Test Index

KEAM 2013 Math Paper

© examsnet.com

Question : 100 of 120

Marks: +1, -0

If the slope of

y=3x^2+ax^3

x=\frac{1}{2},

then the value of

a

is equal to

2
1
$-1$
$-2$
3

Solution:

Given curve is

y=3 x^{2}+a x^{3}

On differentiating w.r.t.

x

, we get

\;\frac{dy}{dx}=6x+3ax^{2}=m(

say

)

....(i)

Now, differentiating Eq. (i) w.r.t.

x

, we get

\;\frac{dm}{dx}=6+6ax

and

\;\frac{d^{2}m}{dx^{2}}=6a

For maximum or minimum of

m

,

put

\;\; \;\frac{dm}{dx}=0 \Rightarrow 6+6ax=0

\Rightarrow \;\; x=-\;\frac{1}{a}<0

0 (max)

But given

m

is maximum at

x=\;\frac{1}{2}

.

\therefore \;\; -\;\frac{1}{a}=\;\frac{1}{2} \Rightarrow a=-2

© examsnet.com

Go to Question:

Prev Question

Next Question