Search any question & find its solution
Question:
Answered & Verified by Expert
Let $f$ be a continuous function defined on $[0,1]$ such that $\int_{0}^{1} f^{2}(x) d x=\left(\int_{0}^{1} f(x) d x\right)^{2}$. Then the range of $f$
Options:
Solution:
2557 Upvotes
Verified Answer
The correct answer is:
is a singleton
By Cauchy Schwarz inequality
$$
\left\{\int_{a}^{b} f(x) g(x) d x\right\}^{2} \leq \int_{a}^{b}(f(x))^{2} d x \int_{a}^{b}(g(x))^{2} d x
$$
Here $g(x)=1$
and equality holds only when $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lambda$
So, $f(x)$ is constant
$$
\left\{\int_{a}^{b} f(x) g(x) d x\right\}^{2} \leq \int_{a}^{b}(f(x))^{2} d x \int_{a}^{b}(g(x))^{2} d x
$$
Here $g(x)=1$
and equality holds only when $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lambda$
So, $f(x)$ is constant
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.