The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+10 votes
667 views
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 $5.$
asked in Calculus by Veteran (59.9k points)
edited by | 667 views
0

(&le; 720)

what is this?

1 Answer

+12 votes
Best answer

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

$f '(x) = -4 + 4x = 0  \implies  x=1$

$f ''(x) = 4$

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

Ans for B:

answered by Loyal (8.1k points)
edited by
0
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
0
thank you.:-)
0
nice explanation
+2
Why have you included 720.....In question in is mentioned numbers less than 720.
0

@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.

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
48,725 questions
52,831 answers
183,518 comments
68,656 users