Maxima Minima (Derivative)- 2 - Free MCQ Practice Test with solutions,

f(x) = sinx.cosx
f’(x) = -sin²x + cos²x
-sin²x + cos²x = 0
sin²x = cos²x
tan²x = 1
tanx = +-1
for tan x = 1, x = π/4
for tan x = -1, x = 3π/4
At x = π/4, f(π/4) = ½
At x = 3π/4, f(3π/4) = -½
At π/4 is the local maxima.

f(x) = log x
f’(x) = 1/x
Putting f’(x) = 0
1/x = 0
x = ∞ (this is not defined for x)
So, f(x) does not have minima or maxima

f(x) = x⁵ - 5x⁴ + 5x³ - 1
On differentiating w.r.t.x, we get 
f'(x) = 5x⁴ - 20x³ + 15x²
For maxima or minima,f'(x) = 0
⇒ 5x⁴ - 20x³ + 15x² = 0
⇒ 5x²(x²-4x+3) = 0
⇒ 5x²(x-1)(x-3) = 0
⇒ x = 0,1,3
So, y has maxima at x = 1 and minima at x = 3
At   x = 0, y has neither maxima nor minima, which is the point of inflection .
f(1) = 1-5+5-1=0,which is local maximum value at x = 1. 

A point C in the domain of a function f at which either f(c) = 0 or f is not differentiable.  
The point f  is called the critical point.
c is called the point of local maxima
If f ′(x) changes sign from positive to negative as x increases through c, that is, if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c.
c is called the point of local minima
If f ′(x) changes sign from negative to positive as x increases through c, that is, if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c.
c is called the point of inflexion
If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima.

two positive numbers x and y are such that x + y = 60.
 x + y = 60
⇒ x = 60 – y  ...(1)
Let P = xy³
∴ P =(60 – y)y³ = 60y³ – y4
Differentiating both sides with respect to y, we get

For maximum or minimum dP/dy = 0
⇒ 180y² - 4y³ = 0
⇒ 4y² (45 - y) = 0
⇒ y = 0 or 45 - y = 0
⇒ y = 0 or y = 45
⇒ y = 45 (∵ y = 0 is not possible)


Thus, the two positive numbers are 15 and 45.