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Principle: Modulus / absolute value behaviour
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Q26
#26
NDA → Mathematics → Probability → Probability via Counting
·
Moderate
Two integers x and y are chosen with replacement from the set
{
0
,
1
,
2
,
…
,
10
}
\{0,1,2,\ldots,10\}
{
0
,
1
,
2
,
…
,
10
}
. The probability that
∣
x
−
y
∣
>
5
|x-y|>5
∣
x
−
y
∣
>
5
is
Add
Lever: Modulus / absolute value behaviour
Concept: Choosing numbers with a property
A
6
11
\frac{6}{11}
11
6
B
35
121
\frac{35}{121}
121
35
C
30
121
\frac{30}{121}
121
30
D
25
121
\frac{25}{121}
121
25
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[Q115 · Sep · 2018]
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Q27
#27
NDA → Mathematics → Quadratic Equations → Special Quadratics — Parametric, Logarithmic, Constructed
·
Moderate
What are the roots of the equation
∣
x
2
−
x
−
6
∣
=
x
+
2
|x^2 - x - 6| = x + 2
∣
x
2
−
x
−
6∣
=
x
+
2
?
Add
Lever: Modulus / absolute value behaviour
A
−
2
,
1
,
4
-2, 1, 4
−
2
,
1
,
4
B
0
,
2
,
4
0, 2, 4
0
,
2
,
4
C
0
,
1
,
4
0, 1, 4
0
,
1
,
4
D
−
2
,
2
,
4
-2, 2, 4
−
2
,
2
,
4
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[Q8 · Apr · 2019]
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Q28
#28
NDA → Mathematics → Quadratic Equations → Special Quadratics — Parametric, Logarithmic, Constructed
·
Easy
The number of real roots for the equation
x
2
+
9
∣
x
∣
+
20
=
0
x^2+9|x|+20=0
x
2
+
9∣
x
∣
+
20
=
0
is
Add
Lever: Modulus / absolute value behaviour
A
Zero
B
One
C
Two
D
Three
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[Q78 · Apr · 2019]
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Q29
#29
NDA → Mathematics → Definite Integration → Integration of Absolute Value, Piecewise, and Greatest Integer Functions
·
Moderate
∫
0
π
/
2
∣
sin
x
−
cos
x
∣
d
x
\displaystyle\int_0^{\pi/2}|\sin x-\cos x|\,dx
∫
0
π
/2
∣
sin
x
−
cos
x
∣
d
x
is equal to
Add
Lever: Modulus / absolute value behaviour
A
0
0
0
B
2
(
2
−
1
)
2(\sqrt{2}-1)
2
(
2
−
1
)
C
2
2
2\sqrt{2}
2
2
D
2
(
2
+
1
)
2(\sqrt{2}+1)
2
(
2
+
1
)
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[Q82 · Apr · 2019]
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Q30
#30
NDA → Mathematics → Linear Inequalities → Linear Systems and Feasible Regions
·
Moderate
If
∣
x
2
−
3
x
+
2
∣
>
x
2
−
3
x
+
2
\left|x^{2}-3x+2\right| > x^{2}-3x+2
x
2
−
3
x
+
2
>
x
2
−
3
x
+
2
, then which one of the following is correct ?
Add
Lever: Modulus / absolute value behaviour
A
x
≤
1
x \leq 1
x
≤
1
or
x
≥
2
x \geq 2
x
≥
2
B
1
≤
x
≤
2
1 \leq x \leq 2
1
≤
x
≤
2
C
1
<
x
<
2
1 < x < 2
1
<
x
<
2
D
x
x
x
is any real value except 3 and 4
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[Q7 · Sep · 2019]
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Q31
#31
NDA → Mathematics → Quadratic Equations → Special Quadratics — Parametric, Logarithmic, Constructed
·
Easy
How many real roots does the equation
x
2
+
3
∣
x
∣
+
2
=
0
x^{2} + 3|x| + 2 = 0
x
2
+
3∣
x
∣
+
2
=
0
have ?
Add
Lever: Modulus / absolute value behaviour
A
Zero
B
One
C
Two
D
Four
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[Q37 · Sep · 2019]
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Q32
#32
NDA → Mathematics → Applications of Integration → Area Bounded by a Curve, Lines, and Axes
·
Easy
What is the area of the region bounded by
∣
x
∣
<
5
|x| < 5
∣
x
∣
<
5
,
y
=
0
y = 0
y
=
0
and
y
=
8
y = 8
y
=
8
?
Add
Lever: Modulus / absolute value behaviour
A
40 square units
B
80 square units
C
120 square units
D
160 square units
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[Q71 · Sep · 2019]
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Q33
#33
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Easy
For what value of
k
k
k
is the function
f
(
x
)
=
{
2
x
+
1
4
,
x
<
0
k
,
x
=
0
(
x
+
1
2
)
2
,
x
>
0
f(x) = \begin{cases} 2x + \dfrac{1}{4}, & x < 0 \\ k, & x = 0 \\ \left(x + \dfrac{1}{2}\right)^{2}, & x > 0 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
2
x
+
4
1
,
k
,
(
x
+
2
1
)
2
,
x
<
0
x
=
0
x
>
0
continuous ?
Add
Lever: Modulus / absolute value behaviour
A
1
4
\dfrac{1}{4}
4
1
B
1
2
\dfrac{1}{2}
2
1
C
1
D
2
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[Q98 · Sep · 2019]
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Q34
#34
NDA → Mathematics → Functions → Domain, Range, and Function Properties
·
Easy
What is the minimum value of
∣
x
−
1
∣
|x-1|
∣
x
−
1∣
, where
x
∈
R
x\in\mathbb{R}
x
∈
R
?
Add
Lever: Modulus / absolute value behaviour
A
0
B
1
C
2
D
−
1
-1
−
1
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[Q89 · Apr · 2020]
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Q35
#35
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Moderate
Consider the following statements for
f
(
x
)
=
e
−
∣
x
∣
f(x)=e^{-|x|}
f
(
x
)
=
e
−
∣
x
∣
: 1. The function is continuous at
x
=
0
x=0
x
=
0
. 2. The function is differentiable at
x
=
0
x=0
x
=
0
. Which of the above statements is/are correct?
Add
Lever: Differentiability at a point
A
1 only
B
2 only
C
Both 1 and 2
D
Neither 1 nor 2
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[Q91 · Apr · 2020]
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Q36
#36
NDA → Mathematics → Differentiation → Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
·
Easy
If
f
(
x
)
=
e
∣
x
∣
f(x)=e^{|x|}
f
(
x
)
=
e
∣
x
∣
, then which one of the following is correct?
Add
Lever: Differentiability at a point
A
f
′
(
0
)
=
1
f'(0)=1
f
′
(
0
)
=
1
B
f
′
(
0
)
=
−
1
f'(0)=-1
f
′
(
0
)
=
−
1
C
f
′
(
0
)
=
0
f'(0)=0
f
′
(
0
)
=
0
D
f
′
(
0
)
f'(0)
f
′
(
0
)
does not exist
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[Q84 · Apr · 2021]
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Q37
#37
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Moderate
If the function
f
(
x
)
=
{
a
+
b
x
,
x
<
1
5
,
x
=
1
b
−
a
x
,
x
>
1
f(x)=\begin{cases}a+bx, & x<1\\5, & x=1\\b-ax, & x>1\end{cases}
f
(
x
)
=
⎩
⎨
⎧
a
+
b
x
,
5
,
b
−
a
x
,
x
<
1
x
=
1
x
>
1
is continuous, then what is the value of
(
a
+
b
)
(a+b)
(
a
+
b
)
?
Add
Lever: Modulus / absolute value behaviour
A
5
B
10
C
15
D
20
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[Q93 · Apr · 2021]
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Q38
#38
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Moderate
Let
f
(
x
)
=
{
1
+
x
2
k
,
0
<
x
<
2
k
x
,
2
≤
x
<
4
f(x)=\begin{cases}1+\frac{x}{2k},&0<x<2\\kx,&2\le x<4\end{cases}
f
(
x
)
=
{
1
+
2
k
x
,
k
x
,
0
<
x
<
2
2
≤
x
<
4
. If
lim
x
→
2
f
(
x
)
\lim_{x\to2}f(x)
lim
x
→
2
f
(
x
)
exists, then what is the value of
k
k
k
?
Add
Lever: Modulus / absolute value behaviour
A
−
2
-2
−
2
B
−
1
-1
−
1
C
0
0
0
D
1
1
1
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[Q75 · Sep · 2021]
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Q39
#39
NDA → Mathematics → Limits & Continuity → Continuity and Differentiability — Piecewise, Modulus, Composed, Oscillatory
·
Easy
Consider the following statements in respect of
f
(
x
)
=
∣
x
∣
−
1
f(x)=|x|-1
f
(
x
)
=
∣
x
∣
−
1
: 1.
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
1
x=1
x
=
1
. 2.
f
(
x
)
f(x)
f
(
x
)
is differentiable at
x
=
0
x=0
x
=
0
. Which of the above statements is/are correct?
Add
Lever: Differentiability at a point
A
1 only
B
2 only
C
Both 1 and 2
D
Neither 1 nor 2
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[Q76 · Sep · 2021]
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Q40
#40
NDA → Mathematics → Limits & Continuity → One-Sided Limits, Greatest Integer, and Absolute Value Limits
·
Easy
If
f
(
x
)
=
[
x
]
∣
x
∣
f(x)=\frac{[x]}{|x|}
f
(
x
)
=
∣
x
∣
[
x
]
,
x
≠
0
x\ne0
x
=
0
, where
[
⋅
]
[\cdot]
[
⋅
]
denotes the greatest integer function, then what is the right-hand limit of
f
(
x
)
f(x)
f
(
x
)
at
x
=
1
x=1
x
=
1
?
Add
Lever: Modulus / absolute value behaviour
A
−
1
-1
−
1
B
0
0
0
C
1
1
1
D
Right-hand limit of
f
(
x
)
f(x)
f
(
x
)
at
x
=
1
x=1
x
=
1
does not exist
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[Q77 · Sep · 2021]
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Q41
#41
NDA → Mathematics → Definite Integration → Integration of Absolute Value, Piecewise, and Greatest Integer Functions
·
Easy
If
∫
−
2
0
f
(
x
)
d
x
=
k
\int_{-2}^{0}f(x)\,dx=k
∫
−
2
0
f
(
x
)
d
x
=
k
, then
∫
−
2
0
∣
f
(
x
)
∣
d
x
\int_{-2}^{0}|f(x)|\,dx
∫
−
2
0
∣
f
(
x
)
∣
d
x
is
Add
Lever: Modulus / absolute value behaviour
A
less than
k
k
k
B
greater than
k
k
k
C
less than or equal to
k
k
k
D
greater than or equal to
k
k
k
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[Q97 · Sep · 2021]
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Q42
#42
NDA → Mathematics → Applications of Integration → Area Bounded by a Curve, Lines, and Axes
·
Moderate
What is the area bounded by
y
=
[
x
]
y=[x]
y
=
[
x
]
, where
[
⋅
]
[\cdot]
[
⋅
]
is the greatest integer function, the
x
x
x
-axis and the lines
x
=
−
1
⋅
5
x=-1\cdot5
x
=
−
1
⋅
5
and
x
=
−
1
⋅
8
x=-1\cdot8
x
=
−
1
⋅
8
?
Add
Lever: Modulus / absolute value behaviour
A
0·3 square unit
B
0·4 square unit
C
0·6 square unit
D
0·8 square unit
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[Q99 · Sep · 2021]
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Q43
#43
NDA → Mathematics → Functions → Greatest Integer Function
·
Moderate
Consider the following statements in respect of the function
y
=
[
x
]
y=[x]
y
=
[
x
]
,
x
∈
(
−
1
,
1
)
x\in(-1,1)
x
∈
(
−
1
,
1
)
where
[
.
]
[.]
[
.
]
is the greatest integer function: 1. Its derivative is 0 at
x
=
0.5
x=0.5
x
=
0.5
2. It is continuous at
x
=
0
x=0
x
=
0
Which of the above statements is/are correct?
Add
Lever: Modulus / absolute value behaviour
A
1 only
B
2 only
C
Both 1 and 2
D
Neither 1 nor 2
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[Q56 · Apr · 2022]
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Q44
#44
NDA → Mathematics → Definite Integration → Integration of Absolute Value, Piecewise, and Greatest Integer Functions
·
Easy
What is
∫
−
2
−
1
x
∣
x
∣
d
x
\int_{-2}^{-1}\frac{x}{|x|}\,dx
∫
−
2
−
1
∣
x
∣
x
d
x
equal to?
Add
Lever: Modulus / absolute value behaviour
A
-2
B
-1
C
1
D
2
Tap an option to check your answer.
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[Q64 · Apr · 2022]
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Q45
#45
NDA → Mathematics → Differentiation → Differentiability of Absolute Value, Piecewise, and Greatest Integer Functions
·
Easy
Let
y
=
[
x
+
1
]
y=[x+1]
y
=
[
x
+
1
]
,
−
4
<
x
<
−
3
-4<x<-3
−
4
<
x
<
−
3
where
[
.
]
[.]
[
.
]
is the greatest integer function. What is the derivative of
y
y
y
with respect to
x
x
x
at
x
=
−
3.5
x=-3.5
x
=
−
3.5
?
Add
Lever: Differentiability at a point
A
-4
B
-3.5
C
-3
D
0
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[Q71 · Apr · 2022]
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Q46
#46
NDA → Mathematics → Functions → Greatest Integer Function
·
Moderate
Let
z
=
[
y
]
z = [y]
z
=
[
y
]
and
y
=
[
x
]
−
x
y = [x] - x
y
=
[
x
]
−
x
, where [.] is the greatest integer function. If x is not an integer but positive, then what is the value of z?
Add
Lever: Modulus / absolute value behaviour
A
−
1
-1
−
1
B
0
0
0
C
1
1
1
D
2
2
2
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[Q71 · Sep · 2022]
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Q47
#47
NDA → Mathematics → Applications of Integration → Area Bounded by a Curve, Lines, and Axes
·
Moderate
What is the area of the region bounded by
x
−
∣
y
∣
=
0
x - |y| = 0
x
−
∣
y
∣
=
0
and
x
−
2
=
0
x - 2 = 0
x
−
2
=
0
?
Add
Lever: Modulus / absolute value behaviour
A
1
B
2
C
4
D
8
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[Q81 · Sep · 2022]
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Q48
#48
NDA → Mathematics → Limits & Continuity → One-Sided Limits, Greatest Integer, and Absolute Value Limits
·
Moderate
If
f
(
x
)
=
x
2
+
x
+
∣
x
∣
x
f(x) = \frac{x^2+x+|x|}{x}
f
(
x
)
=
x
x
2
+
x
+
∣
x
∣
, then what is
lim
x
→
0
f
(
x
)
\lim_{x\to 0} f(x)
lim
x
→
0
f
(
x
)
equal to?
Add
Lever: Modulus / absolute value behaviour
A
0
B
1
C
2
D
lim
x
→
0
f
(
x
)
\lim_{x\to0}f(x)
lim
x
→
0
f
(
x
)
does not exist
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[Q86 · Sep · 2022]
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Q49
#49
NDA → Mathematics → Application of Derivatives → Monotonicity, Extrema, and Critical Points
·
Moderate
Consider the following statements in respect of the function
f
(
x
)
=
{
∣
x
∣
+
1
,
0
<
∣
x
∣
≤
3
1
,
x
=
0
f(x) = \begin{cases}|x|+1, & 0<|x|\leq3\\1, & x=0\end{cases}
f
(
x
)
=
{
∣
x
∣
+
1
,
1
,
0
<
∣
x
∣
≤
3
x
=
0
: 1. The function attains maximum value only at x=3 2. The function attains local minimum only at x=0 Which of the statements given above is/are correct?
Add
Lever: Modulus / absolute value behaviour
A
1 only
B
2 only
C
Both 1 and 2
D
Neither 1 nor 2
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[Q93 · Sep · 2022]
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Q50
#50
NDA → Mathematics → Functions → Domain, Range, and Function Properties
·
Easy
How many real numbers satisfy the equation
∣
x
−
4
∣
+
∣
x
−
7
∣
=
15
|x-4|+|x-7|=15
∣
x
−
4∣
+
∣
x
−
7∣
=
15
?
Add
Lever: Modulus / absolute value behaviour
A
Only one
B
Only two
C
Only three
D
Infinitely many
Tap an option to check your answer.
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[Q21 · Apr · 2023]
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