Let f(x)=(1+b2)x2+2bx+1 and m(b) be the minimum value of f(x). If b can assume different values, then range of m(b) is equal to
[1/2,1]
[0,1]
(0,1]
(0,1/2]
A
(0,1]
B
[0,1]
C
(0,1/2]
D
[1/2,1]
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Solution
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f(x)=(1+b2)x2+2bx+1 It is quadratic expression with coefficient of x2 as (1+b2)>0 Whose minimum value is −D4a ∴m(b)=−{4b2−4(1+b2)}4.(1+b2) m(b)=11+b2 For range of m(b) Let y=11+b2 ⇒b2=1−yy Since, b2≥0 ⇒1−yy≥0 ⇒y>0and1−y≥0 ⇒y>0andy≤1
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