I differentiated:

if we use -5,it will give a very large negative value, so cancel it. the larger the value,the more negative it will keep getting. just check for -\sqrt{3} and 1.

Ans:D

Can you please explain elaborately

In math, the first derivative is used to find max. and min. values of a function. Starting from this point, i differentiated to find the \(x\) where the function takes the greatest value. Hope it's clear.

for simplification assume x^2 = y then 6y-y^2-4 or -y^2+6y-4
rewrite as

-y^2 + 6y - 9 + 5 (to make perfect squares)
-( y^2 – 6y + 9) + 5
5 – (y-3)^2

We know that
(y-3)^2 ≥ 0
Therefore, 5 – (y-3)^2 will always be lesser than or equal to 5
The greatest value of 5 – (y - 3)^2 is hence 5
This happens when
y = 3
Or
x^2 = 3
x = +√3 or x = -√3

Bunuel , chetan2u, yashikaaggarwal can anyone of you help me with this ?

@knombat see, you just have to check for 3 values
Option A and E are out of the constraint given, between which the value of X must lie

now,
put \( X = -√5, -2, -√3\)
in eq \(6x^2-x^4-4\)

\(X = -√5\)
\(6(-√5)^2-(-√5)^4-4\)
\(6*5 - 25 - 4 = 30-25-4 = 1\)

\(X = -√3\)
\(6(-√3)^2-(-√3)^4-4\)
\(6*3 - 9 - 4 = 18-9-4 = 5 \)

\(X = -2\)
\(6(-2)^2-(-2)^4-4\)
\(6*4 - 16 - 4 = 24-16-4 = 4\)

Among all only X = -√3 gives higher value, so that's our answer.
P.S. I saw earlier explanations, Maxima & Minima, Substitution X^2 as Y all works fine, it depends on you what method you can retain till last.
P.P.S - Before solving for question check whether you can eliminate option or not, It will save you some time.

Hi,

Questions where the value of x is asked and the options contain the values, the best way is to substitute these values in the original equation and look for the choice that fits in in the given criteria or criterion.

Here too, the criterion is the GREATEST value, so substitute the values and you will get the answer.
_________________

Another Approach

because we are given this equation i.e. 6x^2-x^4-4
we can transform this equation to -(4-6x^2+4)
= -((x^2-3)^2 -9+4)
= 9-4 - (x^2-3)^2
= 5 - (x^2-3)^2
now we know that Since (x^2-3)^2 is always >=0
the minimum value is 0, which is at -root(3),root(3)
greatest value of the expression is 5,which can be obtained at -root(3),root(3)
hence the answer is D

I did not realize we could use differentiation for this. Thanks for showing how it works.

I have a question, though. We ended up with 3 values for x. Does this mean the value of the equation is maximum when x is either of these numbers?

If x = sqrt(3) or -sqrt(3), the value of the equation is 5.

But when x = 0, value of the equation is -4.

So, what do these 3 values really denote?

These values denote the point where the slope of the equation is 0.
to find which of these values will give the maxima or minima, differentiate second time and plug in the values.
if the resultant is <0 , then the substituted value is the maxima
if the resultant >0, the value is the minima.

I put x = 1, which is giving me 1. Then why any other negative value be the answer?