IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

nitish_iit

_{[x][infinity]} {6a

(x^2+ax+a^2) -

(x^2+a^2)} /{a(6x+2a-1/ax)-xe^1/x}

(a) 6a-1 (b) 6a+1 (c) 1 (d) 0

PLZZZZ..give me ans in all steps....

nitin62225

ans=1

soln:

take x common in numerator nd denominator.and then u get (6a-1)/(6a-1)=1

nik_d_1

smit.chandra

the ans is 1.

after multiplying with ax in denominator we get

{6a[(rootx^2+ax+a^2)] - root(x^2+a^2)}/{6ax+2a^2-1/x-xe^1/x}

=take x common fm num. and denom.

=x^2{6a[root(1+a/x+(a/x)^2)]-root(1+(a/x)^2)/x^2{6a+2a^2/x-1/x^2-e^1/x}

=6a-1/6a-1=1(on applying x tends to infinity)

elessar_iitkgp

Transform the quotient as follows
_x

_infinity [x{6a

(1+ a/x + (a/x)²) -

1+(a/x)²] / [(6 + 2/x -1/x) - x e ^1/x]

Now take the x from the numerator & transform as follows

_x

_infinity [{6a

(1+ a/x + (a/x)²) -

1+(a/x)²] / [{(6 + 2/x -1/x) - x e ^1/x}/x]
_x

_infinity [{6a

(1+ a/x + (a/x)²) -

1+(a/x)²] / [(6/x + 2/x² -1/x²) - e ^1/x]
6a -1

The answer is 6a -1

iitkgp_bipin

Divide nemerator and denominator by x.

L = _[x]

_[infinity] {6a

(1+ a/x + a²/x²) -

(1+a²/x²)} / {a(6 + 2a/x - 1/ax²) - e^1/x}

L = (6a-1)/(6a-1)

L = 1