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function f x x 1 cotx is continuous at x 0 then th
Question:
Function f(x)=(x+1)
cotx
is continuous at x = 0, then the value of f(0) is
CUET (PG) - 2023
CUET (PG)
Updated On:
Mar 12, 2025
1
e
\frac{1}{e}
e
1
0
e
1
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The Correct Option is
C
Solution and Explanation
The correct answer is(C): e
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Top Questions on Functions
The sum of all local minimum values of the function
f
(
x
)
f(x)
f
(
x
)
as defined below is:
f
(
x
)
=
{
1
−
2
x
if
x
<
−
1
,
1
3
(
7
+
2
∣
x
∣
)
if
−
1
≤
x
≤
2
,
11
18
(
x
−
4
)
(
x
−
5
)
if
x
>
2.
f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases}
f
(
x
)
=
⎩
⎨
⎧
1
−
2
x
3
1
(
7
+
2∣
x
∣
)
18
11
(
x
−
4
)
(
x
−
5
)
if
x
<
−
1
,
if
−
1
≤
x
≤
2
,
if
x
>
2.
JEE Main - 2025
Mathematics
Functions
View Solution
In
I
(
m
,
n
)
=
∫
0
1
x
m
−
1
(
1
−
x
)
n
−
1
d
x
I(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx
I
(
m
,
n
)
=
∫
0
1
x
m
−
1
(
1
−
x
)
n
−
1
d
x
, where
m
,
n
>
0
m, n > 0
m
,
n
>
0
, then
I
(
9
,
14
)
+
I
(
10
,
13
)
I(9, 14) + I(10, 13)
I
(
9
,
14
)
+
I
(
10
,
13
)
is:
JEE Main - 2025
Mathematics
Functions
View Solution
Let
f
(
x
)
=
2
x
+
2
+
16
2
2
x
+
1
+
2
x
+
4
+
32
f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32}
f
(
x
)
=
2
2
x
+
1
+
2
x
+
4
+
32
2
x
+
2
+
16
. Then the value of
8
(
f
(
1
15
)
+
f
(
2
15
)
+
⋯
+
f
(
59
15
)
)
8 \left( f\left( \frac{1}{15} \right) + f\left( \frac{2}{15} \right) + \dots + f\left( \frac{59}{15} \right) \right)
8
(
f
(
15
1
)
+
f
(
15
2
)
+
⋯
+
f
(
15
59
)
)
is equal to:
JEE Main - 2025
Mathematics
Functions
View Solution
Let
f
:
R
→
R
f: \mathbb{R} \to \mathbb{R}
f
:
R
→
R
be a function defined by
f
(
x
)
=
(
2
+
3
a
)
x
2
+
(
a
+
2
a
−
1
)
x
+
b
,
a
≠
1
f(x) = \left( 2 + 3a \right)x^2 + \left( \frac{a+2}{a-1} \right)x + b, a \neq 1
f
(
x
)
=
(
2
+
3
a
)
x
2
+
(
a
−
1
a
+
2
)
x
+
b
,
a
=
1
. If
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
+
1
−
2
7
x
y
,
f(x + y) = f(x) + f(y) + 1 - \frac{2}{7}xy,
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
+
1
−
7
2
x
y
,
then the value of
28
∑
i
=
1
5
f
(
i
)
28 \sum_{i=1}^5 f(i)
28
∑
i
=
1
5
f
(
i
)
is:
JEE Main - 2025
Mathematics
Functions
View Solution
The area of the region enclosed by the curves
y
=
e
x
y = e^x
y
=
e
x
,
y
=
∣
e
x
−
1
∣
y = |e^x - 1|
y
=
∣
e
x
−
1∣
, and the y-axis is:
JEE Main - 2025
Mathematics
Functions
View Solution
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