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Mathematics
List of top Mathematics Questions asked in AP EAMCET
Let
D
=
R
−
{
0
,
1
}
D=R-\{0,1\}
D
=
R
−
{
0
,
1
}
and
f
:
D
→
D
,
g
:
D
→
D
f: D \rightarrow D, g: D \rightarrow D
f
:
D
→
D
,
g
:
D
→
D
and
h
:
D
→
D
h: D \rightarrow D
h
:
D
→
D
be three functions defined by
f
(
x
)
=
1
x
;
g
(
x
)
=
1
−
x
f(x)=\frac{1}{x} ; g(x)=1-x
f
(
x
)
=
x
1
;
g
(
x
)
=
1
−
x
and
h
(
x
)
=
1
1
−
x
.
h(x)=\frac{1}{1-x} .
h
(
x
)
=
1
−
x
1
.
If
j
:
D
→
D
j: D \rightarrow D
j
:
D
→
D
is such that
(
g
o
j
o
f
)
(gojof)
(
g
o
j
o
f
)
(
x
)
=
f
(
x
)
(x)=f(x)
(
x
)
=
f
(
x
)
for all
x
∈
D
x \in D
x
∈
D
, then which one of the following is
j
(
x
)
?
j(x) ?
j
(
x
)?
AP EAMCET - 2019
AP EAMCET
Mathematics
Functions
If
y
=
f
(
x
)
y = f(x)
y
=
f
(
x
)
is twice differentiable function such that at a point
P
,
d
y
d
x
=
4
,
d
2
y
d
x
2
=
−
3
P , \frac{dy}{dx} = 4 , \frac{d^2 y}{dx^2} = - 3
P
,
d
x
d
y
=
4
,
d
x
2
d
2
y
=
−
3
, then
(
d
2
x
d
y
2
)
P
=
\left( \frac{d^2 x}{dy^2} \right)_P =
(
d
y
2
d
2
x
)
P
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Derivatives of Functions in Parametric Forms
If the line joining the points
A
(
α
)
A(\alpha)
A
(
α
)
and
B
(
β
)
B(\beta)
B
(
β
)
on the ellipse
x
2
25
+
y
2
9
=
1
\frac{x^2}{25} + \frac{y^2}{9} = 1
25
x
2
+
9
y
2
=
1
is a focal chord, then one possible values of
cot
α
2
.
cot
β
2
\cot \frac{\alpha}{2} . \cot \frac{\beta}{2}
cot
2
α
.
cot
2
β
is
AP EAMCET - 2019
AP EAMCET
Mathematics
Ellipse
If
x
=
3
10
+
3.7
10.15
+
3.7.9
10.15.20
+
x = \frac{3}{10} + \frac{3.7}{10.15} + \frac{3.7.9}{10.15.20} +
x
=
10
3
+
10.15
3.7
+
10.15.20
3.7.9
+
...., then
5
x
+
8
5x + 8
5
x
+
8
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Series
If each line of a pair of lines passing through origin is at a perpendicular distance of
4
4
4
units from the point
(
3
,
4
)
(3, 4)
(
3
,
4
)
, then the equation of the pair of lines is
AP EAMCET - 2019
AP EAMCET
Mathematics
general equation of a line
P
P
P
is a variable point on the ellipse
x
2
a
2
+
y
2
b
2
=
1
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
a
2
x
2
+
b
2
y
2
=
1
with foci
F
1
F_1
F
1
and
F
2
F_2
F
2
. If
A
A
A
is the area of the triangle
P
F
1
F
2
PF_1F_2
P
F
1
F
2
. then the maximum value of
A
A
A
is
AP EAMCET - 2019
AP EAMCET
Mathematics
Ellipse
In a
Δ
A
B
C
,
2
x
+
3
y
+
1
=
0
,
x
+
2
y
−
2
=
0
\Delta ABC, 2x + 3y + 1 = 0 , x + 2y - 2 = 0
Δ
A
BC
,
2
x
+
3
y
+
1
=
0
,
x
+
2
y
−
2
=
0
are the perpendicular bisectors of its sides
A
B
AB
A
B
and
A
C
AC
A
C
respectively and if
A
=
(
3
,
2
)
A = (3,2)
A
=
(
3
,
2
)
, then the equation of the side
B
C
BC
BC
is
AP EAMCET - 2019
AP EAMCET
Mathematics
general equation of a line
If
A
B
C
D
ABCD
A
BC
D
is a cyclic quadrilateral with
A
B
=
6
,
B
C
=
4
,
C
D
=
5
,
D
A
=
3
AB = 6, BC = 4, CD = 5, DA = 3
A
B
=
6
,
BC
=
4
,
C
D
=
5
,
D
A
=
3
and
∠
A
B
C
\angle ABC
∠
A
BC
=
θ
\theta
θ
then
c
o
s
θ
cos\, \theta
cos
θ
=
AP EAMCET - 2019
AP EAMCET
Mathematics
circle
If
′
a
′
'a'
′
a
′
is the middle term in the expansion of
(
2
x
−
3
y
)
8
(2x - 3y)^8
(
2
x
−
3
y
)
8
and
b
,
c
b, c
b
,
c
are the middle terms in the expansion of
(
3
x
+
4
y
)
7
(3x + 4y)^7
(
3
x
+
4
y
)
7
, then the value of
b
+
c
a
\frac{b +c}{a}
a
b
+
c
,when
x
=
2
x = 2
x
=
2
and
y
=
3
y = 3
y
=
3
, is
AP EAMCET - 2019
AP EAMCET
Mathematics
binomial expansion formula
If a random variable
X
X
X
has the probability distribution given by
P
(
X
=
0
)
=
3
C
3
,
P
(
X
=
2
)
=
5
C
−
10
C
2
P(X = 0) = 3C^3, P(X = 2 ) = 5C - 10C^2
P
(
X
=
0
)
=
3
C
3
,
P
(
X
=
2
)
=
5
C
−
10
C
2
and
P
(
X
=
4
)
=
4
C
−
1
P(X = 4) = 4C - 1
P
(
X
=
4
)
=
4
C
−
1
, then the variance of that distribution is
AP EAMCET - 2019
AP EAMCET
Mathematics
Random Experiments
If
y
=
sin
2
(
cot
−
1
1
+
x
1
−
x
)
y = \sin^2 (\cot^{-1} \sqrt{\frac{1+x}{1-x}} )
y
=
sin
2
(
cot
−
1
1
−
x
1
+
x
)
, then
d
y
d
x
\frac{dy}{dx}
d
x
d
y
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Derivatives of Functions in Parametric Forms
Let
A
A
A
and
B
B
B
be finite sets and
P
A
P_{A}
P
A
and
P
B
P_{B}
P
B
respectively denote their power sets. If
P
B
P_{B}
P
B
has
112
112
112
elements more than those in
P
A
′
P_{A^{\prime}}
P
A
′
then the number of functions from
A
A
A
to
B
B
B
which are injective is
AP EAMCET - 2019
AP EAMCET
Mathematics
Relations
The variance of the following continuous frequency distribution is
AP EAMCET - 2019
AP EAMCET
Mathematics
Variance and Standard Deviation
In triangle
Δ
A
B
C
\Delta A B C
Δ
A
BC
, if
b
+
c
9
=
c
+
a
10
=
a
+
b
11
,
\frac{b + c}{9} = \frac{c + a}{10} = \frac{a+b}{11},
9
b
+
c
=
10
c
+
a
=
11
a
+
b
,
then
cos
A
+
cos
B
cos
C
=
\frac{\cos A + \cos B}{\cos C} =
c
o
s
C
c
o
s
A
+
c
o
s
B
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Trigonometric Functions
If the perpendicular bisector of the line segment joining
A
(
α
,
3
)
A(\alpha, 3)
A
(
α
,
3
)
and
B
(
2
,
−
1
)
B (2, -1)
B
(
2
,
−
1
)
has
y
y
y
-intercept
1
1
1
, then
α
\alpha
α
=
AP EAMCET - 2019
AP EAMCET
Mathematics
x-intercepts and y-intercepts
Variable straight lines
y
=
m
x
+
c
y = mx + c
y
=
m
x
+
c
make intercepts on the curve
y
2
−
4
a
x
=
0
y^2 - 4ax = 0
y
2
−
4
a
x
=
0
which subtend a right angle at the origin. Then the point of concurrence of these lines
y
=
m
x
+
c
y = mx + c
y
=
m
x
+
c
is
AP EAMCET - 2019
AP EAMCET
Mathematics
x-intercepts and y-intercepts
The polynomial equation of degree
4
4
4
having real coefficients with three of its roots as
2
±
3
2 \pm \sqrt{3}
2
±
3
and
1
+
2
i
1+2i
1
+
2
i
. is
AP EAMCET - 2019
AP EAMCET
Mathematics
Algebra of Complex Numbers
If
x
4
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
=
x
+
k
+
A
x
−
1
+
B
x
−
2
+
C
x
−
3
\frac{x^{4}}{\left(x-1\right)\left(x-2\right)\left(x-3\right)} =x + k+ \frac{A}{x-1}+\frac{B}{x-2} + \frac{C}{x-3}
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
x
4
=
x
+
k
+
x
−
1
A
+
x
−
2
B
+
x
−
3
C
, then
k
+
A
−
B
+
C
=
k + A - B + C =
k
+
A
−
B
+
C
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Integration by Partial Fractions
If
p
p
p
and
q
q
q
are respectively the global maximum and global minimum of the function
f
(
x
)
=
x
2
e
2
x
f(x) = x^2 e^{2x}
f
(
x
)
=
x
2
e
2
x
on the interval
[
−
2
,
2
]
[-2, 2]
[
−
2
,
2
]
, then
p
e
−
4
+
q
e
4
=
pe^{-4} + qe^4 =
p
e
−
4
+
q
e
4
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Maxima and Minima
If
cos
θ
≠
0
\cos \theta \neq 0
cos
θ
=
0
and
sec
θ
−
1
=
(
2
−
1
)
tan
θ
\sec \theta - 1 = ( \sqrt{2} - 1 ) \tan \theta
sec
θ
−
1
=
(
2
−
1
)
tan
θ
then
θ
\theta
θ
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Trigonometric Equations
If
α
\alpha
α
and
β
\beta
β
are the roots of
x
2
+
7
x
+
3
=
0
x^{2}+7 x+3=0
x
2
+
7
x
+
3
=
0
and
2
α
3
−
4
α
,
2
β
3
−
4
β
\frac{2 \alpha}{3-4 \alpha}, \frac{2 \beta}{3-4 \beta}
3
−
4
α
2
α
,
3
−
4
β
2
β
are the roots of
a
x
2
+
b
x
+
c
=
0
a x^{2}+b x+c=0
a
x
2
+
b
x
+
c
=
0
and
G
C
D
GCD
GC
D
of
a
,
b
,
c
a, b, c
a
,
b
,
c
is
1
1
1
, then
a
+
b
+
c
=
a+b+c=
a
+
b
+
c
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Quadratic Equations
When the coordinate axes are rotated about the origin in the positive direction through an angle
π
4
\frac{\pi}{4}
4
π
, if the equation
25
x
2
+
9
y
2
=
225
25 x^2 + 9y^2 = 225
25
x
2
+
9
y
2
=
225
is transformed to
α
x
2
+
β
x
y
+
γ
y
2
=
δ
\alpha x^2 + \beta xy + \gamma y^2 = \delta
α
x
2
+
β
x
y
+
γ
y
2
=
δ
, then
(
α
+
β
+
γ
−
δ
)
2
(\alpha + \beta + \gamma - \sqrt{\delta})^2
(
α
+
β
+
γ
−
δ
)
2
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Plane
If
α
,
β
,
γ
\alpha , \beta , \gamma
α
,
β
,
γ
are the roots of
x
3
−
6
x
2
+
11
x
−
6
=
0
x^3 - 6x^2 + 11x - 6 = 0
x
3
−
6
x
2
+
11
x
−
6
=
0
, then the equation having the roots
α
2
+
β
2
+
γ
2
\alpha^2 + \beta^2 + \gamma^2
α
2
+
β
2
+
γ
2
and
γ
2
+
α
2
\gamma^2 + \alpha^2
γ
2
+
α
2
is
AP EAMCET - 2019
AP EAMCET
Mathematics
Quadratic Equations
If
∫
cos
x
.
cos
2
x
.
cos
5
x
d
x
=
A
sin
2
x
+
B
sin
4
x
+
C
sin
6
x
+
D
sin
8
x
+
k
\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k
∫
cos
x
.
cos
2
x
.
cos
5
x
d
x
=
A
sin
2
x
+
B
sin
4
x
+
C
sin
6
x
+
D
sin
8
x
+
k
(where
k
k
k
is the arbitrary constant of integration), then
1
B
+
1
C
=
\frac{1}{B} + \frac{1}{C} =
B
1
+
C
1
=
AP EAMCET - 2019
AP EAMCET
Mathematics
Definite Integral
If
α
=
lim
x
→
0
x
.
2
x
−
x
1
−
cos
x
\alpha =\displaystyle\lim_{x\to0} \frac{x.2^{x}-x}{1-\cos x}
α
=
x
→
0
lim
1
−
cos
x
x
.
2
x
−
x
and
β
=
lim
x
→
0
x
.
2
x
−
x
1
+
x
2
−
1
−
x
2
,
\beta =\displaystyle\lim_{x\to0} \frac{x.2^{x}-x}{\sqrt{1+x^{2} } - \sqrt{1-x^{2}} } ,
β
=
x
→
0
lim
1
+
x
2
−
1
−
x
2
x
.
2
x
−
x
,
then
AP EAMCET - 2019
AP EAMCET
Mathematics
Limits
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