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Differential Calculus

Blazing goIITian

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2 Nov 2008 19:39:12 IST

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IIT sample paper qn --

None

f(x+y) = f(x) + f(y) + 2xy - 1 for all x,y.

f is differentiable and f ‘(0) = cos a.

Prove that f(x) > 0 for all x = Real.

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Mirka

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2 Nov 2008 20:08:07 IST

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ANYBODY TRYING ???

santhosh in NUS

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2 Nov 2008 20:41:37 IST

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Re:IIT sample paper qn --

nikhil

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Joined: 5 Oct 2008

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3 Nov 2008 01:07:36 IST

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find derivative using first princile

u will get

f'(x)= lim h>0f(h)-1\h+2x

f'(x)=f'(0)+2x

F(x)=cos^2a=2x^2

which is always greater than zero

plz rateeeeeeeeeeeee

Anant Kumar

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Joined: 10 Jul 2008

Posts: 598

4 Nov 2008 15:53:51 IST

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$f(x+y)=f(x)+f(y)+2xy-1\quad \forall\ x,y\in \mathbb{R}$

Differentiate with respect to $y$ keeping $x$ constant,

$f^\prime(x+y)=f^\prime(y) + 2x$

Set $y=0$ , getting $f^\prime(x)=f^\prime(0)+2x$

$\Rightarrow \ f^\prime(x)=\cos a + 2x\qquad (\since \ f^\prime(0)=\cos a)$

Integerating with respect to $x$ , we get

$f(x)=x\cos a + x^2 +f(0)$

Setting $x=y=0$ in the original functional equation, we obtain $f(0)=1$ . Hence,

$\boxed{f(x)=x\cos a + x^2 +1}$ .

Transform $f$ as

$f(x)=\left(x^2 + 2\cdot x \cdot \frac{\cos a}{2}+\frac{\cos^2a}{4}\right) +1-\frac{\cos^2a}{4} = \left(x+\frac{\cos a}{2}\right)^2 + 1-\frac{\cos^2a}{4}$

Since $\left(x+\frac{\cos a}{2}\right)^2 \geq 0$ and $1-\frac{\cos^2a}{4}\geq \frac{3}{4}\quad \forall\ x\in \mathbb{R}$ .

Accordingly, $f(x)\geq \frac{3}{4}>0\quad \forall\ x\in \mathbb{R}$ .

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