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if f r2 r2 is a function defined as f x y cases x
Question:
If f: R
2
→R
2
is a function defined as
f
(
x
,
y
)
=
{
x
x
2
+
y
2
,
x
≠
0
,
y
≠
0
2
,
x
=
0
,
y
=
0
f(x,y) = \begin{cases} \frac{x}{\sqrt{x^2+y^2}}, & x\neq0,y\neq0\\ 2, & x=0,y=0 \end{cases}
f
(
x
,
y
)
=
{
x
2
+
y
2
x
,
2
,
x
=
0
,
y
=
0
x
=
0
,
y
=
0
then, which of the following is correct?
CUET (PG) - 2023
CUET (PG)
Updated On:
Mar 12, 2025
f(x,y) is continuous at origin
f(x,y) is differentiable at origin
lim
(
x
,
y
)
→
(
0
,
0
)
\lim\limits_{(x,y)\rightarrow(0,0)}
(
x
,
y
)
→
(
0
,
0
)
lim
f(x,y) exists and is equal to 2
f(x,y) is not continuous at origin
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The Correct Option is
D
Solution and Explanation
The correct answer is(D): f(x,y) is not continuous at origin
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Top Questions on Continuity and differentiability
Let
f
:
R
→
R
f : \mathbb{R} \to \mathbb{R}
f
:
R
→
R
be a twice-differentiable function such that
f
(
2
)
=
1
f(2) = 1
f
(
2
)
=
1
. If
F
(
x
)
=
x
f
(
x
)
F(x) = x f(x)
F
(
x
)
=
x
f
(
x
)
for all
x
∈
R
x \in \mathbb{R}
x
∈
R
, and the integrals
∫
0
2
x
F
′
(
x
)
d
x
=
6
\int_0^2 x F'(x) \, dx = 6
∫
0
2
x
F
′
(
x
)
d
x
=
6
and
∫
0
2
x
2
F
′
′
(
x
)
d
x
=
40
\int_0^2 x^2 F''(x) \, dx = 40
∫
0
2
x
2
F
′′
(
x
)
d
x
=
40
, then
F
′
(
2
)
+
∫
0
2
F
(
x
)
d
x
F'(2) + \int_0^2 F(x) \, dx
F
′
(
2
)
+
∫
0
2
F
(
x
)
d
x
is equal to:
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If
f
(
x
)
=
∫
1
x
1
/
4
(
1
+
x
1
/
4
)
d
x
,
f
(
0
)
=
−
6
,
t
h
e
n
f
(
1
)
i
s
e
q
u
a
l
t
o
:
f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6, { then } f(1) { is equal to:}
f
(
x
)
=
∫
x
1/4
(
1
+
x
1/4
)
1
d
x
,
f
(
0
)
=
−
6
,
t
h
e
n
f
(
1
)
i
se
q
u
a
lt
o
:
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View Solution
If
f
(
x
)
,
defined by
f
(
x
)
=
{
k
x
+
1
if
x
≤
π
cos
x
if
x
>
π
is continuous at
x
=
π
,
then the value of
k
is:
\text{ If } f(x), \text{ defined by } f(x) = \begin{cases} kx + 1 & \text{if } x \leq \pi \\ \cos x & \text{if } x > \pi \end{cases} \text{ is continuous at } x = \pi, \text{ then the value of } k \text{ is:}
If
f
(
x
)
,
defined by
f
(
x
)
=
{
k
x
+
1
cos
x
if
x
≤
π
if
x
>
π
is continuous at
x
=
π
,
then the value of
k
is:
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Let [x] denote the greatest integer function. Then match List-I with List-II:
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Let
f
:
R
→
R
f : \mathbb{R} \to \mathbb{R}
f
:
R
→
R
be a function given by
f
(
x
)
=
{
1
−
cos
2
x
x
2
,
x
<
0
α
,
x
=
0
,
where
α
,
β
∈
R
.
β
1
−
cos
x
/
x
,
x
>
0
f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x < 0 \\\alpha, & x = 0, \text{ where } \alpha, \beta \in \mathbb{R}. \\\beta \sqrt{1 - \cos x} / x, & x > 0 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
x
2
1
−
c
o
s
2
x
,
α
,
β
1
−
cos
x
/
x
,
x
<
0
x
=
0
,
where
α
,
β
∈
R
.
x
>
0
If
f
f
f
is continuous at
x
=
0
x = 0
x
=
0
, then
α
2
+
β
2
\alpha^2 + \beta^2
α
2
+
β
2
is equal to:
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Mathematics
Continuity and differentiability
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