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Q.

The range of the function f(x)=log0.5(x42x2+3) is

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a

(,1]

b

[1,)

c

(,)

d

[1,1]

answer is B.

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Detailed Solution

f(x)=log0.5(x42x2+3)
Let x2=t
t22t+3=y
D=b24ac=(2)24×1×3=8
No real roots
y will be always positive
tmin=b2a=(2)2×1=1
ymin=D4a=(8)4×1=2
  Base is less than 1.
Graph will decrease for x > 1.
 f(x)=log0.(2)=log(12) (2)=log21=(1)log22=1
Range(,1]

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