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Question Number 38699 by Tinkutara last updated on 28/Jun/18

Answered by behi83417@gmail.com last updated on 28/Jun/18

$$y^2=a^2+b^2+2\sqrt{(a^2{cos}^{2}x+b^2{sin}^{2}x)(a^2{sin}^{2}x+b^2{cos}^{2}x)}$$$$P=a^2{cos}^{2}x+b^2{sin}^{2}x,Q=a^2{sin}^{2}x+b^2{cos}^{2}x$$$$P+Q=const,PQ,willbemax,when:$$$$P=Q$$$$\Rightarrow a^2{cos}^{2}x+b^2{sin}^{2}x=a^2{sin}^{2}x+b^2{cos}^{2}x$$$$\Rightarrow a^{2}cos2x=b^{2}cos2x\Rightarrow \mathit{a}=\mathit{b}\left(\times \right)\vee cos2x=0$$$$\Rightarrow y^2⩽a^2+b^2+2\sqrt{\frac{a^2+b^2}{2}.\frac{a^2+b^2}{2}}=2(a^2+b^2)$$$$\Rightarrow 0<y⩽\sqrt{2(a^2+b^2)}.◼$$

Commented by behi83417@gmail.com last updated on 28/Jun/18

dear master:mrW3! thank you so much for solving Q#38032. no message recive to mi from app.so i don't see your answer on this Q.anyway thanks.

Commented by MrW3 last updated on 29/Jun/18

$$thankyousir!$$$$shouldtheanswernotbe$$$$\mid a\mid +\mid b\mid ⩽y⩽\sqrt{2(a^2+b^2)}?$$

Commented by behi83417@gmail.com last updated on 28/Jun/18

$$waynot!pleaseconfirm.$$

Commented by MrW3 last updated on 29/Jun/18

$$y^2=a^2+b^2+2\sqrt{(a^2{cos}^{2}x+b^2{sin}^{2}x)(a^2{sin}^{2}x+b^2{cos}^{2}x)}$$$$y^2=a^2+b^2+2\sqrt{(a^4{\sin }^2{xcos}^{2}x+a^2{b}^2{sin}^{4}x+a^2{b}^2{cos}^{4}x+b^4{\sin }^{2}x{\cos }^{2}x}$$$$y^2=a^2+b^2+2\sqrt{(a^4+b^4){\sin }^2{xcos}^{2}x+a^2{b}^2({sin}^{4}x+{cos}^{4}x)}$$$$y^2=a^2+b^2+2\sqrt{(a^4+b^4){\sin }^2{xcos}^{2}x+a^2{b}^2({sin}^{4}x+{cos}^{4}x+{2\sin }^{2}x{\cos }^{2}x)-2a^2{b}^2{\sin }^{2}x{\cos }^{2}x}$$$$y^2=a^2+b^2+2\sqrt{(a^4+b^4-2a^2{b}^2){\sin }^2{xcos}^{2}x+a^2{b}^2{({\sin }^{2}x+{\cos }^{2}x)}^2}$$$$y^2=a^2+b^2+2\sqrt{{(a^2-b^2)}^2{\sin }^2{xcos}^{2}x+a^2{b}^2}$$$$y^2=a^2+b^2+2\sqrt{\frac{1}{4}{(a^2-b^2)}^2{\sin }^{2}2x+a^2{b}^2}$$$$y^2=a^2+b^2+\sqrt{4a^2{b}^2+{(a^2-b^2)}^2{\sin }^{2}2x}$$$$\Rightarrow y=\sqrt{a^2+b^2+\sqrt{4a^2{b}^2+{(a^2-b^2)}^2{\sin }^{2}2x}}$$$$min.yat\sin 2x=0:$$$$y_{min}=\sqrt{a^2+b^2+\sqrt{4a^2{b}^2}}=\sqrt{a^2+b^2+2\mid a\mid \mid b\mid }=\mid a\mid +\mid b\mid$$$$max.yat\sin 2x=\pm 1:$$$$y_{max}=\sqrt{a^2+b^2+\sqrt{4a^2{b}^2+{(a^2-b^2)}^2}}$$$$=\sqrt{a^2+b^2+\sqrt{{(a^2+b^2)}^2}}$$$$=\sqrt{2(a^2+b^2)}$$$$$$$$\Rightarrow \mid a\mid +\mid b\mid ⩽y⩽\sqrt{2(a^2+b^2)}$$

Commented by behi83417@gmail.com last updated on 29/Jun/18

$$perfect!$$

Commented by Tinkutara last updated on 29/Jun/18

$$Ansgivenis[(a+b),\sqrt{2(a^2+b^2)}]$$

Commented by Tinkutara last updated on 29/Jun/18

Thank you very much Sir! I got the answer. ��������

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