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Integral Calculus

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6 Oct 2009 23:16:01 IST

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

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

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Guruprasad Raghavan

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6 Oct 2009 23:17:51 IST

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tell me yr question!!!

Nishank

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6 Oct 2009 23:25:43 IST

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let f(x) is a function in which sighn of f(x) and f''(x) are same for all x belongs R. f(x) =x^2-ax+2. g(x) & h(x) are two functions defined as g(x) = f(x).f'(x) & h(k)= int.lim. (a to 4) |x| dx.1. g(x) isa.inc. function b.dec. function c. have local maximad. hava local minima2.range of h(k) isa. [2,8]b.[4,8]c.[2,4]d. N.O.T

Nishank

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6 Oct 2009 23:26:32 IST

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CAN ANYONE DO THIS

Nishank

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6 Oct 2009 23:32:35 IST

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oh c'mon goiitians.................no one....

Nishank

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6 Oct 2009 23:41:05 IST

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vijay pal ji ..plz solve

Nishank

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6 Oct 2009 23:58:54 IST

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plz ans.

kabi

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7 Oct 2009 01:27:16 IST

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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

Nishank

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7 Oct 2009 11:17:37 IST

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explain

Nishank

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7 Oct 2009 12:33:49 IST

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....

Nishank

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7 Oct 2009 14:16:04 IST

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no one............?????????/

Pritish Chakraborty

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7 Oct 2009 14:49:56 IST

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He wrote f"(x), not f'(x)...

f"(x) in this case = 2.

Thus f(x) > 0.

We have f'(x) = 2x - a, so putting it equal to 0,

x = a/2. Sign changes across this checkpoint. By wavy curve method, for f(x) > 0, x > a/2. Meaning the domain of f(x) is (a/2, infinity).

g(x) = f(x).f'(x) = (x² - ax + 2)(2x - a).

Will solve further. Be back in 20 minutes.

Nishank

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7 Oct 2009 15:26:47 IST

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help

apollo

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7 Oct 2009 15:30:15 IST

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plese post the question a bit more clearly

VINEET KUMAR

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7 Oct 2009 15:44:01 IST

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F"X=2 WHICH IS > 0 AND

SINCE FOR ALL X F(X) AND F"X ARE OF SAME SIGN

SO F(X)>0

IT WILL HAPPEN ONLY WHEN DISCRIMINANT <0

SO BY SOLVING a BELONGS TO [-2ROOT2,2ROOT2]

NOW G(x) =2(X^2 -aX+2) WILL HAVE LOCAL MINIMA

AND H(K) BELONGS TO (4,8) (

Nishank

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7 Oct 2009 15:46:37 IST

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vineet can u plz explain???

Nishank

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7 Oct 2009 15:48:16 IST

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POSTING THE QUESTION IN PIECES...................let f(x) is a function in which sighn of f(x) and f''(x) are same for all x belongs R. f(x) =x^2-ax+2. g(x) & h(x) are two functions defined as g(x) = f(x).f'(x) & h(k)= int.lim. (a to 4) |x| dx.

Nishank

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7 Oct 2009 15:50:04 IST

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1. g(x) is [a].inc. function [b].dec. function[c]. have local maxima[d]. hava local minima

VINEET KUMAR

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7 Oct 2009 15:50:30 IST

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I M SORRY

G(X) IS NOT = 2(X^2-aX+2)

BUT G(X)=(2X-a)((X^2-aX+2)

SO G'(X)=2(X^2-aX+2)+( 2X-a) ^2 WHICH IS >0 AS (X^2-aX+2) >0

SO G(X) IS AN INCREASING FUNCTION (

Nishank

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7 Oct 2009 15:50:49 IST

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2.range of h(k) is[a]. [2,8][b].[4,8][c].[2,4][d]. N.O.T

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