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Find the minimum value of $3-4x+2x^2$.

Determine the number of positive integers $(< 720)$ which are not divisible by any of $2, 3$ or $5.$
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(&le; 720)

what is this?

$f(x) = 3-4x+2x^2$

$f '(x) = -4 + 4x = 0 \implies x=1$

$f ''(x) = 4$

$f ''(1) = 4>0$, therefore at $x=1$ we will get minimum value, which is : $3 - 4(1) + 2(1)^2 = 1.$

Ans for B:

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can you plz further explain this figure, jayendra?
+5
div by 2=720/2=360

by 3=240

by 5=144

by 2 and 3=120

by 2 and 5=72

by 3 and 5=48

by 2&3&5=24

for 2 or 3 or 5=528

so not div=192
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thank you.:-)
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nice explanation
+2
Why have you included 720.....In question in is mentioned numbers less than 720.
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@Dushyant Raut 4

U are correct but as 720 is divisible by 2 it is not included in answer . So including it does not affect ans. But i should not be included.

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