PLZZZZZZZZZ.............. SOLVE THIS QUES.....URGENT.......

Ask iit jee aieee pet cbse icse state board community

Discussion Response Post to: PLZZZZZZZZZ.............. SOLVE THIS QUES.....URGENT.......

Forum Index -> Integral Calculus -> View Full Question

Author

Message

kabi (1665)

Blazing goIITian

total posts: 575
Offline

look man question is confusing me . f (x) is parabola and f'(x) is staight line there is no. value of x for which condition of having same sign satisfyies .

u know f(x) will either meet x-axis at 1 point or it will intersect it at two points where as any values of a f'(x) will cut x-axis at one point only ie. x = a/2 so for all x it can't be that f(x) and f'(x) have same sign .

If the question is that for right part of f(x) then only cond applies then the solution is as follows ::

f(x) =x^2 -ax +2

f'(x) = 2x -a

f'(x) change sign at x = a/2

so x=a/2 should be d only root of f(x)

f (a/2 )=0

this gives a =-2Sqrt(2)

g' (x) = f(x) f'' (x) + ( f(x) ' )2 = 2 x^2 -2ax +4 +4x^2 +a^2 -4ax

= 6x^2-6ax +4 +a^2

D= 36a^2 -24 (4+a^2) putting a = -2Sqrt(2) D= 0

so g'(x) is zero at one point

and g(x) has one min point

The time u guys take to find the derivative of a function or for finding the equilibrium constant of a reaction or for finding the angle of dispersion of prism or for standing from ur seat to congratulate our team after their win almost in that time one kid die because of poverty.

Like 0 people liked this

Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info