JEE Advanced2009MathematicsDefinite IntegrationActual
Let f be a non-negative function defined on the interval [0,1] . If ₀^x 1- f^ (t) ^2 d t= ₀^x f(t) d t, 0 x 1 and f(0)=0 , then
Options
- Af ( 1 2 ) 1 3
- Bf ( 1 2 )> 1 2 and f ( 1 3 )> 1 3
- Cf ( 1 2 ) < 1 2 and f ( 1 3 ) < 1 3
- Df ( 1 2 )> 1 2 and f ( 1 3 ) < 1 3
Correct answer
C. f ( 1 2 ) < 1 2 and f ( 1 3 ) < 1 3
Step-by-step solution
Given ₀^x 1- (f^ (t) )^2 d t= ₀^x f(t) d t , 0 x 1 Applying Leibnitz theorem, we get array rlrl & & 1- (f^ (x) )^2 & =f(x) & & 1- (f^ (x) )^2 & =f^2(x) & & (f^ (x) )^2 & =1-f^2(x) & & f^ (x) & = 1-f^2(x) & & d y d x & = 1-y^2 & where y=f(x) & d y 1-y^2 = d x array On integrating both sides, we get aligned & ⁻¹(y)= x+C & f(0)=0 C=0 y= x & y= x=f(x) given f(x) 0 for & x [0,1] aligned It is known that x < x, x R⁺ ( 1 2 ) < 1 2 f ( 1 2 ) < 1 2 and ( 1 3 ) < 1 3 f ( 1 3 ) < 1 3