Bài 110861

Question

a)Tìm số $n$ nguyên dương sao cho:
$C^{1}_{2n+1}-2.2 C^{2}_{2n+1}+ 3.2^2. C^{3}_{2n+1}-4.2^3 C^{4}_{2n+1}+...+ (2n+1)2^{ 2n} C^{2n+1}_{2n+1}=2007 $
b) Giải phương trình lượng giác: $\sin 5x -\cos 5x =\tan 2x +\cot 2x$
c) Cho $f(x)=\sin x + \frac{1}{3} \sin^3 x -\cos x -\frac{1}{3} \cos^3 x $ và $g(x)=x-\frac{1}{2} \cos2x $.
Tìm $x$ để $f'(x)=g'(x)$

score 0 · Answer 1

a) Xét   $P(x)=(x+1)^{2n+1}$.    Ta có:
              $P(x)= C^{0}_{2n+1}+C^{1}_{2n+1}x+C^{2}_{2n+1}x^2+C^{3}_{2n+1}x^3 +...+ C^{2n+1}_{2n+1}x^{2n+1}     $
            $P'(x)=C^{1}_{2n+1}+2 C^{2}_{2n+1}x+3 C^{3}_{2n+1}x^2 +...+ (2n+1)C^{2n+1}_{2n+1} x^{ 2n}   $
          $P'(-2)=C^{1}_{2n+1}-2.2C^{2}_{2n+1}+3.2^2.C^{3}_{2n+1}+ ...+ (2n+1)2^{2n} C^{2n+1}_{2n+1} =2007 $
Mặt khác       $P(x)=(x+1)^{2n+1}$
           $ \Rightarrow P'(x)=(2n+1)(x+1)^{2n}$
                      $P'(-2)=2n+1=2007 \Rightarrow    n=1003$

b)   VT:    $|\sin 5x - \cos 5x|=\sqrt{2}|\sin (5x-\frac{\pi}{4} )| \leq \sqrt{2}$
      VP:     $|\tan 2x+\cot 2x| \geq 2$
      Do đó phương trình vô nghiệm.

c)    $f'(x)=\cos x + \sin^2 x \cos x + \sin x+\cos^2 x\sin x$
       $g'(x)=1+\sin 2x$
       $f'(x)=g'(x) \Leftrightarrow (1+\sin ^2 x) \cos x+ (1+\cos^2 x) \sin x =1+\sin 2x$
                                   $\Leftrightarrow (\sin x +\cos x)+\sin x \cos x(\sin x+ \cos x)=(\sin x+ \cos x)^2$
                                   $\Leftrightarrow (\sin x+ \cos x)[(\sin x+ \cos x)- \sin x \cos x -1 ]=0$
                                   $\Leftrightarrow \left[ \begin{array}{l}\sin x+\cos x=0                  (a)\\ (1-\sin x)(1-\cos x)            (b)\end{array} \right. $
$(a) \sin x + \cos x =0 \Leftrightarrow    \sqrt{2} \sin (x+\frac{\pi }{4} )=0 \Leftrightarrow     x=-\frac{\pi}{4}+k\pi $
$(b) \Leftrightarrow    \left[ \begin{array}{l}\sin x=1\\\cos x=1\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}x=\frac{\pi}{2}+k2\pi \\x=k2\pi\end{array} \right. $

Bài 110861

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Bài 110861

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