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In acute angled $\Delta ABC$ prove that,

$sin A + sin B + sin C > cos A + cos B + cos C$

In acute angled triangle, $\cos \frac{A}{2} > \sin \frac{A}{2}$

So, $\sin \frac{A}{2}.\cos \frac{A}{2} > \sin \frac{A}{2}.\sin \frac{A}{2}$

Hence, $2\sin \frac{A}{2}\cos \frac{A}{2} > 2{\sin ^2}\frac{A}{2}$

Thus, $\sin A > 1 - \cos A$

Likewise, $\sin B > 1 - \cos B$ & $\sin C > 1 - \cos C$

On addition, $\sin A + \sin B + \sin C > 3 - (\cos A + \cos B + \cos C)$ ........(!)

In any triangle, $\cos A + \cos B + \cos C \le \frac{3}{2}$*

$\therefore 2(\cos A + \cos B + \cos C) \le 3$

$ \Rightarrow \cos A + \cos B + \cos C \le 3 - (\cos A + \cos B + \cos C)$ ........(!!)

From (!) & (!!), $\sin A + \sin B + \sin C > \cos A + \cos B + \cos C$

*Please refer textbook or study material for the proof.

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$