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Question Number 76586 by Master last updated on 28/Dec/19

Commented by Master last updated on 28/Dec/19

$$\mathbf{solve}\mathbf{please}$$

Commented by Tony Lin last updated on 28/Dec/19

$$\left(5\right)$$$${\int }_0^{\frac{\pi }{2}}{sin}^n{xcos}^{n}xdx$$$$=\frac{1}{2^n}{\int }_0^{\frac{\pi }{2}}{sin}^n\left(2x\right)dx$$$$let2x=u,du=2dx$$$$\frac{1}{2^{n-1}}{\int }_0^{\pi }{sin}^{n}udu$$$$=\frac{1}{2^{n-2}}{\int }_0^{\frac{\pi }{2}}{sin}^{n}xdx$$$$=\frac{1}{2^{n-2}}\times \{\begin{array}{l}\frac{(n-1)(n-3)\cdot \cdot \cdot 1}{n(n-2)\cdot \cdot \cdot 2}\times \frac{\pi }{2},ifniseven\\ \frac{(n-1)(n-3)\cdot \cdot \cdot 2}{n(n-2)\cdot \cdot \cdot 1},ifnisodd\end{array}$$

Commented by john santu last updated on 28/Dec/19

$$sorrysirmissinginline2$$$$\frac{1}{2^n}{\left(2\sin x\cos x\right)}^n$$

Commented by Tony Lin last updated on 28/Dec/19

$$\left(1\right)$$$$a_{n+2}=\frac{1}{a_{n+1}}+a_n$$$$\Rightarrow a_{n+2}{a}_{n+1}=a_{n+1}{a}_n+1$$$$\Rightarrow a_{2019}{a}_{2018}$$$$=a_{2018}{a}_{2017}+1$$$$=a_{2017}{a}_{2016}+2$$$$=a_1{a}_2+2017$$$$=2018$$$$\Rightarrow a_{n+1}{a}_n=n$$$$\frac{a_{2019}}{a_{2017}}=\frac{2018}{2017}$$$$\frac{a_{2017}}{a_{2015}}=\frac{2016}{2015}$$$$\frac{a_5}{a_3}=\frac{4}{3}$$$$\frac{a_3}{a_1}=\frac{2}{1}$$$$\Rightarrow a_{2019}=a_1\times \frac{2\times 4\times \cdot \cdot \cdot \times 2018}{1\times 3\times \cdot \cdot \cdot \times 2017}$$$$=\frac{2018!!}{2017!!}$$

Commented by Master last updated on 28/Dec/19

$$2\mathrm{and}3,4\mathrm{solution}\mathrm{plz}$$

Commented by Master last updated on 28/Dec/19

$$\mathrm{thanks}$$

Commented by MJS last updated on 28/Dec/19

$$\left(3\right)\mathrm{all}\mathrm{you}\mathrm{need}\mathrm{is}\mathrm{calculate}\mathrm{the}\mathrm{determinant}...$$$$\mathrm{let}p=2019$$$$-2(x-p)(x-p+1)(x-p-1)(3px+3p^2-1)=0$$$$\Rightarrow$$$$x_1=p-1=2018$$$$x_2=p=2019$$$$x_3=p+1=2020$$$$x_4=\frac{1-3p^2}{3p}=-\frac{12229082}{6057}$$

Answered by benjo 1/2 santuyy last updated on 29/Dec/19

$$\left(4\right)f\left(0\right)=4\to c=4$$$$0⩽f\left(x\right)⩽4for0⩽x⩽3\to$$$${ax}^2+bx+4⩽4,$$$$x(ax+b)⩽0,bmustbenegatif$$$$sowehas0⩽x⩽-\left(b/a\right).$$$$wegetting-\left(b/a\right)=3\to b=-3a,a>0◼$$$$$$

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