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Q.

Let fx =x3x2+10x7, x12x+log2b24, x>1 Then the set of all values of b, for which f(x) has maximum value
at x = 1, is

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a

(2,6)

b

[6,2)(2,6]

c

[6,2)(2,6]

d

(–6, –2)

answer is C.

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Detailed Solution

f1=3  For x<1,f'x=3x22x+10>0 fx is increasing   For x>1,f'x<0fx is decreasing  limx1+fx=2+log2b24  For maximum value at x=1 

32+log2b24log2b245b24>0 and b2432 or b2360b[6,2)(2,6] 

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