CUET Preparation Today
CUET
-- Mathematics - Section B1
Differential Equations
If f(x) is differentiable function in the interval (0,∞) such that f(1)=1 and lim for each x>0, then f\left(\frac{3}{2}\right) is equal to |
\frac{13}{6} \frac{23}{18} \frac{25}{9} \frac{31}{18} |
\frac{31}{18} |
We have, \lim\limits _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1 \Rightarrow \lim\limits _{t \rightarrow x} \frac{t^2\{f(x)-f(t)\}-\left(x^2-t^2\right) f(t)}{t-x}=1 \Rightarrow \lim\limits _{t \rightarrow x}\left\{\frac{f(x)-f(t)}{t-x}\right\} t^2+\lim\limits _{t \rightarrow x}(x+t) f(t)=1 \Rightarrow -x^2 f'(x)+2 x f(x)=1 [∵ f is differentiable and so continuous also] \Rightarrow f'(x)-\frac{2}{x} f(x)=-\frac{1}{x^2} .....(i) This is a linear differential equation with integrating factor e^{-\int \frac{2}{x} d x}=\frac{1}{x^2} Multiplying (i) by integrating factor =\frac{1}{x^2} integrating, we get \frac{f(x)}{x^2}=\frac{1}{3 x^3}+C .....(ii) It is given that f(1)=1. Putting x=1 in (ii), we get 1=\frac{1}{3}+C \Rightarrow C=\frac{2}{3} ∴ \frac{f(x)}{x^2}=\frac{1}{3 x^3}+\frac{2}{3} \Rightarrow f(x)=\frac{1}{3 x}+\frac{2 x^2}{3} \Rightarrow f\left(\frac{3}{2}\right)=\frac{2}{9}+\frac{3}{2}=\frac{31}{18} |