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let f x y ln 1 x2 y2 for x y 2 define matrix p 2f
Question:
Let f(x, y) = ln(1 + x
2
+ y
2
) for (x, y) ∈
R
2
\R^2
R
2
. Define
P
=
∂
2
f
∂
x
2
∣
(
0
,
0
)
Q
=
∂
2
f
∂
x
∂
y
∣
(
0
,
0
)
R
=
∂
2
f
∂
y
∂
x
∣
(
0
,
0
)
S
=
∂
2
f
∂
y
2
∣
(
0
,
0
)
\begin{matrix} P=\frac{∂^2f}{∂x^2}|_{(0,0)} & Q=\frac{∂^2f}{∂x∂y}|_{(0,0)} \\ R=\frac{∂^2f}{∂y∂x}|_{(0,0)} & S=\frac{∂^2f}{∂y^2}|_{(0,0)} \end{matrix}
P
=
∂
x
2
∂
2
f
∣
(
0
,
0
)
R
=
∂
y
∂
x
∂
2
f
∣
(
0
,
0
)
Q
=
∂
x
∂
y
∂
2
f
∣
(
0
,
0
)
S
=
∂
y
2
∂
2
f
∣
(
0
,
0
)
Then
IIT JAM MA - 2023
IIT JAM MA
Updated On:
Oct 1, 2024
PS − QR > 0 and P < 0
PS − QR > 0 and P > 0
PS − QR < 0 and P > 0
PS − QR < 0 and P < 0
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The Correct Option is
B
Solution and Explanation
The correct option is (B) : PS − QR > 0 and P > 0.
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Top Questions on Differential Equations
If
y
=
y
(
x
)
y = y(x)
y
=
y
(
x
)
is the solution of the differential equation,
4
−
x
2
d
y
d
x
=
(
(
sin
−
1
(
x
2
)
)
2
−
y
)
sin
−
1
(
x
2
)
,
\sqrt{4 - x^2} \frac{dy}{dx} = \left( \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2 - y \right) \sin^{-1} \left( \frac{x}{2} \right),
4
−
x
2
d
x
d
y
=
(
(
sin
−
1
(
2
x
)
)
2
−
y
)
sin
−
1
(
2
x
)
,
where
−
2
≤
x
≤
2
-2 \leq x \leq 2
−
2
≤
x
≤
2
, and
y
(
2
)
=
π
2
−
8
4
y(2) = \frac{\pi^2 - 8}{4}
y
(
2
)
=
4
π
2
−
8
, then
y
2
(
0
)
y^2(0)
y
2
(
0
)
is equal to:
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Mathematics
Differential Equations
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Let
y
=
y
(
x
)
y = y(x)
y
=
y
(
x
)
be the solution of the differential equation
(
x
y
−
5
x
2
1
+
x
2
)
d
x
+
(
1
+
x
2
)
d
y
=
0
,
y
(
0
)
=
0.
\left( xy - 5x^2 \sqrt{1 + x^2} \right) dx + (1 + x^2) dy = 0, \quad y(0) = 0.
(
x
y
−
5
x
2
1
+
x
2
)
d
x
+
(
1
+
x
2
)
d
y
=
0
,
y
(
0
)
=
0.
Then
y
(
3
)
y(\sqrt{3})
y
(
3
)
is equal to:
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Mathematics
Differential Equations
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Let a curve
y
=
f
(
x
)
y = f(x)
y
=
f
(
x
)
pass through the points
(
0
,
5
)
(0,5)
(
0
,
5
)
and
(
log
2
,
k
)
(\log 2, k)
(
lo
g
2
,
k
)
. If the curve satisfies the differential equation:
2
(
3
+
y
)
e
2
x
d
x
−
(
7
+
e
2
x
)
d
y
=
0
,
2(3+y)e^{2x}dx - (7+e^{2x})dy = 0,
2
(
3
+
y
)
e
2
x
d
x
−
(
7
+
e
2
x
)
d
y
=
0
,
then
k
k
k
is equal to:
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Mathematics
Differential Equations
View Solution
Let
x
=
x
(
y
)
x = x(y)
x
=
x
(
y
)
be the solution of the differential equation:
y
=
(
x
−
y
d
x
d
y
)
sin
(
x
y
)
,
y
>
0
and
x
(
1
)
=
π
2
.
y = \left( x - y \frac{dx}{dy} \right) \sin\left( \frac{x}{y} \right), \, y > 0 \, \text{and} \, x(1) = \frac{\pi}{2}.
y
=
(
x
−
y
d
y
d
x
)
sin
(
y
x
)
,
y
>
0
and
x
(
1
)
=
2
π
.
Then
cos
(
x
(
2
)
)
\cos(x(2))
cos
(
x
(
2
))
is equal to:
JEE Main - 2025
Mathematics
Differential Equations
View Solution
Let for some function
y
=
f
(
x
)
y = f(x)
y
=
f
(
x
)
,
∫
0
x
t
f
(
t
)
d
t
=
x
2
f
(
x
)
,
x
>
0
\int_0^x t f(t) \, dt = x^2 f(x), x>0
∫
0
x
t
f
(
t
)
d
t
=
x
2
f
(
x
)
,
x
>
0
and
f
(
2
)
=
3
f(2) = 3
f
(
2
)
=
3
. Then
f
(
6
)
f(6)
f
(
6
)
is equal to:
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Mathematics
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S
(
c
)
=
{
x
∈
R
:
f
(
x
)
=
c
}
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(
c
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∈
R
:
f
(
x
)
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}
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D
=
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(
x
,
y
,
z
)
∈
R
3
:
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2
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y
2
+
z
2
≤
a
2
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x
2
+
y
2
≥
b
2
}
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D
=
{
(
x
,
y
,
z
)
∈
R
3
:
x
2
+
y
2
+
z
2
≤
a
2
and
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2
+
y
2
≥
b
2
}
.
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Let P
7
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7
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(
f
(
x
)
)
=
f
(
x
)
+
d
f
(
x
)
d
x
T(f(x))=f(x)+\frac{df(x)}{dx}
T
(
f
(
x
))
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f
(
x
)
+
d
x
df
(
x
)
.
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∈
Z
n
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cd
(
r
,
n
)
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