Maxima and MinimaHard
Question
f(x) =
and g(x) =
f(t) dt, x ∈ [1, 3] then g (x) has
and g(x) =
f(t) dt, x ∈ [1, 3] then g (x) hasOptions
A.local maxima at x = 1 + ln 2 and local minima at x = e
B.local maxima at x = 1 and local minima at x = 2
C.no local maxima
D.no local minima
Solution
g′(x) = f(x) = 
g′(x) = 0, when x = 1 + ln2 and x = e
g′′(x) =
g′′(1 + ln 2) = - eln2 < 0 hence at x = 1 + ln 2, g(x) has a local maximum
g′′(e) = 1 > 0 hence at x = e, g(x) has local minimum.
∵ f(x) is discontinuous at x = 1, then we get local maxima at x = 1 and local minima at x = 2.

g′(x) = 0, when x = 1 + ln2 and x = e
g′′(x) =

g′′(1 + ln 2) = - eln2 < 0 hence at x = 1 + ln 2, g(x) has a local maximum
g′′(e) = 1 > 0 hence at x = e, g(x) has local minimum.
∵ f(x) is discontinuous at x = 1, then we get local maxima at x = 1 and local minima at x = 2.
Create a free account to view solution
View Solution FreeMore Maxima and Minima Questions
Suppose the cube x3 - px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?...f′(x) = (x − a)2n(x − b)2p + 1; n,p ∈ N, then-...The area of the rectangle with given perimeter will be maximum when the ration of its two sides is -...Let f(x) = then :......