Let f be a differentiable function from R to R such that fx-fy≤2x-y^(3/2), for all x,y∈R. If f0=1 then ∫01f^2xdx is equal to

Correct answer: (B) 1. Given functional equality is fx-fy≤2 x-y^(3/2) Put x=y+h, we get fy+h-fy≤2. |h^(3/2) ⇒(fy+h-f(y))/h≤2 h^(1/2) ⇒limh→0⁡(fy+h-fy)/h≤limh→0⁡2 h^(1/2) ⇒f^(′)y≤0 ⇒f^(′)y=0 ⇒f′y=0 ⇒fy=c ⇒fy=1 (since f0=1 ) Now ∫01f^2xdx= ∫011 dx=1.

Q59 / 90 · JEE Main 2019 (09 Jan Shift 2) · Mathematics

Let $f$ be a differentiable function from $R$ to $R$ such that $|f (x) - f (y)| \leq 2 {|x - y|}^{3 / 2},$ for all $x, y \in R .$ If $f (0) = 1$ then $\int_{0}^{1} f^{2} (x) d x$ is equal to

A
$0$
Correct answer
B
$1$
C
$2$
D
$\frac{1}{2}$

Solution

Given functional equality is

|f (x) - f (y)| \leq 2 {|x - y|}^{3 / 2}

Put

x = y + h,

we get

|f (y + h)| - f (y) {|\leq 2 . | h|}^{3 / 2}

\Rightarrow |\frac{f (y + h) - f (y)}{h}| \leq 2 {|h|}^{1 / 2}

\Rightarrow \lim_{h \to 0} |\frac{f (y + h) - f (y)}{h}| \leq \lim_{h \to 0} 2 {|h|}^{1 / 2}

\Rightarrow |f^{'} (y)| \leq 0

\Rightarrow |f^{'} (y)| = 0

\Rightarrow f' (y) = 0

\Rightarrow f (y) = c

\Rightarrow f (y) = 1

(since

f (0) = 1

)

Now

\int_{0}^{1} f^{2} (x) d x = \int_{0}^{1} 1 d x = 1

Hence, the correct answer is option: 'B'.

Let f be a differentiable function from R to R such that fx-fy≤2x-y3/2, for all x,y∈R. If f0=1 then ∫01f2xdx is equal to

Solution

Relevant Questions

Let $f$ be a differentiable function from $R$ to $R$ such that $|f (x) - f (y)| \leq 2 {|x - y|}^{3 / 2},$ for all $x, y \in R .$ If $f (0) = 1$ then $\int_{0}^{1} f^{2} (x) d x$ is equal to