Question

Let f(x)=(1+b^2)x^2+bx+1 and let m(b) be the minimum value of f(x).As b varies the range of m(b) is-

Answer 1

ans. (0,1]

Answer 2

remember the vertex formula for finding minimum of this quadratic function

Answer 3

(-b/2a , -D/4a) ??

Answer 4

yep

Answer 5

now u know that m(b)=-D/4a

Answer 6

yes

Answer 7

well find the formula for m(b)

Answer 8

could you pls solve!

Answer 9

ok

Answer 10

??

Answer 11

sorry its not (0,1]

Answer 12

@dpaInc
help

Answer 13

yes... i'm here.... :)
@shubham.bagrecha, is (0, 1] the answer you got or is that the solution?

Answer 14

thats solution

Answer 15

hmmm. i keep getting (3/4, 1].... lemme try again...

Answer 16

the vertex of the parabola is given by -b/(2a), so in this problem, the vertex is:
\(\large \frac{-b}{2a}\rightarrow \frac{-b}{2(1+b^2)} \)

Answer 17

U r Quite right answer is (3/4 ,1]

Answer 18

go on

Answer 19

the y-coordinate is f(\(\large \frac{-b}{2(1+b^2)} \)) = \(\large \frac{3}{4}+\frac{1}{4(1+b^2)} \)

Answer 20

as b varies, \(\large \frac{1}{4(1+b^2)} \) ranges from (0, 1/4]

Answer 21

so the m(b) varies in the range (3/4,1]

Answer 22

yes... but idk why the solution is (0, 1]...:(

Answer 23

maybe question is f(x)=(1+b^2)x^2+2bx+1 .....

Answer 24

@shubham.bagrecha
if given function is f(x)=(1+b^2)x^2+bx+1 then answer is (3/4,1] not (0,1]