Question

Let f(x+y)=f(x)+f(y)+2xy-1 for all real values of x and y, and f(x) is a differentiable function and f'(0)=sina, prove that f(x)>0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f(x + y) = f(x) + f(y) + 2xy - 1$

and

$\rm :\longmapsto\:f'(0) = sina$

Now,

By First Principal we have,

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

So, can be rewritten as

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(x + y) - f(x)}{y}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(x) + f(y) + 2xy - 1 - f(x)}{y}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(y) + 2xy - 1}{y}$

can be rewritten by rearranging as

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{(f(y) - 1) + 2xy}{y}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(y) - 1}{y} + \displaystyle\lim_{y \to 0} \frac{2xy}{y}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(y) - 1}{y - 0} + \displaystyle\lim_{y \to 0} 2x$

$\bf :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(y) - 1}{y - 0} + 2x - - - - (1)$

Now, from given we have,

$\red{\rm :\longmapsto\:f(x + y) = f(x) + f(y) + 2xy - 1}$

On substituting x = y = 0, we get

$\rm :\longmapsto\:f(0 + 0) = f(0) + f(0) + 0 - 1$

$\rm :\longmapsto\:f(0) = 2f(0) - 1$

$\rm :\longmapsto\:f(0) - 2f(0) = - 1$

$\rm :\longmapsto\: - f(0) = - 1$

$\red{\bf :\longmapsto\: f(0) = 1}$

On substituting this value in equation (1), we get

$\bf :\longmapsto\:f'(x) = \displaystyle\lim_{y \to 0} \frac{f(y) - f(0)}{y - 0} + 2x$

$\rm :\longmapsto\:f'(x) = f'(0) + 2x$

$\rm :\longmapsto\:f'(x) = sina + 2x$

On integrating both sides w. r. t. x, we get

$\rm :\longmapsto\:\displaystyle\int \:f'(x)dx = \displaystyle\int \:(sina + 2x)dx$

We know,

$\boxed{ \bf{ \: \displaystyle\int \:kdx = kx + c}}$

and

$\boxed{ \bf{ \: \displaystyle\int \: {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c}}$

So, using these, we get

$\rm :\longmapsto\:f(x) = x \: sina \: + \: 2 \times \dfrac{ {x}^{2} }{2} + c$

$\rm :\longmapsto\:f(x) = x \: sina \: + \: {x}^{2} + c$

On substituting x = 0, we get

$\rm :\longmapsto\:f(0) = 0 \times \: sina \: + \: {0}^{2} + c$

$\rm :\longmapsto\:1 = c$

Thus,

$\purple{\bf :\longmapsto\:f(x) = x \: sina \: + \: {x}^{2} + 1}$

can be rewritten as

$\rm :\longmapsto\:f(x) = \: {x}^{2} + xsina + 1$

Now, its a quadratic equation and we know that for a quadratic polynomial f(x) = ax² + bx + c,

$\boxed{ \bf{ \: If \: a > 0 \: and \: {b}^{2} - 4ac < 0\bf\implies \:f(x) > 0}}$

$\boxed{ \bf{ \: If \: a < 0 \: and \: {b}^{2} - 4ac < 0\bf\implies \:f(x) < 0}}$

So, here,

$\red{\rm :\longmapsto\:a = 1}$

$\red{\rm :\longmapsto\:b = sina}$

$\red{\rm :\longmapsto\:c = 1}$

So,

$\red{\rm :\longmapsto\: {b}^{2} - 4ac = {sin}^{2}a - 4 < 0 \: \: \: \: \: \: \: \{ - 1 \leqslant sina \leqslant 1 \}}$

and

$\red{\rm :\longmapsto\:a = 1 > 0}$

$\bf\implies \:f(x) > 0$

Hence, Proved