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Differentiation- Limits & Derivatives MCQs for A Level Exam
It covers all Important Questions with answers on Differentiation- Limits & Derivatives for the A Level exam. The questions are based on important topics. Details about the questions:- Topic: Differentiation- Limits & Derivatives
- Type of Questions: MCQs with solutions
- Number of Questions: 50
- You can attempt them on EduRev to score high in A Level exam.
- a)1
- b)1/3
- c)1/2
- d)0
Correct answer is option 'B'. Can you explain this answer?
a)
1
b)
1/3
c)
1/2
d)
0
 | Lohit Matani answered |
lim(x → 0) (tanx-x)/x2 tanx
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x2(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x2 sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x2cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x2cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x2(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x2 sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x2cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x2cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3
The derivative of f(x) = 99x at x = 100 is:- a)9900
- b)0
- c)99
- d)100
Correct answer is option 'C'. Can you explain this answer?
The derivative of f(x) = 99x at x = 100 is:
a)
9900
b)
0
c)
99
d)
100
 | Raghav Bansal answered |
f'(100) = lim(h→0) [f(100+h) - f(100)]/h
= lim(h→0) [99(100+h) - 99(100)]/h
= lim(h→0) [9900 + 99h - 9900]/h
= lim(h→0) 99h/h
= lim(h→0) 99
= 99
= lim(h→0) [99(100+h) - 99(100)]/h
= lim(h→0) [9900 + 99h - 9900]/h
= lim(h→0) 99h/h
= lim(h→0) 99
= 99
- a)e5
- b)e4
- c)e2
- d)e3
Correct answer is option 'A'. Can you explain this answer?
a)
e5
b)
e4
c)
e2
d)
e3
 | Manish Aggarwal answered |
lim (x → 0) [((1-3x)+5x)/(1-3x)]1/x
lim (x → 0) [1 + 5x/(1-3x)]1/x
= elim(x → 0) (1 + 5x/(1-3x) - 1) * (1/x)
= elim(x → 0) (5x/(1-3x)) * (1/x)
= elim(x → 0) (5x/(1-3x))
= e5
lim (x → 0) [1 + 5x/(1-3x)]1/x
= elim(x → 0) (1 + 5x/(1-3x) - 1) * (1/x)
= elim(x → 0) (5x/(1-3x)) * (1/x)
= elim(x → 0) (5x/(1-3x))
= e5
If is a real valued function and c is a point in its domain, then 
is ;- a)f (x)
- b)f’ (x)
- c)f (c)
- d)f’ (c)
Correct answer is option 'D'. Can you explain this answer?
If is a real valued function and c is a point in its domain, then is ;
is ;a)
f (x)
b)
f’ (x)
c)
f (c)
d)
f’ (c)
 | Gaurav Kumar answered |
This is a formula for finding derivative or differentiation which is represented by
but here at the place of x , c is written So this is equal to f'(c)
but here at the place of x , c is written So this is equal to f'(c)Derivative of sum of two functions is sum of the derivatives of the functions. If , f and g be two functions such that their derivatives are defined in ______.- a)Common domain
- b)Their individual domains
- c)Universally
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Derivative of sum of two functions is sum of the derivatives of the functions. If , f and g be two functions such that their derivatives are defined in ______.
a)
Common domain
b)
Their individual domains
c)
Universally
d)
None of the above
| | Gaurav Kumar answered |
- The process of determining the derivative of a function is known as differentiation. It is clearly visible that the basic concept of derivative of a function is closely intertwined with limits. Therefore, it can be expected that the rules of derivatives are similar to that of limits.
- The following rules are a part of algebra of derivatives:
Consider f and g to be two real-valued functions such that the differentiation of these functions is defined in a common domain.
- a)e
- b)1
- c)log 2
- d)2 log 2
Correct answer is option 'D'. Can you explain this answer?
a)
e
b)
1
c)
log 2
d)
2 log 2
| | Manish Aggarwal answered |
lim(x → 0) (x2x - x)/(1-cosx)
lim(x → 0) x2(2x - 1)/x(1-cosx)
By formula : lim(x → 0) (ax - 1)/x = log a
lim(x → 0) x2 log2/(1-cosx)
Differentiate it
lim(x → 0) 2x log2/(sinx)
= 2 log2
lim(x → 0) x2(2x - 1)/x(1-cosx)
By formula : lim(x → 0) (ax - 1)/x = log a
lim(x → 0) x2 log2/(1-cosx)
Differentiate it
lim(x → 0) 2x log2/(sinx)
= 2 log2
- a)∞
- b)e
- c)e1/3
- d)1
Correct answer is option 'C'. Can you explain this answer?
a)
∞
b)
e
c)
e1/3
d)
1
 | Ciel Knowledge answered |
lim(x → 0) (tanx/x)(1/x^2)
= (1)∞
elim(x → 0) (1/x2)(tanx/x - 1)
= elim(x → 0) ((tanx - x)/x3) .....(1)
lim(x → 0) ((tanx - x)/x3)
(0/0) form, Apply L hospital rule
lim(x → 0) [sec2x -1]/3x2
lim(x → 0) [tan2x/3x2]
= 1/3 lim(x → 0) [tan2x/x2]
= 1/3 * 1
= e1/3
= (1)∞
elim(x → 0) (1/x2)(tanx/x - 1)
= elim(x → 0) ((tanx - x)/x3) .....(1)
lim(x → 0) ((tanx - x)/x3)
(0/0) form, Apply L hospital rule
lim(x → 0) [sec2x -1]/3x2
lim(x → 0) [tan2x/3x2]
= 1/3 lim(x → 0) [tan2x/x2]
= 1/3 * 1
= e1/3
Let (tan α) x + (sin α) y = α and (α cosec α) x + (cos α) y = 1 be two variable straight lines, α being the parameter. Let P be the point of intersection of the lines. In the limiting position when α→ 0, the coordinates of P are- a)(2 , 1)
- b) (–2 ,–1)
- c)(–2 , 1)
- d)(2 , –1)
Correct answer is option 'D'. Can you explain this answer?
Let (tan α) x + (sin α) y = α and (α cosec α) x + (cos α) y = 1 be two variable straight lines, α being the parameter. Let P be the point of intersection of the lines. In the limiting position when α→ 0, the coordinates of P are
a)
(2 , 1)
b)
(–2 ,–1)
c)
(–2 , 1)
d)
(2 , –1)
 | Maths Learner answered |
Can anyone explain the question and concept
- a)y + 1
- b)1/y
- c)y
- d)y – 1
Correct answer is option 'C'. Can you explain this answer?
a)
y + 1
b)
1/y
c)
y
d)
y – 1
 | Om Desai answered |
Since the number of terms is finite, just differentiate term by term using the power rule.
dy/dx = 0 + 1 + 2x/2! + 3x2/3! + ... + nxn-1/n!
dy/dx = 1 + x + x2/2! + ... + xn-1/(n -1)!
Compare dy/dx to y and note that the last term of y, which is xn/n!, is not in dy/dx, while all the other terms of y are in dy/dx.
Then dy/dx = y - xn/n!
⇒ dy/dx + xn/n! = y
dy/dx = 0 + 1 + 2x/2! + 3x2/3! + ... + nxn-1/n!
dy/dx = 1 + x + x2/2! + ... + xn-1/(n -1)!
Compare dy/dx to y and note that the last term of y, which is xn/n!, is not in dy/dx, while all the other terms of y are in dy/dx.
Then dy/dx = y - xn/n!
⇒ dy/dx + xn/n! = y
The derivate of x3 - 20x at x = 10 is :- a)280
- b)320
- c)250
- d)180
Correct answer is option 'A'. Can you explain this answer?
The derivate of x3 - 20x at x = 10 is :
a)
280
b)
320
c)
250
d)
180
 | Yash Patel answered |
Y=x3-20x
dy/dx=3x2-20
(dy/dx) at x=10 is
=3×100-20
=280
dy/dx=3x2-20
(dy/dx) at x=10 is
=3×100-20
=280
- a)e
- b)1/e
- c)2e
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
a)
e
b)
1/e
c)
2e
d)
None of these
 | Praveen Kumar answered |
lim(x → 1) (log2 2x)1/log2x
= lim(x →1) (log22 + log2x)1/log2x
As we know that {log ab = log a + log b}
lim(x → 1) {1 + log2x}1/log2x
log2x → 0
Put t = log2x
lim(t → 0) {1 + t}1/t
= e
= lim(x →1) (log22 + log2x)1/log2x
As we know that {log ab = log a + log b}
lim(x → 1) {1 + log2x}1/log2x
log2x → 0
Put t = log2x
lim(t → 0) {1 + t}1/t
= e
- a)sec2 x
- b)sec x. tan x
- c)- cosec x. cot x
- d)- cot2 x
Correct answer is option 'C'. Can you explain this answer?
a)
sec2 x
b)
sec x. tan x
c)
- cosec x. cot x
d)
- cot2 x
 | Krishna Iyer answered |
y = cosecx
= 1/sinx= (sinx)−1
⇒ dy/dx=−(sinx)−2 × d/dx(sinx)
= −cosx/sin2x
= −cosx/sinx ×(1/sinx)
= −cot x cosec x
= 1/sinx= (sinx)−1
⇒ dy/dx=−(sinx)−2 × d/dx(sinx)
= −cosx/sin2x
= −cosx/sinx ×(1/sinx)
= −cot x cosec x
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