Test: Maxima Minima (Derivative) - JEE MCQ

f(x) = sinx + 1/2cos2x  
⇒ f'(x) = cos x – sin2x 
Now, f'(x) = 0 gives cosx – sin2x = 0 
⇒ cos x (1 – 2 sinx) = 0 
⇒ cos x = 0, (1 – 2 sinx) = 0 
⇒ cos x = 0, sinx = 1/2 
⇒ x = π/6 , π/2 
Now, f(0) = 1/2, 
f(π/6) = 1/2 + 1/4 = 3/4, 
f(π/2) = 1 – 1/2 = 1/2 
Therefore, the absolute max value = 3/4 and absolute min = 1/2

f(x) = sin x
f’(x) =cosx 
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0 
Hence, 0 is the maxima.

For every real number (or) valued function f(x), the values of x which satisfies the equation f¹(x)=0 are the point of it's local and global maxima or minima.
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)^x
We will be using the equation, y = (1/x)^x 
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y. 
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating  dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e⁽⁻¹⁾
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e^(1/e).
Hence the maximum value of the function is (e)^1/e

p’(x) = -24 - 36x
p”(x) = -36
Now, p’(x) = 0  ⇒ x = (-24)/36
x = -⅔
Also, p”(-⅔) = -36 < 0
By the second derivative test,  x = -⅔
Therefore, maximum profit = p(-⅔)
= 41 - 24(-⅔) - 18(-⅔)^2 
= 41 +16 - 8  
⇒ 49

f(x) = alog|x| + bx² + x
f’(x) = a/x + 2bx + 1
f’(-1) = - a - 2b + 1
-a - 2b + 1 = 0
a = 1 - 2b
f’(2) = a/2 + 4b + 1 = 0
a + 8b = -2
Put the value of a in eq(1)
(1 - 2b) + 8b = - 2
6b = -3
b = -½, a = 2

f(x)=2x³-24x+107 x ∈ [1, 3]
f'(x) = 6x^2 - 24
To find the points equate f'(x) = 0
in closed interval x= -2 doesn't lies,so discard x = -2
now find the value of function at x = 1, 2 ,3
f(1) =2(1)³-24(1)+107
= 2-24+107 = 85
f(2) =2(2)³-24(2)+107
= 16-48+107 = 75
f(3) =2(3)³-24(3)+107
= 54-72+107 = 89
So, the function has maximum value in close interval at x= 3, Maximum value= 89.
minimum value at x= 2, minimum value = 75