Let \(λ^*\) be the largest value of \(λ\) for which the function \(f_λ(x) = 4λx^3 – 36λx^2 + 36x + 48\) is increasing for all \(x ∈ ℝ\). Then \(f_λ^* (1) + f_λ^* (– 1)\) is equal to :
- 36
- 48
- 64
- 72
The Correct Option is D
Solution and Explanation
As \(f_λ(x) = 4λx^3 – 36λx^2 + 36x + 48\)
Hence, \(f_λ^{'}(x) = 12(λx^2-6λx+3)\)
For \(fλ(x) \)increasing : \((6λ)^2 – 12λ ≤ 0\)
\(∴ λ ∈[0,\frac{1}{3}]\)
\(∴ λ ^∗=\frac{1}{3}\)
Then,
\(fλ^*(x) = \frac{4}{3}x^3−12x^2+36x+48\)
\(∴fλ^*+fλ^*(−1)=73\frac{1}{2}−1\frac{1}{2}\)
= \(72\).
Hence, the correct option is (D): \(72\)
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions