Let
\(f(x) = \begin{cases} x^3 - x^2 + 10x - 7, & x \leq 1 \\ -2x + \log_2(b^2 - 4), & x > 1 \end{cases}\)
Then the set of all values of b, for which f(x) has maximum value at x = 1, is
Let
\(f(x) = \begin{cases} x^3 - x^2 + 10x - 7, & x \leq 1 \\ -2x + \log_2(b^2 - 4), & x > 1 \end{cases}\)
Then the set of all values of b, for which f(x) has maximum value at x = 1, is
- (-6, -2)
- (2,6)
\((-6,-2) ∪ (2, 6)\)
\([-\sqrt6, -2] ∪ (2, \sqrt6]\)
The Correct Option is C
Solution and Explanation
\(f(x) = \begin{cases} x^3 - x^2 + 10x - 7, & x \leq 1 \\ -2x + \log_2(b^2 - 4), &x > 1 \end{cases}\)
If f(x) has maximum value at x = 1 then f(1+) ≤f(1)
–2 + log2(b2 – 4) ≤ 1 – 1 + 10 – 7
log2(b2 – 4) ≤ 5
0 <b2 – 4 ≤ 32
(i) b2–4>0
⇒ b∈(−∞,−2)∪(2,∞)
(ii) b2–36≤0
⇒ b∈[−6,6]
Intersection of above two sets
b∈[−6,−2)∪(2,6]
So, the correct option is (C): [−6,−2)∪(2,6]
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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