Not for JEE

any thing said about f(x)?

We have, for a real \lambda,
\int_0^1f(x)(x+\lambda)\ \mathrm{d}x =1+\lambda
Therefore, by Cauchy-Schwarz, we get
(1+\lambda)^2
=\left(\int_0^1f(x)(x+\lambda)\ \mathrm{d}x\right)^2
\le \int_0^1f^2(x)\ \mathrm{d}x\,\int_0^1(x+\lambda)^2\ \mathrm{d}x
= \dfrac{3\lambda^2 + 3\lambda +1}{3}\int_0^1f^2(x)\ \mathrm{d}x
Hence, we have
\int_0^1f^2(x)\ \mathrm{d}x \geq \dfrac{3(1+\lambda)^2}{3\lambda^2+3\lambda+1}
Since this inequality must hold for all real \lambda, we must have
\int_0^1f^2(x)\ \mathrm{d}x \geq \max_\lambda\left(\dfrac{3(1+\lambda)^2}{3\lambda^2+3\lambda+1}\right)=4
with equality at \lambda = -\dfrac{1}{3} and hence for f(x)=6\left(x-\dfrac{1}{3}\right)

It was a nice one and took me a while to figure it out. However, I enjoyed solving it. :)

Yeah, the Schwarz Bunyakovsky Inequality is called into play here:

I'll write out my solution, which is essentially the same as above, but is a little more straightforward:

\int_0^1 f(x) (1+x) \ dx = 2 \Rightarrow \int_0^1 f^2(x) \ dx \int_0^1 (1+x)^2 \ dx \ge \left(\int_0^1 f(x) (1+x) \ dx \right)^2

But \int_0^1 (1+x)^2 \ dx = 1

Hence the inequality follows

Well, I first did what you did. However, for equality we require f(x) to be proportional to (1+x). So if we took f(x)=k(1+x), then the condition
\int_0^1 f(x)\ \mathrm{d}x=1
give us k = \dfrac{2}{3}. But with this k, our function at equality becomes
f(x) = \dfrac{2}{3}(1+x)
But with this function, the second condition \int_0^1 x f(x)\ \mathrm{d}x=1 does NOT hold. Further, in this case
\int_0^1 f^2(x)\ \mathrm{d}x=\dfrac{28}{27}\neq 4
So, your solution is not entirely correct.

Hmm. I didnt verify that.

Moreover, how did you get \int_0^1(1+x)^2\ \mathrm{d}x=1 ?

the problem, as usual, was: 1. i did this in my head 2. i did not bother to check the solution at the back :D

Apologies and thank you kaymant sir for the correct solution. Otherwise the students would have been misled