Q1.
\mathrm{f(x)=x+\int_{0}^{1}(xy^2+x^2y)f(y)dy} ,\texttt{Then f(x) attains a minimum at} \\ \\ (A)x=\frac{9}{8} \ \ \ \(B)x=\frac{-9}{8} \\ \\ (A)x=0 \ \ \ \(D)x=1
Q2.
\texttt{Let f(x) be a positive , continuous and differentiable on the interval (a,b).}\\ \texttt{If }\mathrm{\lim_{x \rightarrow a^+}f(x)=1} \texttt{ and }\mathrm{\lim_{x \rightarrow b^-}f(x)=3^{\frac{1}{4}}}.\texttt{Also }f'(x)\geq f^3(x)+\frac{1}{f(x)}\texttt{, then} \\ \\ \\ \mathrm{(A)b-a\geq \frac{\pi}{4} \ \ \ \ (B)b-a\leq \frac{\pi}{4} \ \ (C)b-a\leq \frac{\pi}{24} \ \ \ \ (D)\text{None of these}}
for practice
341If its integer it helps because you can set bounds for x. To show you what i mean,
We must have x^3 \ge 100 \Rightarrow x \ge 5
Also we need
x^2+100 > x^3-100 \Rightarrow 200>x^3-x^2 = x^2(x-1) \ge 4x^2 \Rightarrow x^2 \le 50 \Rightarrow x \le 7
So x=5,6, or 7
If its real solutions, then one method available is to prove that f(x) = (x^3-100)^3-(x^2+100)^2 is monotonic in at least [4,8]. then using the above discussion you have that x = 5 is the only root
21prophet sir, i didnt understand wat u said inside the bracket.
aditya, may be i call not very very easy things ''easy'' but that is just to make myself and people around me confident..acc to me self belief and confidence solve 70%(75%?) of any problem.....these days i hv grown a habit of these..infact if i can't do u qstn..after seeing the solution , i start calling it easy...so that is just my nature not anything else [3]..
1yes nasiko i also moved backward ..
i thought solved the differential equation and then applied
Lagrange Mean Value Theorem ..
but this question is from FIITJEE's GMP which is reputed to be of Jee level by the aspirants
anyways ,undoubtedly your problem solving skills are way better than an average jee aspirant
that is why you feel it too be "a very easy one"
341i got that (see my hint. in fact to be lil more rigorous, u hv to state that your are dealing with an improper integral and by using continuity justify the final form of the inequality). I was playing with using AM-GM and after a serious of mishaps concluded that i get a tighter bound
Teaches me not to underestimate problem setters
21yes...:D
btw, this problem is not tricky..
no need to think abt the function tan-1(f(x)2)
just an attempt to solve the differential equation will lead us there...
341that was a miscalculation. sorry. I was trying to do a stunt using AM-GM inequality
341Someone posted this on AOPS and there's a cute solution from one of my favourite problem solvers there:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=420110
341By the way, in Q2 can you try your hand at establishing a tighter bound.
I mean we have got that
b-a \le \frac{\pi}{24} \approx 0.13
In fact with far less manipulation establish that b-a \le 0.104
(which makes the given problem a bit lame actually :D)
1hint for the second question:
think about a function: g(x)= tan-1(f2(x))
1yes nasiko...i did the same way for q1 ...
in case it is integer solution...i think some concept of number theory is used
prophet sir and nasiko please explain
211st is very easy
It's visible that f(x) is of the form ax2+bx
its very simple to prove that f(-98)= minf(x)
1doubt :
http://123iitjee.com/mod/forum/discuss.php?d=2075&parent=5839
how to do this one
wolfram gives 5 as one root
21The given inequality can be written as \frac{2f(x)f^{,}(x)}{1+ (f(x)^{4})}\geq 2
Which means g^{,}(x)\geq 2 if we let g(x)= \tan^{-1} (f(x)^{2})
On integration g(b)-g(a)\geq 2(b-a)
Which gives \frac{\pi}{24}\leq b-a
341Since aditya has given a hint, a further hint: write the eqn as 2f(x) f'(x)1+f4(x)≥2
BTW, nice and useful questions
