IIT JEE , AIEEE Forum - Mathematics , Algebra

sankydreams

Please post detailed replies of following ques for appropriate rates-

1) If x = 2 + 2^2/3 + 2^1/3,then find the value of x³ - 6x² + 6x.

2)Both roots of the eqn (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0 are always :

a)positive

b)negative

c)real

d)none of these

3)If a+b+c = 0, then the quadratic eqn 3ax²+2bx+c=0 has

a)atleast one root in [0,1]

b)one root in [2,3] and the other in [-2,-1]

c)imaginary roots

d)none of these

4)For real x, the function (x-a)(x-b)/(x-c) will assume all real values provided

a)a>b>c

b)a<b<c

c)a>c>b

d)a<c<b

kj

1 ans1 is 2

ans2 is positive

ans3 is a

sankydreams

dude the first & third ans is correct but second is wrong. Please give detailed explanations man if u can.....

lazycol

Q2:

eqn simplyfies 2, x² - 2(a+b+c)x + ab+bc+ca = 0

discriminant = 4(a+b+c)²- 4(ab+bc+ca)

= 4 (a² + b² + c² + ab + bc + ca )

= 2{ (a+b)²+ (b+c)²+ (c+a)²}

discriminant >= 0 ie roots r real

now put a=b=c=0 gives x=0

thus roots can b +ve or -ve

(c) is crt

shobna

is d ans for d 4 th question 'd'?

lazycol

let y = (x-a)(x-b)/(x-c)

simplify, x² - (a+b+y)x + ab + cy = 0

x

R ie D>0 so (a+b+y)²- 4(ab+cy) > 0

y² - (4c-2c-2b)y +(a-b)²> 0

quadratic fnt > 0 ie D < 0

(4c - 2a -2b)²- 4(a-b)²< 0

(c-a)(c-b) < 0

ie a< c< b or b<c<a

(c) or (d) is crt becoz if a&b r interchanged noting is going 2 happen

irlmaks

Q NO. 1

i am giving an eloberate proof since i am asked for!

x-2=2^1/3+2^2/3

cubing on both sides , we get x³-6x²+12x-8=2+3.2^1/3+2/3(2^1/3+2^2/3)+4

since once again 2^1/3+2^2/3 is x-2, now the above equation reduses to

x³-6x²+12x-8=6+6.(x-2)

hence x³-6x²+6x=2

magiclko

x = 2 + 2^2/3 + 2^1/3

=> x-2=2^1/3+2^2/3

cubing both sides,

x³-6x²+12x-8 = 2+3.2 (2^1/3+2^2/3)+4

= 6 + 6(x-2)

=> x³-6x²+6x=2

irlmaks

Q NO. 2

the given equation is equivalent to

3x²-2(a+b+c)x+(ab+bc+ca)

D²=4[(a+b+c)²-3(ab+bc+ca)]

=2[(a-b)²+(b-c)²+(c-a)²]

0 ,equality occuring iff a=b=c

Hence the roots are real

The question is not all over

it is required to check whether all triad of (a,b,c) yield both the roots of given equation are either positive or negative or one +ve and other -ve or one triad yields one result

now here the argument is intresting

product of the roots is (ab+bc+ca)/3

obviously there are triads of (a,b,c) for which the product is >0 , <0 , =0

i.e, not for all triads of (a,b,c) the product is positive or negative

hence both the roots cannot be +ve -ve for all triads of (a,b,c)

hence (A),(B) cannot be true for all triads of (a,b,c)

akku

Q2) Simplifying

3x^2-2x(a+b+c)+ab+bc+ca=0

Discriminant=D=B^2-4AC=4(a+b+c)^2-12(ab+bc+ca)

D=4(a^2+b^2+c^2)-4(ab+bc+ca)

Considering nos a^2,b^2;b^2,c^2;c^2,a^2

Arithmetic mean>Geometric mean

a^2+b^2/2>ab

b^2+c^2/2>bc

c^2+a^2/2>ac ADDING THESE INEQ a^2+b^2+^2>ab+bc+ca

--->D>0 AND=0 when a=b=c

=>ROOTS R REAL c)

Q3) assume a function f(x)=ax^3+bx^2+cx+d

f(x) being a polynomial fn is both continuous n diff for all real nos

f(0)=d

f(1)=a+b+c+d=d (since a+b+c=0)

f(0)=f(1)

as all the condiions of rolles theorem r verified there must exist a 'g' in(0,1)

such that f '(g)=0

ie (3ag^2+2bg+c)=0

HENCE 3ax^2+2bx+c=0 has at least one root in [0,1] a)

PLS RATE MY EFFORTS !

irlmaks

Q NO . 3

the given equation is the differential of the equation ax³+bx²+cx=0

i.e, x(ax²+bx+c)=0 Since a+b+c=0 , the cubic has 0 , 1 as solutions

If f(x)=ax³+bx²+cx then , f(1)=f(0)=0 and f is differentiable on [0,1]

hence there exists a c in [0,1] such that [f(1)-f(0)]=(1-0)f'(c)

i.e,there exists a c in [0,1] such that f'(c)=0

i.e,there exists a c in [0,1] such that 3ax²+2bx+c=0

the proof is now complete

I HOPE THAT U RATE FOR THIS SOL.

sankydreams

Hey irlmaks ur solutions r good but since i havent as yet studied Differentiation of JEE level, i dint get ur proof posted at 23:42. Thanx man

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