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Principle: Modulus / absolute value behaviour
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Q1
#1
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Moderate
Let
f
(
x
)
=
{
2
x
+
1
,
−
3
<
x
<
−
2
x
−
1
,
−
2
≤
x
<
0
x
+
2
,
0
≤
x
<
1
f(x)=\begin{cases}2x+1, & -3<x<-2\\x-1, & -2\leq x<0\\x+2, & 0\leq x<1\end{cases}
f
(
x
)
=
⎩
⎨
⎧
2
x
+
1
,
x
−
1
,
x
+
2
,
−
3
<
x
<
−
2
−
2
≤
x
<
0
0
≤
x
<
1
. Which one of the following statements is correct in respect of the above function?
Add
Lever: Differentiability at a point
A
It is discontinuous at
x
=
−
2
x=-2
x
=
−
2
but continuous at every other point.
B
It is continuous only in the interval
(
−
3
,
−
2
)
(-3,-2)
(
−
3
,
−
2
)
.
C
It is discontinuous at
x
=
0
x=0
x
=
0
but continuous at every other point.
D
It is discontinuous at every point.
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[Q92 · Apr · 2017]
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Q2
#2
NDA → Mathematics → Differentiation → Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
Suppose the function
f
(
x
)
=
x
n
f(x)=x^n
f
(
x
)
=
x
n
,
n
≠
0
n\neq0
n
=
0
is differentiable for all
x
x
x
. Then
n
n
n
can be any element of the interval
Add
Lever: Differentiability at a point
A
[
1
,
∞
)
[1,\infty)
[
1
,
∞
)
B
(
0
,
∞
)
(0,\infty)
(
0
,
∞
)
C
(
1
2
,
∞
)
\left(\dfrac{1}{2},\infty\right)
(
2
1
,
∞
)
D
None of the above
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[Q99 · Apr · 2017]
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Q3
#3
NDA → Mathematics → Definite Integration → Integration of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
What is
∫
e
−
1
e
2
∣
ln
x
x
∣
d
x
\int_{e^{-1}}^{e^2}\left|\dfrac{\ln x}{x}\right|\,dx
∫
e
−
1
e
2
x
ln
x
d
x
equal to?
Add
Lever: Modulus / absolute value behaviour
A
3
2
\dfrac{3}{2}
2
3
B
5
2
\dfrac{5}{2}
2
5
C
3
3
3
D
4
4
4
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[Q100 · Apr · 2017]
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Q4
#4
NDA → Mathematics → Sets & Relations → Relations — Properties, Cartesian Product, and Counting
·
Easy
If
∣
a
∣
|a|
∣
a
∣
denotes the absolute value of an integer, then which of the following are correct? 1.
∣
a
b
∣
=
∣
a
∣
∣
b
∣
|ab|=|a||b|
∣
ab
∣
=
∣
a
∣∣
b
∣
2.
∣
a
+
b
∣
≤
∣
a
∣
+
∣
b
∣
|a+b|\leq|a|+|b|
∣
a
+
b
∣
≤
∣
a
∣
+
∣
b
∣
3.
∣
a
−
b
∣
≥
∣
∣
a
∣
−
∣
b
∣
∣
|a-b|\geq||a|-|b||
∣
a
−
b
∣
≥
∣∣
a
∣
−
∣
b
∣∣
Select the correct answer using the code given below.
Add
Lever: Modulus / absolute value behaviour
A
1 and 2 only
B
2 and 3 only
C
1 and 3 only
D
1, 2 and 3
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[Q9 · Sep · 2017]
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Q5
#5
NDA → Mathematics → Quadratic Equations → Special Quadratics — Parametric, Logarithmic, Constructed
·
Easy
The sum of all real roots of the equation
∣
x
−
3
∣
2
+
∣
x
−
3
∣
−
2
=
0
|x-3|^2 + |x-3| - 2 = 0
∣
x
−
3
∣
2
+
∣
x
−
3∣
−
2
=
0
is
Add
Lever: Modulus / absolute value behaviour
A
2
B
3
C
4
D
6
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[Q12 · Sep · 2017]
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Q6
#6
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Easy
A function is defined as follows:
f
(
x
)
=
{
−
x
x
2
,
x
≠
0
0
,
x
=
0
f(x) = \begin{cases} -\dfrac{x}{\sqrt{x^2}}, & x \neq 0 \\ 0, & x = 0 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
−
x
2
x
,
0
,
x
=
0
x
=
0
. Which one of the following is correct in respect of the above function?
Add
Lever: Differentiability at a point
A
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
0
x = 0
x
=
0
but not differentiable at
x
=
0
x = 0
x
=
0
B
f
(
x
)
f(x)
f
(
x
)
is continuous as well as differentiable at
x
=
0
x = 0
x
=
0
C
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
0
x = 0
x
=
0
D
None of the above
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[Q62 · Sep · 2017]
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Q7
#7
NDA → Mathematics → Differentiation → Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
A function is defined in
(
0
,
∞
)
(0, \infty)
(
0
,
∞
)
by
f
(
x
)
=
{
1
−
x
2
for
0
<
x
≤
1
ln
x
for
1
<
x
≤
2
ln
2
−
1
+
0.5
x
for
2
<
x
<
∞
f(x) = \begin{cases}1 - x^2 & \text{for } 0 < x \leq 1 \\ \ln x & \text{for } 1 < x \leq 2 \\ \ln 2 - 1 + 0.5x & \text{for } 2 < x < \infty\end{cases}
f
(
x
)
=
⎩
⎨
⎧
1
−
x
2
ln
x
ln
2
−
1
+
0.5
x
for
0
<
x
≤
1
for
1
<
x
≤
2
for
2
<
x
<
∞
. Which one of the following is correct in respect of the derivative of the function, i.e.,
f
′
(
x
)
f'(x)
f
′
(
x
)
?
Add
Lever: Differentiability at a point
A
f
′
(
x
)
=
2
x
f'(x) = 2x
f
′
(
x
)
=
2
x
for
0
<
x
≤
1
0 < x \leq 1
0
<
x
≤
1
B
f
′
(
x
)
=
−
2
x
f'(x) = -2x
f
′
(
x
)
=
−
2
x
for
0
<
x
≤
1
0 < x \leq 1
0
<
x
≤
1
C
f
′
(
x
)
=
−
2
x
f'(x) = -2x
f
′
(
x
)
=
−
2
x
for
0
<
x
<
1
0 < x < 1
0
<
x
<
1
D
f
′
(
x
)
=
0
f'(x) = 0
f
′
(
x
)
=
0
for
0
<
x
<
∞
0 < x < \infty
0
<
x
<
∞
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[Q68 · Sep · 2017]
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Q8
#8
NDA → Mathematics → Functions → Domain, Range, and Function Properties
·
Easy
The function
f
(
x
)
=
∣
x
∣
−
x
3
f(x) = |x| - x^3
f
(
x
)
=
∣
x
∣
−
x
3
is
Add
Lever: Modulus / absolute value behaviour
A
odd
B
even
C
both even and odd
D
neither even nor odd
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[Q72 · Sep · 2017]
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Q9
#9
NDA → Mathematics → Applications of Integration → Area Bounded by a Curve, Lines, and Axes
·
Easy
The area bounded by the curve
∣
x
∣
+
∣
y
∣
=
1
|x| + |y| = 1
∣
x
∣
+
∣
y
∣
=
1
is
Add
Lever: Modulus / absolute value behaviour
A
1 square unit
B
2
2
2\sqrt{2}
2
2
square units
C
2 square units
D
2
3
2\sqrt{3}
2
3
square units
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[Q77 · Sep · 2017]
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Q10
#10
NDA → Mathematics → Limits & Continuity → One-Sided Limits, Greatest Integer, and Absolute Value Limits
·
Hard
The left-hand derivative of
f
(
x
)
=
[
x
]
sin
(
π
x
)
f(x) = [x]\sin(\pi x)
f
(
x
)
=
[
x
]
sin
(
π
x
)
at
x
=
k
x = k
x
=
k
, where
k
k
k
is an integer and
[
x
]
[x]
[
x
]
is the greatest integer function, is
Add
Lever: Differentiability at a point
A
(
−
1
)
k
(
k
−
1
)
π
(-1)^k(k-1)\pi
(
−
1
)
k
(
k
−
1
)
π
B
(
−
1
)
k
−
1
(
k
−
1
)
π
(-1)^{k-1}(k-1)\pi
(
−
1
)
k
−
1
(
k
−
1
)
π
C
(
−
1
)
k
k
π
(-1)^k k\pi
(
−
1
)
k
k
π
D
(
−
1
)
k
−
1
k
π
(-1)^{k-1}k\pi
(
−
1
)
k
−
1
k
π
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[Q79 · Sep · 2017]
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Q11
#11
NDA → Mathematics → Differentiation → Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
·
Hard
The set of all points, where the function
f
(
x
)
=
1
−
e
−
x
2
f(x) = \sqrt{1 - e^{-x^2}}
f
(
x
)
=
1
−
e
−
x
2
is differentiable, is
Add
Lever: Differentiability at a point
A
(
0
,
∞
)
(0, \infty)
(
0
,
∞
)
B
(
−
∞
,
∞
)
(-\infty, \infty)
(
−
∞
,
∞
)
C
(
−
∞
,
0
)
∪
(
0
,
∞
)
(-\infty, 0) \cup (0, \infty)
(
−
∞
,
0
)
∪
(
0
,
∞
)
D
(
−
1
,
∞
)
(-1, \infty)
(
−
1
,
∞
)
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[Q83 · Sep · 2017]
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Q12
#12
NDA → Mathematics → Functions → Domain, Range, and Function Properties
·
Easy
Which one of the following graphs represents the function
f
(
x
)
=
x
∣
x
∣
,
x
≠
0
f(x) = \dfrac{x}{|x|},\ x \neq 0
f
(
x
)
=
∣
x
∣
x
,
x
=
0
?
Add
Lever: Modulus / absolute value behaviour
A
Graph (a): horizontal line at y=1
B
Graph (b): line at +1 for x>0, line at -1 for x<0 (step)
C
Graph (c): line at +1 for x>0, touching origin
D
None of the above
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[Q86 · Sep · 2017]
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Q13
#13
NDA → Mathematics → Functions → Greatest Integer Function
·
Moderate
Let
f
(
n
)
=
[
1
4
+
n
1000
]
f(n) = \left[\dfrac{1}{4} + \dfrac{n}{1000}\right]
f
(
n
)
=
[
4
1
+
1000
n
]
, where
[
x
]
[x]
[
x
]
denotes the integral part of
x
x
x
. Then the value of
∑
n
=
1
1000
f
(
n
)
\displaystyle\sum_{n=1}^{1000} f(n)
n
=
1
∑
1000
f
(
n
)
is
Add
Lever: Modulus / absolute value behaviour
A
251
B
250
C
1
D
0
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[Q87 · Sep · 2017]
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Q14
#14
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Hard
Let
g
g
g
be the greatest integer function. Then the function
f
(
x
)
=
(
g
(
x
)
)
2
−
g
(
x
2
)
f(x) = (g(x))^2 - g(x^2)
f
(
x
)
=
(
g
(
x
)
)
2
−
g
(
x
2
)
is discontinuous at
Add
Lever: Modulus / absolute value behaviour
A
all integers
B
all integers except 0 and 1
C
all integers except 0
D
all integers except 1
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[Q91 · Sep · 2017]
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Q15
#15
NDA → Mathematics → Functions → Greatest Integer Function
·
Hard
Let [x] denote the greatest integer function. What is the number of solutions of
x
2
−
4
x
+
[
x
]
=
0
x^2 - 4x + [x] = 0
x
2
−
4
x
+
[
x
]
=
0
in
[
0
,
2
]
[0, 2]
[
0
,
2
]
?
Add
Lever: Modulus / absolute value behaviour
A
Zero (No solution)
B
One
C
Two
D
Three
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[Q29 · Apr · 2018]
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Q16
#16
NDA → Mathematics → Functions → Domain, Range, and Function Properties
·
Moderate
Which is correct for
f
:
R
→
R
+
f:\mathbb{R}\to\mathbb{R}^+
f
:
R
→
R
+
defined as
f
(
x
)
=
∣
x
+
1
∣
f(x)=|x+1|
f
(
x
)
=
∣
x
+
1∣
?
Add
Lever: Modulus / absolute value behaviour
A
f
(
x
2
)
=
[
f
(
x
)
]
2
f(x^2)=[f(x)]^2
f
(
x
2
)
=
[
f
(
x
)
]
2
B
f
(
∣
x
∣
)
=
∣
f
(
x
)
∣
f(|x|)=|f(x)|
f
(
∣
x
∣
)
=
∣
f
(
x
)
∣
C
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
f(x+y)=f(x)+f(y)
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
D
None of the above
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[Q71 · Apr · 2018]
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Q17
#17
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Easy
If
f
(
x
)
=
∣
x
∣
+
∣
x
−
1
∣
f(x)=|x|+|x-1|
f
(
x
)
=
∣
x
∣
+
∣
x
−
1∣
, which is correct?
Add
Lever: Differentiability at a point
A
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
0
x=0
x
=
0
and
x
=
1
x=1
x
=
1
B
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
0
x=0
x
=
0
but not at
x
=
1
x=1
x
=
1
C
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
1
x=1
x
=
1
but not at
x
=
0
x=0
x
=
0
D
f
(
x
)
f(x)
f
(
x
)
is neither continuous at
x
=
0
x=0
x
=
0
nor at
x
=
1
x=1
x
=
1
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[Q73 · Apr · 2018]
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Q18
#18
NDA → Mathematics → Differentiation → Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
For
f
(
x
)
=
x
2
ln
∣
x
∣
f(x)=x^2\ln|x|
f
(
x
)
=
x
2
ln
∣
x
∣
(
x
≠
0
x\ne0
x
=
0
),
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
, what is
f
′
(
0
)
f'(0)
f
′
(
0
)
?
Add
Lever: Differentiability at a point
A
0
B
1
C
-1
D
It does not exist
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[Q74 · Apr · 2018]
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Q19
#19
NDA → Mathematics → Definite Integration → Integration of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
What is
∫
0
2
[
x
2
]
d
x
\int_{0}^{\sqrt{2}}[x^2]\,dx
∫
0
2
[
x
2
]
d
x
equal to (where [.] is greatest integer function)?
Add
Lever: Modulus / absolute value behaviour
A
2
−
1
\sqrt{2}-1
2
−
1
B
1
−
2
1-\sqrt{2}
1
−
2
C
2
(
2
−
1
)
2(\sqrt{2}-1)
2
(
2
−
1
)
D
3
−
1
\sqrt{3}-1
3
−
1
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[Q82 · Apr · 2018]
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Q20
#20
NDA → Mathematics → Functions → Domain, Range, and Function Properties
·
Moderate
For
f
(
x
)
=
1
∣
x
∣
−
x
f(x)=\frac{1}{\sqrt{|x|-x}}
f
(
x
)
=
∣
x
∣
−
x
1
, what is the domain of f?
Add
Lever: Modulus / absolute value behaviour
A
(
−
∞
,
0
)
(-\infty,0)
(
−
∞
,
0
)
B
(
0
,
∞
)
(0,\infty)
(
0
,
∞
)
C
(
−
∞
,
∞
)
(-\infty,\infty)
(
−
∞
,
∞
)
D
(
−
∞
,
0
]
(-\infty,0]
(
−
∞
,
0
]
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[Q85 · Apr · 2018]
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Q21
#21
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Easy
For the function
f
(
x
)
=
∣
x
−
3
∣
f(x)=|x-3|
f
(
x
)
=
∣
x
−
3∣
, which one of the following is
not
\textbf{\text{not}}
not
correct?
Add
Lever: Differentiability at a point
A
The function is not continuous at x = -3
B
The function is continuous at x = 3
C
The function is differentiable at x = 0
D
The function is differentiable at x = -3
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[Q74 · Sep · 2018]
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Q22
#22
NDA → Mathematics → Definite Integration → Integration of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
What is
∫
a
b
[
x
]
d
x
+
∫
a
b
[
−
x
]
d
x
\int_a^b [x]\,dx + \int_a^b [-x]\,dx
∫
a
b
[
x
]
d
x
+
∫
a
b
[
−
x
]
d
x
equal to, where
[
.
]
[.]
[
.
]
is the greatest integer function?
Add
Lever: Modulus / absolute value behaviour
A
b
−
a
b-a
b
−
a
B
a
−
b
a-b
a
−
b
C
0
0
0
D
2
(
b
−
a
)
2(b-a)
2
(
b
−
a
)
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[Q81 · Sep · 2018]
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Q23
#23
NDA → Mathematics → Definite Integration → Integration of Absolute Value, Piecewise, and Greatest Integer Functions
·
Easy
What is
∫
2
8
∣
x
−
5
∣
d
x
\int_2^8 |x-5|\,dx
∫
2
8
∣
x
−
5∣
d
x
equal to?
Add
Lever: Modulus / absolute value behaviour
A
2
B
3
C
4
D
9
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[Q82 · Sep · 2018]
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Q24
#24
NDA → Mathematics → Differentiation → Differentiation Techniques — Chain Rule, Logarithmic, Composite Functions
·
Hard
If
y
=
∣
sin
x
∣
∣
x
∣
y=|\sin x|^{|x|}
y
=
∣
sin
x
∣
∣
x
∣
, then what is the value of
d
y
d
x
\frac{dy}{dx}
d
x
d
y
at
x
=
−
π
6
x=-\frac{\pi}{6}
x
=
−
6
π
?
Add
Lever: Modulus / absolute value behaviour
A
2
−
π
6
(
6
ln
2
−
3
π
)
6
\frac{2^{-\frac{\pi}{6}}(6\ln 2-\sqrt{3}\pi)}{6}
6
2
−
6
π
(
6
l
n
2
−
3
π
)
B
2
π
6
(
6
ln
2
+
3
π
)
6
\frac{2^{\frac{\pi}{6}}(6\ln 2+\sqrt{3}\pi)}{6}
6
2
6
π
(
6
l
n
2
+
3
π
)
C
2
−
π
6
(
6
ln
2
+
3
π
)
6
\frac{2^{-\frac{\pi}{6}}(6\ln 2+\sqrt{3}\pi)}{6}
6
2
−
6
π
(
6
l
n
2
+
3
π
)
D
KaTeX parse error: Extra } at position 18: …rac{2^{\frac{6}}̲(6\ln 2-\sqrt{3…
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[Q95 · Sep · 2018]
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Q25
#25
NDA → Mathematics → Differentiation → Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
What is
d
1
−
sin
2
x
d
x
\frac{d\sqrt{1-\sin 2x}}{dx}
d
x
d
1
−
s
i
n
2
x
equal to, where
π
4
<
x
<
π
2
\frac{\pi}{4}<x<\frac{\pi}{2}
4
π
<
x
<
2
π
?
Add
Lever: Modulus / absolute value behaviour
A
cos
x
+
sin
x
\cos x+\sin x
cos
x
+
sin
x
B
−
(
cos
x
+
sin
x
)
-(\cos x+\sin x)
−
(
cos
x
+
sin
x
)
C
±
(
cos
x
+
sin
x
)
\pm(\cos x+\sin x)
±
(
cos
x
+
sin
x
)
D
None of the above
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[Q96 · Sep · 2018]
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