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if f x and g x are two probability density functio
Question:
If
f
(
x
)
and
g
(
x
)
are two probability density functions,
f
(
x
)
=
{
x
a
+
1
:
−
a
≤
x
<
0
−
x
a
+
1
0
≤
x
≤
a
0
otherwise
g
(
x
)
=
{
−
x
a
:
−
a
≤
x
≤
0
x
a
:
0
≤
x
≤
a
0
:
otherewise
Which one of the following statements is true?
MHT CET
Updated On:
Aug 16, 2024
(A) Mean of
f
(
x
)
and
g
(
x
)
are same; Variance of
f
(
x
)
and
g
(
x
)
are same
(B) Mean of
f
(
x
)
and
g
(
x
)
are same; Variance of
f
(
x
)
and
g
(
x
)
are different
(C) Mean of
f
(
x
)
and
g
(
x
)
are different; Variance of
f
(
x
)
and
g
(
x
)
are same
(D) Mean of
f
(
x
)
and
g
(
x
)
are different; Variance of
f
(
x
)
and
g
(
x
)
are different
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The Correct Option is
B
Solution and Explanation
Explanation:
Mean of
f
(
x
)
:
E
(
x
)
=
∫
−
a
0
X
(
X
a
+
1
)
d
x
+
∫
0
a
X
(
−
X
a
+
1
)
d
x
=
(
X
3
3
a
+
X
2
2
)
−
a
0
+
(
−
X
3
3
a
+
X
3
3
)
0
a
=
0
=
∫
−
a
0
X
2
(
X
a
+
1
)
d
x
+
∫
0
a
X
2
(
−
X
a
+
1
)
d
x
(
X
4
4
a
+
X
3
3
)
−
a
0
+
(
−
X
4
4
a
+
X
3
3
)
0
a
=
a
3
6
⇒
Variance
=
a
3
6
Mean of
g
(
x
)
:
E
(
x
)
=
∫
−
a
0
x
(
−
x
a
)
d
x
+
∫
0
a
x
×
(
X
a
)
d
x
=
0
Variance of
g
(
x
)
is
E
(
x
2
)
−
{
E
(
X
)
}
2
, Where
E
(
X
2
)
=
∫
−
a
0
X
2
(
−
X
a
)
d
X
+
∫
0
a
X
2
(
X
a
)
d
x
=
a
3
2
⇒
Variance
=
a
3
2
∴
Mean of
f
(
x
)
and
g
(
x
)
are same but variance of
f
(
x
)
and
g
(
x
)
are different.Hence, the correct option is (B).
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Top Questions on Vectors
Let
a
⃗
=
i
^
+
2
j
^
+
3
k
^
,
b
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=
3
i
^
+
j
^
−
k
^
\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \, \vec{b} = 3\hat{i} + \hat{j} - \hat{k}
a
=
i
^
+
2
j
^
+
3
k
^
,
b
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3
i
^
+
j
^
−
k
^
and
c
⃗
\vec{c}
c
be three vectors such that
c
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c
is coplanar with
a
⃗
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a
and
b
⃗
\vec{b}
b
. If the vector
c
⃗
\vec{c}
c
is perpendicular to
b
⃗
\vec{b}
b
and
a
⃗
⋅
c
⃗
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5
\vec{a} \cdot \vec{c} = 5
a
⋅
c
=
5
, then
∣
c
⃗
∣
|\vec{c}|
∣
c
∣
is equal to:
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View Solution
Let
a
⃗
=
i
+
j
+
k
\vec{a} = i + j + k
a
=
i
+
j
+
k
,
b
⃗
=
2
i
+
2
j
+
k
\vec{b} = 2i + 2j + k
b
=
2
i
+
2
j
+
k
and
d
⃗
=
a
⃗
×
b
⃗
\vec{d} = \vec{a} \times \vec{b}
d
=
a
×
b
. If
c
⃗
\vec{c}
c
is a vector such that
a
⃗
⋅
c
⃗
=
∣
c
⃗
∣
\vec{a} \cdot \vec{c} = |\vec{c}|
a
⋅
c
=
∣
c
∣
,
∥
c
⃗
−
2
d
⃗
∥
=
8
\|\vec{c} - 2\vec{d}\| = 8
∥
c
−
2
d
∥
=
8
and the angle between
d
⃗
\vec{d}
d
and
c
⃗
\vec{c}
c
is
π
4
\frac{\pi}{4}
4
π
, then
∣
10
−
3
b
⃗
⋅
c
⃗
+
∣
d
⃗
∣
∣
2
\left|10 - 3\vec{b} \cdot \vec{c} + |\vec{d}|\right|^2
10
−
3
b
⋅
c
+
∣
d
∣
2
is equal to:
JEE Main - 2025
Mathematics
Vectors
View Solution
If the components of
a
⃗
=
α
i
^
+
β
j
^
+
γ
k
^
\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}
a
=
α
i
^
+
β
j
^
+
γ
k
^
along and perpendicular to
b
⃗
=
3
i
^
+
j
^
−
k
^
\vec{b} = 3\hat{i} + \hat{j} - \hat{k}
b
=
3
i
^
+
j
^
−
k
^
respectively are
16
11
(
3
i
^
+
j
^
−
k
^
)
\frac{16}{11} (3\hat{i} + \hat{j} - \hat{k})
11
16
(
3
i
^
+
j
^
−
k
^
)
and
1
11
(
−
4
i
^
−
5
j
^
−
17
k
^
)
\frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k})
11
1
(
−
4
i
^
−
5
j
^
−
17
k
^
)
, then
α
2
+
β
2
+
γ
2
\alpha^2 + \beta^2 + \gamma^2
α
2
+
β
2
+
γ
2
is equal to:
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Mathematics
Vectors
View Solution
Number of functions
f
:
{
1
,
2
,
…
,
100
}
→
{
0
,
1
}
f: \{1, 2, \dots, 100\} \to \{0, 1\}
f
:
{
1
,
2
,
…
,
100
}
→
{
0
,
1
}
, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to:
JEE Main - 2025
Mathematics
Vectors
View Solution
Let
a
⃗
=
i
+
j
+
k
\vec{a} = i + j + k
a
=
i
+
j
+
k
,
b
⃗
=
2
i
+
2
j
+
k
\vec{b} = 2i + 2j + k
b
=
2
i
+
2
j
+
k
and
d
⃗
=
a
⃗
×
b
⃗
\vec{d} = \vec{a} \times \vec{b}
d
=
a
×
b
. If
c
⃗
\vec{c}
c
is a vector such that
a
⃗
⋅
c
⃗
=
∣
c
⃗
∣
\vec{a} \cdot \vec{c} = |\vec{c}|
a
⋅
c
=
∣
c
∣
,
∥
c
⃗
−
2
d
⃗
∥
=
8
\|\vec{c} - 2\vec{d}\| = 8
∥
c
−
2
d
∥
=
8
and the angle between
d
⃗
\vec{d}
d
and
c
⃗
\vec{c}
c
is
π
4
\frac{\pi}{4}
4
π
, then
∣
10
−
3
b
⃗
⋅
c
⃗
+
∣
d
⃗
∣
∣
2
\left|10 - 3\vec{b} \cdot \vec{c} + |\vec{d}|\right|^2
10
−
3
b
⋅
c
+
∣
d
∣
2
is equal to:
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Mathematics
Vectors
View Solution
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