A differentiable function f is satisfying the relation f(x + y) = f(x) + f(y) + 2xy(x + y) - R and , then value of is ( greatest integer function.)

JEE MainMathematicsModerateLimits

A differentiable function f is satisfying the relation f(x + y)
= f(x) + f(y) + 2xy(x + y) - $\frac{1}{3}\forall x,y \in$ R and
$\lim_{h \rightarrow 0}\mspace{2mu}\frac{3f(h) - 1}{6h} = \frac{2}{3}$ ,
then value of is ( greatest integer function.)

Show answer and solution

f^{'}(x) = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(x + h) - f(x)}{h}

= \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(x) + f(h) + 2x(x + h)h - \frac{1}{3} - \left( f(x) + f(0) - \frac{1}{3} \right)}{h}

= \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(h) - f(0)}{h} + 2x^{2} = f^{'}(0) + 2x^{2}

\lim_{h \rightarrow 0}\mspace{2mu}\frac{3(f(h)) - 1}{6h} = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(h) - f(0)}{2h} = \frac{f^{'}(0)}{2} = \frac{2}{3}

f^{'}(0) = \frac{4}{3}

f^{'}(x) = \frac{4}{3} + 2x^{2}

f(x) = \lambda + \frac{4}{3}x + \frac{2x^{3}}{3}

now

f(0) = \lambda = \frac{1}{3}

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

Scan any doubt from your notes and get a whiteboard explanation.