GATE CSE 1995 | Question: 25a

Question

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

Comments

(≤ 720)

what is this?

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2nd part of this question ….

Determine the number of positive integer(<=720)

      which are not divisible by any of 2,3, or 5.

solution:

total number of positive integers (<=720)=720 

no. of integers divisible by 2=floor(720/2)=360

no. of integers divisible by 3=floor(720/3)=240

no. of integers divisible by 5=floor(720/5)=144

Here we are going to see the formula for (A U B U C).

n(AUBUC)=n(A)+n(B)+n(C)-n(AnB)-n(BnC)-n(CnA)+n(AnBnC)

so number of positive integers(<720) which are  divisible by any of 2,3,or 5:

     =floor(720/2)+floor(720/3)+floor(720/5) – floor(720/6) – floor(720/15) –         floor(720/10).+floor(720/30)

      = 360+240+144 -120 -48-72+24

      = 528  positive integers numbers are there which are divisible by 2 ,3 or 5. between 1 to 720.

SO answer will be 720-528=192 numbers which are not divisible by  2 ,3 or 5.

 

 

 

Answer 1

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

$f '(x) = -4 + 4x = 0  \implies  x=1$

$f ''(x) = 4$

$f ''(1) = 4>0$

 Therefore at $x=1$ we will get minimum value, which is : $3 - 4(1) + 2(1)^2 = 1.$

Answer 2

Step 1: Identify aa and bb

2x2−4x+32x^2-4x+3

So,

a=2,b=−4a=2,\qquad b=-4

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Step 2: Find the value of xx where the minimum occurs

x=−(−4)2(2)=44=1x=\frac{-(-4)}{2(2)} =\frac{4}{4} =1

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Step 3: Substitute x=1x=1

f(1)=3−4(1)+2(1)2f(1)=3-4(1)+2(1)^2 =3−4+2=3-4+2 $$=1$$