OH!! yeah MAK!!!
I never realised that they applied L hosp. in the 1st step, i kept on thinkin THEY HAD USED THE APPROXIMATION as in comp.
but its only use was for a pre-requisite to apply LH ruke
OH!! yeah MAK!!!
I never realised that they applied L hosp. in the 1st step, i kept on thinkin THEY HAD USED THE APPROXIMATION as in comp.
but its only use was for a pre-requisite to apply LH ruke
@tapan... d solution is perfectly correct... [1]
what did u not understand in d solution...? [7]
see, first of all, apply L' Hospital rule to d given limit 2 tyms... finally u'll get lim(h→0) f''(a+h) = 0... which implies f''(x) = 0 for xεR... this is possible only if f(x) has a maximum degree of 1... since if f(x) is of second or higher degree... f(x) may or may not b zero... (eg: f(x)=x2 => f''(x)=2)... got it...!!!
well tapan, isn't d solution given... 'coz i too feel d answer must b 3...
agar solution diya ho tho post kar dena... let us discuss upon it...
oh the answer is not staring.. :P :D
Graph wont work.. i thought for a moment it will.. but abhee it has stopped staring in my face.. for a moment i had thought ki ho gaya :()
if v apply approximation Formula 1
then to our numerator gets cancelled only [2]
if v use Formula 2
then num is sumthing lyk : (t-a) {f(t) + ...... }
--------------------------
(t-a)3
then v get {f(t) + ........}
-------------------
(t-a)2
aage Kuch STARE nahi kar raha [2] [17]
lim t→a \int_{a}^{t}{f(t)dt-\frac{t-a}{2}(f(t)+f(a))}/(t-a)^{3}
its too tiring to write the equation
but answer is b