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Mathematics Question on Functions

Let
f(x)={x3−x2+10x−7,x≤1 −2x+log⁡2(b2−4),x>1f(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

A

(-6, -2)

B

(2,6)

C

(−6,−2)∪(2,6)(-6,-2) ∪ (2, 6)

D

[−6,−2]∪(2,6][-\sqrt6, -2] ∪ (2, \sqrt6]

Answer

(−6,−2)∪(2,6)(-6,-2) ∪ (2, 6)

Explanation

Solution

f(x)={x3−x2+10x−7,x≤1 −2x+log⁡2(b2−4),x>1f(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(b 2 – 4) ≤ 1 – 1 + 10 – 7
log2(b 2 – 4) ≤ 5
0 <b 2 – 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]