IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - a sum from LIMITS....PLZZZ TELL ME THE SOLN......!!!!

nayan_dbb

if f(9) = 9 and f ' (9)=1 then

_{[ x]}

_{[ 9]} { 3 - f(x) } / {3 - ^{[2 ]}

x } is equal to ???

Please let me know the solution.

The above question is from the Pathfinder study material/

nishant

see here f(x)=x symbolises tht the function is a constant function..so f(3) will be equal to 3...therefore,the limit on substituting the f(x) value and x value becomes 0/0 form..apply l'hospital rule in it to land up with the answer..cheers!!

iitkgp_bipin

Sorry.....I posted the wrong solution.

The limit is in constant/0 form which does not exist.

Actually the question should be

L = _{[ x]}

_{[ 9]} {3 -

f(x)}/{3 -

x} which is in 0/0 form.

Apply L'Hospital rule :

L = _[x]

_[9] {d(3 -

f(x))/dx}/{d(3 -

x)/dx}

L = _[x]

_[9] {f'(x)/2

f'(x)}/{1/2

x}

L = {f'(9).

9}/(

f(9)} = 1

fwks_phoenix

if f(9) = 9 and f ' (9)=1 then

_{[ x]}

_{[ 9]} { 3 - f(x) } / {3 - ^{[2 ]}

x } is equal to ???

---------------------------------

differentiate the numerator n the denominator...
u'll get " - differentiation of f(x) " in the numerator n " - differentiation of x to the power 1/2" in the denominator

now u sub. the value for x as 9..
u'll get " -1 / - 1 / 2 X 3" = " +6 "
so i think the ans qil b " + 6 "

aditya_arora04

Hi, Modify your question as :

_{[ ]}

_{[ ]} [ 9 - f(x) ] / [ 3 - (x)^1/2 ] - _{[ ]}

_{[ ]}6 / [ 3 - (x)^1/2]

Applying L'Hospital rule in first part, we get :

_{[ ]}

_{[ ]} [ - f '(x) ] / [ 1/2(x)^-1/2 ] - _{[ ]}

_{[ ]}6 / [ 3 - (x)^1/2]

Now, put the limits in first part,

For, second part, rationalize the denominator with 3 + x^(1/2) and put the limits.

Hope you got it.

Ask & Discuss Questions with Community & Experts