MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
CBSE
Class 11 Maths
Limits And Derivatives
Quiz 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
The value of the limit Lim
x→0
(cos x)
cot2 x
is
0%
1
0%
e
0%
e
1/2
0%
e
-1/2
Explanation
e
-1/2
Hint:
Given, Lim
x→0
(cos x)
cot² x
= Lim
x→0
(1 + cos x – 1)
cot² x
= eLim
x→0
(cos x – 1) × cot² x
= eLim
x→0
(cos x – 1)/tan² x
= e
-1/2
The value of limit Lim
x→0
{sin (a + x) – sin (a – x)}/x is
0%
0
0%
1
0%
2 cos a
0%
2 sin a
Explanation
2 cos a
Hint:
Given, Lim
x→0
{sin (a + x) – sin (a – x)}/x
= Lim
x→0
{2 × cos a × sin x}/x
= 2 × cos a × Lim
x→0
sin x/x
= 2 cos a
Lim
x→-1
[1 + x + x² + ……….+ x
10
] is
0%
0
0%
1
0%
-1
0%
2
Explanation
1
Hint:
Given, Lim
x→-1
[1 + x + x² + ……….+ x
10
]
= 1 + (-1) + (-1)² + ……….+ (-1)
10
= 1 – 1 + 1 – ……. + 1
= 1
The value of Lim
x→01
(1/x) × sin
-1
{2x/(1 + x²) is
0%
0
0%
1
0%
2
0%
-2
Explanation
2
Hint:
Given, Lim
x→0
(1/x) × sin
-1
{2x/(1 + x²)
= Lim
x→0
(2× tan
-1
x)/x
= 2 × 1
= 2
Lim
x→0
log(1 – x) is equals to
0%
0
0%
1
0%
1/2
0%
None of these
Explanation
0
Hint:
We know that
log(1 – x) = -x – x²/2 – x³/3 – ……..
Now,
Lim
x→0
log(1 – x) = Lim
x→0
{-x – x²/2 – x³/3 – ……..}
⇒ Lim
x→0
log(1 – x) = Lim
x→0
{-x} – Lim
x→0
{x²/2} – Lim
x→0
{x³/3} – ……..
⇒ Lim
x→0
log(1 – x) = 0
Lim
x→0
{(a
x
– b
x
)/ x} is equal to
0%
log a
0%
log b
0%
log (a/b)
0%
log (a×b)
Explanation
log (a/b)
Hint:
Given, Lim
x→0
{(a
x
– b
x
)/ x}
= Lim
x→0
{(a
x
– b
x
– 1 + 1)/ x}
= Lim
x→0
{(a
x
– 1) – (b
x
– 1)}/ x
= Lim
x→0
{(a
x
– 1)/x – (b
x
– 1)/x}
= Lim
x→0
(a
x
– 1)/x – Lim
x→0
(b
x
– 1)/x
= log a – log b
= log (a/b)
The value of lim
y→0
{(x + y) × sec (x + y) – x × sec x}/y is
0%
x × tan x × sec x
0%
x × tan x × sec x + x × sec x
0%
tan x × sec x + sec x
0%
x × tan x × sec x + sec x
Explanation
x × tan x × sec x + sec x
Hint:
Given, lim
y→0
{(x + y) × sec (x + y) – x×sec x}/y
= lim
y→0
{x sec (x + y) + y sec (x + y) – x×sec x}/y
= lim
y→0
[x{ sec (x + y) – sec x} + y sec (x + y)]/y
= lim
y→0
x{ sec (x + y) – sec x}/y + lim
y→0
{y sec (x + y)}/y
= lim
y→0
x{1/cos (x + y) – 1/cos x}/y + lim
y→0
{y sec (x + y)}/y
= lim
y→0
[{cos x – cos (x + y)} × x/{y×cos (x + y)×cos x}] + lim
y→0
{y sec (x + y)}/y
= lim
y→0
[{2sin (x + y/2) × sin(y/2)} × 2x/{2y×cos (x + y)×cos x}] + lim
y→0
{y sec (x + y)}/y
= lim
y→0
{sin (x + y/2) × lim
y→0
{sin(y/2)/(2y/2)} × lim
y→0
{ x/{y×cos (x + y)×cos x}] + sec x
= sin x × 1 × x/cos² x + sec x
= x × tan x × sec x + sec x
So, lim
y→0
{(x + y) × sec (x + y) – x×sec x}/y = x × tan x × sec x + sec x
Lim
y→∞
{(x + 6)/(x + 1)}
(x+4)
equals
0%
e
0%
e³
0%
e
5
0%
e
6
Explanation
e
5
Hint:
Given, Lim
y→∞
{(x + 6)/(x + 1)}(x + 4)
= Lim
y→∞
{1 + 5/(x + 1)}(x + 4)
= e
Lim
y→∞
5(x + 4)/(x + 1)
= e
Lim
y→∞
5(1 + 4/x)/(1 + 1/x)
= e
5(1 + 4/∞)/(1 + 1/∞)
= e
5/(1 + 0)
= e
5
The expansion of log(1 – x) is
0%
x – x²/2 + x³/3 – ……..
0%
x + x²/2 + x³/3 + ……..
0%
-x + x²/2 – x³/3 + ……..
0%
-x – x²/2 – x³/3 – ……..
Explanation
-x – x²/2 – x³/3 – ……..
Hint:
log(1 – x) = -x – x²/2 – x³/3 – ……..
If f(x) = x × sin(1/x), x ≠ 0, then Lim
x→0
f(x) is
0%
1
0%
0
0%
-1
0%
does not exist
Explanation
0
Hint:
Given, f(x) = x × sin(1/x)
Now, Lim
x→0
f(x) = Lim
x→0
x × sin(1/x)
⇒ Lim
x→0
f(x) = 0
The value of Lim
n→∞
{1² + 2² + 3² + …… + n²}/n³ is
0%
0
0%
1
0%
-1
0%
n
Explanation
0
Hint:
Given, Lim
n→∞
{1² + 2² + 3² + …… + n²}/n³
= Lim
n→∞
[{n×(n + 1)×(2n + 1)}/6]/{n(n + 1)/2}²
= Lim
n→∞
[{n×n×n ×(1 + 1/n)×(2 + 1/n)}/6]/{n × n ×(1 + 1/n)/2}²
= Lim
n→∞
[{n³ ×(1 + 1/n)×(2 + 1/n)}/6]/{n² ×(1 + 1/n)/2}²
= Lim
n→∞
[{(1 + 1/n)×(2 + 1/n)}/6]/[n
4
× {(1 + 1/n)/2}²]
⇒ Lim
n→∞
[{(1 + 1/n)×(2 + 1/n)}/6]/[n × {(1 + 1/n)/2}²]
= [{(1 + 1/∞)×(2 + 1/∞)}/6]/[∞×{(1 + 1/∞)/2}²
= [{(1 + 0)×(2 + 0)}/6]/∞ {since 1/∞ = 0}
= {(1 × 2)/6}/∞
= (2/6)/∞
= (1/3)/∞
= 0
So, Lim
n→∞
{1² + 2² + 3² + …… + n²}/n³ = 0
The value of Lim
n→∞
(sin x/x) is
0%
0
0%
1
0%
-1
0%
None of these
Explanation
0
Hint:
Lim
n→∞
(sin x/x) = Lim
y→0
{y × sin (1/y)} = 0
The value of Lim
x→0
ax is
0%
0
0%
1
0%
1/2
0%
3/2
Explanation
1
Hint:
We know that
a
x
= 1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + ………..
Now,
Lim
x→0
a
x
= Lim
x→0
{1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + …}
⇒ Lim
x→0
a
x
= Lim
x→0
1 + Lim
x→0
{x/1! × (log a)} + Lim
x→0
{x² /2! × (log a)²}+ ………
⇒ Lim
x→0
a
x
= 1
Let f(x) = cos x, when x ≥ 0 and f(x) = x + k, when x < 0 Find the value of k given that Lim
x→0
f(x) exists.
0%
0
0%
1
0%
-1
0%
None of these
Explanation
1
Hint:
Given, Lim
x→0
f(x) exists
⇒ Lim
x→0
– f(x) = Lim
x→0
+ f(x)
⇒ Lim
x→0
(x + k) = Lim
x→0
cos x
⇒ k = cos 0
⇒ k = 1
Lim
x→0
sin (ax)/bx is
0%
0
0%
1
0%
a/b
0%
b/a
Explanation
a/b
Hint:
Given, Lim
x→0
sin (ax)/bx
= Lim
x→0
[{sin (ax)/ax} × (ax/bx)]
⇒ (a/b) Lim
x→0
sin (ax)/ax
= a/b
The value of the limit Lim
x→0
{log(1 + ax)}/x is
0%
0
0%
1
0%
a
0%
1/a
Explanation
a
Hint:
Given, Lim
x→0
{log(1 + ax)}/x
= Lim
x→0
{ax – (ax)² /2 + (ax)³ /3 – (ax)
4
/4 + …….}/x
= Lim
x→0
{ax – a² x² /2 + a³ x³ /3 – a
4
x
4
/4 + …….}/x
= Lim
x→0
{a – a² x /2 + a³ x² /3 – a
4
x³ /4 + …….}
= a – 0
= a
If f(x) = (x + 1)/x then df(x)/dx is
0%
1/x
0%
-1/x
0%
-1/x²
0%
1/x²
Explanation
-1/x²
Hint:
Given, f(x) = (x + 1)/x
Now, df(x)/dx = d{(x + 1)/x}/dx
= {1 × x – (x + 1)×1}/x²
= (x – x – 1)/x²
= -1/x²
Lim
x→0
(e
x²
– cos x)/x² is equals to
0%
0
0%
1
0%
2/3
0%
3/2
Explanation
3/2
Hint:
Given, Lim
x→0
(e
x²
– cos x)/x²
= Lim
x→0
(e
x²
– cos x -1 + 1)/x²
= Lim
x→0
{(e
x²
– 1)/x² + (1 – cos x)}/x²
= Lim
x→0
{(e
x²
– 1)/x² + Lim
x→0
(1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page