IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

srichaitanya

_{[ ]}

_N

_{[ ]}

_R=1¹⁰⁰ (N+R)¹⁰/N¹⁰+10¹⁰

nitish_iit

can you make the question a more clear.....

i can't understand iam confused....

iitkgp_bipin

L = _[x]

_[0] (36^x - 9^x - 4^x + 1)/{

2 -

(1+cosx)} which is in the form of 0/0 .

Numerator , N = (36^x - 9^x - 4^x + 1) = (9^x.4^x - 9^x - 4^x + 1) = (9^x - 1)(4^x - 1)

Denominator, D = {

2 -

(1+cosx)}.{

2 +

(1+cosx)}/{

2 +

(1+cosx)} (rationalize)

= (1 - cosx)/{

2 +

(1+cosx)}

= {2sin²(x/2)}/{

2 +

(1+cosx)}

As x tends to zero sin²(x/2) tends to (x/2)²

Hence _[x]

_[0] D = 2(x/2)²}/{

2 +

(1+1)} = x²/4

2

Hence L = _[x]

_[0] N/D

L = _[x]

_[0] (9^x - 1)(4^x - 1)/{x²/4

2}

L = _[x]

_[0] {(9^x - 1)/x}.{(4^x - 1)/x}.(4

2)

Now use _[x]

_[0] (a^x - 1)/x = loga

L = (4

2).(log4).(log9)

L = (16

2).(log2).(log3)