Question

The least integral value of 'm' for which the expression
mx^2 -4x +3m +1 is positive for every
x belongs to R is

Answer 1

Is there any algebraic method to this

Answer 2

There are several methods. One of which requires calculus.

Answer 3

well +ve with zero ?

Answer 4

Put, f(x) = mx^2 - 4x + 3m + 1

Find minima of f(x) and put it >0

Answer 5

You would get a very simple answer.

Answer 6

That's one method.

Answer 7

Ok and the next method

Answer 8

Change f(x) into complete square form.

Answer 9

(A) 1
(B) -2
(C) -1
(D) 2
ARe the options

Answer 10

there another algebraic method if u only ask for +ve -{0}

Answer 11

@ikram002p i don't know if zero is included

Answer 12

D < 0

Answer 13

I know finding max and min but haven't learnt it in math as such

Answer 14

Yeah that's one method probably @rsadhvika

Answer 15

Here is the second method:

\(\large{f(x) = mx^2 - 4x + 3m + 1}\)

\(\large{f(x) = m(x^2 - \cfrac{4}{m}x + 3 + \cfrac{1}{m})}\)

\(\large{f(x) = m[(x-\cfrac{2}{m})^2 - \cfrac{4}{m^2} + 3 + \cfrac{1}{m}]}\)

Answer 16

Now, this must be positive. Solve it

Answer 17

umm
why not try b^2-4ac >0

Answer 18

Other method is the one @rsadhvika mentioned. But, for that remember that m must be +ve.

Answer 19

No @aajugdar that will not work :



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