JEE Maths Zone - 2018-09-20

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3:13 PM

> Let $a,b,c \in \mathbb R$ such that no two of them are equal and
> satisfy $\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0$,
> then the equation $24ax^2 + 4bx +c=0$ has:
>
> a) atleast one root in $[0,\frac 12]$
>
> b) at least one root in $[-\frac 12, 0)$
>
> c) at least one root in $[1,2]$
>
> d) none of these

@AvnishKabaj @Jasmine Its from package did you attempt it

Avnish Kabaj

3:33 PM

@Abcd Cayley Hamilton se nahi ho Raha?

Abcd

@AvnishKabaj WTH is that!

Anonymous

4:01 PM

(I don't know, but isn't that a circular determinant?)

Anonymous

(Oh, is that a matrix lol? I thought det meant determinant)

Anonymous

(A quick Google search tells me that det indeed means determinant. I am acting so dumb?)

Anonymous

(And if it is a circular determinant, that would give us $2a= -(b+c)$ meaning option (1) is correct?)

Anonymous

(I haven't studied determinants yet, so I might be speaking a little gibberish.)

Anonymous

(And that's why I am putting everything in these brackets)

Anonymous

4:14 PM

@Abcd ^

Jasmine

@Abcd circular determinant so either $(2a+b+c)=0$ or $2a=b=c$ the later condition will always give imaginary roots

Anonymous

Oh yeah, I was correct. I am happy.

Jasmine

@Abcd Now if you take the first condition $(2a+b+c) =0$ and accordingly plot the values of function in the intervals as in option 1 you Will get opposite signs at $x=0$ and $x=1/2$

By IMVT. One real root is for sure

@IceInkberry det of matrix means corresponding determinant value of the matrix

You didn't mention how did you get option 1 did you use the same IMVT logic I used

@IceInkberry ?

Jasmine

4:33 PM

I got confused

Shouldn't the answer be D

It says at lease one root

But if the 2a=b=c was chosen then no real roots in any Condition

Abcd

@Jasmine it's given a b c not equal

Jasmine

@Abcd oh I didn't read

Oh thank you! I am crazy..

Abcd

@Jasmine what do you mean here?

@IceInkberry why?

I get 2a = -(b+c) but how does that make option a correct

Jasmine

At x=0 the value of the equation is $-(2a+b)$ and at x=1/2 value =$(4a+b)$

Anonymous

@Jasmine IMVT means?

Anonymous

4:40 PM

Oh ha. Got it.

Anonymous

Like at $0$ and $1/2$ the expression's value has opposite signs.

Jasmine

If $b>-4a$ so $b>-2a$ @Abcd

Anonymous

And since I was really lazy. I used desmos to graph it and verify my answer.

Jasmine

I feel so slow and irritated my phone is just autocorrecting any thing I don't know how to turn it off

Anonymous

@Jasmine Settings > Language and Input > Keyboard or it may be just on that page

Jasmine

4:43 PM

@IceInkberry input option is not available

Anonymous

Search it in settings.

Anonymous

In the search bar.

Anonymous

It's in General management.

Anonymous

Or anything like 'General stuff'

Jasmine

I have tried so many times it's no use I don't remember myself I tuned auto correct on and now I want to off it but don't know how should I anyways leave it

Anonymous

4:50 PM

@Jasmine Please use commas and full stops in your sentences :p

Anonymous

I always get confused haha

Jasmine

@IceInkberry Intermediate Mean Value Theorem if I am not wrong

Anonymous

10 mins ago, by Ice Inkberry

Oh ha. Got it.

Jasmine

@IceInkberry Ok!