Given: f(x+y)=f(x)+f(y)+2xy-1and f'(0)=sinϕWe know,f'(x)=limh→0fx+h-f(x)h=limh→0f(x)+f(h)+2xh-1-f(x)h using given relation=limh→02x+f(h)-1h=limh→02x+f(h)-f(0)hPutting x=0=y in the given relation we findf(0)=f(0)+f(0)+0-1⇒f(0)=1∴f'(x)=2x+f'(0)⇒f'(x)=2x+sinϕIntegrating both sides, we getf(x)=x2+xsinϕ+CPut x=0, we getf(0)=C⇒C=1So, f(x)=x2+xsinϕ+1>0 ∀x∈R

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Solve this: 46 Let + y) = —l forall x, y e R. If is differentiable and f z sin"rnprove that V x e R.

G i v e n : f (x + y) = f (x) + f (y) + 2 x y - 1 a n d f' (0) = \sin ϕ W e k n o w, f' (x) = \lim_{h \to 0} f \frac{(x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\{f (x) + f (h) + 2 x h - 1\} - f (x)}{h} [u \sin g g i v e n r e l a t i o n] = \lim_{h \to 0} 2 x + \frac{f (h) - 1}{h} = \lim_{h \to 0} 2 x + \frac{f (h) - f (0)}{h} [P u t t i n g x = 0 = y i n t h e g i v e n r e l a t i o n w e f i n d f (0) = f (0) + f (0) + 0 - 1 \Rightarrow f (0) = 1] ∴ f' (x) = 2 x + f' (0) \Rightarrow f' (x) = 2 x + \sin ϕ I n t e g r a t i n g b o t h s i d e s, w e g e t f (x) = x^{2} + x \sin ϕ + C P u t x = 0, w e g e t f (0) = C \Rightarrow C = 1 S o, f (x) = x^{2} + x \sin ϕ + 1 > 0 \forall x \in R

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Solve this: Solve this: 46 Let + y) = —l forall x, y e R. If is differentiable and f z sin"rnprove that V x e R.

Solve this:

Solve this: 46 Let + y) = —l forall x, y e R. If is differentiable and f z sin"rnprove that V x e R.