Let f be a continuous function defined on [0, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to the Cauchy-Schwarz inequality,for continuous functions $f(x)$ and $g(x)$,we have $(\int_a^b f(x)g(x) dx)^2 \leq (\int_a^b f^2(x) dx)(

Vedclass · Accepted Answer

is a singleton

Vedclass · Answer

(D) According to the Cauchy-Schwarz inequality,for continuous functions $f(x)$ and $g(x)$,we have $(\int_a^b f(x)g(x) dx)^2 \leq (\int_a^b f^2(x) dx)(\int_a^b g^2(x) dx)$.By setting $g(x) = 1$,we get $(\int_0^1 f(x) dx)^2 \leq (\int_0^1 f^2(x) dx)(\int_0^1 1^2 dx) = \int_0^1 f^2(x) dx$.Since it is given that $\int_0^1 f^2(x) dx = (\int_0^1 f(x) dx)^2$,the equality holds if and only if $f(x)$ is proportional to $g(x) = 1$,which means $f(x) = c$ (a constant).Therefore,$f(x)$ is a constant function,which implies that its range is a singleton set.

Let $f$ be a continuous function defined on $[0,1]$ such that $\int_0^1 f^2(x) dx = (\int_0^1 f(x) dx)^2$. Then,the range of $f$

Similar Questions

$\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x=$

$\int\limits_{\frac{1}{2}}^{3\frac{1}{2}} {\left\{ {\frac{1}{2}\,\left( {|x - 3| + |1 - x| - 4} \right)} \right\}\,dx} $ equals: Where $\{*\}$ denotes the fractional part function.

$\int_{0}^{\pi} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x$ is equal to

If $I_{n} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \, dx$,where $n$ is a positive integer,then $I_{10} + I_{8}$ is equal to

If $I$ is the greatest of the definite integrals
${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module