If
A.  2x^{2} 
B.  √x 
C.  0 
D.  1 
Option: A Explanation :
Click on Discuss to view users comments. Sagnik Bhawal said: (4:56pm 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: (7:48am on Saturday 5th December 2015)
it should be 2x^2
Waqar said: (4:57pm on Thursday 4th May 2017)
Dear, Its Answer Should be 2x square not 2x square. how can you exchange limits?

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(x^{2}4x+3) f ''(x)=6(x)12 Click on Discuss to view users comments. abc said: (12:22pm on Friday 5th April 2013)
how x=1 and x=3 arrive??
Akem said: (4:02pm 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: (6:14pm on Friday 27th 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: (5:38am 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.

The value of a = is
A.  >0 
B.  2 
C.  0  1 + 100  10 + 1 
D.  undefined 
Option: A Explanation :
Click on Discuss to view users comments. kiran said: (4:23am on Friday 2nd December 2016)
how cos5p=1

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. 
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. 