NCERT Solutions
Class 12 Maths
Application of Derivatives
Ex.6.5 Q.2
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = |x + 2| − 1
(ii) g(x) = − |x + 1| + 3
(iii) h(x) = sin(2x) + 5
(iv) f(x) = |sin 4x + 3|
(v) h(x) = x + 4, x ∈ (−1, 1)
(i) Given, f(x) = |x + 2| - 1
We know that |x + 2| ≥ 0 for every x ∈ R.
Therefore, f(x) = |x + 2| - 1 ≥ -1 for every x ∈ R.
The minimum value of f is attained when |x + 2| = 0
⟹ |x + 2| = 0
⟹ x = -2
So, minimum value of f = f(−2) = |-2 + 2| - 1 = -1
Hence, function f does not have a maximum value.
(ii) Given, g(x) = -|x + 1| + 3
We know that -|x + 1| ≤ 0 for every x ∈ R.
Therefore, g(x) = -|x + 1| + 3 ≤ 3 for every x ∈ R.
The maximum value of g is attained when |x + 1| = 0
⟹ |x + 1| = 0
⟹ x = -1
So, maximum value of g = g(−1) = -|-1 + 1| + 3 = 3
Hence, function g does not have a minimum value.
(iii) Given, h(x) = sin2x + 5
We know that − 1 ≤ sin 2x ≤ 1
⟹ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5
⟹ 4 ≤ sin 2x + 5 ≤ 6
Hence, the maximum and minimum values of h are 6 and 4 respectively.
(iv) Given, f(x) = |sin 4x + 3|
We know that −1 ≤ sin 4x ≤ 1
⟹ 2 ≤ sin 4x + 3 ≤ 4
⟹ 2 ≤ |sin 4x + 3| ≤ 4
Hence, the maximum and minimum values of f are 4 and 2 respectively.
(v) Given, h(x) = x + 1, x ∈ (−1, 1)
Here, if a point x0 is closest to −1, then we find x0/2 + 1 < x0 < 1 for all x0 ∈ (−1, 1).
Also, if x1 is closest to 1, then (x1 + 1 < (x1 + 1)/2 + 1< 1 for all x ∈ (−1, 1).
Hence, function h(x) has neither maximum nor minimum value in (−1, 1).