
+
s
i
n
t
)
2
with the
x
−
a
x
i
s
at any point on it is
Report Question
0%
1
4
(
π
+
2
t
)
0%
1
−
sin
t
cos
t
0%
1
4
(
2
t
−
π
)
0%
1
+
sin
t
cos
2
t
Explanation
x
=
a
(
t
+
s
i
n
t
c
o
s
t
)
d
x
d
t
=
2
a
c
o
s
2
t
y
=
a
(
1
+
s
i
n
t
)
2
d
y
d
t
=
2
a
(
c
o
s
t
+
c
o
s
t
s
i
n
t
)
⇒
d
y
d
x
=
1
+
s
i
n
t
c
o
s
t
=
(
c
o
s
t
2
+
s
i
n
t
2
)
2
c
o
s
2
t
2
−
s
i
n
2
t
2
=
1
+
t
a
n
t
2
1
−
t
a
n
t
2
=
t
a
n
(
π
4
+
t
2
)
Slope of tangent to the curve
=
t
a
n
(
π
4
+
t
2
)
So, the angle made by the tangent to the curve with the x-axis
=
π
4
+
t
2
or
1
4
(
π
+
2
t
)
Consider the function
f
(
x
)
=
{
x
sin
π
x
,
f
o
r
x
>
0
0
,
f
o
r
x
=
0
. Then, the number of points in
(
0
,
1
)
where the derivative
f
′
(
x
)
vanishes is
Report Question
0%
0
0%
1
0%
2
0%
infinite
The abscissas of points
P
and
Q
on the curve
y
=
e
x
+
e
−
x
such that tangents at
P
and
Q
make
60
∘
with the
x
-axis are
Report Question
0%
ln
(
√
3
+
√
7
7
)
and
ln
(
√
3
+
√
5
2
)
0%
ln
(
√
3
+
√
7
2
)
0%
ln
(
√
7
−
√
3
7
)
0%
±
ln
(
√
3
+
√
7
2
)
Explanation
y
=
e
x
+
e
−
x
d
y
d
x
=
e
x
−
e
−
x
=
e
2
x
−
1
e
x
Let P
(
x
1
,
y
1
)
and Q
(
x
2
,
y
2
)
be the points on the given curve.
Slope of tangent at P is
m
1
=
e
2
x
1
−
1
e
x
1
⇒
√
3
=
e
2
x
1
−
1
e
x
1
⇒
e
2
x
1
−
√
3
e
x
−
1
=
0
⇒
(
e
x
1
−
√
3
2
)
2
=
7
4
⇒
e
x
1
=
±
√
7
2
+
√
3
2
⇒
x
1
=
l
n
(
√
7
2
+
√
3
2
)
as logarithm of negative numbers is not defined.
Similarly,
⇒
x
2
=
l
n
(
√
7
2
+
√
3
2
)
as it makes same angle,
60
0
with x-axis.
The graphs
y
=
2
x
3
−
4
x
+
2
and
y
=
x
3
+
2
x
−
1
intersect at exactly 3 distinct points. The slope of the line passing through two of these points
Report Question
0%
is equal to 4
0%
is equal to 6
0%
is equal to 8
0%
is not unique
Explanation
Given equation of curves
y
=
2
x
3
−
4
x
+
2
y
=
x
3
+
2
x
−
1
Let
(
x
1
,
y
1
)
be a point of intersection of the curves.
⇒
y
1
=
2
x
3
1
−
4
x
1
+
2
And
y
1
=
x
3
1
+
2
x
1
−
1
⇒
y
1
=
8
x
1
−
4
Let
(
x
2
,
y
2
)
be another point of intersection of the curves.
⇒
y
2
=
2
x
3
2
−
4
x
2
+
2
And
y
2
=
x
3
2
+
2
x
2
−
1
⇒
y
2
=
8
x
2
−
4
Now, slope
=
y
2
−
y
1
x
2
−
x
1
⇒
m
=
8
(
x
2
−
x
1
)
x
2
−
x
1
⇒
m
=
8
If the curve represented parametrically by the equations
x
=
2
ln
cot
t
+
1
and
y
=
tan
t
+
cot
t
Report Question
0%
tangent 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
Explanation
x
=
2
ln
cot
t
+
1
d
x
d
t
=
−
2
c
o
s
e
c
2
t
cot
t
=
−
2
sin
t
cos
t
y
=
tan
t
+
cot
t
d
y
d
t
=
s
e
c
2
t
−
c
o
s
e
c
2
t
=
s
i
n
2
t
−
c
o
s
2
t
s
i
n
2
t
c
o
s
2
t
⇒
d
y
d
x
=
−
s
i
n
2
t
−
c
o
s
2
t
s
i
n
2
t
⇒
d
y
d
x
=
cot
2
t
Slope of tangent at
t
=
π
4
is 0.
Hence, tangent is parallel to x-axis.
Let
S
be a square with sides of length
x
. If we approximate the change in size of the area of
S
by
h
.
d
A
d
x
|
x
=
x
0
, when the sides are changed from
x
0
to
x
o
+
h
, then the absolute value of the error in our approximation, is
Report Question
0%
h
2
0%
2
h
x
0
0%
x
2
0
0%
h
Explanation
We have S be a square with side length x
Since, area of square
A
=
x
2
Differentiate both sides with respect to x
d
A
d
x
=
2
x
Given , approximate change in size of area
=
Δ
A
=
h
.
d
A
d
x
|
x
=
x
0
⇒
Δ
A
=
2
x
0
h
Also,
(
A
+
Δ
A
)
−
A
=
(
x
0
+
h
)
2
−
x
0
2
=
2
x
0
h
+
h
2
Hence, absolute value of error in our approximation is
h
2
Hence, Option (A) is correct.
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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