Classical

Calculus - Calculus MCQ

1:  

If calculus mcq

A.

2x2

B.

√x

C.

0

D.

1

 
 

Option: A

Explanation :

Calculus questions

Click on Discuss to view users comments.

Sagnik Bhawal said: (11:56pm on Wednesday 5th August 2015)
while integrating how are the upper and lower limits getting changed? coz the integration doesn't result a negative sign. In absence of a negative sign how are the limits getting interchanged? i mean the answer should have been -2x^2 instead of 2x^2
ravi said: (2:48pm on Saturday 5th December 2015)
it should be -2x^2
Waqar said: (11:57pm on Thursday 4th May 2017)
Dear, Its Answer Should be -2x square not 2x square. how can you exchange limits?

Write your comments here:



2:  

The function f(x) = x3 - 6x2 + 9x + 25 has

A.

a maxima at x= 1 and a minima at x = 3

B.

a maxima at x = 3 and a minima at x = 1

C.

no maxima, but a minima at x = 1

D.

a maxima at x = 1, but no minima

 
 

Option: A

Explanation :

f '(x) = 3(x2-4x+3)    f ''(x)=6(x)-12
at  x=1,  f "(1)  
at  x=3, f ''(3) = +ve

Click on Discuss to view users comments.

abc said: (7:22pm on Friday 5th April 2013)
how x=1 and x=3 arrive??
Akem said: (11:02pm on Tuesday 8th September 2015)
i taught maxima at x=3 because is positive and minima at x = 1 because it is negative
Jay Vyas said: (1:14am on Saturday 28th November 2015)
at X = 1 We got minimum value and at X= 3 we got maximum valueSo in my opinion option B is right
Samrah said: (12:38pm on Wednesday 13th July 2016)
Take 1st derivative of the function and put it equal to 0 to find the value of "x".Now take 2nd derivative and put the 2 values of X previously found in equation to get a result respectively. If the final answer is ve so it is a Minima. If the answer is -ve it is a maxima.

Write your comments here:



3:  

The value of a = is 

A.

>0

B.

2

C.

0 - 1 + 100 - 10 + 1

D.

undefined

 
 

Option: A

Explanation :

Calculus questions answers

Click on Discuss to view users comments.

kiran said: (11:23am on Friday 2nd December 2016)
how cos5p=-1

Write your comments here:



4:  

The interval in which the Lagrange's theorem is applicable for the function f(x) = 1/x is

A.

[-3, 3]

B.

[-2, 2]

C.

[2, 3]

D.

[-1, 1]

 
 

Option: C

Explanation :

Since f(x) = (1/x) is not continuous in [- 3, 3] [- 4, 2] or [- 1, 1], The point of discontinuity is '0'. Only in [2, 3] the function is continuous, and differentiable hence mean value theorem is applicable in [2, 3].

Click on Discuss to view users comments.

Write your comments here:



5:  

 If f(x) = | x | , then for interval [-1, 1] ,f(x)

 

 

A.

satisied all the conditions of Rolle's Theorem

B.

satisfied all the conditions of Mean Value Theorem

C.

does not satisied the -conditions of Mean Value Theorem

D.

None of these

 
 

Option: C

Explanation :

Since f(x) = |x| is continuous in [-1, 1] but  it is not differentiable at x = 0 ε (-1, 1)

Click on Discuss to view users comments.

Write your comments here: