NCERT Solutions
Class 12 Maths
Application of Derivatives
Ex.6.5 Q.1
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = (2x − 1)2 + 3
(ii) f(x) = 9x2 + 12x + 2
(iii) f(x) = −(x − 1)2 + 10
(iv) g(x) = x3 + 1
(i) The given function is f(x) = (2x − 1)2 + 3
It can be observed that (2x − 1)2 ≥ 0 for every x ∈ R.
Therefore, f(x) = (2x − 1)2 + 3 ≥ 3 for every x ∈ R.
The minimum value of f is attained when 2x − 1 = 0
(ii) The given function is f(x) = 9x2 + 12x + 2 = (3x + 2)2 − 2
It can be observed that (3x + 2)2 ≥ 0 for every x ∈ R.
Therefore, f(x) = (3x + 2)2 − 2 ≥ −2 for every x ∈ R.
The minimum value of f is attained when 3x + 2 = 0
⟹ 3x + 2 = 0
⟹ x = -23
So, minimum value of f = {3 (-23 )+ 2}2 – 2 = -2
Hence, function f does not have a maximum value.
(iii) The given function is f(x) = − (x − 1)2 + 10
It can be observed that (x − 1)2 ≥ 0 for every x ∈ R.
Therefore, f(x) = − (x − 1)2 + 10 ≤ 10 for every x ∈ R.
The maximum value of f is attained when (x − 1) = 0
⟹ x − 1 = 0
⟹ x = 1
So, maximum value of f = f(1) = − (1 − 1)2 + 10 = 10
Hence, function f does not have a minimum value.
(iv) The given function is g(x) = x3 + 1
Hence, function g neither has a maximum value nor a minimum value.
⟹ 2x − 1 = 0
⟹ x = 1/2
So, minimum value of f(1/2) = (2 * 1/2 − 1)2 + 3 = 3
Hence, function f does not have a maximum value.