Question 24 - Week 8 Practice | Prasnya
Prasnya
Continue with Google
Practice path
Dashboard
Exam
Quiz 2
Go to subject
Mathematics 1
Go to chapter
Week 8
Question 24
00:00
Est. 3 min
Marks:
+5.00
0.00
Passage
Consider the following functions
f
1
:
D
1
→
R
f_1:D_1\to\mathbb{R}
,
f
2
:
D
2
→
R
f_2:D_2\to\mathbb{R}
,
f
3
:
D
3
→
R
f_3:D_3\to\mathbb{R}
and
g
:
D
→
R
g:D\to\mathbb{R}
defined as:
f
1
(
x
)
=
sin
2
x
f_1(x)=\sin 2x
,
f
2
(
x
)
=
ln
(
x
2
−
6
x
+
8
)
f_2(x)=\ln(x^2-6x+8)
,
f
3
(
x
)
=
e
3
x
+
5
f_3(x)=e^{3x}+5
, and
g
(
x
)
=
f
1
(
x
)
+
f
2
(
x
)
+
f
3
(
x
)
g(x)=f_1(x)+f_2(x)+f_3(x)
. Let
D
1
,
D
2
,
D
3
D_1,D_2,D_3
and
D
D
be the largest domains of
f
1
,
f
2
,
f
3
f_1,f_2,f_3
and
g
g
, respectively. Use this information to answer the subquestions.
Question
Which of the following options is/are true?
Comprehension
Medium Difficulty
06 August 2023
A
The function
g
(
x
)
g(x)
is not differentiable in its domain.
B
The function
g
(
x
)
g(x)
is continuous in its domain.
C
If
g
(
x
)
g(x)
is differentiable in its domain, then
g
′
(
x
)
=
2
cos
2
x
+
2
x
−
6
x
2
−
6
x
+
8
+
3
e
3
x
g'(x)=2\cos 2x+\frac{2x-6}{x^2-6x+8}+3e^{3x}
.
D
If
g
(
x
)
g(x)
is differentiable in its domain, then
g
′
(
x
)
=
2
cos
2
x
+
1
x
2
−
6
x
+
8
+
3
e
3
x
g'(x)=2\cos 2x+\frac{1}{x^2-6x+8}+3e^{3x}
.